Algebra

Algebra Mark Steinberger The University at Albany State University of New York Preface The intent of this book is to ...

Author: M. Steinberger

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Algebra Mark Steinberger The University at Albany State University of New York

Preface The intent of this book is to introduce the student to algebra from a point of view that stresses examples and classification. Thus, whenever possible, the main theorems are treated as tools that may be used to construct and analyze specific types of groups, rings, fields, modules, etc. Sample constructions and classifications are given in both text and exercises. It is also important to note that many beginning graduate students have not taken a deep, senior-level undergraduate algebra course. For this reason, I have not assumed a great deal of sophistication on the part of the reader in the introductory portions of the book. Indeed, it is hoped that the reader may acquire a sophistication in algebraic techniques by reading the presentations and working the exercises. In this spirit, I have attempted to avoid any semblance of proofs by intimidation. The intent is that the exercises should provide sufficient opportunity for the student to rise to mathematical challenges. The first chapter gives a summary of the basic set theory we shall use throughout the text. Other prerequisites for the main body of the text are trigonometry and the differential calculus of one variable. We also presume that the student has seen matrix multiplication at least once, though the requisite definitions are provided. Chapter 2 introduces groups and homomorphisms, and gives some examples that will be used as illustrations throughout the material on group theory. Chapter 3 develops symmetric groups and the theory of G-sets, giving some useful counting arguments for classifying low order groups. We emphasize actions of one group on another via automorphisms, with conjugation as the initial example. Chapter 4 studies consequences of normality, beginning with the Noether Isomorphism Theorems. We study simple groups and composition series, and then classify all finite abelian groups via the Fundamental Theorem of Finite Abelian Groups. The automorphism group of a cyclic group is calculated. Then, semidirect products are introduced, using the calculations of automorphism groups to construct numerous examples that will be essential in classifying low order groups. We then study extensions of groups, developing some useful classification tools. In Chapter 5, we develop some additional tools, such as the Sylow Theorems, and apply the theory we’ve developed to the classification of groups of many of the orders ≤ 63, in both text and exercises. The methods developed are sufficient to classify the rest of the orders in that range. We also study solvable and nilpotent groups. Chapter 6 is an introduction to basic category theoretic notions, with examples drawn from the earlier material. The concepts developed here are useful in understanding rings and modules, though they are used sparingly in the text on that material. Pushouts of groups are constructed and the technique of generators and relations is given. Chapter 7 provides a general introduction to the theory of rings and modules. Exam p , the cyclotomic ples are introduced, including the quaternions, H, the p-adic integers, Z ix

Preface

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integers and rational numbers, Z[ζn ] and Q[ζn ], polynomials, group rings, and more. Free modules and chain conditions are studied, and the elementary theory of vector spaces and matrices is developed. The chapter closes with the study of rings and modules of fractions, which we shall apply to the study of P.I.D.s and fields. Chapter 8 develops the theory of P.I.D.s and covers applications to field theory. The exercises treat prime factorization in some particular Euclidean number rings. The basic theory of algebraic and transcendental field extensions is given, including many of the basic tools to be used in Galois theory. The cyclotomic polynomials over Q are calculated, and the chapter closes with a presentation of the Fundamental Theorem of Finitely Generated Modules over a P.I.D. Chapter 9 gives some of the basic tools of ring and module theory: Nakayama’s Lemma, primary decomposition, tensor products, extension of rings, projectives, and the exactness properties of tensor and hom. In Section 9.3, Hilbert’s Nullstellensatz is proven, using many of the tools so far developed, and is used to define algebraic varieties. In Section 9.5, extension of rings is used to extend some of the standard results about injections and surjections between finite dimensional vector spaces to the study of maps between finitely generated free modules over more general rings. Also, algebraic K-theory is introduced with the study of K0 . Chapter 10 gives a formal development of linear algebra, culminating in the classification of matrices via canonical forms. This is followed by some more general material about general linear groups, followed by an introduction to K1 . Chapter 11 is devoted to Galois theory. The usual applications are given, e.g. the Primitive Element Theorem, the Fundamental Theorem of Algebra, and the classification of finite fields. Sample calculations of Galois groups are given in text and exercises, particularly for splitting fields of polynomials of the form X n − a. The insolvability of polynomials of degree ≥ 5 is treated. Chapter 12 gives the theory of hereditary and semisimple rings with an emphasis on Dedekind domains and group algebras. Stable and unstable classifications of projective modules are given for Dedekind domains and semisimple rings. The major dependencies between chapters are as follows, where A → B means that B depends significantly on material in A. 1−4

/ 7

5 ? 8 ?? ?? ?? ?? ? 6 11

/ 9

/ 10

/ 12

Additionally there are some minor dependencies. For instance, Chapter 9 assumes an understanding of the basic definitions of categories and functors, as given in the first two sections in Chapter 6. Some categorical notions would also be useful in Chapter 7, but are optional there. Also, the last section in Chapter 5 assumes an understanding of some of the material on linear algebra from Chapter 10. Some material on linear algebra is also used in the closing section of Chapter 11. And optional references are made in Chapter 10 to the material on exterior algebras from Chapter 9. Other than the references to linear algebra in Chapter 5, the chapters may be read in order.

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A solid box symbol ( ) is used to indicate either the end of a proof or that the proof is left to the reader. Those exercises that are mentioned explicitly in the body of the text are labeled with a dagger symbol (†) in the margin. This does not necessarily indicate that the reference is essential to the main flow of discussion. Similarly, a double dagger symbol (‡) indicates that the exercise is explicitly mentioned in another exercise. Of course, many more exercises will be used implicitly in other exercises.

Acknowledgments Let me first acknowledge a debt to Dock Rim, who first introduced me to much of this material. His love for the subject was obvious and catching. I also owe thanks to Hara Charalambous, SUNY at Albany, and David Webb, Dartmouth College, who have used preliminary versions of the book in the classroom and made some valuable suggestions. I am also indebted to Bill Hammond, SUNY at Albany, for his insights. I’ve used this material in the classroom myself, and would like to thank the students for many useful suggestions regarding the style of presentation and for their assistance in catching typos. Mark Steinberger Ballston Lake, NY

Contents 1 A Little Set Theory 1.1 Properties of Functions . . . . . . . 1.2 Factorizations of Functions . . . . . 1.3 Relations . . . . . . . . . . . . . . . 1.4 Equivalence Relations . . . . . . . . 1.5 Generating an Equivalence Relation 1.6 Cartesian Products . . . . . . . . . . 1.7 Formalities about Functions . . . . .

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1 1 3 6 7 10 11 13

2 Groups: Basic Definitions and Examples 2.1 Groups and Monoids . . . . . . . . . . . . 2.2 Subgroups . . . . . . . . . . . . . . . . . . 2.3 The Subgroups of the Integers . . . . . . . 2.4 Finite Cyclic Groups: Modular Arithmetic 2.5 Homomorphisms and Isomorphisms . . . . 2.6 The Classification Problem . . . . . . . . 2.7 The Group of Rotations of the Plane . . . 2.8 The Dihedral Groups . . . . . . . . . . . . 2.9 Quaternions . . . . . . . . . . . . . . . . . 2.10 Direct Products . . . . . . . . . . . . . . .

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15 15 19 24 27 29 35 37 38 40 42

3 G-sets and Counting 3.1 Symmetric Groups: Cayley’s Theorem . 3.2 Cosets and Index: Lagrange’s Theorem 3.3 G-sets and Orbits . . . . . . . . . . . . . 3.4 Supports of Permutations . . . . . . . . 3.5 Cycle Structure . . . . . . . . . . . . . . 3.6 Conjugation and Other Automorphisms 3.7 Conjugating Subgroups: Normality . . .

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46 47 51 55 64 66 71 77

4 Normality and Factor Groups 4.1 The Noether Isomorphism Theorems . . . . 4.2 Simple Groups . . . . . . . . . . . . . . . . 4.3 The Jordan–H¨ older Theorem . . . . . . . . 4.4 Abelian Groups: the Fundamental Theorem 4.5 The Automorphisms of a Cyclic Group . . . 4.6 Semidirect Products . . . . . . . . . . . . . 4.7 Extensions . . . . . . . . . . . . . . . . . . .

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82 . 83 . 90 . 96 . 98 . 105 . 111 . 119

xii

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CONTENTS

xiii

5 Sylow Theory, Solvability, and Classification 5.1 Cauchy’s Theorem . . . . . . . . . . . . . . . 5.2 p-Groups . . . . . . . . . . . . . . . . . . . . 5.3 Sylow Subgroups . . . . . . . . . . . . . . . . 5.4 Commutator Subgroups and Abelianization . 5.5 Solvable Groups . . . . . . . . . . . . . . . . 5.6 Hall’s Theorem . . . . . . . . . . . . . . . . . 5.7 Nilpotent Groups . . . . . . . . . . . . . . . . 5.8 Matrix Groups . . . . . . . . . . . . . . . . .

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134 136 137 141 149 150 153 156 158

6 Categories in Group Theory 6.1 Categories . . . . . . . . . . . . . . . . . 6.2 Functors . . . . . . . . . . . . . . . . . . 6.3 Universal Mapping Properties: Products 6.4 Pushouts and Pullbacks . . . . . . . . . 6.5 Infinite Products and Coproducts . . . . 6.6 Free Functors . . . . . . . . . . . . . . . 6.7 Generators and Relations . . . . . . . . 6.8 Direct and Inverse Limits . . . . . . . . 6.9 Natural Transformations and Adjoints . 6.10 General Limits and Colimits . . . . . . .

. . . . . . . . . . . . . . . . . . . . and Coproducts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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163 164 167 171 176 183 186 189 191 195 198

7 Rings and Modules 7.1 Rings . . . . . . . . . . . . . . 7.2 Ideals . . . . . . . . . . . . . . 7.3 Polynomials . . . . . . . . . . . 7.4 Symmetry of Polynomials . . . 7.5 Group Rings and Monoid Rings 7.6 Ideals in Commutative Rings . 7.7 Modules . . . . . . . . . . . . . 7.8 Chain Conditions . . . . . . . . 7.9 Vector Spaces . . . . . . . . . . 7.10 Matrices and Transformations . 7.11 Rings of Fractions . . . . . . .

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201 202 215 222 234 239 245 250 268 273 277 284

8 P.I.D.s and Field Extensions 8.1 Euclidean Rings, P.I.D.s, and U.F.D.s 8.2 Algebraic Extensions . . . . . . . . . . 8.3 Transcendence Degree . . . . . . . . . 8.4 Algebraic Closures . . . . . . . . . . . 8.5 Criteria for Irreducibility . . . . . . . . 8.6 The Frobenius . . . . . . . . . . . . . 8.7 Repeated Roots . . . . . . . . . . . . . 8.8 Cyclotomic Polynomials . . . . . . . . 8.9 Modules over P.I.D.s . . . . . . . . . .

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295 296 308 314 317 320 324 325 327 332

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CONTENTS

xiv

9 Radicals, Tensor Products, and Exactness 9.1 Radicals . . . . . . . . . . . . . . . . . . . . . 9.2 Primary Decomposition . . . . . . . . . . . . 9.3 The Nullstellensatz and the Prime Spectrum 9.4 Tensor Products . . . . . . . . . . . . . . . . 9.5 Tensor Products and Exactness . . . . . . . . 9.6 Tensor Products of Algebras . . . . . . . . . . 9.7 The Hom Functors . . . . . . . . . . . . . . . 9.8 Projective Modules . . . . . . . . . . . . . . . 9.9 The Grothendieck Construction: K0 . . . . . 9.10 Tensor Algebras and Their Relatives . . . . .

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341 342 346 351 365 377 383 386 392 398 406

10 Linear algebra 10.1 Traces . . . . . . . . . . . . . 10.2 Multilinear alternating forms 10.3 Properties of determinants . . 10.4 The characteristic polynomial 10.5 Eigenvalues and eigenvectors 10.6 The classification of matrices 10.7 Jordan canonical form . . . . 10.8 Generators for matrix groups 10.9 K1 . . . . . . . . . . . . . . .

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416 416 418 424 429 432 434 440 443 447

11 Galois Theory 11.1 Embeddings of Fields . . . . 11.2 Normal Extensions . . . . . . 11.3 Finite Fields . . . . . . . . . 11.4 Separable Extensions . . . . . 11.5 Galois Theory . . . . . . . . . 11.6 The Fundamental Theorem of 11.7 Cyclotomic Extensions . . . . 11.8 n-th Roots . . . . . . . . . . 11.9 Cyclic Extensions . . . . . . . 11.10Kummer Theory . . . . . . . 11.11Solvable Extensions . . . . . . 11.12The General Equation . . . . 11.13Normal Bases . . . . . . . . . 11.14Norms and Traces . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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450 451 454 458 460 465 474 476 480 485 488 493 498 500 502

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506 507 512 521 526 529 531 539

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12 Hereditary and Semisimple Rings 12.1 Maschke’s Theorem and Projectives 12.2 Semisimple Rings . . . . . . . . . . . 12.3 Jacobson Semisimplicity . . . . . . . 12.4 Homological Dimension . . . . . . . 12.5 Hereditary Rings . . . . . . . . . . . 12.6 Dedekind Domains . . . . . . . . . . 12.7 Integral Dependence . . . . . . . . . Bibliography

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550

Chapter 1

A Little Set Theory Here, we discuss the basic properties of functions, relations and cartesian products. The first section discusses injective and surjective functions and their behavior under composition. This is followed in Section 1.2 by a discussion of commutative diagrams and factorization. Here, duality is introduced, and injective and surjective maps are characterized in terms of their properties with respect to factorization. In the latter case, the role of the Axiom of Choice is discussed. Some introductory remarks about the axiom are given, to be continued in Chapter 6, where infinite products are introduced. In the next three sections, we discuss relations, including equivalence relations and partial orderings. Particular attention is given to equivalence relations. For an equivalence relation ∼, the set of equivalence classes, X/∼, is studied, along with the canonical map π : X → X/∼. In Section 1.5, we analyze the process by which an arbitrary relation may be used to generate an equivalence relation. We shall make use of this in constructing free products of groups. We then describe the basic properties of cartesian products, showing that they satisfy the universal mapping property for the categorical product. We close with some applications to the theory of functions and to the basic structure of the category of sets. With the exception of the discussion of the Axiom of Choice, we shall not delve into axiomatics here. We shall assume a familiarity with basic naive set theory, and shall dig deeper only occasionally, to elucidate particular points.

1.1

Properties of Functions

We use the notation f : X → Y to denote that f is a function from X to Y , and say that X is its domain, and Y its codomain. We shall assume an intuitive understanding of functions at this point in the discussion, deferring a more formal discussion to Section 1.7. We shall use the word “map” as a synonym for function. Definitions 1.1.1. Let X be a set. The identity map 1X : X → X is the function defined by 1X (x) = x for all x ∈ X. If Y is a subset of X, then the inclusion map i : Y → X is obtained by setting i(y) = y for all y ∈ Y . We give names for some important properties of functions: Definitions 1.1.2. A function f : X → Y is one-to-one, or injective, if f (x) = f (x ) for x = x . An injective function is called an injection. 1

CHAPTER 1. A LITTLE SET THEORY

2

A function f : X → Y is onto, or surjective, if for each y ∈ Y , there is at least one x ∈ X with f (x) = y. A surjective function is called a surjection.1 A function is bijective if it is both one-to-one and onto. Such a function may also be called a one-to-one correspondence, or a bijection. Composition of functions is useful in studying the relationships between sets. The reader should verify the following properties of composition. Lemma 1.1.3. Suppose given functions f : X → Y , g : Y → Z, and h : Z → W . Then 1. (h ◦ g) ◦ f = h ◦ (g ◦ f ). 2. f ◦ 1X = f . 3. 1Y ◦ f = f . Here are some more interesting facts to verify regarding composition. Lemma 1.1.4. Suppose given functions f : X → Y and g : Y → Z. Then the following relationships hold. 1. If f and g are injective, so is g ◦ f . 2. If f and g are surjective, so is g ◦ f . 3. If g ◦ f is injective, so is f . 4. If g ◦ f is surjective, so is g. Two sets which are in one-to-one correspondence may be thought of as being identical as far as set theory is concerned. One way of seeing this is in terms of inverse functions. Definition 1.1.5. Let f : X → Y . We say that a function g : Y → X is an inverse function for f if the composite f ◦ g is the identity map of Y and the composite g ◦ f is the identity map of X. Proposition 1.1.6. A function f : X → Y has an inverse if and only if it is bijective. If f does have an inverse, the inverse is unique, and is defined by setting f −1 (y) to be the unique element x ∈ X such that f (x) = y. Proof If f is bijective, then for each y ∈ Y , there is a unique x ∈ X such that f (x) = y. Thus, there is a function f −1 : Y → X defined by setting f −1 (y) to be equal to this unique x. The reader may now easily verify that f −1 is an inverse function for f . Conversely, suppose that f has an inverse function g. Note that identity maps are always bijections. Thus, f ◦g = 1Y is surjective, and hence f is surjective by Lemma 1.1.4. Also, g ◦ f = 1X is injective, so that f is injective as well. Finally, since f (g(y)) = y for all y, we see that g(y) is indeed the unique x ∈ X such that f (x) = y. Definition 1.1.7. Let f : X → Y be a function. Then the image of f , written im f , is the following subset of Y : im f = {y ∈ Y | y = f (x) for some x ∈ X}. 1 “One-to-one” and “onto” are the terms native to the development of mathematical language in English, while “injection” and “surjection” come from French mathematics.


3

Thus, the image of f is the set that classically would be called the range of f . Since the values f (x) of the function f : X → Y all lie in im f , f defines a function, which we shall call by the same name, f : X → im f . Notice that by the definition of the image, f : X → im f is surjective. Notice also that f factors as the composite f

X

/ im f ⊂ Y,

where the second map is the inclusion of im f in Y . Since inclusion maps are always injections, the next lemma is immediate. Lemma 1.1.8. Every function may be written as a composite g ◦ f , where g is injective and f is surjective. Exercises 1.1.9. 1. Give an example of a pair of functions f : X → Y and g : Y → X, where X and Y are finite sets, such that g ◦ f = 1X , but f is not surjective, and g is not injective. 2. Show that if X and Y are finite sets with the same number of elements, then any injection f : X → Y is a bijection. Show also that any surjection f : X → Y is a bijection. 3. Give an example of a function f : X → X which is injective but not bijective. 4. Give an example of a function f : X → X which is surjective but not bijective.

1.2

Factorizations of Functions

Commutative diagrams are a useful way to summarize the relationships between different sets and functions. Definition 1.2.1. Suppose given a diagram of sets and functions. X@ @@ @@ f @@

h

Y

/ Z. > } }} } }g }}

We say that the diagram commutes if h = g ◦ f . Similarly, given a diagram X

f

g

g

Z

/ Y

f

/ W

we say the diagram commutes if g ◦ f = f ◦ g. More generally, given any diagram of maps, we say the diagram commutes if, when we follow two different paths of arrows which start and end at the same points (following the directions of the arrows, of course) and compose the maps along these paths, then the compositions from the two paths agree. Commutative diagrams are frequently used to model the factorization of functions.


4

Definitions 1.2.2. Let f : X → Y be a function. We say that a function g : X → Z factors through f if there is a function g : Y → Z such that the following diagram commutes.

X@ @@ @@ f @@

g

Y

/Z ? ~ ~ ~ ~~ ~~ g

We say that g factors uniquely through f if there is exactly one function g : Y → Z that makes the diagram commute. The language of factorization can be confusing, as there is a totally different meaning to the words “factors through,” which is also used frequently: Definitions 1.2.3. Let f : X → Y be a function. We say that a function g : Z → Y factors through f if there is a function g : Z → X such that the following diagram commutes. X? ??f ? ?? / Y Z g g

We say that g factors uniquely through f if there is exactly one function g : Z → X that makes the diagram commute. Notice that despite the fact that there are two possible meanings to the statement that a function factors through f : X → Y , the meaning in any given context is almost always unique. Thus, if we say that a map g : X → Z factors through f : X → Y , then the only possible interpretation is that g = h ◦ f for some h : Y → Z. Similarly, if we say that a map g : Z → Y factors through f : X → Y , then the only possible interpretation is that g = f ◦ h for some h : Z → X. Indeed, the only way that ambiguity can enter is if we ask whether a function g : X → X factors through another function f : X → X. In this last case, one must be precise as to which type of factorization is being discussed. Factorization properties may be used to characterize injective and surjective maps. Proposition 1.2.4. Let f : X → Y . Then the following two statements are equivalent. 1. The map f is injective. 2. Any function g : Z → Y with the property that im g ⊂ im f factors uniquely through f. Proof Suppose that f is injective. Let g : Z → Y , and suppose that im g ⊂ im f . Then for each z ∈ Z there is an x ∈ X such that g(z) = f (x). Since f is injective, this x is unique, and hence there is a function g : Z → X defined by setting g (z) equal to the unique x ∈ X such that g(z) = f (x). But then f (g (z)) = g(z), and hence f ◦ g = g, so that g factors through f . The factorization is unique, since if f ◦ h = g, we have f (h(z)) = g(z), and hence f (h(z)) = f (g (z)) for all z ∈ Z. Since f is injective, this says that h(z) = g (z) for all z ∈ Z, and hence h = g .


5

Conversely, suppose that any function g : Z → Y with im g ⊂ im f factors uniquely through f . Let Z be the set with one element: Z = {z}. Then if f (x) = f (x ), define g : Z → Y by g(z) = f (x). Then if we define h, h : Z → X by h(z) = x and h (z) = x , respectively, we have f ◦ h = f ◦ h = g. By uniqueness of factorization, h = h , and hence x = x . Thus, f is injective. The uniqueness of the factorization is the key in the above result. There is an entirely different characterization of injective functions in terms of factorizations, where the factorization is not unique. Proposition 1.2.5. Let f : X → Y with X = ∅. Then the following two statements are equivalent. 1. The map f is injective. 2. Any function g : X → Z factors through f . Proof Suppose that f is injective and that g : X → Z. Choose any z ∈ Z,2 and define g : Y → Z by g(x) if y = f (x) g (y) = z if y ∈ im f . Since f is injective, g is a well defined function. By construction, g ◦ f = g, and hence g provides the desired factorization of g. (Note that if Z has more than one element, then g is not unique. For instance, we could choose a different element z ∈ Z.) Conversely, suppose that every map g : X → Z factors through f . We apply this by setting g equal to the identity map of X. Thus, there is a function g : Y → X such that g ◦ f = 1X . But then f is injective by Lemma 1.1.4. There is a notion in mathematics called dualization. The dual of a statement is the statement obtained by turning around all of the arrows in it. It is often interesting to see what happens when you dualize a statement, but one should bear in mind that the dual of a true statement is not always true. The dual of Proposition 1.2.4 is true, but its proof is best deferred to our discussion of equivalence relations below. The dual of Proposition 1.2.5 is also true. In fact, it turns out to be equivalent to the famous Axiom of Choice. Axiom of Choice (first form) Let f : X → Y be a surjective function. Then there is a function s : Y → X such that f ◦ s = 1Y . In this form, the Axiom of Choice seems to be obviously true. In order to define s, all we have to do is choose, for each y ∈ Y , an element x ∈ X with f (x) = y. Since such an x is known to exist for each y ∈ Y (because f is surjective), intuition says we should be able to make these choices. Sometimes the desired function s : Y → X may be constructed by techniques that depend only on the other axioms of set theory. One example of this is when f is a bijection. Here, s is the inverse function of f , and is constructed in the preceding section. Alternatively, if Y is finite, we may construct s by induction on the number of elements. Other examples exist where we may construct s by a particular formula. 2 Since

X = ∅ and g : X → Z, we must have Z = ∅ as well. (See Proposition 1.7.2.)


6

Nevertheless, intuition aside, for a generic surjection f : X → Y , we cannot construct the desired function s via the other axioms, and, if we wish to be able to construct s in all cases, we must simply accept the Axiom of Choice as part of the foundations of mathematics. Most mathematicians are happy to do this without question, but some people prefer models of set theory in which the Axiom of Choice does not hold. Thus, while we shall use it freely, we shall give some discussion of the ways in which it enters our arguments.3 We now give the dual of Proposition 1.2.5. Proposition 1.2.6. Let f : X → Y . Then the following two statements are equivalent. 1. The map f is surjective. 2. Any function g : Z → Y factors through f . Proof Suppose that f : X → Y is surjective. Then the Axiom of Choice provides a function s : Y → X with f ◦ s = 1Y . But then if g : Z → Y is any map, s ◦ g : Z → X gives the desired factorization of g through f . Conversely, suppose that any function g : Z → Y factors through f . Applying this for g = 1Y , we obtain a function s : Y → X with f ◦ s = 1Y . But then f is surjective by Lemma 1.1.4.

1.3

Relations

We shall make extensive use of both equivalence relations and order relations in our work. We first consider the general notion of a relation on a set X. Definitions 1.3.1. A relation R on a set X is a subset R ⊂ X × X of the cartesian product of X with itself. We shall use x R y as a shorthand for the statement that the pair (x, y) lies in the relation R. We shall often discuss a relation entirely in terms of the expressions x R y, and dispense with discussion of the associated subset of the product. This is an example of what’s known as “abuse of notation”: we can think and talk about a relation as a property x R y which holds for certain pairs (x, y) of elements of X. We can then, if we choose, define an associated subset R ⊂ X × X by letting R equal the set of all pairs such that x R y. For instance, there is a relation on the real numbers R, called ≤, which is defined in the usual way: we say that x ≤ y if y − x is a non-negative number. We can discuss all the properties of the relation ≤ without once referring to the subset of R × R which it defines. Not all relations on a set X have a whole lot of value to us. We shall enumerate some properties which could hold in a relation that might make it interesting for certain purposes. 3 The desired functions s : Y → X often exist for reasons that do not require the assumption of the choice axiom. In the study of the particular examples of groups, rings, fields, topological spaces, etc., that mathematicians tend to be most interested in, the theorems that we know and love tend to be true without assuming the axiom. However, the proofs of these theorems are often less appealing without the choice axiom. Moreover, if we assume the axiom, we can prove very general theorems, which then may be shown to hold without the axiom, for different reasons in different special cases, in the cases of greatest interest. Thus, the Axiom of Choice permits us to look at mathematics from a broader and more conceptual viewpoint than we could without it. And indeed, in the opinion of the author, a world without the Axiom of Choice is a poorer one, with less functions and narrower horizons.


7

Definitions 1.3.2. Let R be a relation on a set X. If x R x for each x ∈ X, then we say that R is reflexive. Let R be a relation on a set X. Suppose that y R x if and only if x R y. Then we say that R is symmetric. Alternatively, we say that R is antisymmetric if x R y and y R x implies that x = y. Let R be a relation on a set X. Then we say that R is transitive if x R y and y R z together imply that x R z. These definitions can be used to define some useful kinds of relations. Definitions 1.3.3. A relation R on X is an equivalence relation if it is reflexive, symmetric, and transitive. A relation R on X is a partial ordering if it is reflexive, antisymmetric, and transitive. A set X together with a partial ordering R on X is called a partially ordered set. A relation R on X is a total ordering if it is a partial ordering with the additional property that for all x, y ∈ X, either x R y or y R x. A set X together with a total ordering R on X is called a totally ordered set. Examples 1.3.4. 1. The usual ordering ≤ on R is easily seen to be a total ordering. 2. There is a partial ordering, ≤, on R × R obtained by setting (x, y) ≤ (z, w) if both x ≤ z and y ≤ w. Note that this example differs from the last one in that it is not a total ordering. 3. There is a different partial ordering, called the lexicographic ordering, on R × R obtained by setting (x, y) ≤ (z, w) if either x < z, or both x ≤ z and y ≤ w. (Thus, both (1, 5) and (2, 3) are ≤ (2, 4).) Note that in this case, we do get a total ordering. 4. There is an equivalence relation, ≡, on the integers, Z, obtained by setting m ≡ n if m − n is even. 5. For any set X, there is an equivalence relation, ∼, on X defined by setting x ∼ y for all x, y ∈ X (i.e., the subset of X × X in question is all of X × X). 6. For any set X, there is exactly one relation which is both a partial ordering and an equivalence relation: here x R y if and only if x = y. It is customary to use symbols such as ≤ and for a partial ordering, and to use symbols such as ∼, , and ≡ for an equivalence relation.

1.4

Equivalence Relations

We now look at equivalence relations, as defined in the preceding section, in greater depth. Definition 1.4.1. Let ∼ be an equivalence relation on X and let x ∈ X. The equivalence class of x in X (under the relation ∼), denoted E(x), is defined by E(x) = {y ∈ X | x ∼ y}. In words, E(x) is the set of all y ∈ X which are equivalent to x under the relation ∼.


8

Lemma 1.4.2. Let ∼ be an equivalence relation on X and let x, y ∈ X. Then y is in the equivalence class E(x) of x if and only if E(x) = E(y). Proof Suppose y ∈ E(x). This just says x ∼ y. But if z ∈ E(y), then y ∼ z, and hence x ∼ z by the transitivity of ∼. Therefore, E(y) ⊂ E(x). But equivalence relations are also symmetric. Thus, x ∼ y if and only if y ∼ x, and hence y ∈ E(x) if and only if x ∈ E(y). In particular, the preceding argument then shows that x ∼ y also implies that E(x) ⊂ E(y), and hence E(x) = E(y). Conversely, if E(x) = E(y), then reflexivity shows that y ∈ E(y) = E(x). In consequence, we see that equivalence classes partition the set X into distinct subsets: Corollary 1.4.3. Let ∼ be an equivalence relation on X and let x, y ∈ X. Then the equivalence classes of x and y have a common element if and only if they are equal. In symbols, E(x) ∩ E(y) = ∅

if and only if

E(x) = E(y).

Proof If z ∈ E(x) ∩ E(y), then both E(x) and E(y) are equal to E(z) by Lemma 1.4.2. The global picture is now as follows. Corollary 1.4.4. Let ∼ be an equivalence relation on X. Then every element of X belongs to exactly one equivalence class under ∼. Proof By Corollary 1.4.3, every element belongs to at most one equivalence class. But reflexivity shows that x ∈ E(x) for each x ∈ X, and the result follows. Given a set with an equivalence relation, the preceding observation allows us to define a new set from the equivalence classes. Definitions 1.4.5. Let ∼ be an equivalence relation on a set X. By the set of equivalence classes under ∼, written X/∼, we mean the set whose elements are the equivalence classes under ∼ in X. In particular, each element of X/∼ is a subset of X. We define π : X → X/∼ by π(x) = E(x). Thus, π takes each element of X to its equivalence class. We call π the canonical map from X to X/∼. Example 1.4.6. Let X = {1, 2, 3, 4} and let ∼ be the equivalence relation on X in which 1 ∼ 3, 2 ∼ 4, and the other related pairs are given by reflexivity and symmetry. Then there are two elements in X/∼: {1, 3} and {2, 4}. We have π(1) = π(3) = {1, 3} and π(2) = π(4) = {2, 4}. We shall show that the canonical map π characterizes the equivalence relation completely. Lemma 1.4.7. Let ∼ be an equivalence relation on X and let π : X → X/∼ be the canonical map. Then π is surjective, and satisfies the property that π(x) = π(y) if and only if x ∼ y. Proof Surjectivity of π follows from the fact that every equivalence class has the form E(x) for some x ∈ X, and E(x) = π(x). Now π(x) = π(y) if and only if E(x) = E(y). By Lemma 1.4.2, the latter condition is equivalent to the statement that y ∈ E(x), which is defined to mean that x ∼ y.


9

Intuitively, this says that X/∼ is the set obtained from X by identifying two elements if they lie in the same equivalence class. The canonical map has an important universal property.4 Proposition 1.4.8. Let ∼ be an equivalence relation on X. Then a function g : X → Y factors through the canonical map π : X → X/∼ if and only if x ∼ x implies that g(x) = g(x ) for all x, x ∈ X. Moreover, if g factors through π, then the factorization is unique. Proof Suppose that g = g ◦ π for some g : X/∼ → Y . If x ∼ x ∈ X, then π(x) = π(x ), and hence g (π(x)) = g (π(x )). Since g = g ◦ π, this gives g(x) = g(x ), as desired. Also, since π is surjective, every element of X/∼ has the form π(x) for some x ∈ X. Since g (π(x)) = g(x), the effect of g on the elements of X/∼ is determined by g. In other words, the factorization g , if it exists, is unique. Conversely, suppose given a function g : X → Y with the property that x ∼ x implies that g(x) = g(x ). Now define g : X/∼ → Y by setting g (π(x)) = g(x) for all x ∈ X. Since π(x) = π(x ) if and only if x ∼ x , this is well defined by the property assumed for the function g. By the construction of g , g ◦ π = g, and hence g gives the desired factorization of g through π. We have seen that an equivalence relation determines a surjective function: the canonical map. We shall give a converse to this, and show that specifying an equivalence relation on X is essentially equivalent to specifying a surjective function out of X. Definition 1.4.9. Let f : X → Y be any function. By the equivalence relation determined by f , we mean the relation ∼ on X defined by setting x ∼ x if f (x) = f (x ). The reader may easily verify that the equivalence relation determined by f is indeed an equivalence relation. We now show that if ∼ is the equivalence relation determined by f : X → Y , then there is an important relationship between the canonical map π : X → X/∼ and f . Proposition 1.4.10. Let f : X → Y be any function and let ∼ be the equivalence relation determined by f . Then f factors uniquely through the canonical map π : X → X/∼, via a commutative diagram f / Y. XD DD z= z DD zz D zzf π DD z z ! X/∼

Moreover, the map f : X/∼ → Y gives a bijection from X/∼ to the image of f . In particular, if f is surjective, then f gives a bijection from X/∼ to Y . Proof Recall that the definition of ∼ was that x ∼ x if and only if f (x) = f (x ). Thus, the universal property of the canonical map (i.e., Proposition 1.4.8) provides the unique factorization of f through π: just set f (π(x)) = f (x) for all x ∈ X. Since every element of X/∼ has the form π(x), the image of f is the same as the image of f . Thus, it suffices to show that f is injective. Suppose that f (π(x)) = f (π(x )). Then f (x) = f (x ), so by the definition of ∼, x ∼ x . But then π(x) = π(x ), and hence f is injective as claimed. 4 The expression “universal property” has a technical meaning which we shall explore in greater depth in Chapter 6 on category theory.


10

In particular, if f : X → Y is surjective, and if ∼ is the equivalence relation determined by f , then we may identify Y with X/∼ and identify f with π. In particular, the dual of Proposition 1.2.4 is now almost immediate from the universal property of the canonical map π. Proposition 1.4.11. Let f : X → Y . Then the following two statements are equivalent. 1. The map f is surjective. 2. Let g : X → Z be such that f (x) = f (x ) implies that g(x) = g(x ) for all x, x ∈ X. Then g factors uniquely through f . Proof Suppose that f is surjective, and let ∼ be the equivalence relation determined by f . Then we may identify f with the canonical map π : X → X/∼. But under this identification, the second statement is precisely the universal property of π. Conversely, suppose that the second condition holds. Then the hypothesis says that there is a unique map g : Y → Y such that the following diagram commutes. X@ @@ @@ f @@

f

Y

/Y ~? ~ ~ ~~ ~~ g

Of course, g = 1Y will do, but the key is the uniqueness: if f were not surjective, we could alter g on those points of Y which are not in im f without changing the fact that the diagram commutes, contradicting the uniqueness of the factorization. Exercises 1.4.12. † 1. Let n be a positive integer. Define a relation ≡ on the integers, Z, by setting i ≡ j if j − i is a multiple of n. Show that ≡ is an equivalence relation. (We call it congruence modulo n.) 2. Let ∼ be any equivalence relation. Show that the equivalence relation determined by the canonical map π : X → X/∼ is precisely ∼.

1.5

Generating an Equivalence Relation

There are many situations in mathematics where we are given a relation R which is not an equivalence relation, and asked to form the “smallest” equivalence relation, ∼, with the property that x R y implies that x ∼ y. Here, “smallest” means that if ≡ is another equivalence relation with the property that x R y implies that x ≡ y, then x ∼ y also implies that x ≡ y. In other words, the equivalence classes of ∼ are subsets of the equivalence classes of ≡. This smallest equivalence relation is called the equivalence relation generated by R, and its canonical map π : X → X/∼ will be seen to have a useful universal property with respect to R. Of course, it is not immediately obvious that such an equivalence relation exists, so we shall have to construct it. Definition 1.5.1. Let R be any relation on X. The equivalence relation, ∼, generated by R is the relation obtained by setting x ∼ x if there is a sequence x1 , . . . , xn ∈ X such that x1 = x, xn = x , and for each i with 1 ≤ i < n, either xi = xi+1 , or xi R xi+1 , or xi+1 R xi .


11

The intuition is that we take the relation R to be a collection of “basic equivalences,” and declare that x ∼ x if they may be connected by a chain of basic equivalences, some in one direction and some in the other. Example 1.5.2. Let n be a positive integer, and define a relation R on the integers, Z, by setting i R j if j = i + n. Note then that if ∼ is the equivalence relation generated by R and if i ∼ j, then j = i + kn for some k ∈ Z. But the converse is also clear: in fact, i ∼ j if and only if j = i + kn for k ∈ Z. Thus, ∼ is the equivalence relation known as “congruence modulo n,” which was given in Problem 1 of Exercises 1.4.12. We now verify that the equivalence relation generated by R has the desired properties. Proposition 1.5.3. Let R be any relation and let ∼ be the equivalence relation generated by R. Then ∼ is, in fact, an equivalence relation, and has the property that x R x implies that x ∼ x . Moreover, if ≡ is another equivalence relation with the property that x R x implies that x ≡ x , then x ∼ x also implies that x ≡ x . Proof If x R x , then the sequence x1 = x, x2 = x shows that x ∼ x , so that x R x implies that x ∼ x , as desired. Similarly, we see that ∼ is reflexive. Symmetry of ∼ may be obtained by reversing the sequence x1 , . . . , xn that’s used to show that x ∼ x , while transitivity is obtained by concatenating two such sequences. Now let ≡ be another equivalence relation with the property that x R x implies that x ≡ x , and suppose that x ∼ x . Thus, we are given a sequence x1 , . . . , xn ∈ X such that x1 = x, x2 = x , and for each i with 1 ≤ i < n, either xi = xi+1 , or xi R xi+1 , or xi+1 R xi . But in any of the three situations, this says that xi ≡ xi+1 , so that x = x1 ≡ x2 ≡ · · · ≡ xn = x . Since ≡ is transitive, this gives x ≡ x , as desired. The fact that ∼ is the smallest equivalence relation containing R translates almost immediately into a statement about the canonical map π : X → X/∼. Corollary 1.5.4. Let R be a relation on X and let ∼ be the equivalence relation generated by R. Then a map f : X → Y factors through the canonical map π : X → X/∼ if and only if x R x implies that f (x) = f (x ) for all x, x ∈ X. The factorization, if it exists, is unique. Proof Suppose that f : X → Y has the property that x R x implies that f (x) = f (x ) for all x, x ∈ X. Let be the equivalence relation induced by f (i.e., x x if and only if f (x) = f (x )). Then x R x implies that x x . Thus, by Proposition 1.5.3, x ∼ x implies that x x . By the definition of , this just says that x ∼ x implies that f (x) = f (x ). But now the universal property of the canonical map π : X → X/∼ (Proposition 1.4.8) shows that f factors uniquely through π. For the converse, if f factors through π, then Proposition 1.4.8 shows that x ∼ x implies that f (x) = f (x ). But x R x implies that x ∼ x , so the result follows.

1.6

Cartesian Products

The next definition should be familiar. Definition 1.6.1. Let X and Y be sets. The cartesian product X × Y is the set of all ordered pairs (x, y) with x ∈ X and y ∈ Y .


12

When describing a function f : X × Y → Z, it is cumbersome to use multiple parentheses, as in f ((x, y)). Instead, we shall write f (x, y) for the image of the ordered pair (x, y) under f . Definition 1.6.2. There are projection maps π1 : X × Y → X and π2 : X × Y → Y defined by π1 (x, y) = x, and π2 (x, y) = y, respectively. An understanding of products with the null set is useful for the theory of functions. Lemma 1.6.3. Let X be any set and let ∅ be the null set. Then X × ∅ = ∅ × X = ∅. Proof If one of the factors of a product is empty, then there are no ordered pairs: one cannot choose an element from the empty factor. We shall make extensive use of products in studying groups and modules. The next proposition is useful for studying finite groups. Here, if X is finite, we write |X| for the number of elements in X. Proposition 1.6.4. Let X and Y be finite sets. Then X × Y is finite, and |X × Y | = |X| · |Y |, i.e., the number of elements in X × Y is the product of the number of elements in X and the number of elements in Y . Proof If either X or Y is empty, the result follows from Lemma 1.6.3. Thus, we assume that both X and Y are nonempty. Consider the projection map π1 : X ×Y → X. For each x0 ∈ X, write π1−1 (x0 ) for the set of ordered pairs that are mapped onto x0 by π1 . Thus, π1−1 (x0 ) = {(x0 , y) | y ∈ Y }.5 For each x ∈ X, there is an obvious bijection between π1−1 (x) and Y , so that each π1−1 (x) has |Y | elements. Now π1−1 (x) ∩ π1−1 (x ) = ∅ for x = x , and X × Y = −1 x∈X π1 (x). Thus, X × Y is the union of |X| disjoint subsets, each of which has |Y | elements, and hence |X × Y | = |X| · |Y | as claimed. Products have an important universal property. Let f : Z → X × Y be a map, and write f (z) = (f1 (z), f2 (z)) for z ∈ Z. In the usual parlance, f1 : Z → X and f2 : Z → Y are the component functions of f . Note that if π1 : X × Y → X and π2 : X × Y → Y are the projection maps, then fi = πi ◦ f , for i = 1, 2. Since a pair (x, y) ∈ X × Y is determined by its coordinates, we see that a function f : Z → X×Y is determined by its component functions. In other words, if g : Z → X×Y is a map, and if g1 : Z → X and g2 : Z → Y are its component functions, then f = g if and only if f1 = g1 and f2 = g2 . Finally, if we are given a pair of functions f1 : Z → X and f2 : Z → Y , then we can define a function f : Z → X × Y by setting f (z) = (f1 (z), f2 (z)) for z ∈ Z. In other words, we have shown the following. Proposition 1.6.5. For each pair of functions f1 : Z → X and f2 : Z → Y , there is a unique function f : Z → X × Y such that fi = πi ◦ f for i = 1, 2. Here, π1 : X × Y → X and π2 : X × Y → Y are the projection maps. that the subsets π1−1 (x), as x ranges over the elements of X, are precisely the equivalence classes for the equivalence relation determined by π1 . 5 Note


1.7

13

Formalities about Functions

Intuitively, a function f : X → Y is a correspondence which assigns to each element x ∈ X an element f (x) ∈ Y . The formal definition of function is generally given in terms of its graph. Definitions 1.7.1. A subset f ⊂ X × Y is the graph of a function from X to Y if for each x ∈ X there is exactly one ordered pair in f whose first coordinate is x. If f ⊂ X × Y is the graph of a function and if (x, y) ∈ f , we write y = f (x). Thus, f = {(x, f (x)) | x ∈ X}. By the function determined by f we mean the correspondence f : X → Y that assigns to each x ∈ X the element f (x) ∈ Y . Two functions from X to Y are equal if and only if their graphs are equal. Thus, every function is determined uniquely by its graph. We obtain some immediate consequences. Proposition 1.7.2. Let Y be any set. Then there is one and only one function from the null set, ∅, to Y . On the other hand, if X is a set with X = ∅, then there are no functions from X to ∅. Proof For a set Y , we have ∅ × Y = ∅. Thus, ∅ × Y has exactly one subset, and hence there is only one candidate for a function. The subset in question is the graph of a function, because, since the null set has no elements, the condition that must be satisfied is automatically true. On the other hand, if X = ∅, then the unique subset of X × ∅ is not a function: for any given x ∈ X, there is no ordered pair in this subset whose first coordinate is x. The next proposition is important in the foundations of category theory. Proposition 1.7.3. Let X and Y be sets. Then the collection of all functions from X to Y forms a set. Proof The collection of all functions from X to Y is in one-to-one correspondence with the collection of their graphs. But the latter is a subcollection of the collection of all subsets of X × Y . The fact that the collection of all subsets of a set, W , is a set (called the power set of W ) is one of the standard axioms of set theory. An important special case of this is when X is finite. For simplicity of discussion, let us establish a notation. Notation 1.7.4. Let X and Y be sets. We write F (X, Y ) for the set of functions from X to Y . Proposition 1.7.5. Let X1 and X2 be subsets of X such that X = X1 ∪ X2 and X1 ∩ X2 = ∅. Write ι1 : X1 ⊂ X and ι2 : X2 ⊂ X for the inclusions of these two subsets in X. Let Y be any set, and define α : F (X, Y ) → F (X1 , Y ) × F (X2 , Y ) by α(f ) = (f ◦ ι1 , f ◦ ι2 ). Then α is a bijection.6 6 In the language of category theory, this says that the disjoint union of sets is the coproduct in the category of sets.


14

Proof Suppose that α(f ) = α(g). Then f ◦ ι1 = g ◦ ι1 , and hence f and g have the same effect on the elements of X1 . Similarly, f ◦ ι2 = g ◦ ι2 , and hence f and g have the same effect on the elements of X2 . Since a function is determined by its effect on the elements of its domain, f = g, and hence α is injective. Now suppose given an element (f1 , f2 ) ∈ F (X1 , Y ) × F (X2 , Y ). Thus, f1 : X1 → Y and f2 : X2 → Y are functions. Define f : X → Y by f1 (x) if x ∈ X1 f (x) = f2 (x) if x ∈ X2 . Then f ◦ ιi = fi for i = 1, 2, and hence α(f ) = (f1 , f2 ). Thus, α is surjective. As a result, we can count the number of distinct functions between two finite sets. Corollary 1.7.6. Let X and Y be finite. Then the set of functions from X to Y is finite. Specifically, if X has n elements and Y has |Y | elements, then the set F (X, Y ) of n functions from X to Y has |Y | elements. (Here, we take 00 to be equal to 1.) Proof If Y = ∅, the result follows from Proposition 1.7.2. Thus, we assume that Y = ∅, and argue by induction on n. If n = 0, then the result is given by Proposition 1.7.2. For n = 1, say X = {x}, there is a one-to-one correspondence ν : F (X, Y ) → Y given by ν(f ) = f (x). Thus, we assume n > 1. Write X = X ∪ {x}, with x ∈ X . Then we have a bijection α : F (X, Y ) → n−1 elements, while F ({x}, Y ) has F (X , Y ) × F ({x}, Y ). By induction, F (X , Y ) has |Y | |Y | elements by the case n = 1, above. Thus, the result follows from Proposition 1.6.4.

Chapter 2

Groups: Basic Definitions and Examples In the first four sections, we define groups, monoids, and subgroups, and develop the most basic properties of the integers and the cyclic groups Zn . We then discuss homomorphisms and classification. The homomorphisms out of cyclic groups are determined, and the abstract cyclic groups classified. Classification is emphasized strongly in this volume, and we formalize the notion as soon as possible. We then discuss isomorphism classes and show that there are only finitely many isomorphism classes of groups of any given finite order. We then give some examples of groups, in particular the dihedral and quaternionic groups, which we shall use throughout the book as illustrations of various concepts. We close with a section on cartesian products, giving a way to create new groups out of old ones.

2.1

Groups and Monoids

A binary operation on a set X is a function μ : X × X → X. That is, μ assigns to each ordered pair (x, y) of elements of X an element μ(x, y) ∈ X. We think of these as multiplications, and generally write x · y, or just xy (or sometimes x + y), instead of μ(x, y). Note that the order of x and y is important. Generally x · y and y · x are different elements of X. Binary operations without additional properties, while they do arise in various situations, are too general to study in any depth. The condition we shall find most valuable is the associative property: our binary operation is associative if (x · y) · z = x · (y · z) for every x, y, and z in X. Another important property is the existence of an identity element. An identity element for the operation is an element e ∈ X such that e·x=x=x·e for every x in X. The next result is basic. 15

CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES

16

Lemma 2.1.1. A binary operation can have at most one identity element. Proof Suppose that e and e are both identity elements for the binary operation ·. Since e is an identity element, we have e · e = e . But the fact that e is also an identity element shows that e · e = e. Thus, e = e . Definition 2.1.2. A monoid is a set with an associative binary operation that has an identity element. Examples 2.1.3. 1. Let Z+ be the set of positive integers: Z+ = {1, 2, . . . }. Then ordinary multiplication gives Z+ the structure of a monoid, with identity element 1. 2. Let N ⊂ Z be the set of non-negative integers: N = {0, 1, 2, . . . }. Then ordinary addition gives N the structure of a monoid, with identity element 0. Monoids occur frequently in mathematics. They are sufficiently general that their structure theory is extremely difficult. Definition 2.1.4. Let X be a monoid. We say that an element x ∈ X is invertible if there is an element y ∈ X with x · y = e = y · x, where e is the identity. We call y the inverse for x and write y = x−1 . The next result shows that the above definition uniquely defines y. Lemma 2.1.5. Let X be a monoid. Then any element x ∈ X which is invertible has a unique inverse. Proof Suppose that y and z are both inverses for x. Then x · y = e = y · x, and x · z = e = z · x. Thus, y = y · e = y · (x · z) = (y · x) · z = e · z = z.

Note that the statement that x · y = e = y · x, used in defining inverses, is symmetric in x and y, establishing the following lemma. Lemma 2.1.6. Let X be a monoid, and let x ∈ X be invertible, with inverse y. Then y is also invertible, and y −1 = x. Finally, we come to the object of our study. Definition 2.1.7. A group is a monoid in which every element is invertible. Notice that the monoid, Z+ , of positive integers under multiplication is not a group, as, in fact, no element other than 1 is invertible: if n ∈ Z+ is not equal to 1, then n > 1. Thus, for any m ∈ Z+ , n · m > 1 · m, and hence n · m > 1. The monoid, N, of the non-negative integers under addition also fails to be a group, as the sum of two positive integers is always positive. We shall generally write G for a group, rather than X. We do have numerous examples of groups which come from numbers.


17

Examples 2.1.8. The simplest example of a group is the group with one element, e. There is only one binary operation possible: e · e = e. We call this group the trivial group and denote the group itself by e. Let Z be the integers: Z = {0, ±1, ±2, . . . }. Then addition of numbers gives Z the structure of a group: the identity element is 0, and the inverse of n ∈ Z is −n. Similarly, the rational numbers, Q, and the real numbers, R, form groups under addition. We also know examples of groups obtained from multiplication of numbers. Write Q× and R× for the nonzero numbers in Q and R, respectively. Since the product of two nonzero real numbers is nonzero, multiplication provides Q× and R× with an associative operation with identity element 1. And if x ∈ R× , the reciprocal, 1/x, provides an inverse to x in R× . Moreover, if x = m/n ∈ Q× , with m, n ∈ Z, then 1/x = n/m ∈ Q× , and hence x has an inverse in Q× . × Now write Q× + and R+ for the positive rational numbers and the positive real numbers, respectively. Since the product of two positive numbers is positive and the reciprocal × of a positive number is positive, the above arguments show that Q× + and R+ are groups under multiplication. The examples above are more than just groups. They also satisfy the commutative law : x · y = y · x for all x and y in G. In fact, the ubiquity of the commutative law in our background can exert a pernicious influence on our understanding of groups, as most groups don’t satisfy it. The ones that do merit a special name: Definitions 2.1.9. A group or monoid is abelian if it satisfies the commutative law. A group or monoid in which the commutative law is known to fail is said to be nonabelian. We will often use + for the operation in an abelian group. In fact, we shall restrict the use of + to these groups. Notice that the groups in the examples above all have infinitely many elements. Our main focus here will be on finite groups: those with finitely many elements. Definition 2.1.10. If G is a finite group, then the order of G, written |G|, is the number of elements in G. If G is infinite, we say |G| = ∞. Let X be a monoid. Write Inv(X) ⊂ X for the collection of invertible elements in X. Lemma 2.1.11. Inv(X) is a group under the multiplication inherited from that of X. Proof Let x and y be elements of Inv(X). Thus, x has an inverse, x−1 , and y has an inverse y −1 . Now the element y −1 x−1 may easily be seen to be an inverse for xy, and hence xy ∈ Inv(X), also. Since the operation on Inv(X) is inherited from that on X, it must be associative. Since e−1 = e, e is an element of Inv(X), and acts as an identity element there. And for x ∈ Inv(X), x−1 is invertible, with inverse x, and hence x−1 ∈ Inv(X), and is the inverse of x in Inv(X). In particular, every element of Inv(X) is invertible, so that Inv(X) is a group. A primary source for nonabelian groups is from matrices. We assume familiarity with the basic properties of matrix multiplication shown in a beginning course in linear algebra. We shall state the properties we are to use, and leave them as exercises to the reader.


18

Definitions 2.1.12. We write Mn (R) for the set of n×n matrices over the real numbers, R. We shall make use of the binary operation of matrix multiplication, which is defined as follows. If ⎛ ⎞ ⎛ ⎞ a11 . . . a1n b11 . . . b1n ⎜ ⎟ ⎜ ⎟ .. .. A=⎝ and B=⎝ ⎠ ⎠, . . an1

...

ann

bn1

...

bnn

then the product AB is the matrix whose ij-th coordinate is cij =

n

aik bkj = ai1 b1j + · · · + ain bnj .

k=1

We write I = In for the n × n identity matrix: the matrix whose diagonal entries are all equal to 1 and whose off-diagonal entries are all equal to 0. 1 0 0 Thus, I2 = 10 01 , I3 = 0 1 0 , etc. 001 The properties we need here regarding matrix multiplication are few. We shall discuss matrices in greater detail in Section 7.10 and Chapter 10. Lemma 2.1.13. Multiplication of n × n matrices gives an associative binary operation on Mn (R). Moreover, In is an identity element for this operation. Thus, Mn (R) is a monoid under matrix multiplication. The invertible elements in this monoid structure on Mn (R) are precisely the invertible matrices in the usual sense. As such, they merit a name. Definition 2.1.14. We write Gln (R) for Inv(Mn (R)), the group of invertible elements of Mn (R). We call it the n-th general linear group of R. Later in this chapter, we shall construct two infinite families of finite nonabelian groups, the dihedral groups and the quaternionic groups, as explicit subgroups of Gln (R) for n = 2 and 4, respectively. We shall show in Chapter 10 that every finite group is a subgroup of Gln (R) for some value of n. It is sometimes useful to study partial inverses in a monoid. Definitions 2.1.15. Let X be a monoid with identity element e. If x, y ∈ X with xy = e, then we say that x is a left inverse for y and that y is a right inverse for x. Exercises 2.1.16. 1. Let M be the set of all nonzero integers. Then M is a monoid under multiplication. What is Inv(M )? † 2. Let G be a group and let x, y ∈ G. Show that x and y commute if and only if x2 y 2 = (xy)2 . † 3. Let G be a group such that x2 = e for all x ∈ G. Show that G is abelian. −1 4. Let G be a group and let x1 , . . . , xk ∈ G. Show that (x1 . . . xk )−1 = x−1 k . . . x1 .

5. Let G be a group and let x, y ∈ G. Show that x and y commute if and only if (xy)−1 = x−1 y −1 . 6. Let G be a group and let x, y ∈ G. Suppose there are three consecutive integers n such that xn y n = (xy)n . Show that x and y commute.


19

7. Let G be an abelian group and let x, y ∈ G. Show that xn y n = (xy)n for all n ∈ Z. 8. Verify that multiplication of n × n matrices is associative. 9. Show that Gl1 (R) is isomorphic to R× , the group of non-zero real numbers under multiplication. 10. Show that Gln (R) is nonabelian for n ≥ 2. 11. Let

G=

a b 0 c

b ∈ R, a, c ∈ Q and ac = 0 .

Show that G is a group under matrix multiplication. Is G abelian? 12. Let X be a monoid. Suppose that x ∈ X has a left inverse, y, and a right inverse, z. Show that y = z, and that x is invertible with inverse y. † 13. Let X be a monoid with the property that every element of X has a left inverse. Show that X is a group. 14. Let X be a finite monoid with the left cancellation property: if xy = xz, then y = z. Show that X is a group. 15. Let X be a finite set with an associative binary operation. Suppose this operation has both the left and the right cancellation properties. Show that X is a group.

2.2

Subgroups

One can tell quite a bit about about a group by knowing its subgroups. Definitions 2.2.1. A subset S of a group G is said to be closed under multiplication if for x and y in S, the product xy is also in S. A subset H of G is said to be a subgroup if it is nonempty and closed under multiplication, and if for each x ∈ H, the inverse element x−1 is also in H. Examples 2.2.2. 1. The groups Z and Q are subgroups of R. 2. The inclusions below are all inclusions of subgroups. Q× + ∩ R× +

⊂ Q× ∩ ⊂ R×

3. Z+ ⊂ R× + is closed under multiplication, but is not a subgroup, because the inverses in R× + of the non-identity elements of Z+ do not lie in Z+ . 4. Any group G is a subgroup of itself. 5. For any group G, consider the subset {e} ⊂ G, consisting of the identity element alone. Because e · e = e and e−1 = e, {e} is a subgroup of G, called the trivial subgroup, or identity subgroup. By abuse of notation (i.e., for convenience), we shall generally write e in place of {e} for this subgroup. (When the identity element is called 1 or 0, we shall write 1 or 0 for the trivial subgroup as well.)


20

The next lemma is immediate from the definitions. Lemma 2.2.3. Let H ⊂ G be a subgroup. Then H is a group under the operation inherited from that of G. We also have what is sometimes called Gertrude Stein’s Theorem.1 We leave the proof as an exercise to the reader. Lemma 2.2.4. A subgroup of a subgroup is a subgroup. (I.e., if H is a subgroup of G and K is a subgroup of H, then K is a subgroup of G.) Identifying the subgroups of a given group is a deep and interesting activity, and the result says a lot about the group. As we shall see presently, the easiest subgroups to find are the smallest ones. Definition 2.2.5. Let G be a group and let g ∈ G. For any integer n, we define the n-th power of g, g n , as follows. For n = 0, 1, g n is given by e and g, respectively. For n > 1, we define g n by induction on n: g n = g n−1 · g. For negative exponents, we require that g −1 be the inverse of g, and if n is positive, then g −n = (g −1 )n . If G is abelian and written in additive notation, we write n · g, or ng for the n-th power of G in the above sense. The power laws for a group are fundamental: Lemma 2.2.6. Let G be a group and let g ∈ G. Then for any integers m and n, gm · gn m n

(g )

= g m+n , = g

mn

and

.

If the group operation in G is written additively, this translates to the following: mg + ng = (m + n)g, n(mg) = (nm)g.

and

Proof We use the multiplicative notation. Consider the first statement first. For n = 0, it reduces to g m · e = g m which is certainly true. For n = 1, the statement depends on the value of m: if m > 0 it just gives the definition of g m+1 , and hence is true. For m = 0, it says e · g = g. For m < 0, we have m = −k, say, and g m = (g −1 )k = (g −1 )k−1 · g −1 . Thus, g m · g = (g −1 )k−1 (g −1 · g) = (g −1 )k−1 = g −(k−1) = g m+1 , because −k = m. For n > 1, we argue by induction on n. We have g n = g n−1 · g, so that g m · g n = m (g · g n−1 ) · g. By induction, this is g m+n−1 · g, which is g m+n by the case n = 1 above. The case of n < 0 follows from that of n > 0 by exchanging the roles of g and g −1 . We also use induction for the second statement. As with the first, we may assume n ≥ 0. For n = 0, 1 it says nothing more than e = e, or g m = g m , respectively. For n > 1, we have (g m )n = (g m )n−1 · g m . By induction, this is g mn−m · g m , which is g mn by an application of the first statement. 1 This

usage was coined by Arunas Liulevicius.


21

Notice that there’s a potential conflict between the additive notation for powers and the notation for ordinary multiplication if G = Z. This is resolved by the following lemma. Lemma 2.2.7. Let m and n be integers. Then mn (as defined by ordinary multiplication of integers) is the m-th power of n in the group Z. Proof First note the definition of multiplication in Z: if m and n are both positive, then mn is the sum of m copies of n and (−1)k m · (−1)l n = (−1)k+l mn. The sum of m copies of n is by definition the m-th power of n. The remaining cases follow from Lemma 2.2.6. Lemma 2.2.6 gives us a convenient source of subgroups. Definition 2.2.8. Let G be a group and let g ∈ G. Define g ⊂ G to be the set of all integer powers of g: g = {g n | n ∈ Z}. Our next result will show that g is a subgroup of G. We call it the cyclic subgroup generated by g. Note that if the group operation in G is written additively, then g is defined by g = {ng | n ∈ Z}. In particular, in the case G = Z, for n ∈ Z, n = {nk | k ∈ Z} is the set of all integer multiples of n. In this case, there are a couple of alternate notations that come out of ring theory: for n ∈ Z we may write either (n) or nZ for n. Lemma 2.2.9. Let G be a group and let g ∈ G. Then g is a subgroup of G. Moreover, g has the property that it is the smallest subgroup of G which contains g. In other words, if H ⊂ G is a subgroup, and if g ∈ H, then g ⊂ H. Proof Since g m · g n = g m+n , g is closed under multiplication in G. Also, the inverse of g m is g −m ∈ g, so that g is a subgroup of G. If g ∈ H ⊂ G, with H a subgroup of G, then g n must lie in H for all n ∈ Z by the closure properties of a subgroup. Thus, g ⊂ H. Indeed, viewing g as an element of H, g is the subgroup of H generated by g. We shall see in the next section that the cyclic subgroups n ⊂ Z give all of the subgroups of Z. Cyclic subgroups have a very special kind of group structure. They are cyclic groups: Definition 2.2.10. A group G is cyclic if there is an element g ∈ G such that G = g. Cyclic groups are simplest kind of group there is. For this reason, we can answer some very complicated questions about them, questions we could not answer for more complicated groups. Nevertheless, there are many deep and interesting questions about cyclic groups (some of them connected with algebraic number theory) which have yet to be answered. We can also talk about the subgroup generated by a family of elements.


22

Definitions 2.2.11. Let G be a group. For any subset S ⊂ G, the subgroup, S, generated by S is the set of all products (of arbitrary length) of powers of elements of S: S = {g1n1 . . . gknk | k ≥ 1, gi ∈ S

and ni ∈ Z for 1 ≤ i ≤ k}.

Warning: In the products g1n1 . . . gknk , we do not, and cannot, assume that the g1 , . . . , gk are distinct. Indeed, if g, h ∈ S do not commute, then, for instance, g 2 h2 and ghgh are distinct elements of S, by Problem 2 of Exercises 2.1.16. If S is finite, say S = {x1 , . . . , xn }, then by abuse of notation, we shall write x1 , . . . , xn for S. If S = G, we say that G is generated by S. If G is generated by a finite set, we say that G is finitely generated. The following lemma is straightforward. Lemma 2.2.12. S is the smallest subgroup of G which contains S. We now give a method for detecting finite subgroups. Proposition 2.2.13. Let G be a group and let H be a nonempty finite subset of G which is closed under multiplication. Then H is a subgroup of G. Proof We must show H is closed under inverses. Let x ∈ H and let S = {xk | 1 ≤ k ∈ Z}. Then S is contained in H, and hence S must be finite. Thus, the elements xk with k ≥ 1 cannot all be distinct. We must have xk = xl for some k = l. Say l > k. But then multiplying both sides by x−k , we get e = xl−k = xl−k−1 · x. Thus, xl−k−1 = x−1 . But l − k − 1 ≥ 0 since l > k. If l − k − 1 = 0, we have x = e = x−1 ∈ H. Otherwise, xl−k−1 ∈ S ⊂ H, and hence H is closed under inverses. We can give one useful operation on subgroups at this point. The proof is left to the reader. Lemma 2.2.14. Let H and K be subgroups of G. Then H ∩ K is also a subgroup of G. Warning: The union of two subgroups is almost never a subgroup. The collection of all subgroups of a group G is easily seen to form a partially ordered set as defined in Definition 1.3.3. Here the order relation is given by inclusion as subsets of G; explicitly, H is less than or equal to K if H ⊂ K. Definition 2.2.15. Let G be a group. The partially ordered set of subgroups of G is known as the lattice of subgroups of G. It is often useful to diagram the lattice of subgroups of a group G. Right now we have neither enough examples of groups nor enough theory about subgroups to be able to give a good example and verify its accuracy. However, we shall give one anyway, for a group we shall encounter in this chapter. The reader should reconsider the following example as the theory develops.


23

Example 2.2.16. We write D6 for the dihedral group of order 6. As will become clear later, the following is the lattice of subgroups of D6 . D6 ?? oo o o ?? oo o ?? o o o ? oo o o o b O a ab ab2 OOO ? ? OOO OOO ??? OOO ?? OOO? e Here, b has order 3, while a, ab, and ab2 have order 2. The upward-slanted lines represent the inclusions of subgroups. Because the elements of monoids don’t necessarily have inverses, the definition of a submonoid will have to be different from that of a subgroup. Definition 2.2.17. A submonoid of a monoid M is a subset which is closed under multiplication and contains the identity element. Exercises 2.2.18. 1. Show that in the real numbers R, the cyclic subgroup generated by 1 is the integers. In particular, Z is cyclic. 2. In Z, show that n = Z if and only if n = ±1. 3. Consider the group, Q× + , of positive rational numbers under multiplication. What are the elements of 2 ⊂ Q× +? 4. Consider the group Q× of nonzero rational numbers under multiplication. What are the elements of −1 ⊂ Q× ? What are the elements of −2? † 5. Let M be a monoid. We can define the positive powers of the elements in M in exactly the same way that positive powers in a group are defined. We have m1 = m for all m ∈ M , and the higher powers are defined by induction: mk = mk−1 m. Show that for m ∈ M and for i, j ≥ 1, we have (a) mi · mj = mi+j , and (b) (mi )j = mij . 6. In Z, show that 2, 3 = Z. 7. In Z, show that 3n, 5n = n for any n ∈ Z. 8. In the group, Q× + , of positive rational numbers under multiplication, show that 2, 3 is not a cyclic subgroup. In other words, there is no rational number q such that 2, 3 = q. 9. Show that Q× + is generated by the set of all prime numbers. 10. Show that the group Q× of nonzero rational numbers is generated by the set consisting of −1 and all of the prime numbers.


24

11. Let G be a group and let a, b ∈ G such that aba−1 ∈ b. Show that H = {ai bj | i, j ∈ Z} is a subgroup of G. Deduce that H = a, b. 12. In Z, show that 2 ∩ 3 = 6. 13. In Z, show that m ∩ n = k, where k is the least common multiple of m and n. 14. In the group Q× + of positive rational numbers under multiplication, show that 2 ∩ 3 = 1. Here, 1 is the trivial subgroup of Q× +. 15. Let M be a monoid. Show that the group of invertible elements Inv(M ) is a submonoid of M . 16. Show that not every submonoid of a group is a group. 17. Show that every submonoid of a finite group is a group.

2.3

The Subgroups of the Integers

One of the simplest yet most powerful results in mathematics is the Euclidean Algorithm. We shall use it here to identify all subgroups of Z and to derive the properties of prime decomposition in Z. Theorem 2.3.1. (The Euclidean Algorithm2 ) Let m and n be integers, with n > 0. Then there are integers q and r, with 0 ≤ r < n, such that m = qn + r. Proof First, we assume that m ≥ 0, and argue by induction on m. If m < n, we may take q = 0 and r = m. If m = n, we take q = 1 and r = 0. Thus, we may assume that m > n and that the result holds for all non-negative integers less than m. In particular, the induction hypothesis gives m − 1 = q n + r , for integers q and r with 0 ≤ r < n. If r < n − 1, we may take q = q and r = r + 1 for the desired result. Otherwise, m = (q + 1)n, and the proof for m ≥ 0 is complete. If m < 0, then −m is positive, and hence −m = q n + r with 0 ≤ r < n. If r = 0, this gives m = −q n. Otherwise, we have m = −q n − r . Subtracting and adding a copy of n on the right of the equation gives m = (−q − 1)n + (n − r ), and since 0 < r < n, we have 0 < n − r < n as well. We shall use this to characterize the subgroups of Z. The subgroups we already know are the cyclic ones: n = {qn | q ∈ Z}. We shall next show that these are all the subgroups of Z. First, note that if H is a nonzero subgroup of Z, then there must be a nonzero element in it, and hence, by closure under inverses, a positive element. Since the set of positive integers less than or equal to a given one is finite, there is a unique smallest positive element in it. 2 The terminology that we’ve chosen here is not universal. There are some mathematicians who refer to Theorem 2.3.1 as the Division Algorithm, and use the term Euclidean Algorithm for the procedure for calculating greatest common divisors, outlined in Problem 3 of Exercises 2.3.18.


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Proposition 2.3.2. Let H ⊂ Z be a subgroup, and suppose that H = 0. Let n be the smallest positive element of H. Then H = n. Proof Since n ∈ H, n ⊂ H. Thus, it suffices to show that H ⊂ n. For m ∈ H, write m = qn + r with q and r integers and 0 ≤ r < n. Now m is in H, as is qn. Thus, m − qn = r must be in H. But r is non-negative and is less than n. Since n is the smallest positive integer in H, we must have r = 0. Thus, m = qn ∈ n. The language of divisibility is useful for discussing subgroups of Z: Definition 2.3.3. Let m and n be integers. We say n divides m (or that m is divisible by n) if there is an integer q such that m = nq. The next lemma is now immediate from the fact that n is the set of all multiples of n. Lemma 2.3.4. Let m and n be integers. Then m ∈ n if and only if n divides m. By Lemma 2.2.9, Lemma 2.3.4 may be restated this way. Lemma 2.3.5. Let m and n be integers. Then n divides m if and only if m ⊂ n. We shall also need a uniqueness statement about generators of cyclic subgroups. Proposition 2.3.6. Let m and n be integers such that m = n. Then m = ±n. Proof Clearly, if either of m and n is 0, so is the other. By passage to negatives, if necessary, we may assume that m and n are both positive. By Lemma 2.3.5, they must divide each other. Say m = qn and n = rm. But then q and r are positive. If q > 1, then m > n, and hence n = rm > rn ≥ n, i.e., n > n, which is impossible. Thus, q = 1, and hence m = n. Corollary 2.3.7. Let H be a subgroup of Z. Then there is a unique non-negative integer which generates H. Definition 2.3.8. Let m and n be integers. We write m + n = {rm + sn | r, s ∈ Z}. Lemma 2.3.9. m + n is a subgroup of Z. Proof The inverse of rn + sm is (−r)m + (−s)n. The result follows since (rm + sn) + (r m + s n) = (r + r )m + (s + s )n ∈ m + n. In fact, it is easy to see that m + n is the subgroup of Z generated by the set {m, n}. Definition 2.3.10. Let m, n ∈ Z. We write (m, n) for the unique non-negative integer which generates the subgroup m + n of Z: m + n = (m, n). We shall refer to (m, n) as the greatest common divisor, or g.c.d., of m and n.


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We shall now derive the familiar properties of the g.c.d. and of prime decomposition from the above results. Proposition 2.3.11. Let m and n be integers. Then an integer k divides both m and n if and only if it divides (m, n). In particular, if m and n are not both 0, then (m, n) is the largest integer which divides both m and n. In addition, there exist integers r and s with rm + sn = (m, n). Proof The last statement follows because (m, n) ∈ m + n = {rm + sn | r, s ∈ Z}. Now let k divide both m and n. Then it must divide any number of the form rm+sn, including (m, n). Conversely, to show that the divisors of (m, n) must divide both m and n, it suffices to show that (m, n) divides both m and n. But m = 1 · m + 0 · n and n = 0 · m + 1 · n both lie in (m, n), so the result follows from Lemma 2.3.4. Definition 2.3.12. We say that the integers m and n are relatively prime if (m, n) = 1. Note that if 1 ∈ m + n, then m + n = Z. The following lemma is immediate. Lemma 2.3.13. The integers m and n are relatively prime if and only if there exist integers r and s with rm + sn = 1. Definition 2.3.14. We say that an integer p > 1 is prime if its only divisors are ±1 and ±p. By Lemma 2.3.5, this says that p ⊂ k if and only if either k = p or k = Z. Lemma 2.3.15. Let p and a be integers, with p prime. Then p and a are relatively prime if and only if p does not divide a. Proof The greatest common divisor of p and a must be either p or 1. In the former case, p divides a. In the latter, it does not. Proposition 2.3.16. Let p be a prime number. Then p divides a product ab if and only if it divides at least one of a and b. Proof Clearly, if p divides a or b, it must divide their product. Conversely, suppose p divides ab, but does not divide a. We shall show that it must divide b. Since p does not divide a, we have (p, a) = 1, so that qp + ra = 1 for some integers q and r. But then qpb + rab = b. Since p divides both qpb and rab, it must divide b. We can now give the fundamental theorem regarding prime factorization. Theorem 2.3.17. Every integer m > 1 has a unique prime decomposition: there is a factorization m = pr11 . . . prkk , where pi is prime and ri > 0 for 1 ≤ i ≤ k, and pi < pj for i < j. Uniqueness means that if m = q1s1 . . . qlsl with qi prime and si > 0 for 1 ≤ i ≤ l and qi < qj for i < j, then we must have k = l and have pi = qi and ri = si for 1 ≤ i ≤ k.


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Proof The existence of a prime decomposition amounts to showing that any integer m > 1 is a product of primes. Clearly, this is true for m = 2. We argue by induction on m, assuming that any integer k with 1 < k < m has a prime decomposition. If m is prime, we are done. Otherwise, m = kl, where both k and l are greater than 1 and less than m. But then k and l are products of primes by induction, and hence so is m. For uniqueness, suppose given decompositions pr11 . . . prkk and q1s1 . . . qlsl of m as above. We argue by induction on t = r1 + · · · + rk . If t = 1, then m = p1 , a prime. But then p1 is divisible by q1 , and since both are prime, they must be equal. We then have p1 (1 − q1s1 −1 . . . qlsl ) = 0, so that (1 − q1s1 −1 . . . qlsl ) = 0, and hence q1s1 −1 . . . qlsl = 1. Since the qi are prime and since no positive number divides 1 other than 1 itself, we must have l = 1 and s1 = 1. Suppose t > 1. Since p1 divides m, Proposition 2.3.16, together with an induction on s1 + · · · + sl , shows that p1 must divide qi for some i. But since p1 and qi are prime, we must have p1 = qi . A similar argument shows that q1 = pj for some j, and our order assumption then shows that i = j = 1, so that p1 = q1 . Now p1 (p1r1 −1 . . . prkk − q1s1 −1 . . . qlsl ) = 0, so that pr11 −1 . . . prkk = q1s1 −1 . . . qlsl . But our inductive hypothesis applies here, and the resulting equality is exactly what we wanted. Exercises 2.3.18. 1. Let p1 , . . . , pk be distinct primes and let ri ≥ 0 for 1 ≤ i ≤ k. Show that an integer n divides pr11 . . . prkk if and only if n = ±ps11 . . . pskk with 0 ≤ si ≤ ri for 1 ≤ i ≤ k. 2. Show that there are infinitely many prime numbers. † 3. Let n = qm + r with 0 ≤ r < m. Show that (m, n) = (r, m). Use this to obtain an iterative procedure to calculate (m, n). 4. Implement the iterative procedure above by a computer program to calculate greatest common divisors. † 5. A subgroup G ⊂ R is called discrete if for each g ∈ G there is an open interval (a, b) ⊂ R such that (a, b) ∩ G = {g}. Show that every discrete subgroup of R is cyclic.

2.4

Finite Cyclic Groups: Modular Arithmetic

Here, we develop Zn , the group of integers modulo n. Definition 2.4.1. Let n > 0 be a positive integer. We say that integers k and l are congruent modulo n, written k ≡ l mod n, or sometimes k ≡ l (n), if k − l is divisible by n (i.e., k − l ∈ n). Lemma 2.4.2. Congruence modulo n is an equivalence relation. Proof Clearly, k − k ∈ n for any n, so congruence modulo n is reflexive. It is also symmetric, since l − k = −(k − l) is contained in any subgroup containing k − l. If k ≡ l mod n and l ≡ m mod n, then both k − l and l − m are in n, and hence so is k − m = (k − l) + (l − m). Thus, the relation is transitive as well.


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Definition 2.4.3. We write Zn for the set of equivalence classes of integers modulo n, and write m ∈ Zn for the equivalence class containing m ∈ Z. 3 We shall also use 0 to denote 0. Zn is called the group of integers modulo n. Lemma 2.4.4. There are exactly n elements in Zn , represented by the elements 0 = 0, 1, . . . , n − 1. Proof The elements listed are precisely the classes r for 0 ≤ r < n. For m ∈ Z, the Euclidean algorithm provides an equation m = qn + r, with 0 ≤ r < n. But then m = r, and hence m is in the stated list of elements. To see that these classes are all distinct, suppose that 0 ≤ k < l < n. Then 0 < l − k < n, so that l − k is not divisible by n, and hence k = l in Zn . There are two important binary operations on Zn . Definitions 2.4.5. The operations of addition and multiplication in Zn are defined by setting k + l = k + l and k · l = kl, respectively, for all k, l ∈ Z. We must show that these operations are well defined. Lemma 2.4.6. The operations of addition and multiplication are well defined binary operations on Zn . Proof Suppose that k = q and l = r in Zn . In other words, k ≡ q mod n and l ≡ r mod n. We must show that k + l ≡ q + r mod n and kl ≡ qr mod n. Addition is easy: k − q ∈ n and l − r ∈ n. Now, (k + l) − (q + r) = (k − q) + (l − r). This lies in n, as n closed under addition. For multiplication, we have kl − qr = kl − kr + kr − qr = k(l − r) + r(k − q). This last is divisible by n because l − r and k − q are. The chief properties of addition and multiplication are inherited from those in the integers: Proposition 2.4.7. Addition gives Zn the structure of an abelian group with identity 0. Multiplication makes Zn into an abelian monoid with identity 1. The two operations satisfy the distributive law: a(b + c) = ab + a c. Proof (a + b) + c = a + b + c = a + b + c. Since a + (b + c) has the same expansion, addition is associative. The other assertions follow by similar arguments. In the language of abstract algebra, Proposition 2.4.7 says, simply, that the operations of addition and multiplication give Zn the structure of a commutative ring. When we refer to Zn as a group, we, of course, mean with respect to addition. By Lemma 2.4.4, Zn has order n. Corollary 2.4.8. Let n be a positive integer. Then there are groups of order n. 3 In

the notation of Section 1.4, Zn = Z/ ≡, where ≡ denotes equivalence modulo n.


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Definition 2.4.9. The canonical map π : Z → Zn is defined by setting π(k) = k for all k ∈ Z. Given that Zn is the set of equivalence classes of integers with respect to the equivalence relation of congruence modulo n, and that π takes each integer to its equivalence class, we see that this coincides with the canonical map defined in Section 1.4 for an arbitrary equivalence relation. We shall see in the next section that the canonical map π : Z → Zn is what’s known as a homomorphism of groups. Exercises 2.4.10. 1. Show that Zn is generated as a group by 1. Thus, Zn is a cyclic group. 2. Show that 2 does not generate Z4 . Deduce that there are finite cyclic groups that possess nontrivial proper subgroups. 3. Let p be a prime. Show that the nonzero elements of Zp form a group under multiplication modulo p. 4. If n is not prime, show that the nonzero elements of Zn do not form a group under multiplication modulo n.

2.5

Homomorphisms and Isomorphisms

There are many ways to construct a finite set, but the thing that describes the essence of the structure of such a set is the number of elements in it. Two sets with the same number of elements may be placed in one-to-one correspondence, at which point their properties as sets are indistinguishable. In the case of groups, the structure is too complicated to be captured by the number of elements alone. We shall see examples of finite groups with the same order whose group properties are quite different. For instance, one may be abelian and the other not. But the idea of a one-to-one correspondence makes a good start at deciding when two groups may be identified, provided that the correspondence respects the multiplications of the groups. Definition 2.5.1. An isomorphism between two groups is a function f : G → G which is one-to-one and onto, such that f (x · y) = f (x) · f (y) for all x, y ∈ G. Here, · is used for the multiplications of both G and G . We say that the groups G and G are isomorphic if there exists an isomorphism from G to G , and write G ∼ = G . This turns out to be enough to capture the full structure of the groups in question. A glance at the definition shows that there is a weaker notion which is part of it: Definition 2.5.2. Let G and G be groups. A homomorphism from G to G is a function f : G → G such that f (x · y) = f (x) · f (y) for all x, y ∈ G. Thus, an isomorphism is a homomorphism which is one-to-one and onto. Recall that one-to-one functions are sometimes called injective functions, or injections, and that onto functions are known as surjective functions, or surjections. Also, a function which is both one-to-one and onto is called bijective. Clearly, an understanding of homomorphisms will contribute to our understanding of isomorphisms, but in fact, homomorphisms will turn out to say a lot more about the


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structure of a group than may be apparent at this point. Before we dive into theory, let’s consider some examples. Examples 2.5.3. 1. For any two groups, G and G , there is a homomorphism f : G → G obtained by setting f (g) = e for all g ∈ G. We call this the trivial homomorphism from G to G . 2. Let H be a subgroup of the group G. Then the inclusion i : H ⊂ G is a homomorphism, because the multiplication in H is inherited from that of G. 3. Recall that the canonical map π : Z → Zn is defined by π(k) = k for all k ∈ Z. Thus, π(k + l) = k + l = k + l = π(k) + π(l), and hence π is a homomorphism. For some pairs G, G of groups, the trivial homomorphism is the only homomorphism from G to G . The relationship between a homomorphism and the power laws in a group is an important one: Lemma 2.5.4. Let f : G → G be a homomorphism of groups and let g ∈ G. Then f (g n ) = f (g)n for all n ∈ Z. In particular, f (e) = e (recall that e = g 0 ) and f (g −1 ) = f (g)−1 . Proof For n > 0, this is a quick induction from the definition of powers. For n = 0, we have f (e) = f (e · e) = f (e) · f (e). Multiplying both sides of the resulting equation f (e) = f (e) · f (e) by f (e)−1 , we get e = f (e). (Any group element which is its own square must be the identity element of the group.) For n < 0, we have e = f (e) = f (g n g −n ) = f (g n )f (g −n ) = f (g n )f (g)−n by the two cases above, so that f (g n ) is the inverse of f (g)−n . Since the canonical map π : Z → Zn is a homomorphism, Lemma 2.2.7 gives us the next corollary. Corollary 2.5.5. Let m ∈ Zn and let k be any integer. Then the k-th power of m with respect to the group operation in Zn is km. Another consequence of Lemma 2.5.4 is a determination of all of the homomorphisms from Z to an arbitrary group G. Proposition 2.5.6. Let G be a group and let g ∈ G. Then there is a unique homomorphism fg : Z → G with fg (1) = g. Proof Define fg by fg (n) = g n for all n ∈ Z. This is a homomorphism by the power laws of Lemma 2.2.6. For uniqueness, suppose that f : Z → G is an arbitrary homomorphism with f (1) = g. We’d like to show that f (n) = g n . Since n is the n-th power of 1 with respect to the group operation in Z, this follows from Lemma 2.5.4. The most basic properties of homomorphisms are the following. We leave the proof to the reader. Lemma 2.5.7. Let f : G → G and g : G → G be homomorphisms. Then so is the composite g ◦ f : G → G . If f and g are both isomorphisms, then so is g ◦ f . If f is an isomorphism, then so is the inverse map f −1 : G → G.


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There are two subgroups associated with a homomorphism that can be useful in detecting its deviation from being an isomorphism. One of them is the kernel: Definition 2.5.8. Let f : G → G be a homomorphism. Then the kernel of f , written ker f , is given by ker f = {x ∈ G | f (x) = e}. Recall that the image of a function f : X → Y is the subset im f ⊂ Y given by im f = {y ∈ Y | y = f (x) for some x ∈ X}. An easy check, using Lemma 2.5.4, shows that the kernel and image of a homomorphism f : G → G are subgroups of G and G , respectively. Lemma 2.5.9. Let f : G → G be a homomorphism. Then ker f detects the deviation of f from being injective: we have f (x) = f (x ) if and only if x−1 x ∈ ker f . In particular, f is injective if and only if the kernel of f is the trivial subgroup, e, of G. The image detects whether f is onto: it is onto if and only if im f = G . Proof Suppose that f (x) = f (x ). Then f (x−1 x ) = f (x)−1 f (x ) = e, and hence x−1 x ∈ ker f . Conversely, if x−1 x ∈ ker f , then f (x ) = f (xx−1 x ) = f (x)f (x−1 x ) = f (x) · e, and hence f (x ) = f (x). Suppose that f is injective. Since f (e) = e, no element of G other than e may be carried to e by f , and hence ker f = e. Conversely, if ker f = e, then x−1 x ∈ ker f implies that x−1 x = e, so that x = x . But then x = x whenever f (x) = f (x ), and hence f is injective. The statement regarding im f is immediate. We shall refer to an injective homomorphism as an embedding. The following lemma is immediate: Lemma 2.5.10. Let f : G → G be an embedding. Then f induces an isomorphism of G onto im f . Definitions 2.5.11. Let G be a group and let n ∈ Z. We say that g ∈ G has exponent n if g n = e. If there is a positive integer n which is an exponent for g, we say that g has finite order, and define the order, o(g), of g to be the smallest positive integer which is an exponent for g. If g does not have a positive exponent, we say that o(g) = ∞. If one knows the order of an element, one can determine all possible exponents for it. Lemma 2.5.12. Let G be a group and let g ∈ G be an element of finite order. Let n be any integer. Then g has exponent n if and only if o(g) divides n. Proof Of course, g mo(g) = (g o(g) )m = em = e, so if o(g) divides n, then g has exponent n. Conversely, if g has order k and exponent n, write n = qk + r with 0 ≤ r < k. Then e = g n = g qk g r = g r , since both n and qk are exponents for g. But this says g has exponent r. If r were positive, this would contradict the fact that k is the smallest positive exponent for g. Thus, r = 0, and n = qk, and hence k divides n, as claimed.


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Information about exponents sometimes leads to a determination that a particular element must be the identity. Corollary 2.5.13. Let m and n be integers that are relatively prime. Suppose that m and n are both exponents for an element g ∈ G. Then g = e. Proof The order of g has to divide both m and n, and hence it must also divide their g.c.d. Since the order of an element is positive, g has order 1, and hence g 1 = e. Let fg : Z → G be the unique homomorphism with fg (1) = g. Then fg (n) = g n for all n ∈ Z. Thus, an integer n is in the kernel of fg if and only if g has exponent n. Also, im fg is precisely g. The next result now follows immediately from Proposition 2.3.2. It includes, in part, a slicker argument for Lemma 2.5.12. Proposition 2.5.14. Let G be a group and let g ∈ G. Let fg : Z → G be the unique homomorphism with fg (1) = g. Then the kernel of fg is o(g), if o(g) is finite, and is 0 otherwise.4 The image of fg is g, the cyclic subgroup of G generated by g. Lemma 2.5.10 now gives us the following corollary. Corollary 2.5.15. If o(g) = ∞, then fg : Z → g is an isomorphism. Corollary 2.5.16. Every nonzero subgroup of Z is isomorphic to Z. Proof Since mn = 0 if and only if either m or n is 0, no nonzero element of Z has finite order. But every subgroup of Z is cyclic. Next, we study the homomorphisms out of Zn . We shall make use of the notion of commutative diagrams, as defined in Definition 1.2.1. Proposition 2.5.17. Let G be a group and let g ∈ G. Then there is a homomorphism hg : Zn → G with hg (1) = g if and only if g has exponent n. If g does have exponent n, then there is only one such homomorphism, and it has the property that the following diagram commutes fg

/ G, Z@ > @@ } @@ }} } @ π @@ }} }} hg Zn meaning that hg ◦ π = fg . Here π : Z → Zn is the canonical map, while fg : Z → G is the unique homomorphism that takes 1 to g. The homomorphism hg must satisfy the formula hg (m) = g m for all m ∈ Zn . If the homomorphism hg exists, then its image is g. Moreover, the induced map hg : Zn → g is an isomorphism if and only if o(g) = n. In particular, if g ∈ G has finite order, then g ∼ = Zo(g) . 4 Recall that, since we are using additive notation for Z, we write 0 for the trivial subgroup consisting of the identity element alone.


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Proof Suppose there exists a homomorphism hg : Zn → G with hg (1) = g. Then hg ◦ π is a homomorphism from Z to G which takes 1 to hg (π(1)) = hg (1) = g. Thus, hg ◦ π = fg , and hence hg (m) = hg (π(m)) = fg (m) = g m for all m ∈ Zn . Thus, hg is unique if it exists. Moreover, the existence of hg gives e = hg (0) = hg (n) = g n , since n = 0 in Zn , and hence g has exponent n. Clearly, im hg = g. Conversely, if g has exponent n, we wish to show that hg (m) = g m gives a well defined function from Zn to G. If m = l, then m−l = sn for some integer s, and hence m = l+sn. But then g m = g l (g n )s = g l es = g l by the rules of exponents. Thus, hg is well defined. But hg (q + r) = g q+r = g q g r = hg (q)hg (r), and hence hg is a homomorphism. It suffices to show that hg is injective if and only if o(g) = n. Suppose first that o(g) = n. Since g has exponent n, o(g) divides n. Thus, 0 < o(g) < n, and hence 0 = o(g) ∈ Zn . Since o(g) ∈ ker hg , hg is not injective. On the other hand, if o(g) = n, let k ∈ ker hg . But this says that g k = e, and hence g has exponent k. But that means that k is divisible by o(g) = n, and hence k = 0 in Zn . In particular, ker hg = 0, as desired. Since Zn has order n, we obtain the following corollary. Corollary 2.5.18. Let G be a group and let g ∈ G. Then the order of g is equal to the order of the group g. Corollary 2.5.19. Let f : G → H be a homomorphism, and suppose that g ∈ G has finite order. Then the order of f (g) divides the order of g. Proof Let hg : Zo(g) → G be the homomorphism which carries 1 to g. Then f ◦ hg : Zo(g) → H is a homomorphism that takes 1 to f (g). By Proposition 2.5.17, f (g) has exponent o(g), so the order of f (g) divides o(g). Because the elements of monoids do not all have inverses, the definition of homomorphism must be modified for it to be useful for monoids. Definitions 2.5.20. A homomorphism of monoids is a function f : M → M such that 1. f carries the identity element of M to that of M . 2. For m, m ∈ M , f (mm ) = f (m)f (m ). An isomorphism of monoids is a homomorphism f : M → M which is bijective. Exercises 2.5.21. 1. Show that any group with one element is isomorphic to the trivial group. 2. Show that any group with two elements is isomorphic to Z2 . 3. List the homomorphisms from Z9 to Z6 . 4. List the homomorphisms from Z5 to Z6 . 5. Show that the order of m in Zn is n/(m, n). Deduce that the order of each element divides the order of Zn . Deduce that every non-identity element of Zp has order p, for any prime p. † 6. Let G be a group and let x ∈ G. Show that o(x−1 ) = o(x). (Hint: We may assume G = x.)


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7. Show that a group G is cyclic if and only if there is a surjective homomorphism f : Z → G. † 8. Let f : G → G be a homomorphism. (a) Let H ⊂ G be a subgroup. Define f −1 (H ) ⊂ G by f −1 (H ) = {x ∈ G | f (x) ∈ H }. Show that f −1 (H ) is a subgroup of G which contains ker f . (b) Let H ⊂ G be a subgroup. Define f (H) ⊂ G by f (H) = {f (x) | x ∈ H}. Show that f (H) is a subgroup of G . If ker f ⊂ H, show that f −1 (f (H)) = H. (c) Suppose that H ⊂ im f . Show that f (f −1 (H )) = H . Deduce that there is a one-to-one correspondence between the subgroups of im f and those subgroups of G that contain ker f . What does this tell us when f is onto? 9. Let f : G → G be a homomorphism and let x ∈ G. Show that f (x) = f (x). 10. Show that every subgroup of Zn is cyclic. 11. Show that Zn has exactly one subgroup of order d for each d dividing n, and has no other subgroups. (In particular, among many other important implications, this shows that the order of every subgroup of Zn divides the order of Zn .) Deduce the following consequences. (a) Show that a group containing a subgroup isomorphic to Zk × Zk for k > 1 cannot be cyclic. (b) Show that if a prime p divides n, then Zn has exactly p − 1 elements of order p. (c) Let H and K be subgroups of Zn . Show that H ⊂ K if and only if |H| divides |K|. (d) Show that if d divides n, then Zn has exactly d elements of exponent d. 12. Let H and K be subgroups of Zpr , where p is prime. Show that either H ⊂ K or K ⊂ H. 13. Let p and q be distinct prime numbers and let n be divisible by both p and q. Find a pair of subgroups H, K ⊂ Zn such that neither H nor K is contained in the other. 14. Show that a group G is abelian if and only if the function f : G → G given by f (g) = g 2 is a homomorphism. 15. Show that a group G is abelian if and only if the function f : G → G given by f (g) = g −1 is a homomorphism. 16. Let G be an abelian group and let n be an integer. Show that the function f : G → G given by f (g) = g n is a homomorphism. 17. Show that the exponential function exp(x) = ex gives an isomorphism from the additive group of real numbers to R× +.


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18. Show that a group is finite if and only if it has finitely many subgroups. 19. Let i : Gln (R) → Mn+1 (R) be defined by setting i(A) to be the matrix whose first n rows consist of the matrix A with a column of 0’s added on the right, and whose last row consists of n 0’s followed by a 1, for all A ∈ Gln (R). Pictorially, ⎞ ⎛ 0 ⎜ .. ⎟ ⎜ . ⎟ A i(A) = ⎜ ⎟. ⎝ 0 ⎠ 0 ... 0 1 Show that i(A) lies in Gln+1 (R), and that i defines an embedding from Gln (R) into Gln+1 (R). 20. Show that there is a monoid with two elements which is not a group. Show that any other monoid with two elements which is not a group is isomorphic to this one. 21. Let f : M → M be a homomorphism of monoids. Show that f restricts to a group homomorphism f : Inv(M ) → Inv(M ) between their groups of invertible elements. In particular, if m ∈ M is invertible, then so is f (m). 22. Give an example to show that condition 1 in the definition of monoid homomorphism does not follow from condition 2.

2.6

The Classification Problem

Definition 2.6.1. By a classification of the groups with a certain property, we mean the identification of a set of groups {Gi | i ∈ I}5 such that the following conditions hold. 1. Each group Gi has the stated property. 2. If Gi is isomorphic to Gj , then i = j. 3. Each group with the stated property is isomorphic to one of the Gi . A classification of groups with a given property is often referred to as a classification “up to isomorphism” of the groups in question. The property by which we collect the groups to be classified is often their order. This is a natural criterion, as isomorphic groups must have the same order. However, other criteria are often used. For instance, the Fundamental Theorem of Finite Abelian Groups, which we give in Section 4.4, gives a classification of all finite abelian groups. Examples 2.6.2. 1. Problem 2 of Exercises 2.5.21 shows that every group of order 2 is isomorphic to Z2 . This classifies the groups of order 2. 2. Corollary 2.5.15 and Proposition 2.5.17 show that every cyclic group is isomorphic to exactly one of the groups in {Z, e, Zn | n ≥ 2}. Thus, we obtain a classification of the cyclic groups. 5 Here

the set I is called the indexing set for the set {Gi | i ∈ I} of groups.


36

Classification can be phrased in more general terms using the language of category theory. (See Chapter 6.) In general, questions of classification are the most fundamental and important questions that can be asked in mathematics. It may come as a surprise to some readers that the finite groups have not yet been classified. It is often astounding that questions which may be so easily phrased can be so difficult to answer. Modern mathematics abounds with such questions. A beginning student of mathematics has no conception of the number of important, yet really basic, mathematical questions that have not yet been solved, despite the number of astute mathematicians that have been working on them over the years. The surprise that comes later is how many interesting open questions are amenable to solutions, given the right mixture of ideas, luck, and persistence. The main goal of the group theory portion of this book is to be able to obtain as much information as possible about the classification of groups of relatively small order. In particular, for a given integer n, say with n ≤ 63, we’d like to classify all groups of order n. There is some language we can use to make our discussion clearer. Definition 2.6.3. We say that two groups are in the same isomorphism class if the groups are isomorphic to one another. Isomorphism classes are analogous to equivalence classes, where the equivalence relation used is that of isomorphism. The point is that the collection of all groups of a given order is not a set, as it has too many elements. (There are too many groups of order 1, for instance, to form a set, even though there is only one isomorphism class of groups of order 1.) Thus, we are looking at an analogue for classes (as opposed to sets) of an equivalence relation. The interesting thing is that classification, as defined above, can only occur in situations where the isomorphism classes themselves form a set. Indeed, the next lemma is an immediate consequence of the definition of a classification. Lemma 2.6.4. The set {Gi | i ∈ I} classifies the collection of groups with a given property if and only if each isomorphism class of the groups with the stated property includes exactly one of the groups Gi . In particular, the collection of isomorphism classes of the groups in question forms a set which is in one-to-one correspondence with {Gi | i ∈ I}. Notice that if X is a finite set, then the set of possible group multiplications we could put on X is a subset of the set of functions from X × X to X. But the set of all functions from one finite set to another is finite, by Corollary 1.7.6. Thus, there are finitely many possible group multiplications on a finite set. Different multiplications will in many cases give rise to isomorphic groups, but the number of isomorphism classes of groups with |X| elements must be finite. Proposition 2.6.5. Let n > 0. Then there are finitely many isomorphism classes of groups of order n. A countable union of finite sets is countable. Corollary 2.6.6. The set of isomorphism classes of finite groups is countable. Exercises 2.6.7. 1. Let X = {x, y} be a set with two elements. Which of the functions from X × X to X are group multiplications? Which are monoid multiplications? 2. Classify the groups of order 3.


2.7

37

The Group of Rotations of the Plane

Here, we describe an important subgroup of Gl2 (R), the group, SO(2), of rotations of the plane. Definition 2.7.1. For θ ∈ R, we define the rotation matrix Rθ by cos θ − sin θ Rθ = . sin θ cos θ Here, the sine and cosine are taken with respect to radian measure. Elements of M2 (R) act as transformations of the plane in the usual way. We identify the plane with the space of column vectors v = ( xy ). A matrix A = ac db acts on the plane by the usual multiplication of matrices on columns: ax + by A · v = . cx + dy To display the geometric effect of the transformation induced by a rotation matrix, we make use of polar coordinates and the sum formulæ for sines and cosines, as developed in calculus. Proposition 2.7.2. The transformation of the plane induced by the rotation matrix Rθ is the counterclockwise rotation of the plane through the angle θ. Proof We write the vector v = ( xy ) in polar coordinates, setting x = r cos φ and y = r sin φ, where r = x2 + y 2 is the length of v and φ is the angle that starts at the positive x-axis and ends at v . The product formula now gives r(cos θ cos φ − sin θ sin φ) r cos(θ + φ) Rθ · v = = , r(cos θ sin φ + sin θ cos φ) r sin(θ + φ) by the sum formulæ for the sine and cosine. Thus, the length of Rθ ·v is the same as that of v , but the angle that it makes with the positive x-axis has been increased by θ. As geometric intuition would suggest, the product (composite) of two rotations is again a rotation: Lemma 2.7.3. For any real numbers θ and φ, we have Rθ · Rφ = Rθ+φ . Proof The definition of matrix multiplication gives cos θ cos φ − sin θ sin φ − cos θ sin φ − sin θ cos φ Rθ · R φ = . cos θ sin φ + sin θ cos φ cos θ cos φ − sin θ sin φ Once again, the result follows from the sum formulæ for the sine and cosine. Definition 2.7.4. Let SO(2) ⊂ Gl2 (R) be the set of all rotation matrices Rθ , as θ ranges over all real numbers. We call it the second special orthogonal group. By Lemma 2.7.3, SO(2) is closed under multiplication in Gl2 (R). Moreover, since cos 0 = 1 and sin 0 = 0, we have R0 = I2 , the identity element of Gl2 (R). Since Rθ · R−θ = R0 , R−θ is the inverse of Rθ .


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Lemma 2.7.5. SO(2) is a subgroup of Gl2 (R). Notice that since addition of real numbers is commutative, Lemma 2.7.3 shows that Rθ ·Rφ = Rφ ·Rθ . Thus, unlike Gl2 (R), SO(2) is an abelian group. Lemma 2.7.3 actually shows more: Proposition 2.7.6. There is a homomorphism, exp, from R onto SO(2), defined by exp(θ) = Rθ . The kernel of exp is 2π = {2πk | k ∈ Z}. Thus, Rθ = Rφ if and only if θ − φ = 2πk for some k ∈ Z. Proof That exp is a homomorphism is the content of Lemma 2.7.3. Suppose that θ ∈ ker exp. Since the identity element of SO(2) is the identity matrix, the definition of Rθ forces cos θ to be 1 and sin θ to be 0. From calculus, we know this happens if and only if θ is a multiple of 2π.6 By Lemma 2.5.9, we have that exp θ = exp φ if and only if −φ + θ ∈ ker(exp), so the result follows. We conclude with a calculation of the order of the elements of SO(2). Proposition 2.7.7. The rotation Rθ has finite order if and only if θ = 2πk/n for some integers k and n. The order of R2πk/n is n/(k, n). Proof Suppose that Rθ has finite order. Then (Rθ )n = Rnθ = I2 for some integer n. Then Proposition 2.7.6 shows that nθ = 2πk for some integer k, so that θ = 2πk/n. If θ = 2πk/n, the order of Rθ will be the smallest positive integer m for which mθ is an integer multiple of 2π. Thus, m is the smallest positive integer such that mk is divisible by n. But a glance at the prime decompositions of n and k shows that this smallest positive integer is precisely n/(n, k). In particular, R2π/n has order n for all n ≥ 1. Corollary 2.7.8. There are embeddings of Zn in SO(2) for all n ≥ 1. Since SO(2) gives transformations of the plane, this says that Zn may be used to produce geometric transformations of a familiar space. A certain school of thought would summarize this by saying that Zn “exists in nature.”

2.8

The Dihedral Groups

Let n > 0, we construct a group, D2n , called the dihedral group of order 2n. These groups are nonabelian for n > 2, and provide our first examples of nonisomorphic groups of the same order. We give a definition here via matrices, which relies on calculus. A direct definition as an abstract group is also possible, but is a little awkward notationally without some further discussion. We shall give such a derivation in Section 4.6. Let b, or bn , if more than one dihedral group is being discussed, be the rotation of the plane through the angle 2π/n: b = R2π/n ∈ SO(2) in the notation of the last section. Then by Proposition 2.7.7, b has order n in Gl2 (R). Let a ∈ Gl2 (R) be the matrix 1 0 a= . 0 −1 6 Actually, one of the nicer ways to prove this fact from calculus is to use derivatives to show that the kernel of exp is a discrete subgroup of R and apply Problem 5 of Exercises 2.3.18. One may then define 2π as the smallest positive element in ker(exp) and use numerical methods to compute its value. Here, we may take the sine and cosine to be defined by their Taylor series, and derive their basic properties, such as their derivatives and their sum formulæ, by manipulating the series.


39

The matrix a is said to reflect the plane through the x-axis, and we say that a matrix is a reflection matrix if it is a product a · c with c ∈ SO(2). (It is possible to give a direct geometric definition of reflection using dot products, and then to prove that these geometric operations are induced by precisely the matrices just stated. We shall treat this approach in exercises below, and content ourselves here with the algebraic definition just given.) The most important piece of information about rotations and reflections is the way they interrelate. They don’t commute with each other, but they come close to it: Lemma 2.8.1. For θ ∈ R, we have Rθ · a = a · R−θ . Thus, for b = R2π/n as above, bk · a = a · b−k for any integer k. Proof Multiplying matrices, we get Rθ · a =

cos θ sin θ

sin θ − cos θ

.

This is equal to the matrix product a · R−θ , since cos(−θ) = cos θ and sin(−θ) = − sin θ. The statement about powers of b follows since b±k = R±k·2π/n . The following lemma may be verified by direct computation. Lemma 2.8.2. The matrix a has order 2. Definition 2.8.3. Let n ≥ 1. Then the dihedral group of order 2n, D2n ⊂ Gl2 (R), is given as follows: D2n = {bk , abk | 0 ≤ k < n}, with a and b = bn as above. Of course, this definition needs a bit of justification: Proposition 2.8.4. The dihedral group D2n is a subgroup of Gl2 (R). Moreover, the elements {bk , abk | 0 ≤ k < n} are all distinct, so that D2n has order 2n. Proof We first show that the 2n elements bk and abk with 0 ≤ k < n are all distinct. If abk = bl , then multiplication by b−k on the right gives a = bl−k , a rotation matrix. But rotation matrices all commute with one another, and Lemma 2.8.1 shows that a fails to commute with most rotations. Thus, abk = bl . If abk = abl , then multiplying both sides on the left by a−1 gives bk = bl , which implies that k ≡ l mod n by Proposition 2.5.17. With 0 ≤ k, l < n, this gives k = l by Lemma 2.4.4. We now show that D2n is closed under multiplication in Gl2 (R). By Lemma 2.8.1, we have abk · abl = a2 bl−k = bl−k , as a has order 2. Since b has order n, any power of b is equal to one in the list. Similarly, bk · abl = abl−k is in D2n . The remaining cases are covered by the fact that b is closed under multiplication. The fact that D2n is closed under multiplication implies that it is a subgroup of Gl2 (R), since D2n is finite (Proposition 2.2.13).


40

The dihedral groups have geometric meaning as groups of transformations of regular polygons in the plane. In particular, for n ≥ 3, there is a regular n-gon whose vertices are the points cos(2πk/n) vk = sin(2πk/n) for 1 ≤ k ≤ n. This means that the n-gon is the set of convex of the vertices, combinations n where a convexcombination of v1 , . . . , vn is a vector sum i=1 aivi with 0 ≤ ai ∈ R for n 1 ≤ i ≤ n and i=1 ai = 1. If c is a matrix in D2n , then it is easy to see that for each i with 1 ≤ i ≤ n, the matrix product c · vi is equal to vj for some j with 1 ≤ j ≤ n. In other words, multiplication by c carries each vertex of our n-gon to another vertex of the n-gon. But the linearity of matrix multiplication now shows that multiplication by c carries convex combinations of the vertices to convex combinations of the vertices, and hence c gives a transformation of the n-gon. In fact, it can be shown that the elements of D2n give all possible linear transformations from the n-gon onto itself. Exercises 2.8.5. 1. What is the order of abk ∈ D2n ? 2. Show that D4 and Z4 are non-isomorphic abelian groups of order 4. 3. Show that D2n is nonabelian if and only if n ≥ 3. Deduce that D2n and Z2n are non-isomorphic groups of order 2n for all n ≥ 3. 4. Show that there is a subgroup of Gl2 (R) consisting of all rotation matrices Rθ , together with the matrices a · Rθ . (This subgroup is generally called O(2), the second orthogonal group.) What is the order of a · Rθ ? 5. Show that if k divides n, then there is a subgroup of D2n which is isomorphic to D2k . † 6. Let G be a group. Show that there is a one-to-one correspondence between the homomorphisms from D2n to G and pairs of elements x, y ∈ G such that o(x) divides 2, o(y) divides n, and x−1 yx = y −1 . 7. Show that if k divides n, then there is a homomorphism of D2n onto D2k . 8. Let be a line through the origin in the plane. Standard vector geometry shows there is a vector u of length 1 such that is the set of all vectors perpendicular to u. (Note this does not define u uniquely: −u would do as well.) Then the reflection through is the function f defined by f (w) =w − 2(w · u)u, where · is the dot product. Show that the reflection through is the transformation induced by the matrix aRθ for some θ. Show also that the transformation induced by aRθ is the reflection through some line for any θ ∈ R.

2.9

Quaternions

We construct yet another family of groups via matrices. They are the quaternionic groups Q4n ⊂ Gl4 (R) of order 4n. First, we give a generalization of the rotation matrices. For

CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES θ ∈ R, let

⎛

cos θ − sin θ ⎜ sin θ cos θ Sθ = ⎜ ⎝ 0 0 0 0

0 0 cos θ sin θ

41

⎞ 0 0 ⎟ ⎟. − sin θ ⎠ cos θ

The geometric effect of Sθ on an element of the space R4 of column vectors is to simultaneously rotate the first two coordinates and rotate the last two coordinates, treating these pairs of coordinates as independent planes. The following lemma may be verified directly by multiplying the matrices. Lemma 2.9.1. Let θ and φ be real numbers. Then Sθ · Sφ = Sθ+φ . Thus, there is a homomorphism f : R → Gl4 (R) defined by f (θ) = Sθ . Moreover, as in the case of rotations of the plane, Sθ = Sφ if and only if θ − φ = 2πk for some k ∈ Z. Once again, we shall give a matrix a that will interact with these “rotations” in an interesting way. It is given by ⎛

0 ⎜ 0 a=⎜ ⎝ −1 0

0 1 0 0 0 0 1 0

⎞ 0 −1 ⎟ ⎟. 0 ⎠ 0

Lemma 2.9.2. The matrix a has order 4. The lower powers of ⎛ ⎞ ⎛ −1 0 0 0 0 ⎜ 0 −1 ⎟ ⎜ 0 0 0 2 3 ⎟ and a = ⎜ a = −I4 = ⎜ ⎝ 0 ⎝ 1 0 −1 0 ⎠ 0 0 0 −1 0

a are given by ⎞ 0 −1 0 0 0 1 ⎟ ⎟. 0 0 0 ⎠ −1 0 0

Proof The square of a is as stated by direct calculation. We then have that a3 = a2 a = −a, which is the displayed matrix. Clearly, this is not the identity matrix, but (−I4 )2 = a4 is the identity matrix. Thus, the 4-th power of a is the smallest that will produce the identity, so that a has order 4 and a3 = a−1 . Once again, there is a deviation from commutativity between the Sθ and a. The following lemma may be verified by direct computation. Lemma 2.9.3. We have Sθ · a = a · S−θ . Because Sθ and Sφ commute for any real numbers θ and φ, we obtain the following corollary. Corollary 2.9.4. The matrix a is not equal to Sθ for any θ. Write b (or b2n if more than one quaternionic group is under discussion) for S2π/2n . Then b has order 2n. The quaternionic group of order 4n is the subgroup of Gl4 (R) generated by a and b: Definition 2.9.5. The quaternionic group of order 4n, Q4n ⊂ Gl4 (R), is given as follows: Q4n = {bk , abk | 0 ≤ k < 2n}, with a and b as above.


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Once again we need some justification. Proposition 2.9.6. For each n ≥ 1, Q4n is a subgroup of Gl4 (R). Moreover, the matrices {bk , abk | 0 ≤ k < 2n} are all distinct, so that Q4n does indeed have order 4n. Proof The key here is that a2 = −I4 = Sπ = bn , which is why b was chosen to have even order. Thus, abk · abl = a2 bl−k = bn bl−k = bn+l−k . Since bk · abl = abl−k and since b has order 2n, Q4n is closed under multiplication. Since it is finite, it is a subgroup by Proposition 2.2.13. It remains to show that the stated elements are all distinct. As in the dihedral case, it suffices to show that abk = bl , and hence that a = bk−l for any k and l. But once again this follows from the fact that a cannot equal Sθ for any θ. Exercises 2.9.7. 1. What is the order of abk ∈ Q4n ? 2. Show that Q4 is isomorphic to Z4 . 3. For n ≥ 2, show that Q4n is isomorphic to neither D4n nor Z4n . 4. Show that if k divides n, then there is a subgroup of Q4n which is isomorphic to Q4k . † 5. Let G be a group. Show that there is a one-to-one correspondence between the homomorphisms from Q4n to G and pairs of elements x, y ∈ G such that o(x) divides 4, o(y) divides 2n, x2 = y n , and x−1 yx = y −1 . 6. Show that there is a homomorphism of Q4n onto D2n . 7. Show that if n is an odd multiple of k, then there is a homomorphism of Q4n onto Q4k . 8. Show that every subgroup of Q8 other than e contains a2 . (Hint: The cyclic subgroups are always the smallest ones in a group.) 9. Let n be any power of 2. Show that every subgroup of Q4n other than e contains a2 .

2.10

Direct Products

We give a way of constructing new groups from old. It satisfies an important universal property with respect to homomorphisms. Definition 2.10.1. Let G and H be groups. Then the product (or direct product) of G and H is the group structure on the cartesian product G × H which is given by the binary operation (g1 , h1 ) · (g2 , h2 ) = (g1 g2 , h1 h2 ). We denote this group by G × H. Once again, we must justify our definition: Lemma 2.10.2. The direct product G × H is a group.


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Proof Associativity follows directly from the associativity of G and H. The identity element is the ordered pair (e, e). The inverse of (g, h) is (g −1 , h−1 ). The reader should note that while this is not the only interesting group structure we can put on the product of G and H (we shall discuss semidirect products in Section 4.6), it is the only one we shall refer to as G × H. The next result is immediate from Proposition 1.6.4. Corollary 2.10.3. Let G and H be finite groups. Then G × H is finite, and |G × H| = |G| · |H|.

The reader should supply the verification for the next lemma. Lemma 2.10.4. There are subgroups G × e = {(g, e) | g ∈ G} and e × H = {(e, h) | h ∈ H} of G × H. Any element of G × e commutes with any element of e × H. There are homomorphisms ι1 : G → G × H and ι2 : H → G × H defined by ι1 (g) = (g, e) and ι2 (h) = (e, h). These induce isomorphisms G ∼ = G × e and H ∼ = e × H. There are also homomorphisms π1 : G × H → G and π2 : G × H → H defined by π1 (g, h) = g and π2 (g, h) = h. The homomorphisms ιi are called the canonical inclusions of G and H in G × H. The homomorphisms π1 and π2 are called the projections of G × H onto the first and second factors, respectively. Recall that a function into a product is defined uniquely by its component functions: f : X → G × H is defined by f (x) = (f1 (x), f2 (x)), where f1 : X → G and f2 : X → H. In fact, the component functions are nothing other than the composites fi = πi ◦ f , where the πi are the projections in the above lemma. The group product in G × H is the unique product that makes the following statement true: Lemma 2.10.5. Let G, H, and K be groups. A function f : K → G × H is a homomorphism if and only if its component functions are homomorphisms. We can also study homomorphisms out of a product. If f : G × H → K is a homomorphism, so are f ◦ ι1 : G → K and f ◦ ι2 : H → K. Proposition 2.10.6. Let f : G × H → K be a homomorphism. Then every element of the image of f ◦ ι1 commutes with every element of the image of f ◦ ι2 . Moreover, f is uniquely determined by its restrictions to G × e and e × H. Conversely, given homomorphisms g1 : G → K and g2 : H → K such that each element in the image of g1 commutes with each element in the image of g2 , then there is a unique homomorphism f : G × H → K with f ◦ ιi = gi for i = 1, 2. Proof Since (g, h) = (g, e) · (e, h), f (g, h) = f ◦ ι1 (g) · f ◦ ι2 (h). Thus, f is determined by the f ◦ ιi , which are determined by the restriction of f to the stated subgroups. The commutativity assertion follows since if two elements commute, so must their images under any homomorphism. Given the gi as above, define f : G × H → K by f (g, h) = g1 (g) · g2 (h). Then f (g, h) · f (g , h ) = g1 (g)g2 (h)g1 (g )g2 (h ) = g1 (g)g1 (g )g2 (h)g2 (h ) = g1 (gg )g2 (hh ), with the key step being given by the fact that g1 (g ) commutes with g2 (h). But the last term is clearly f ((g, h) · (g , h )).


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It is valuable to know when a group breaks up as the product of two of its subgroups. Definition 2.10.7. Let H and K be subgroups of the group G. We say that G is the internal direct product of H and K if there is an isomorphism μ : H × K → G such that μ ◦ ι1 and μ ◦ ι2 are the inclusions of H and K, respectively, as subgroups of G. Explicitly, this says μ(h, k) = hk. We now give a characterization of internal direct products. Proposition 2.10.8. Let H and K be subgroups of the group G. Then G is the internal direct product of H and K if and only if the following conditions hold. 1. H ∩ K = e. 2. If h ∈ H and k ∈ K, then h and k commute in G.7 3. G is generated by the elements of H and K. Proof Suppose that G is the internal direct product of H and K. Then the function μ : H × K → G defined by μ(h, k) = hk is an isomorphism of groups. Since μ is a homomorphism and since μ ◦ ι1 (h) = h and μ ◦ ι2 (k) = k, each h ∈ H must commute with each k ∈ K by Proposition 2.10.6. Since μ is onto, each element of G must have the form hk, with h ∈ H and k ∈ K, so the elements of H and K must generate G. Finally, let x ∈ H ∩ K. Then (x, x−1 ) ∈ H × K, and μ(x, x−1 ) = xx−1 = e. Since μ is injective, this says that (x, x−1 ) is the identity element of H × K, which is (e, e). Thus, x = e, and hence H ∩ K = e. Conversely, suppose that the three stated conditions hold. Since the elements of H commute with those of K, Proposition 2.10.6 shows that the function μ : H × K → G given by μ((h, k)) = hk is a homomorphism. Suppose that (h, k) is in the kernel of μ. Then hk = e, and hence h = k −1 is in H ∩ K. But H ∩ K = e, and hence h = e, and hence k = h−1 = e also. Thus, (h, k) = (e, e), and hence μ is injective. Since H and K generate G, any element in G may be written as a product g1 . . . gn where gi is an element of either H or K for i = 1, . . . , n. Since elements of H commute with those of K, we can rewrite this as a product h1 . . . hr k1 . . . ks with hi ∈ H for i = 1, . . . , r and ki ∈ K for i = 1, . . . , s, by moving the elements in H past those in K. But this is just μ(h1 . . . hr , k1 . . . ks ), so μ is onto. We shall see in Proposition 3.7.12 that the second condition of Proposition 2.10.8 may be replaced with the statement that both H and K are what’s known as normal subgroups (to be defined in Section 3.7) of G. Now let us generalize the material in this section to the study of products of more than two groups. Definition 2.10.9. Suppose given groups G1 , . . . , Gk , for k ≥ 3. The product (or direct product), G1 × · · · × Gk , of the groups G1 , . . . , Gk is the set of all k-tuples (g1 , . . . , gk ) with gi ∈ Gi for 1 ≤ i ≤ k, with the following multiplication: (g1 , . . . , gk ) · (g1 , . . . , gk ) = (g1 g1 , . . . , gk gk ). The reader should be able to supply the proof of the following proposition. 7 Of

course, neither H nor K need be abelian. Thus, if h, h ∈ H, then h and h must commute with every element of K, but they need not commute with each other.


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Proposition 2.10.10. The direct product G1 × · · · × Gk is a group under the stated multiplication. There are homomorphisms πi : G1 × · · · × Gk → Gi and ιi : Gi → G1 × · · · × Gk for 1 ≤ i ≤ k, defined by πi (g1 , . . . , gk ) = gi , and ιi (g) = (e, . . . , e, g, e, . . . , e), where g appears in the i-th coordinate. Given homomorphisms fi : H → Gi for 1 ≤ i ≤ k, there is a unique homomorphism f : H → G1 × · · · × Gk such that πi ◦ f = fi for 1 ≤ i ≤ k. Similarly, for each collection hi : Gi → H of homomorphisms such that each element of the image of hi commutes with each element of the image of hj whenever i = j, there is a unique homomorphism h : G1 × · · · × Gk → H with h ◦ ιi = hi for 1 ≤ i ≤ k. Similarly, we can discuss internal direct products of several subgroups. Definition 2.10.11. Let H1 , . . . , Hk be subgroups of G. We say that G is the internal direct product of H1 , . . . , Hk if the function μ : H1 × · · · × Hk → G defined by μ(h1 , . . . , hk ) = h1 . . . hk is an isomorphism of groups. We shall not give a characterization here of the internal direct product of multiple subgroups. (I.e., we shall not give a generalization of Proposition 2.10.8 to the case of more than two subgroups.) The reader may wish to formulate one after reading the proof of Proposition 4.4.9. Exercises 2.10.12. 1. Show that the order of (g, h) ∈ G × H is the least common multiple of the orders of g and h. † 2. More generally, show that the order of (g1 , . . . , gk ) ∈ G1 × · · · × Gk is the least common multiple of the orders of g1 , . . . , gk . 3. Show that Z6 is isomorphic to Z3 × Z2 . † 4. Let m and n be relatively prime. Show that Zmn is isomorphic to Zm × Zn . 5. Let n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes. Show that Zn ∼ = Zpr11 × · · · × Zprkk . 6. Show that if m and n are not relatively prime, then Zm × Zn is not cyclic. 7. Show that D4 is isomorphic to Z2 × Z2 . 8. Show that D12 is isomorphic to D6 × Z2 . 9. Show that if k is odd, then D4k is isomorphic to D2k × Z2 . 10. Show that G × H ∼ = H × G. 11. Show that (G × H) × K ∼ = G × (H × K) ∼ = G × H × K. 12. Show that G × e ∼ = G, where e is the trivial group. 13. Show that the group multiplication μ : G × G → G is a homomorphism if and only if G is abelian. 14. Let fi : Gi → Hi be a homomorphism for 1 ≤ i ≤ k. Define f1 × · · · × fk : G1 × · · · × Gk → H1 × · · · × Hk by f1 × · · · × fk (g1 , . . . , gk ) = (f1 (g1 ), . . . , fk (gk )). Show that f1 × · · · × fk is a homomorphism. In addition, show that f1 × · · · × fk is an isomorphism if and only if each fi is an isomorphism. 15. Show that not every homomorphism f : G1 × G2 → H1 × H2 has the form f1 × f2 .

Chapter 3

G-sets and Counting Many of the examples in Chapter 2 were obtained from matrices, and hence give symmetries of n-dimensional Euclidean space, Rn . Here, we study the symmetries of sets, obtaining a more combinatorial view of symmetry. In the first section, we define the symmetric group on n letters, Sn , and prove Cayley’s Theorem, which shows that every finite group is a subgroup of Sn for some n. Later, in Chapter 10, we will show that Sn embeds in Gln (R). In this manner, we obtain the assertion of the last chapter that every finite group is a subgroup of Gln (R). The symmetric groups are interesting in their own right as examples of groups. Via the technique of cycle decomposition, which we give in Section 3.5, one can say quite a bit about its structure. Also, there are natural subgroups An ⊂ Sn , called the alternating groups, which are the easiest examples of nonabelian simple groups. We define An in Section 3.5 in terms of the sign homomorphism ε : Sn → {±1} to be defined there. We shall define simple groups in Section 4.2, and show that An is simple for n ≥ 5. We may then deduce in Chapter 5 that Sn is not what’s known as a solvable group for n ≥ 5. The non-solvability of Sn will allow us, in Chapter 11, to prove Galois’s famous theorem that there is no formula for solving polynomials of degree ≥ 5. This should be enough to arouse one’s interest in the symmetric groups. We also introduce a very important concept in group theory: that of the set G/H of left cosets of a subgroup H in G. One of the first consequences of the study of cosets is Lagrange’s Theorem: if G is finite, then the order of any subgroup of G must divide the order of G. We give Lagrange’s Theorem in Section 3.2, and give many more applications of the study of cosets throughout the material on group theory. The sets of cosets G/H form an important example of the notion of a group acting on a set. We formulate the notion of G-sets in Section 3.3 and give a generalization of Cayley’s Theorem in which a one-to-one correspondence is established between the actions of G on a set X and the homomorphisms from G to the group of symmetries of X. This allows us to associate to a finite G-set (such as G/H, if G is finite) a homomorphism from G into Sn for appropriate n. These homomorphisms are quite useful in determining the structure of a group G, and will figure prominently in the classification results we get in Chapter 5. Section 3.3 also contains a couple of important counting arguments, including what’s known as the G-set Counting Formula. In the final two sections of the chapter, we study actions of one group on another. The point is that not all symmetries of a group H preserve the group structure. The ones that do are the automorphisms of H, which are defined to be the group isomorphisms

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CHAPTER 3. G-SETS AND COUNTING

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H → H. These form a subgroup, Aut(H), of the group of symmetries of H. We study the actions of G on H such that the action of each g ∈ G induces an automorphism of H. Actions through automorphisms play an important role in constructing and classifying the extensions of one group by another (Section 4.7), and therefore, as we shall see, in the classification of groups. Closer at hand, there is a very important action of a group G on itself through automorphisms, called conjugation. Many of the important results in group theory follow from the study of conjugation, including the Noether Isomorphism Theorems of Chapter 4 and the Sylow Theorems of Chapter 5. We shall begin that study in this chapter.

3.1

Symmetric Groups: Cayley’s Theorem

Here, we show that every finite group embeds in the group of symmetries of a finite set. Recall that a function is bijective if it is one-to-one and onto. Definitions 3.1.1. Let X be a set. By a permutation of X we mean a bijective function σ : X → X. We write S(X) for the set of all permutations of X. Note that S(X) has a binary operation given by composition of functions: for σ, τ ∈ S(X), σ · τ is the composite σ ◦ τ . As we shall see shortly, S(X) is a group under this operation, called the group of symmetries of X. Recall that the identity function 1X of a set X is the function defined by 1X (x) = x for all x ∈ X. Recall from Proposition 1.1.6 that every bijective function σ : X → Y has an inverse function σ −1 : Y → X, with the property that σ −1 ◦ σ = 1X and σ ◦ σ −1 = 1Y . Here, the inverse function σ −1 is defined by setting σ −1 (y) to be the unique x ∈ X such that σ(x) = y. Lemma 3.1.2. Let X be a set. Then S(X) forms a group under the operation of composition of functions. The identity element of S(X) is the identity function 1X , and the inverse of an element σ ∈ S(X) is its inverse function, σ −1 . We shall see below that if X is infinite, then not only is S(X) an infinite group, it also contains elements of infinite order. On the other hand, if X is finite, then there are only finitely many elements in S(X), as, in fact, there are only finitely many functions from X to X by Corollary 1.7.6. We shall calculate the order of S(X), for X finite, in Corollary 3.3.11. Definition 3.1.3. Let f : X → Y be a bijection of sets. Define the induced map f∗ : S(X) → S(Y ) by f∗ (σ) = f ◦ σ ◦ f −1 for σ ∈ S(X). Since two sets are considered equivalent if they may be placed in bijective correspondence, the following lemma should come as no surprise. Lemma 3.1.4. Let f : X → Y be a bijection of sets. Then the induced map f∗ : S(X) → S(Y ) is an isomorphism of groups.


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Proof It is easy to see that f∗ gives a group homomorphism from S(X) to S(Y ). Moreover, (f −1 )∗ : S(Y ) → S(X) gives an inverse function for f∗ , so f∗ is an isomorphism. Note that there is no preferred isomorphism between S(X) and S(Y ), but rather an isomorphism for each choice of bijection from X to Y . Of special interest here is the case when X is finite. Our preferred model for a set with n elements is {1, . . . , n}. Definition 3.1.5. We write Sn for S({1, . . . , n}), and call it the symmetric group on n letters. Fixed points are an important feature of a permutation: Definition 3.1.6. Let σ be a permutation of X. Then σ is said to fix an element x ∈ X if σ(x) = x. We call such an x a fixed point of σ. We should now construct some examples of elements of Sn . Recall that if Y is a subset of X, then the complement of Y in X is the set of all elements of X that are not in Y . Definition 3.1.7. Let i1 , . . . , ik be k distinct elements of {1, . . . , n}, k ≥ 2. We write (i1 i2 · · · ik ) for the permutation of {1, . . . , n} which carries ij to ij+1 for 1 ≤ j < k, carries ik to i1 , and is the identity map on the complement of {i1 , . . . , ik } in {1, . . . , n}. We call (i1 i2 · · · ik ) a k-cycle, or a cycle of length k. Note that a 2-cycle (i j) just interchanges i and j and fixes all other elements of {1, . . . , n}. Another name for a 2-cycle is a transposition. Some people consider a 1-cycle (i) to be another way of writing the identity map. While this is consistent with the notation above, our cycles will all have length ≥ 2. Note that the permutations (i1 i2 · · · ik ) and (ik i1 · · · ik−1 ) are equal, as they have the same effect on all elements of {1, . . . , n}. Thus, there is more than one way to denote a given k-cycle. We can calculate the order and inverse of a k-cycle. Lemma 3.1.8. A k-cycle (i1 i2 · · · ik ) has order k. Its inverse is (ik · · · i2 i1 ). Proof Let σ = (i1 i2 · · · ik ). An easy induction shows that for 1 ≤ j < k, σ j (i1 ) = ij+1 = i1 , while σ k does fix i1 . Thus, σ j = e for 1 ≤ j < k. But since σ may also be written as (i2 · · · ik i1 ), σ k fixes i2 by the preceding argument. Another induction shows that σ k fixes each of the ij . Since it also fixes the complement of {i1 , . . . , ik }, we have σ k = e. The permutation (ik · · · i1 ) is easily seen to reverse the effect of σ. One of the reasons for studying the symmetric groups is that every finite group is isomorphic to a subgroup of Sn for some n. Recall that an embedding of groups is an injective homomorphism. Theorem 3.1.9. (Cayley’s Theorem) Let G be a group. For g ∈ G write g : G → G for the function g (g ) = gg . Then g is a permutation of G. Moreover, if we define Λ : G → S(G) by Λ(g) = g , then Λ is an embedding. In consequence, if G is finite, then it is isomorphic to a subgroup of S|G| .


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Proof To show that g is injective, suppose g (g ) = g (g ), and hence gg = gg . Multiplying both sides on the left by g −1 and cancelling, we get g = g . And g is surjective because g = g (g −1 g ) for all g ∈ G. Thus, g is a permutation of G. Now Λ(g) · Λ(g ) is the permutation g ◦ g . We wish to show this coincides with Λ(gg ), which is the permutation gg . We show they have the same value on each g ∈ G. ( g ◦ g )(g ) = g(g g ) = (gg )g = gg (g ). Thus, Λ is a homomorphism. To show it is an injection, we consider its kernel. Suppose that g = e. Thus, g (g ) = g for all g ∈ G, i.e., gg = g for all g ∈ G. But multiplying both sides of the equation on the right by (g )−1 , we get g = e. Thus, ker Λ = e, the trivial subgroup, so Λ is an embedding by Lemma 2.5.9. Recall that a set is countably infinite if it may be put in one-to-one correspondence with the positive integers, Z+ . It is easy to show that Z is countably infinite. Since every nonzero element of Z has infinite order, the next corollary follows from Cayley’s Theorem. Corollary 3.1.10. Let X be a countably infinite set. Then S(X) contains elements of infinite order. Cayley’s Theorem shows that any finite group G is isomorphic to a subgroup of S|G| . But most groups are actually isomorphic to a group of permutations of a smaller number of letters. As we shall see in a moment, there is a natural choice of embedding of Sn in Sn+1 . Thus, if G embeds in Sn , then it also embeds in Sn+1 . In fact, if Y is a subset of X, there is a natural embedding of S(Y ) in S(X): Definition 3.1.11. Let Y be a subset of X, and let i : Y ⊂ X be the inclusion map. Define i∗ : S(Y ) → S(X) by σ(x) if x ∈ Y (i∗ (σ))(x) = x otherwise. We shall refer to i∗ as the canonical inclusion of S(Y ) in S(X). Lemma 3.1.12. Let i : Y ⊂ X be the inclusion of a subset, Y , of a set X. Then the canonical inclusion i∗ : S(Y ) → S(X) is an embedding onto the subgroup of S(X) consisting of those elements of S(X) which fix all elements of the complement of Y in X. Proof Write X − Y for the complement of Y in X. It is easy to see that i∗ is an embedding and that elements in the image of i∗ fix all elements of Y − X. Thus, it suffices to show that any element of S(X) that fixes the elements of Y − X lies in the image of i∗ . Thus, suppose that τ ∈ S(X) fixes the elements of of X − Y . Note that it suffices to show that τ restricts to a bijection from Y to itself, as then this restriction defines an element of S(Y ), which we shall denote by τ . Since τ is the identity on X − Y and agrees with τ on Y , we obtain that i∗ (τ ) = τ , as desired. We first show that τ (Y ) ⊂ Y . Suppose not. Then we can find y ∈ Y such that τ (y) = x ∈ Y − X. But since τ fixes Y − X, we have τ (x) = x, so x and y have the same image under τ . This is impossible, as τ is injective.


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Thus, τ restricts to a function from Y to Y , which is injective, because restrictions of injective functions are always injective. But note that if x ∈ X − Y , then τ (x) cannot lie in Y . Thus, since τ is onto, we must have τ (Y ) = Y . Thus, τ restricts to a bijection from Y to itself, and the result follows. Taking i to be the inclusion map {1, . . . , n − 1} ⊂ {1, . . . , n}, we get the canonical inclusion Sn−1 ⊂ Sn . The next corollary is now immediate from Lemma 3.1.12. Corollary 3.1.13. The canonical inclusion of Sn−1 in Sn identifies Sn−1 with the subgroup of Sn consisting of those permutations that fix n. Note that the inclusion of {1, . . . , n} in the positive integers, Z+ , induces a canonical inclusion Sn ⊂ S(Z+ ) which is compatible with the inclusion of Sn in Sn+1 . Definition 3.1.14. Write i∗ : Sn ⊂ S(Z+ ) for the canonical inclusion, and define S∞ ⊂ S(Z+ ) by S∞ = i∗ (Sn ). n≥1

Lemma 3.1.15. S∞ is a subgroup of S(Z+ ).1 Proof Since each (i∗ (x))−1 = i∗ (x−1 ), S∞ is closed under inverses, and it suffices to show that S∞ is closed under multiplication. For x, y ∈ S∞ , there are m, n ≥ 1 such that x ∈ i∗ (Sm ) and y ∈ i∗ (Sn ). Suppose, then, that m ≤ n. Since the canonical inclusions of the Sk in S(Z+ ) are compatible with their canonical inclusions in one another, we have i∗ (Sm ) ⊂ i∗ (Sn ), and hence x, y ∈ i∗ (Sn ). But i∗ (Sn ) is a subgroup of S(Z+ ), and hence xy ∈ i∗ (Sn ) ⊂ S∞ . In practice, we identify each Sn with its image under i∗ , and treat Sn as a subgroup of S∞ (and hence also of S(Z+ )). Exercises 3.1.16. 1. Show that S2 is isomorphic to Z2 . 2. Show that S(X) is nonabelian whenever X has more than two elements. 3. Show that the square of a 4-cycle is not a 4-cycle. 4. What is the order of (1 2) · (3 4) in S4 ? 5. What is the order of (1 2 3) · (4 5) in S5 ? 6. Show that there is an embedding f : D6 → S3 induced by setting f (a) = (1 3) and f (b) = (1 2 3). We shall show in Section 3.3 that S3 has order 6. We may then deduce that f is an isomorphism. 7. Define x, y ∈ S4 by x = (1 4) · (2 3) and y = (1 2 3 4). Show that there is an embedding f : D8 → S4 induced by setting f (a) = x and f (b) = y. 8. Show that D12 embeds in S5 . 9. Show that D2n embeds in Sn for all n ≥ 3. 1 Those

who are acquainted with category theory should note that we can identify S∞ with the direct limit lim −→ Sn .


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10. Show that Z2 × Z2 embeds in S4 . 11. Show that Zm × Zn embeds in Sm+n . 12. Show that the subgroup S∞ ⊂ S(Z+ ) is an infinite group whose elements all have finite order. 13. Show that an element of S(Z+ ) lies in S∞ if and only if it fixes all but finitely many of the elements of Z+ . 14. Let G be a group and let g ∈ G. Define rg : G → G by rg (x) = xg. Show that rg ∈ S(G). Show that the function R : G → S(G) defined by R(g) = rg−1 is an embedding. ‡ 15. Let f : Y → Z be any injective function, and define f∗ : S(Y ) → S(Z) by f (σ(y)) if z = f (y) (f∗ (σ))(z) = z if z is not in the image of f . (a) Show that f∗ is an embedding onto the subgroup of S(Z) consisting of those permutations that fix all elements of Z which are not in f (Y ). (b) If f is a bijection, show that this f∗ coincides with the isomorphism f∗ given in Definition 3.1.3. (c) If f is the inclusion of a subset Y ⊂ Z, show that this f∗ coincides with the embedding f∗ given in Definition 3.1.11. (d) Let g : Z → W be another injection. Show that g∗ ◦ f∗ = (g ◦ f )∗ . (e) If X is finite, and if f : X → Z+ is injective, show that the image of f∗ lies in S∞ . Thus, there is an induced embedding f∗ : S(X) → S∞ . 16. Let X be a set. Show that the set of all functions f : X → X forms a monoid under the operation of composition of functions. Show that S(X) is its group of invertible elements. 17. Show that the set of functions from {1, 2} to itself which take 1 to itself gives, via composition of functions, the unique monoid with two elements that is not a group.

3.2

Cosets and Index: Lagrange’s Theorem

We here begin the study of a very important concept in algebra: the left cosets of a subgroup H ⊂ G. One of the consequences of this consideration is the famous theorem of Lagrange: If H is a subgroup of the finite group G, then the order of H divides the order of G. Lagrange’s Theorem is the very first step in trying to classify groups up to isomorphism. Definition 3.2.1. Let H be a subgroup of G and let x ∈ G. Then the left coset xH of H in G is the following subset of G: xH = {xh | h ∈ H}.


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Thus, as x ranges over the elements of G, we get a collection of different subsets xH ⊂ G, which are the left cosets of H in G. Note that eH is just H ⊂ G. The coset H is special in that it is the only left coset of H in G which is a subgroup. (The next lemma shows that e ∈ xH if and only if xH = H.) We study the cosets to understand how H relates to the other elements of G. Example 3.2.2. Let G be a cyclic group of order 4, written multiplicatively, with generator b: G = {e, b, b2 , b3 }. Let H be the cyclic subgroup generated by b2 : H = b2 = {e, b2 }. Then an easy inspection shows that eH = b2 H = {e, b2 }, while bH = b3 H = {b, b3 }. Since we have inspected xH for each element x ∈ G, we see that these are all of the left cosets of H in G. Recall that cosets are defined to be subsets of G. Thus, since bH and b3 H have the same elements, they are equal as cosets. In particular, for this choice of H and G, there are exactly two left cosets of H in G. We now show that two left cosets that have a single element in common must be equal. Lemma 3.2.3. Let H be a subgroup of G and let x, y ∈ G. Then the following statements are equivalent. 1. xH = yH. 2. xH ∩ yH = ∅. 3. x−1 y ∈ H. 4. y −1 x ∈ H. Proof Clearly, the first statement implies the second. Suppose, then, that the second holds. Let z ∈ xH ∩ yH. Then there are h, h ∈ H with z = xh = yh . But then x−1 y = h(h )−1 ∈ H. Thus, the second statement implies the third. The third and fourth statements are equivalent, as y −1 x is the inverse of x−1 y. Suppose, then, that the fourth statement holds. Thus, y −1 x = k ∈ H, and hence x = yk. But then xh = ykh ∈ yH for all h ∈ H, and hence xH ⊂ yH. But also, y = xk −1 , so that yh = xk −1 h for all h ∈ H, and hence yH ⊂ xH. Thus, xH = yH, and we see that the fourth statement implies the first. Since x = x · e we have x ∈ xH for each x ∈ G. But any two left cosets of H that have an element in common must be equal by Lemma 3.2.3. Corollary 3.2.4. Let H be a subgroup of G. Then each element x ∈ G is contained in exactly one left coset of H, and that coset is xH. In particular, the elements x and y lie in the same left coset of H if and only if xH = yH. If the operation of G is written additively, we have an additive notation for cosets. Notation 3.2.5. Suppose that G is abelian and that we use additive notation for the group operation in G. Then we also use additive notation for the left cosets of a subgroup H ⊂ G. For x ∈ G, we write x + H for the coset which contains x: x + H = {x + h | h ∈ H}. It is very useful in group theory to consider the set of all left cosets of H in G.


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Definitions 3.2.6. Let H be a subgroup of G. Write G/H for the set whose elements are the cosets xH of H in G. Thus, if xH = yH as subsets of G, then xH and yH are equal as elements of G/H. If the number of elements of G/H (which is equal to the number of distinct left cosets of H in G) is finite, we say that H has finite index in G, and write [G : H], the index of H in G, for the number of elements in G/H. If the set G/H is infinite, we write [G : H] = ∞. We define the canonical map, π : G → G/H, by π(x) = xH for x ∈ G. Remarks 3.2.7. Let H be a subgroup of G. We put a relation ∼ on G by setting x ∼ y if x−1 y ∈ H. Then ∼ is an equivalence relation: it is reflexive because (x−1 y)−1 = y −1 x, and is transitive because x−1 y · y −1 z = x−1 z. But x−1 y ∈ H if and only if y ∈ xH, so the equivalence class of x under ∼ is precisely xH. Thus, G/H is the set of equivalence classes of G under ∼, and the canonical map π : G → G/H, as defined above, takes each x ∈ H to its equivalence class under ∼. Thus, π agrees with the canonical map of Definition 1.4.5. The following corollary is immediate from Corollary 3.2.4. Corollary 3.2.8. Let H be a subgroup of G and let π : G → G/H be the canonical map. Then π(x) = π(y) if and only if x and y are in the same left coset of H. Proposition 3.2.9. (Lagrange’s Theorem) Let H be a subgroup of G and suppose that H is finite. Then every left coset of H has |H| elements. Given that H is finite, if either of G or [G : H] is finite, then so is the other, and we have |G| = [G : H] · |H|. In particular, if G is a finite group and H is a subgroup of G, then the order of H divides the order of G. Proof For x ∈ G, the permutation x defined in the last section (by x (g) = xg) restricts to a bijection from H to xH (with inverse induced by x−1 ). Thus, if H is finite, each coset xH has the same number of elements as H. Since each element of G lies in exactly one left coset of H, the order of G, if finite, is equal to the sum of the “orders” of the distinct left cosets of H. Since each left coset has |H| elements, this says that |G| must be equal to the number of left cosets of H times the order of H. Of course, if both H and [G : H] are finite, then G is the union of a finite number of the finite sets xH, and hence G is finite as well. Recall that the order of a given element of a group is equal to the order of the cyclic subgroup generated by that element (Corollary 2.5.18). Corollary 3.2.10. Let G be a finite group. Then the order of any element of G must divide the order of G. We should point out that the converse of Lagrange’s Theorem is false: if k divides the order of G, it is not necessarily the case that G will have a subgroup of order k. For instance, in Problem 2 of Exercises 4.2.18, we shall see that the alternating group A4 , which has order 12, has no subgroup of order 6. We shall spend some time investigating partial converses to Lagrange’s Theorem, as they are quite useful in obtaining classification results. In Cauchy’s Theorem, we will


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show that if a prime p divides the order of a group, then G has a subgroup of order p. Then, in the Sylow Theorems, we will show that G has a subgroup whose order is the highest power of p that divides the order of G. We can get even better partial converses to Lagrange’s Theorem if we put extra assumptions on the group G. For instance, Hall’s Theorem (Theorem 5.6.2) gives a result that holds for what’s known as solvable groups: if G is solvable and |G| = nk with (n, k) = 1, then G has a subgroup of order n. Even here, we do not get a full converse, as the alternating group A4 is solvable. But the full converse does hold in finite abelian groups, or, more generally, in the finite nilpotent groups, which we shall study in Section 5.7. Lagrange’s Theorem allows us to classify the groups of order 4. Corollary 3.2.11. Let G be a group of order 4 which is not cyclic. Then G is isomorphic to Z2 × Z2 . Proof Let a = b be elements of G, neither of which is the identity element. Neither a nor b can have order 4, as then G would be cyclic. Thus, both have order 2. Now ab = a, as that would imply that b = e, and ab = b, because then a would be the identity element. And ab = e, as then a and b would be inverse to one another. But an element of order 2 is its own inverse, so this would mean a = b. Thus, the four elements of G are e, a, b, and ab. A similar argument shows that ba cannot equal e, a, or b. But this forces ab = ba, so that a and b commute. But now Proposition 2.10.8 shows that G is the internal direct product of a and b. There are a couple of very important consequences of Lagrange’s Theorem. Corollary 3.2.12. Let H and K be subgroups of G such that |H| and |K| are relatively prime. Then H ∩ K = e. Proof Let x ∈ H ∩ K. Then Corollary 3.2.10 shows that o(x) must divide both |H| and |K|. Since |H| and |K| are relatively prime, this forces o(x) = 1. Corollary 3.2.13. Let G and G be groups such that |G| and |G | are relatively prime. Then the only homomorphism from G to G is the trivial homomorphism, which sends each x ∈ G to e ∈ G . Proof Let x ∈ G. Then o(f (x)) divides o(x) (Corollary 2.5.19), which in turn divides |G|. But o(f (x)) also divides |G |, as f (x) ∈ G . Since |G| and |G | are relatively prime, we must have o(f (x)) = 1. So far, we have studied the left cosets of a subgroup H of G. There is an analogous notion of right cosets, which we shall study in the exercises below: Definition 3.2.14. Let H be a subgroup of the group G. The right cosets of H in G are the subsets Hx = {hx | h ∈ H} for x ∈ G. Clearly, if G is abelian, then xH = Hx, and the right and left cosets of H coincide. For some nonabelian groups, the left and right cosets of a given subgroup can be totally different subsets of G. However, we shall see that the left and right cosets of H agree if H is what’s known as a normal subgroup of G.


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Exercises 3.2.15. 1. Let p be a prime number. Show that any group of order p is isomorphic to Zp . 2. For a prime number p, show that the only subgroups of Zp are e and Zp itself. 3. Let H be a subgroup of the finite group G and let K be a subgroup of H. Show that [G : K] = [G : H] · [H : K]. † 4. Generalize the preceding problem to the case where G is infinite, but [G : K] is finite. 5. State and prove the analogue of Lemma 3.2.3 for right cosets. 6. Show that every subgroup of Q8 other than Q8 itself is cyclic. List all of the subgroups. 7. Find all of the left cosets of a ⊂ D6 . Now find all of the right cosets and show that they are different subsets of D6 . 8. Repeat the preceding exercise with D8 in place of D6 . 9. Repeat the preceding exercise for D2n with n arbitrary. 10. What are the right cosets of b ⊂ D2n ? 11. Find all of the left cosets of a ⊂ Q12 . 12. Let H be a subgroup of G. Show that there is a one-to-one correspondence between the left cosets of H and the right cosets of H. Deduce that H has finite index in G if and only if there are finitely many distinct right cosets of H in G, and that the number of right cosets, if finite, is [G : H]. 13. Show that the order of a submonoid of a finite monoid need not divide the order of the larger monoid.

3.3

G-sets and Orbits

Definitions 3.3.1. Let G be a group. An action of G on a set X is a rule that assigns to each ordered pair (g, x) with g ∈ G and x ∈ X an element g · x ∈ X,2 such that 1. e · x = x for all x ∈ X 2. The following analogue of the associative law holds: g1 · (g2 · x) = (g1 g2 ) · x for all g1 , g2 ∈ G and x ∈ X. A set X together with an action of G on X is called a G-set. If X and Y are G-sets, then a G-map, or G-equivariant map, f : X → Y is a function f from X to Y such that f (g · x) = g · f (x) for all g ∈ G and x ∈ X. A G-isomorphism is a G-map which is a bijection. 2 In other words, the passage from the ordered pair (g, x) to the product g · x defines a function from G × X to X.


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We will often abbreviate g · x by gx. We now note that we may identify two G-sets which are G-isomorphic. Lemma 3.3.2. Let f : X → Y be a G-isomorphism. Then so is the inverse function f −1 : Y → X. Examples 3.3.3. 1. G itself is a G-set, under the action obtained by setting g · g equal to the product gg of g and g in G. We call this the standard action of G on G. 2. Let X be a set with one element: X = {x}. Then there is a unique action of G on X: g · x = x for all g ∈ G. 3. Let H be a subgroup of G. We let G act on G/H by g · xH = (gx)H for g, x ∈ G. As the reader may verify, this gives a well defined action. Moreover, the canonical map π : G → G/H is easily seen to be a G-map, where G is given the action of the first example. The G-sets G/H are sometimes called “homogeneous spaces” of G. Here, “spaces” comes from the fact that a similar construction is studied for topological groups G, where H is a subgroup of G which is also a closed subspace with respect to the topology of G. 4. Let G be any group and let X be any set. Then there is at least one action of G on X: just set g · x = x for all g ∈ G and x ∈ X. We call this the trivial action of G on X. Clearly, it does not tell us very much about either G or X. 5. Let X be a set. Then S(X) acts on X by σ · x = σ(x). Indeed, the facts that 1X is the identity of S(X) and that the group product in S(X) is composition of functions are precisely what is needed to show that this specifies an action of S(X) on X. We give a generalization of this example in Proposition 3.3.16. 6. Let X be any set. Let G be a cyclic group of order 2, written multiplicatively: G = {e, a}, where a has order 2. Then there is an action of G on X × X, obtained by setting a · (x, y) = (y, x) for each ordered pair (x, y) ∈ X × X. As Example 3 suggests, the study of G-sets encodes information about the subgroups of G. We will use G-sets as one of our main tools in understanding the structure of groups. Definitions 3.3.4. Let X be a G-set and let x ∈ X. Then the isotropy subgroup of x, denoted Gx , is the set of elements of G which fix x: Gx = {g ∈ G | g · x = x}. The orbit of x, denoted G · x, is the set of all elements obtained from x via multiplication by elements of G: G · x = {g · x | g ∈ G} ⊂ X. We say that G acts transitively on X if X = G · x for some x ∈ X. The next lemma is needed to justify the preceding definitions. Lemma 3.3.5. Let X be a G-set and let x ∈ X. Then the isotropy subgroup Gx is a subgroup of G. The orbit, G · x, of x is the smallest sub-G-set of X which contains x.


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Proof Clearly, Gx is a submonoid of G, and we must show that Gx is closed under inverses. But if g ∈ Gx , then g −1 · x = g −1 · (g · x) = (g −1 g) · x = x, and hence g −1 ∈ Gx . We leave the rest to the reader. As a first example of these concepts, let us consider their application to the “homogeneous spaces” G/H. Lemma 3.3.6. In the G-set G/H, the isotropy subgroup of eH is H. Also, the orbit of eH is all of G/H, and hence G/H is a transitive G-set. Proof Here g · eH = gH, so that every element in G/H is in the orbit of eH. Also, g is in the isotropy subgroup of eH if and only if gH = eH. By Lemma 3.2.3, this is the case if and only if e−1 g ∈ H. Thus, the isotropy subgroup of eH is H, as claimed. G-maps are of course important in the understanding of G-sets. The reader should verify the following lemma. Lemma 3.3.7. Let f : X → Y be a G-map and let x ∈ X. Then we have an inclusion of isotropy subgroups Gx ⊂ Gf (x) . If, in addition, f is injective, then Gx = Gf (x) . In the study of G-sets, orbits play a role similar to (but simpler than) the role played by cyclic subgroups in the study of groups. Just as the cyclic groups Zn form models for the cyclic subgroups, every orbit turns out to be G-isomorphic to one of the G-sets G/H: Lemma 3.3.8. Let X be a G-set and let x ∈ X. Let H be a subgroup of G. Then there is a G-map fx : G/H → X with fx (eH) = x if and only if H ⊂ Gx . If fx exists, it is unique, and has image equal to the orbit G · x. Moreover, fx induces a G-isomorphism from G/H to G · x if and only if H = Gx . In particular, for X any G-set and x ∈ X, G · x is G-isomorphic to G/Gx . Thus, a G-set is transitive if and only if it is G-isomorphic to G/H for some subgroup H ⊂ G. Proof Suppose given a G-map fx : G/H → X with fx (eH) = x. Then since H is the isotropy subgroup of eH in G/H, it must be contained in the isotropy subgroup of fx (eH) = x. In other words, we must have H ⊂ Gx . Also, since gH = g · eH, fx (gH) = g · fx (eH) = gx, so fx is uniquely determined by its effect on eH. Clearly, its image is G · x. Conversely, if H ⊂ Gx , we use the above formula to define fx : G/H → X: fx (gH) = gx. We must show this is well defined: i.e. if gH = g H, we must show that gx = g x. By Lemma 3.2.3, gH = g H if and only if g = g h for some h ∈ H. But then gx = g hx = g x, since hx = x. Thus, fx is a well defined function. To see that fx is a G-map, note that fx (g · gH) = fx ((g g)H) = g gx = g fx (gH). If fx is a G-isomorphism, then the isotropy subgroups of eH and fx (eH) must be equal. But the isotropy subgroup of eH in G/H is H, while the isotropy subgroup of x is Gx , so if fx is a G-isomorphism, then H = Gx . Conversely, suppose that H = Gx . To show that fx induces an isomorphism of G/Gx onto G · x, it suffices to show that fx is injective. Suppose that fx (gH) = fx (g H), and hence that gx = g x. Multiplying both sides by g −1 and cancelling, we see that x = g −1 g x, so that g −1 g is in the isotropy subgroup of x. Thus g −1 g ∈ Gx = H, so that gH = g H by Lemma 3.2.3. The next corollary now follows from Lagrange’s Theorem.


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Corollary 3.3.9. Let X be a finite G-set and let x ∈ X. Then the number of elements in the orbit G · x is equal to [G : Gx ]. In particular, since [G : Gx ] is finite, if either G or Gx is finite, so is the other, and G · x has |G|/|Gx | elements. As noted in Example 5 of Examples 3.3.3, S(X) acts on X for any set X, via σ · x = σ(x), for σ ∈ S(X) and x ∈ X. Thus, we may use the notions of orbit and isotropy to obtain information about S(X). We are primarily concerned with the case where X is finite. Recall from Corollary 3.1.13 that the canonical inclusion of Sn−1 in Sn identifies Sn−1 with the subgroup of Sn consisting of those elements that fix n. We restate this in terms of isotropy groups: Lemma 3.3.10. Consider the standard action of Sn on {1, . . . , n}. Under this action, the isotropy subgroup of n is the image of the canonical inclusion Sn−1 ⊂ Sn . Corollary 3.3.9 now allows us to give an inductive calculation of the order of Sn . Corollary 3.3.11. The order of Sn is n! (n-factorial). Proof Since the 2-cycle (i n) takes n to i for each 1 ≤ i < n, each element of {1, . . . , n} is in the orbit of n under the standard action of Sn . In other words, Sn acts transitively on {1, . . . , n}. Since this orbit has n elements, and since the isotropy subgroup of n is Sn−1 , we have n = |Sn |/|Sn−1 | by Corollary 3.3.9. The result follows by induction on n, starting with the obvious fact that S1 is the trivial group. It is often desirable to count the elements in a G-set X by adding up the numbers of elements in its orbits. To do that, we shall need the following lemma. Lemma 3.3.12. Let X be a G-set and let x, y ∈ X. Then y ∈ G · x if and only if G · y = G · x. In consequence, if x, z ∈ X, then G · x = G · z if and only if G · x ∩ G · z = ∅. Proof Since y ∈ G · x, y = gx for some g ∈ G. Thus, g y = (g g)x ∈ G · x for all g ∈ G, so G · y ⊂ G · x. But x = g −1 y, so G · x ⊂ G · y by the same reasoning. Notice that the analogy between the orbits in a G-set and the cyclic subgroups of a group has just been left behind, as we can find lots of examples (e.g., in cyclic groups) of inclusions K ⊂ H of cyclic subgroups which are not equal. Definition 3.3.13. Let X be a G-set. Then a set of orbit representatives in X is a subset Y ⊂ X such that 1. If x, y ∈ Y with x = y, then G · x ∩ G · y = ∅. 2. X = x∈Y G · x. For a general G-set, existence of a set of orbit representatives turns out to be an important special case of the Axiom of Choice. But the case for finite G-sets is much easier. Lemma 3.3.14. Any finite G-set has a set of orbit representatives. Proof Let X be our G-set. We argue by induction on the number of elements in X. Let x ∈ X. If G · x = X, then {x} is the desired set. Otherwise, let X be the complement of G · x in X. By Lemma 3.3.12, if g ∈ G and x ∈ X , gx must lie in X . Thus, X is a sub-G-set of X. By induction, X has a set, Y , of orbit representatives. But then Y = Y ∪ {x} is a set of orbit representatives for X.


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For infinite sets, one can prove the existence of a set of orbit representatives by using Zorn’s Lemma (a form of the Axiom of Choice) to perform what amounts to an infinite induction argument. We shall use similar techniques in the material on ring theory in Chapters 7 through 12. The following corollary is now immediate from Corollary 3.3.9. Corollary 3.3.15. (G-set Counting Formula) Let Y be a set of orbit representatives for the finite G-set X. Then the number, |X|, of elements in X is given by [G : Gx ]. |X| = x∈Y

An understanding of G-sets is vital to an understanding of permutation groups because of the following generalization of Cayley’s Theorem. Proposition 3.3.16. Let X be a G-set. For g ∈ G, define g : X → X by g (x) = gx. Then g is a permutation of X. Moreover, the function Λ : G → S(X) defined by Λ(g) = g is a group homomorphism, whose kernel is the intersection of the isotropy subgroups of the elements of X: ker Λ = Gx . x∈X

Conversely, if Λ : G → S(X) is any group homomorphism, then X is a G-set under the action defined by g · x = Λ(g)(x). In fact, the correspondence which takes a G-set X to the homomorphism Λ : G → S(X) gives a one-to-one correspondence between the actions of G on X and the homomorphisms from G to S(X). Proof Let X be a G-set and let g ∈ G. We wish to show that g is a permutation of X. It is onto, because x = g (g −1 x) for all x ∈ X. It is one-to-one, since if g (x) = g (y), then gx = gy, and, multiplying both sides by g −1 , we see that x = y. Now Λ is a homomorphism if and only if g ◦ g = gg for all g, g ∈ G. But clearly, ( g ◦ g )(x) = gg x = gg (x) for all x ∈ X, so Λ is indeed a homomorphism. The kernel of Λ is the set of elements in G which act on X by the identity map, and hence fix each element in X. These then are the group elements which lie in the isotropy group of every element of X. Conversely, given the homomorphism Λ, and defining g · x to be equal to Λ(g)(x), we have that ex = x because Λ(e) is the identity permutation, while g1 (g2 x) = Λ(g1 )(Λ(g2 )(x)) = (Λ(g1 ) ◦ Λ(g2 ))(x) = Λ(g1 g2 )(x) = (g1 g2 )x. Finally, the passage from the G-set X to the homomorphism Λ is easily seen to be inverse to the passage from a homomorphism Λ : G → S(X) to an action on X. Note that if the G-set X is G (with the usual action), then the map Λ : G → S(G) given above is precisely the one given in Cayley’s Theorem. Definitions 3.3.17. Let X be a G-set. 1. We say that the action of G on X is effective, or that X is an effective G-set, if x∈X Gx = e. In words, G acts effectively on X if the only element of G which fixes every element of X is e.


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2. We say that the action of G on X is free, or that X is a free G-set, if the isotropy subgroup of every element of X is e. In other words, X is free if for any g ∈ G and x ∈ X, gx = x implies that g = e. Clearly, any free action is effective. However, there are many effective actions that are not free. Examples 3.3.18. 1. The standard action of G on G is free. Indeed, if gx = x, then right multiplication by x−1 gives g = e. 2. The standard action of S(X) on X is effective, since the only function X → X which fixes every element of X is the identity function. Note, however, that if X has more than two elements, then this action is not free. For instance, a 2-cycle (i j) fixes every element other than i and j. Via the map Λ : G → S(X) of Proposition 3.3.16, we may interpret effective actions in terms of embeddings into symmetric groups. Corollary 3.3.19. An action of a group G on a set X is effective if and only if the induced map Λ : G → S(X) is an embedding. Thus, there is an embedding of G into Sn if and only if there is an effective action of G on a set with exactly n elements. Proof For any G-set X, Proposition 3.3.16 tells us that the kernel of theinduced map Λ : G → S(X) is x∈X Gx . Since effective actions are those for which x∈X Gx = e, they are precisely those actions for which Λ is an embedding. Thus, the key to discovering whether a given group embeds in Sn lies in understanding the isotropy groups that appear in the various orbits of G-sets. We shall take an ad hoc approach to these questions at this time, but the reader may want to come back to this material after reading Section 3.7. Definition 3.3.20. By a minimal subgroup of G we mean a subgroup H = e with the property that the only subgroups of H are e and H. This last is, of course, an abuse of language. Such a subgroup should more properly be called a minimal nontrivial subgroup. But the trivial subgroup is rather uninteresting, so we make the above definition for the purposes of economy of language. (Similarly, in our discussion of commutative rings, we shall use the expression “maximal ideal” for an ideal which is maximal in the collection of all ideals which are not equal to the ring itself.) Note that not all groups have minimal subgroups. For instance, it is easy to see that every nontrivial subgroup of Z contains a subgroup which is strictly smaller, but nontrivial. But this cannot happen in a finite group. Lemma 3.3.21. Every nontrivial finite group G has a minimal subgroup. Proof We argue by induction on the order of G. If G has no subgroups other than e and itself (e.g. if |G| = 2), then G is a minimal subgroup of itself. Otherwise, there is a nontrivial subgroup H = G. But then |H| < |G|, so that H has a minimal subgroup by induction. However, a minimal subgroup of H is also minimal in G.


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Most finite groups have several different minimal subgroups, which are not necessarily isomorphic to each other. However, in the case that there is a unique minimal subgroup, we obtain a strong consequence for the embedding problem. Proposition 3.3.22. Let G be a finite group with exactly one minimal subgroup. Then any effective G-set has at least |G| elements. In particular, n = |G| is the smallest integer such that G embeds in Sn . Proof Let H be the minimal subgroup of G. We claim that every nontrivial subgroup of G contains H. To see this, let K be a nontrivial subgroup of G. Then there is a minimal subgroup K ⊂ K. But any minimal subgroup of K is also a minimal subgroup of G. Since G has only one minimal subgroup, we must have K = H, and hence H ⊂ K. Now suppose given an effective G-set X. We claim there must be at least one x ∈ X with Gx = e. Suppose not. Then each Gx is a nontrivial subgroup, and hence H ⊂ Gx for each x. But then H ⊂ x∈X Gx , and hence X is not effective. Thus, there is an x ∈ X with Gx = e. But then G · x is G-isomorphic to G/e, and hence X contains an orbit which has |G| elements in it. In particular, X has at least |G| elements. Exercises 3.3.23. 1. Show that the standard action on G is G-isomorphic to G/e and that the one-point G-set is G-isomorphic to G/G. ‡ 2. Let G = D6 . Show that G acts effectively (but not freely) on G/a, and hence that G embeds in S3 . Deduce from Corollary 3.3.11 that this embedding is an isomorphism. 3. Generalize the preceding problem to obtain an embedding from D2n into Sn . Since the order of the two groups is different for n > 3, this will not be an isomorphism for these values of n. 4. Let p be a prime and let r ≥ 1. Show that Zpr cannot embed in Sn for n < pr . 5. Show that Q8 cannot embed in Sn for n < 8. 6. Show that Q2r cannot embed in Sn for n < 2r . 7. Find the minimal value of n such that D2r embeds in Sn . Do the same for D2pr , where p is an odd prime. 8. Let X and Y be G-sets. By X × Y , we mean the G-set given by the action of G on the cartesian product of X and Y in which g · (x, y) = (gx, gy), for all g ∈ G, x ∈ X, and y ∈ Y . What is the isotropy subgroup of (x, y) ∈ X × Y ? 9. Let G = Z6 . Let H, K ⊂ G be the subgroups of order 2 and 3, respectively. Show that G/H × G/K is a free G-set. 10. Show that the action of G on X×Y need not be transitive, even if G acts transitively on both X and Y . 11. Show that if G acts freely on X, then it also acts freely on X × Y for any Y .


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12. There is a construction in set theory called the disjoint union. The disjoint union of X and Y, written X Y , is a set containing both X and Y in the following way: in X Y we have X ∩ Y = ∅, and X ∪ Y = X Y . We shall give a formal construction of the disjoint union in Chapter 6. Show that if X and Y are G-sets, then G acts on X Y in such a way that X and Y are G-subsets. 13. Let G = Z6 . Let H, K ⊂ G be the subgroups of order 2 and 3, respectively. Show that G acts effectively (but not freely) on G/H G/K. Deduce that Z6 embeds in S5 . 14. Let m and n be relatively prime. Show that Zmn embeds in Sm+n . 15. Show that Q12 embeds in S7 . 16. What is the isotropy subgroup of gH ∈ G/H for an arbitrary g ∈ G? 17. Let X be a G-set and let H be a subgroup of G. The set of H-fixed points of X, written X H , is defined to be the set of x ∈ X such that hx = x for all h ∈ H. Show that x ∈ X H if and only if H ⊂ Gx . ‡ 18. Let X be a G-set and let H be a subgroup of G. Show that there is a one-to-one correspondence between the G-maps from G/H to X and the elements of X H . ‡ 19. Let X be a G-set. Show that there is an equivalence relation on X defined by setting x ∼ y if y = gx for some g ∈ G. Show that the equivalence classes of this relation are precisely the orbits G · x for x ∈ X. The set of equivalence classes thus obtained is called the orbit space of X, and denoted X/G. (Thus, the elements of X/G are the orbits G · x in X.) We write π : X → X/G for the canonical map, which takes x to its equivalence class. 20. A G-map f : X → Y is said to be isovariant if Gx = Gf (x) for each x ∈ X. Show that a G-map f : X → Y is isovariant if and only if its restriction to each orbit is injective. 21. Recall that a G-set Y has the trivial action if gy = y for all y ∈ Y and g ∈ G. Show that if X is any G-set, and if we give the orbit space X/G the trivial action, then the canonical map π : X → X/G is a G-map. Show also that if Y is any G-set with the trivial action, and if f : X → Y is a G-map, then there is a unique G-map f : X/G → Y such that f ◦ π = f . 22. Show that if G acts trivially on X and if Y is any G-set, then the G-maps from G × X to Y are in one-to-one correspondence with the functions from X to Y . (Here, G in the above product is the standard free G-set G/e.) In this context, we call G × X the free G-set on the set X. ‡ 23. Let X be any set and write X n for the product X × · · · × X of n copies of X. Show that X n is an Sn -set via σ · (x1 , . . . , xn ) = (xσ−1 (1) , . . . , xσ−1 (n) ), for σ ∈ Sn and (x1 , . . . , xn ) ∈ X n . † 24. Let G be a finite group whose order is divisible by a prime number p. In this exercise, we shall prove Cauchy’s Theorem (also given by a different proof as Theorem 5.1.1), which says that such a group G must have an element of order p.


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(a) Define σ : Gp → Gp by σ(g1 , . . . , gp ) = (gp , g1 , . . . , gp−1 ). Show that σ gives a permutation of Gp of order p (i.e., σ ∈ S(Gp ) with σ p = e). Deduce that there is an action of Zp on Gp , given by k · (g1 , . . . , gp ) = σ k (g1 , . . . , gp ) for all k ∈ Zp . (In fact, if we identify Zp with the subgroup of Sn generated by the p-cycle (1 2 · · · p), this action is just the restriction to Zp of the Sn action given in Problem 23.) (b) Define X ⊂ Gp by X = {(g1 , . . . , gp ) | g1 · · · gp = e}. In other words, the product (in order) of the gi is the identity element of G. Show that X is a sub-Zp -set (but not a subgroup, unless G is abelian) of Gp . (c) Show that the isotropy group of (g1 , . . . , gp ) ∈ X is the trivial subgroup of Zp unless g1 = g2 = · · · = gp = g for some g ∈ G with g p = e. Deduce that there is a one-to-one correspondence between the fixed-point set X Zp and the set {g ∈ G | g p = e}. Deduce that G has an element of order p if and only if X Zp has more than one element. (d) Note that if x ∈ X is not in X Zp , then the orbit Zp · x has p elements. Since two orbits are either disjoint or equal, X is the disjoint union of X Zp with the orbits which lie outside of X Zp . Deduce that |X| = |X Zp | + some multiple of p. Show that |X| = |G| , so that |X| is divisible by p. Deduce that |X Zp | is divisible by p, and hence G has an element of order p. p−1

25. We say that G acts on the right on a set X (or that X is a right G-set) if for all x ∈ X and g ∈ G there is an element xg ∈ X such that (xg)g = x(gg )

and

xe = x

for all x ∈ X and all g, g ∈ G. Show that if X is a right G-set, then there is an induced left action of G on X given by g · x = xg −1 . Deduce that there is a one-to-one correspondence between the left G-sets and the right G-sets. 26. Let H be a subgroup of G. Then H acts on the right of G by ordinary multiplication. Show that we may identify the set of left cosets G/H with the orbit space (see Problem 19) of this action. 27. Let G act on the right on X and act on the left on Y . Then G acts on the left on X × Y by g · (x, y) = (xg −1 , gy). We call the orbit space of this action the balanced product of X and Y , which we denote X ×G Y . Let H be a subgroup of G and let X be a (left) H-set. Then using the right action of H on G by ordinary multiplication, we can form the balanced product G ×H X. Show that the standard left action of G on itself induces a left action of G on G ×H X. Write [g, x] for the element of G×H X represented by the ordered pair (g, x) ∈ G×X. (In other words, [g, x] is the orbit of (g, x) under the above action of H on G × X.) Show that the isotropy subgroup G[e,x] of [e, x] under the above G-action coincides with the isotropy subgroup Hx of x under the original action of H on X.


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28. Let H be a subgroup of G and let X be a (left) H-set. Let Y be a (left) G-set. Then H acts on Y by restricting the action of G. Show that there is a one-to-one correspondence between the H-maps from X to Y and the G-maps from G ×H X to Y . 29. Let M be a monoid. Define an M -set by substituting M for G in the definition of G-set. Find a monoid M and an M -set X such that the function m : X → X defined by m (x) = m · x need not be a permutation for all values of m. 30. Show that if you delete the requirement that e · x = x from the definition of G-set, then the functions g need not be permutations.

3.4

Supports of Permutations

We isolate here some of the more elementary considerations for the study of cycle structure in Section 3.5. We give some consequences in the exercises. Definition 3.4.1. Let σ ∈ S(X). The support of σ, Supp(σ), is the following subset of X: Supp(σ) = {x ∈ X | σ(x) = x} In particular, σ is the identity outside of its support. It is easy to calculate the support of a k-cycle. Lemma 3.4.2. The support of the k-cycle (i1 · · · ik ) is {i1 , . . . , ik }. Here is another way to think of supports. Lemma 3.4.3. Let Y be a subset of X and let σ ∈ S(X). Then Supp(σ) ⊂ Y if and only if σ is the identity on the complement of Y . In particular, if i : Y → X is the inclusion of Y in X, then Supp(σ) ⊂ Y if and only if σ is in the image of the canonical inclusion i∗ : S(Y ) ⊂ S(X). Proof The support is the complement of the subset of points fixed by σ, hence the first statement. For the second, recall from Lemma 3.1.12 that the image of i∗ is the set of permutations that fix the complement of Y . Applying Lemma 3.4.3 to Y = Supp(σ), we obtain the following corollary. Corollary 3.4.4. Let σ be a permutation of the set X. Then σ restricts to a bijection σ : Supp(σ) → Supp(σ). Definition 3.4.5. We say the permutations σ, τ ∈ S(X) are disjoint if Supp(σ) ∩ Supp(τ ) = ∅. Disjoint permutations interact well with one another: Proposition 3.4.6. Let σ and τ be disjoint permutations of X. Then σ and τ commute with each other. Moreover, the product στ is given as follows: ⎧ ⎨ σ(x) if x ∈ Supp(σ) τ (x) if x ∈ Supp(τ ) (στ )(x) = ⎩ x otherwise.


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In particular, the support of στ is the union Supp(σ) ∪ Supp(τ ). Finally, στ has finite order if and only if both σ and τ have finite order, in which case the order of στ is the least common multiple of the orders of σ and τ . Proof We wish to show that both στ and τ σ have the effect specified by the displayed formula. Away from Supp(σ) ∪ Supp(τ ), both σ and τ are the identity, and hence both στ and τ σ are also. On Supp(σ), τ is the identity, as Supp(σ) ∩ Supp(τ ) = ∅. Thus, for x ∈ Supp(σ), (στ )(x) = σ(x). Since x ∈ Supp(σ), σ(x) is also in Supp(σ) by Corollary 3.4.4, and hence (τ σ)(x) = σ(x) as well. A similar argument shows that both στ and τ σ act as τ on Supp(τ ). Thus, στ = τ σ, and the displayed formula holds. An easy induction based on the displayed formula now shows: ⎧ k ⎨ σ (x) if x ∈ Supp(σ). τ k (x) if x ∈ Supp(τ ). For k > 0, (στ )k (x) = ⎩ x otherwise. Thus, (στ )k = e if and only if both σ k = e and τ k = e, and the result follows. Exercises 3.4.7. 1. Show that an element σ ∈ S(Z+ ) lies in S∞ if and only if Supp(σ) is finite. ‡ 2. Suppose given subsets Y, Z ⊂ X, with Y ∩ Z = ∅. Show that there is an embedding μ : S(Y )×S(Z) → S(X) whose restrictions to S(Y ) = S(Y )×e and S(Z) = e×S(Z) are the canonical inclusions of S(Y ) and S(Z), respectively, in S(X). Show that the image of μ is the set of permutations σ ∈ S(X) such that the following two properties hold: (a) Supp(σ) ⊂ Y ∪ Z (b) σ(Y ) = Y . ‡ 3. Combinatorics Consider the action of Sn on {0, 1}n obtained by permuting the factors (see Problem 23 of Exercises 3.3.23). We can use it to derive some of the standard properties of the binomial coefficients. First, for 0 ≤ k ≤ n, let us write xk = (1, . . . , 1, 0, . . . , 0 ) ∈ {0, 1}n . k times n−k times (a) Show that the number of elements in the orbit Sn · xk is equal to the number of ways to choose k things out of n things (which we may identify with the number of k-element subsets of an n-element set), which we shall denote by the binomial coefficient nk . (b) Show that the isotropy subgroup of xk is isomorphic to the direct product Sk × Sn−k . Deduce that n n! = , k k! · (n − k)! and that the right-hand side must therefore be an integer.


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(c) Show that the elements x0 , . . . , xn form a set of orbit representatives for the Sn action on {0, 1}n . Deduce that n

2 =

n n k=0

k

.

(d) Deduce that there are exactly 2n subsets of an n-element set. (e) Show that the number of ordered k-element subsets of an n-element set is k! times the number of unordered k-element subsets of that set. Deduce that the number of ordered k-element subsets is n!/(n − k)!. 4. It can be shown that if X is any infinite set, then the disjoint union X X may be placed in bijective correspondence with X. Deduce from Problem 2 that for any infinite set X, we may embed S(X) × S(X) in S(X). 5. Reconsider the preceding problem for X = Z+ . Show that for any choice of injection Z+ Z+ → Z+ , the induced embedding S(Z+ ) × S(Z+ ) ⊂ S(Z+ ) restricts to an embedding of S∞ × S∞ in S∞ . (These embeddings turn out to be important in studying the cohomology of the group S∞ , an endeavor which falls under the heading of homological algebra, but has applications to topology.) 6. In this problem we assume a familiarity with infinite disjoint unions and with infinite products, as discussed in Chapter 6. (a) Suppose given a ! family of sets {X i | i ∈ I}. Show that we may embed the infinite product i∈I S(Xi ) in S( i∈I Xi ). Deduce that there is a permutation of Z+ which may be thought of as the disjoint product of infinitely many cycles, with one k-cycle being given for each k ≥ 2. Deduce that there are permutations in S(Z+ ) whose orbits are all finite, but which have infinite order. (b) Construct a permutation of Z+ which is a product of infinitely many cycles of the same order. Deduce that there are elements of S(Z+ ) which have finite order but do not lie in S∞ .

3.5

Cycle Structure

Here, using the ideas of G-sets and orbits, we develop the notion of cycle decompositions for permutations. This gives a powerful tool for studying permutation groups that has some analogies to prime decompositions in Z. Using cycle decomposition, we shall construct the sign homomorphism ε : Sn → {±1} and define the alternating groups, An . We shall encounter further applications of cycle decomposition in all of our subsequent study of permutation groups, particularly in Section 4.2. Recall that the permutations σ and τ are disjoint if Supp(σ) ∩ Supp(τ ) = ∅, where Supp(σ) is the complement of the set of points fixed by σ. Definition 3.5.1. By a product of disjoint permutations, we mean a product σ1 · · · σk , such that σi and σj are disjoint for i = j. We are particularly interested in products of disjoint cycles: i.e., products σ1 · · · σk of disjoint permutations in Sn , such that each σi is a cycle.


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We now generalize Proposition 3.4.6. Corollary 3.5.2. Let σ1 · · · σk ∈ S(X) be a product of disjoint permutations. Then the effect of σ1 · · · σk is given as follows: on Supp(σi ), the effect of σ1 · · · σk is the same k as that of σi for 1 ≤ i ≤ k. Away from i=1 Supp(σi ), σ1 · · · σk is the identity. In k particular, Supp(σ1 · · · σk ) = i=1 Supp(σi ). In addition, if each σi has finite order, then the order of σ1 · · · σk is the least common multiple of the orders of the σi . Proof We argue by induction on k. The case k = 2 is given k−1by Proposition 3.4.6. For k > 2, the induction hypothesis gives Supp(σ1 · · · σk−1 ) = i=1 Supp(σi ), and hence σ1 · · · σk−1 is disjoint from σk . The result now follows by applying Proposition 3.4.6 (and the induction hypothesis) to the product (σ1 · · · σk−1 ) · σk . The following proposition gives the cycle decomposition for a permutation, σ. Proposition 3.5.3. Let σ ∈ Sn for n ≥ 2, with σ = e. Then σ may be written uniquely as a product of disjoint cycles: σ = σ1 · · · σk . Here, uniqueness means that if σ may also be written as the product τ1 · · · τl of disjoint cycles, then k = l, and there is a permutation φ ∈ Sk such that σi = τφ(i) for 1 ≤ i ≤ k. Proof Let G = σ, the cyclic subgroup of Sn generated by σ. We wish to understand X = {1, . . . , n} as a G-set. Note that if the orbit of x ∈ X has only one element, then the isotropy subgroup of x has index 1 in G by Corollary 3.3.9. This means that the isotropy subgroup of x must be all of G. If this were true for all x ∈ X, then G would act as the identity on X. Since σ = e, this cannot be the case. Suppose, then, that y ∈ X has an orbit with more than one element. Let r be the smallest positive integer with the property that σ r (y) = y. (Such an integer exists, because σ o(σ) (y) = y.) Note that r > 1, as otherwise y is fixed under σ and all of its powers, and hence the orbit of y under G would have only one element. We claim first that the elements y, σ(y), . . . , σ r−1 (y) are all distinct. To see this, note that otherwise there are integers s and t with 0 ≤ s < t < r such that σ s (y) = σ t (y) = σ s (σ t−s (y)). Thus, y and σ t−s (y) have the same image under σ s . Since σ s is injective, this gives σ t−s (y) = y. Since 0 < t − s < r, this contradicts the minimality of r. Thus, there is an r-cycle σ1 = (y σ(y) · · · σ r−1 (y)). Let Y1 = Supp(σ1 ) = {y, σ(y), . . . , σ r−1 (y)}. Then, by inspection, we see that σ has the same effect as σ1 on each element of Y1 . Thus, σ restricts to a bijection Y1 → Y1 . By passage to powers, we see that Y1 is a sub-G-set of X. Since G acts transitively on Y1 , we see that Y1 is the orbit of y under the action of G. Now let Y1 , . . . , Yk be the set of all distinct orbits in X with more than one element. By Lemma 3.3.12, Yi ∩ Yj = ∅ for i = j. Repeating the argument above, we can find cycles σi for 2 ≤ i ≤ k such that σi has the same effect on Yi is σ does, and Supp(σi ) = Yi . In particular, σ1 · · · σk is a product of disjoint cycles. We claim that σ = σ1 · · · σk . By Corollary 3.5.2, the effect of σ1 · · · σk on Yi = Supp(σi ) is the same as that of σi . But σi was chosen to have the same effect on Yi as k σ. Thus, σ agrees with σ1 · · · σk on i=1 Yi = Supp(σ1 · · · σk ). Since σ is the identity on k all one-element orbits, it is the identity off i=1 Yi , and σ = σ1 · · · σk , as claimed. We show uniqueness by reversing the above argument. In a product τ = τ1 . . . τl of disjoint cycles, the effect of τ agrees with that of τi on Supp(τi ). As the cycle notation shows, τi acts transitively on Supp(τi ), so τ acts transitively on Supp(τi ) as well.


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Thus, there are l distinct orbits under the action of τ which have more than one element. Thus, if τ is equal to the permutation σ, above, then k = l and, equating the orbits of σ and τ , we may relabel the τi so that Supp(τi ) = Supp(σi ) for i = 1, . . . , k. But since σ = σi and τ = τi on this common support, τi and σi are equal as permutations. Remarks 3.5.4. Note the concluding statement of the last proof. A cycle is determined by its effect as a permutation, not by its notation. However, it’s easy to see that there are precisely r different ways to write an r-cycle in cycle notation. It depends solely on which element of its support is written first. Proposition 3.5.3 shows that every permutation is a product of cycle permutations. But we can actually write a permutation as a product of even simpler terms as long as we no longer insist that these terms be disjoint. Recall that a transposition is another word for a 2-cycle. The following lemma may be verified by checking the effect of both sides of the proposed equality on the elements {i1 , . . . , ik }. Here, we should emphasize the fact that if τ, τ ∈ Sn and j ∈ {1, . . . , n}, then (τ τ )(j) = τ (τ (j)). Lemma 3.5.5. Let σ be the k-cycle σ = (i1 · · · ik ). Then σ is the product of k − 1 transpositions: σ = (i1 i2 )(i2 i3 ) · · · (ik−1 ik ).

Since the cycle permutations generate Sn , we see that every element of Sn is a product of transpositions. Corollary 3.5.6. Sn is generated by the transpositions. We wish to define the sign of a permutation to be ±1 in such a way that the sign will be +1 if the permutation may be written as a product of an even number of transpositions, and is −1 otherwise. Our initial definition will be more rigid than this. Definition 3.5.7. The sign of a k-cycle is (−1)k−1 . Thus, the sign is −1 if k is even and is +1 if k is odd. The sign of a product of disjoint cycles is the product of the signs of the cycles. The sign of the identity map is +1. Thus, the sign of every permutation is defined, via Proposition 3.5.3. Notice the relationship between the sign of a cycle and the number of transpositions in the product given in Lemma 3.5.5. Multiplying out the signs over a product of disjoint cycles, we see that if the sign of a permutation, σ, is +1, then there is at least one way to write σ as the product of an even number of transpositions. If the sign of σ is −1, then there is at least one way to write σ as the product of an odd number of transpositions. The next proposition is the key to the utility of signs. Notice that {±1} forms a group under multiplication. Since it has two elements, it is isomorphic to Z2 . Proposition 3.5.8. For any σ, τ ∈ Sn , n ≥ 2, the sign of στ is the product of the signs of σ and τ . Thus, there is a homomorphism ε : Sn → {±1} obtained by setting ε(σ) equal to the sign of σ. In particular, ε has the property that ε(τ ) = −1 for any transposition τ .


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Proof Since we may write τ as a product of m transpositions in such a way that the sign of τ is (−1)m , we may argue by induction, assuming that τ is a transposition, say τ = (i j). Write σ as a product of disjoint cycles: σ = σ1 · · · σk . There are four cases to consider, where a non-trivial orbit of σ is one which has more than one element: 1. Neither i nor j lies in a non-trivial orbit of σ. 2. j lies in a non-trivial orbit of σ and i does not. 3. i and j lie in different non-trivial orbits of σ. 4. Both i and j lie in the same non-trivial orbit of σ. We analyze the decomposition of στ as a product of disjoint cycles in each of these cases. The first case is simplest, as then σ and τ are disjoint. Thus, τ is one of the cycles in the decomposition of στ , and the others are those from the decomposition of σ. The result follows in this case from the definition of sign. In case 2, we may assume that σ is an r-cycle, σ = (i1 · · · ir ), with ir = j. Then στ = (i i1 · · · ir ), an (r + 1)-cycle. In case 3, we may assume that σ is the product of two disjoint cycles, σ1 = (i1 · · · ir ), with ir = i, and σ2 = (j1 · · · js ), with js = j. Here, we see that στ is the (r + s)-cycle στ = (i1 · · · ir j1 · · · js ). The signs work out correctly once again. Case 4 is the most complicated. We may assume that σ is a single cycle. The analysis falls naturally into sub-cases. In the first, σ = τ = (i j), and here στ = e, as transpositions have order 2. The next case is where σ is an r-cycle with r ≥ 3 and i and j appear next to each other in the cycle representation of σ, say σ = (i1 · · · ir ) with ir−1 = i and ir = j. Here στ is the (r − 1)-cycle στ = (i1 · · · ir−1 ). In the final case, i and j cannot be written next to each other in the cycle representation of σ. Thus, we may write σ = (i1 · · · ir j1 · · · js ), with ir = i, js = j, and with r and s both greater than or equal to 2. Here, στ is given by στ = (i1 · · · ir )(j1 · · · js ). The sign homomorphism gives rise to an important subgroup of Sn . Definitions 3.5.9. The n-th alternating group, An , is the kernel of the sign homomorphism ε : Sn → {±1}. We say that a permutation is even if it lies in An (i.e., if its sign is 1). A permutation is said to be odd if its sign is −1. The fact that ε is a homomorphism allows us to determine the sign of a permutation from any representation of it as a product of transpositions. Corollary 3.5.10. Let σ ∈ Sn . Then σ ∈ An if and only if σ may be written as a product of an even number of transpositions. If σ ∈ An , then whenever σ is written as a product of transpositions, the number of transpositions must be even. Proof The homomorphism ε carries any product of an odd number of transpositions to −1 and carries any product of an even number of transpositions to 1. Thus, given any two presentations of a permutation σ as a product of transpositions, the number of transpositions in the two presentations must be congruent mod 2. We have made use of the fact that Sn is generated by the transpositions. Using Corollary 3.5.10, we may obtain a similar result for An .


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Corollary 3.5.11. For n ≥ 3, the alternating group An is generated by 3-cycles. Thus, since each 3-cycle lies in An , a permutation σ ∈ Sn lies in An if and only if σ is a product of 3-cycles. Proof Corollary 3.5.10 shows that a permutation is in An if and only if it is the product of an even number of transpositions. Thus, it suffices to show that the product of any two transpositions is a product of 3-cycles. If the two transpositions are not disjoint, then either they are equal, in which case their product is the identity, or the product has the form (a b) · (b c) = (a b c), a 3-cycle. The remaining case is the product of two disjoint cycles, (a b) and (c d). But then we have (a b) · (c d) = (a b c) · (b c d), the product of two 3-cycles. Exercises 3.5.12. 1. Write the product (1 2 3 4) · (3 4 5) as a product of disjoint cycles. What is the order of this product? 2. Consider the function from Z7 to itself induced by multiplication by 2. (Recall that multiplication mod n gives a well-defined binary operation on Zn .) Identifying the elements of Z7 with {1, . . . , 7} (by identifying k with k for 1 ≤ k ≤ 7), show that multiplication by 2 induces a permutation of {1, . . . , 7}. Write down its cycle structure. 3. Repeat the preceding problem using multiplication by 3 in place of multiplication by 2. 4. What are the elements of A3 ? 5. What are the elements of A4 ? 6. Recall from Problem 2 of Exercises 3.3.23 that S3 is isomorphic to the dihedral group D6 . Which subgroup of D6 corresponds to A3 under this isomorphism? 7. Let p be prime and let r ≥ 1. Show that Sk has an element of order pr if and only if k ≥ pr . (Hint: Write the element in question as a product of disjoint cycles. What can you say about the length of the cycles?) 8. For a given n, what is the smallest k such that Zn embeds in Sk ? (Hint: The answer depends on the prime decomposition of n.) 9. What is the largest value of n such that Zn embeds in S8 ? ‡ 10. Expanding on Remarks 3.5.4, show that the number of k-cycles in Sn is 1/k times the number of ordered k-element subsets of {1, . . . , n}. Deduce from Problem 3 of n! Exercises 3.4.7 that there are k·(n−k)! k-cycles in Sn .


3.6

71

Conjugation and Other Automorphisms

In addition to the study of actions of the group G on sets, it is useful to study actions of other groups on G. Of course, S(G) acts on G, but we have seen that the structure of a symmetric group S(X) depends only on the size of the set X. In particular, the action of S(G) on G loses all the information about the group structure of G, and captures only its size. To capture the group structure on G, a group should act on it through permutations σ : G → G which are group homomorphisms. These are precisely the group isomorphisms from G to itself. Definitions 3.6.1. An automorphism of a group G is a group isomorphism from G to itself. We write Aut(G) ⊂ S(G) for the set of all automorphisms of G. The reader may check the details of the following examples, using Problem 6 of Exercises 2.8.5 for the examples involving dihedral groups, and Problem 5 of Exercises 2.9.7 for the examples involving quaternionic groups. Examples 3.6.2. 1. Let G be an abelian group. Then there is an automorphism f : G → G given by f (g) = g −1 for all g ∈ G. 2. Let 0 ≤ k < n. Then there is an automorphism f ∈ Aut(D2n ) determined by f (b) = b, f (a) = abk . 3. There is an automorphism f ∈ Aut(D2n ) determined by f (b) = b−1 , f (a) = a. 4. Let 0 ≤ k < 2n. Then there is an automorphism f ∈ Aut(Q4n ) determined by f (b) = b, f (a) = abk . 5. There is an automorphism f ∈ Aut(Q4n ) determined by f (b) = b−1 , f (a) = a. We now state an immediate consequence of Lemma 2.5.7. Lemma 3.6.3. Let G be a group. Then Aut(G) is a subgroup of S(G). Definition 3.6.4. We say that the group K acts on G via automorphisms if G is a K-set in such a way that the action of each k ∈ K, k : G → G (defined as usual by

k (g) = k · g), is an automorphism of G. The proof of the following proposition is identical to that of Proposition 3.3.16. Proposition 3.6.5. There is a one-to-one correspondence between the actions of K on G via automorphisms and the homomorphisms Λ : K → Aut(G). Here, a given action corresponds to the homomorphism defined by Λ(k) = k . This formalism is useful, as we are about to define a very important action of G on itself through automorphisms, called conjugation. But we cannot denote the resulting G-set by “G,” as we reserve this symbol for the standard action induced by left multiplication. So it is useful to be able to refer to conjugation in terms of a homomorphism from G to Aut(G). Definition 3.6.6. Let G be a group and let x ∈ G. By conjugation by x, we mean the function cx : G → G specified by cx (g) = xgx−1 .


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Lemma 3.6.7. Conjugation by x ∈ G is an automorphism of G. There is an action of G on itself by automorphisms given by the homomorphism Γ : G → Aut(G) defined by Γ(x) = cx . Proof To see that cx is a homomorphism, we have cx (g)cx (g ) = xgx−1 xg x−1 = xgg x−1 = cx (gg ). It is a bijection for the same reason that left or right multiplication by any element of G is a bijection: There is an inverse function given by an appropriate multiplication. In this case the inverse function is conjugation by x−1 . Thus, there is a well-defined function Γ : G → Aut(G) with Γ(x) = cx . To see that Γ is a group homomorphism, we have (cx ◦ cy )(g) = cx (ygy −1 ) = xygy −1 x−1 = (xy)g(xy)−1 = cxy (g), because (xy)−1 = y −1 x−1 . The elements in the image of Γ : G → Aut(G) have a special name. Definitions 3.6.8. An automorphism f ∈ Aut(G) is said to be an inner automorphism of G if f = cx , the conjugation by x, for some x ∈ G. We write Inn(G) ⊂ Aut(G) for the set of inner automorphisms. Note that since Inn(G) = im Γ, Inn(G) is a subgroup of Aut(G). The isotropy subgroup of y ∈ G under the conjugation action is by definition the set {x ∈ G | xyx−1 = y}. It is an important enough subgroup that it merits a name: Definition 3.6.9. Let y be an element of the group G. Then the centralizer of y in G is the subgroup CG (y) = {x ∈ G | xyx−1 = y}. Of course, if x ∈ H ⊂ G, then we write CH (x) for the centralizer of x in H. When there’s no ambiguity, we shall simply refer to CG (x) as the centralizer of x. The next lemma is useful in understanding the effect of conjugation. The proof is left to the reader. Lemma 3.6.10. Let x, y ∈ G. Then the following are equivalent. 1. x and y commute. 2. x ∈ CG (y), i.e., xyx−1 = y. 3. xyx−1 y −1 = e. We now give two useful consequences of Lemma 3.6.10. Corollary 3.6.11. Suppose that the elements x, y ∈ G commute with each other. Then every element of x commutes with every element of y. Proof Since x ∈ CG (y) and since CG (y) is a subgroup of G, x ⊂ CG (y). Thus, y commutes with xk for all k ∈ Z, and hence y ∈ CG (xk ) as well. So y ⊂ CG (xk ), and hence y l commutes with xk for all k, l ∈ Z.


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Corollary 3.6.12. The element x ∈ G is in the kernel of the conjugation-induced homomorphism Γ : G → Aut(G) if and only if x commutes with every element of G. Proof We have x ∈ ker Γ if and only if x is in the isotropy subgroup of every y ∈ G. But this says that x ∈ CG (y) for every y ∈ G, and hence x commutes with every element of G. We shall refer to the kernel of Γ as the center, Z(G), of G:3 Definition 3.6.13. The center, Z(G), of a group G is given as follows: Z(G) = {x ∈ G | xy = yx for all y ∈ G}. Definition 3.6.14. The conjugacy class of g ∈ G is {xgx−1 | x ∈ G}, which is the orbit of g under the action of G on itself by conjugation. Its elements are called the conjugates of g. Since the isotropy subgroup of g is CG (g), the next result follows from Corollary 3.3.9. Corollary 3.6.15. The number of conjugates of g ∈ G is equal to [G : CG (g)], the index in G of the centralizer, CG (g), of g in G. The fact that CG (g) is the set of all elements that commute with g gives a characterization of Z(G) in terms of conjugacy classes. Corollary 3.6.16. The conjugacy class of g ∈ G consists of a single element if and only if g ∈ Z(G). Proof [G : CG (g)] = 1 if and only if CG (g) = G. The G-set Counting Formula (Corollary 3.3.15) for the conjugation action on G is extremely useful. We shall give a variant on it, called the Class Formula. We shall need some setup for it first. Definitions 3.6.17. We say that a conjugacy class of G is nontrivial if it has more than one element. By a set of class representatives for the nontrivial conjugacy classes we mean a subset X ⊂ G such that 1. The conjugacy class of each x ∈ X is nontrivial. 2. If x, y ∈ X with x = y, then the conjugacy classes of x and y intersect in the null set. 3. Every element of G that is not in Z(G) lies in the conjugacy class of some x ∈ X. The desired Class Formula is now immediate from the G-set Counting Formula. Corollary 3.6.18. (Class Formula) Let G be a finite group and let X be a set of class representatives for the nontrivial conjugacy classes in G. Then [G : CG (x)]. |G| = |Z(G)| + x∈X

3 The

German word for center is Zentrum.


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We can compute the conjugates and the centralizers of the elements in the dihedral and quaternionic groups by hand, using the explicit information we have about these groups. We shall leave these calculations for the next set of exercises. The next result computes the conjugates of a k-cycle in Sn . Lemma 3.6.19. Let σ be any permutation and let (i1 · · · ik ) be any k-cycle in Sn . Then σ · (i1 · · · ik ) · σ −1 is a k-cycle: σ · (i1 · · · ik ) · σ −1 = (σ(i1 ) · · · σ(ik )). Proof If j is not in {σ(i1 ), . . . , σ(ik )}, then σ −1 (j) is not in {i1 , . . . , ik }, and hence (i1 · · · ik ) acts as the identity on σ −1 (j). But then σ · (i1 · · · ik ) · σ −1 (j) = j, so that σ · (i1 · · · ik ) · σ −1 acts as the identity outside {σ(i1 ), . . . , σ(ik )}. For 1 ≤ j < k, σ · (i1 · · · ik ) · σ −1 (σ(ij )) = σ · (i1 · · · ik )(ij ) = σ(ij+1 ). Similarly, σ · (i1 · · · ik ) · σ −1 (σ(ik )) = σ(i1 ), so that σ · (i1 · · · ik ) · σ −1 = (σ(i1 ) · · · σ(ik )), as claimed. Lemma 3.6.19 gives important information about the behavior of the symmetric groups. Here is a sample. Corollary 3.6.20. Suppose that σ ∈ Sn lies in the centralizer of the k-cycle (i1 · · · ik ). Then σ({i1 , . . . , ik }) = {i1 , . . . , ik }. Proof Let σ ∈ Sn Then σ · (i1 · · · ik ) · σ −1 = (σ(i1 ) · · · σ(ik )). Thus, Supp(σ · (i1 · · · ik ) · σ −1 )

= {σ(i1 ), . . . , σ(ik )} = σ({i1 , . . . , ik }).

Thus, if σ · (i1 · · · ik ) · σ −1 = (i1 · · · ik ), then σ({i1 , . . . , ik }) = Supp((i1 · · · ik )) = {i1 , . . . , ik }.

There is an important action related to conjugation, in which G acts on the set of homomorphisms from another group into G. We shall use a notation consistent with that used in our discussion of category theory in Chapter 6. Notation 3.6.21. Let K and G be groups. We write MorGp (K, G) for the set of group homomorphisms from K to G. Definition 3.6.22. Let K and G be groups. For g ∈ G and f ∈ MorGp (K, G), the conjugate of f by g, written g · f , is the homomorphism (g · f ) ∈ MorGp (K, G) defined by (g · f )(k) = gf (k)g −1 for k ∈ K. Two homomorphisms f, f ∈ MorGp (K, G) are said to be conjugate if f = g · f for some g ∈ G. Note that g · f is equal to the composite cg ◦ f , where cg : G → G is conjugation by g. Thus, g · f is a homomorphism, as claimed. The following lemma is now immediate.


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Lemma 3.6.23. Let K and G be groups. Then there is an action of G on MorGp (K, G) obtained by setting the action of g ∈ G on f ∈ MorGp (K, G) to be equal to the conjugate homomorphism g · f defined above. Conjugate homomorphisms are important in a number of questions, including the classification of semidirect products and the cohomology of groups. Exercises 3.6.24. 1. What are all the conjugacy classes of elements of D6 ? Write down all the elements in each class. 2. What are all the conjugacy classes of elements of D8 ? Write down all the elements in each class. 3. What are all the conjugacy classes of elements of Q8 ? Write down all the elements in each class. 4. What are all the conjugacy classes of elements of D2n for n ≥ 3? (Hint: The answer depends in a crucial way on the value of n mod 2.) 5. What are all the conjugacy classes of elements of Q4n ? 6. Find the centralizers of each of the elements of Q8 . 7. Find the centralizers of the elements a, b ∈ D2n . 8. Find the centralizers of the elements a, b ∈ Q4n . 9. Let H be a subgroup of G and let x ∈ H. Show that CH (x) = H ∩ CG (x). 10. Suppose that Sn has an element of order k. Show that D2k embeds in Sn . 11. Consider the 2-cycle (n − 1 n) ∈ Sn . Show that the centralizer of (n − 1 n) in Sn is isomorphic to Sn−2 × Z2 . ‡ 12. Let a, b, c, d be distinct elements of {1, . . . , n}. Show that any element of Sn which centralizes both (a b c) and (b c d) must fix both a and d. Deduce that if n ≥ 4, then the only element which centralizes every 3-cycle is the identity. 13. Show that the centralizer of any k-cycle in Sn is isomorphic to Sn−k × Zk . We shall see in Section 4.2 (using a straightforward application of Lemma 3.6.19 above) that n! any two k-cycles in Sn are conjugate. Assuming this, deduce that there are k·(n−k)! k-cycles in Sn . (See Problem 10 of Exercises 3.5.12 for a different line of argument.) 14. What is the centralizer of (1 2 3) in A4 ? How many conjugacy classes of 3-cycles are there in A4 ? 15. Let σ, τ ∈ S(X). Show that Supp(τ στ −1 ) = τ (Supp(σ)). 16. What is the center of D8 ? 17. What is the center of D2n for n ≥ 3? (Hint: The answer depends in a crucial way on the value of n mod 2.)


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18. What is the center of Q8 ? 19. What is the center of Q4n ? 20. What is the center of Sn ? What is the center of An ? ‡ 21. Let H be any group and write H n for the product H × · · · × H of n copies of H. Show that Sn acts on H n through automorphisms via σ · (h1 , . . . , hn ) = (hσ−1 (1) , . . . , hσ−1 (n) ). 22. Verify that the specifications given in Examples 3.6.2 do in fact give automorphisms of the groups in question. 23. Show that the only automorphism of Z4 other than the identity is the one which takes each element to its inverse. 24. Let G be a finite abelian group and let f : G → G be given by f (g) = g n for all g ∈ G, where n is an integer that is relatively prime to |G|. Show that f is an automorphism of G. If G is cyclic of order 7 and n = 2, what is the order of f in Aut(G)? 25. Show that not every automorphism of D8 is an inner automorphism. 26. Show that every automorphism of D6 is an inner automorphism. (Hint: The composite of two inner automorphisms is an inner automorphism.) Deduce that Aut(D6 ) ∼ = D6 . 27. Show that any two nontrivial homomorphisms from Z2 to D6 are conjugate. 28. Show that any two nontrivial homomorphisms from Z3 to D6 are conjugate. 29. What are the conjugacy classes of homomorphisms from Z4 to D8 ? 30. Let G be a group and let f : G → Sn be a homomorphism. Write {1, . . . , n}f for the G-set obtained from the action on {1, . . . , n} induced by f under the correspondence of Proposition 3.3.16: for g ∈ G and i ∈ {1, . . . , n}f , g · i = f (g)(i). Let σ ∈ Sn and let f = σ · f , the conjugate homomorphism obtained by conjugating f by σ. Show that σ induces a G-isomorphism of G-sets, σ : {1, . . . , n}f → {1, . . . , n}f . 31. Let G be a group. Show that there is a one-to-one correspondence between the conjugacy classes of homomorphisms from G to Sn and the G-isomorphism classes of the G-sets with n elements. Here, two G-sets are in the same G-isomorphism class if and only if they are G-isomorphic. 32. Let X and Y be sets, with X finite. Show that if i : X → Y and j : X → Y are any two injective functions, then the induced homomorphisms i∗ : S(X) → S(Y ) and j∗ : S(X) → S(Y ) (as defined in Problem 15 of Exercises 3.1.16) are conjugate. (Hint: In the case that Y is infinite, set theory shows that Y may be put in one-to-one correspondence with the complement of any of its finite subsets.) 33. Let i and j be injective functions from {1, . . . , n} to Z+ . Show that the induced homomorphisms i∗ , j∗ : Sn → S∞ are conjugate.


3.7

77

Conjugating Subgroups: Normality

We haven’t yet made full use of the fact that conjugation is an action by automorphisms. The key point is that if f is an automorphism of G and if H ⊂ G is a subgroup, then f (H) = {f (h) | h ∈ H} is also a subgroup. We get an induced action of Aut(G) on the set Sub(G) of all subgroups of G, via f · H = f (H). The following lemma is immediate. Lemma 3.7.1. If K acts on G via automorphisms, then there is an induced action of K on Sub(G), the set of all subgroups of G. Let’s consider the action of G on Sub(G) induced by conjugation. Here, for x ∈ G and H ∈ Sub(G), the subgroup obtained by letting x act on H is xHx−1 = {xhx−1 | h ∈ H}: Definition 3.7.2. Let H be a subgroup of G, and let x ∈ G. The conjugate of H by x is xHx−1 = {xhx−1 | h ∈ H}, the subgroup obtained by letting x act on H by conjugation. Of course, the set of all conjugates of H is the orbit of H under the action of G on Sub(G) by conjugation. The isotropy subgroup of H ∈ Sub(G) under the conjugation action is called NG (H), the normalizer of H in G: Definition 3.7.3. Let H be a subgroup of G. The normalizer in G of H, NG (H), is given as NG (H) = {x ∈ G | xHx−1 = H}. It is the isotropy subgroup of H under the action of G on Sub(G) by conjugation. Our study of G-sets now allows us to calculate the size of the orbit of H under conjugation. Corollary 3.7.4. The number of conjugates of H ⊂ G is [G : NG (H)], the index of the normalizer of H in G. Of course, if x ∈ H, then conjugation by x gives an automorphism of H, and hence xHx−1 = H. Corollary 3.7.5. Let H be a subgroup of G. Then H ⊂ NG (H). In particular, the number of conjugates of H in G divides [G : H]. Proof We just saw that xHx−1 = H for x ∈ H, so H ⊂ NG (H), as claimed. The last statement follows from Problem 4 of Exercises 3.2.15: If [G : H] is finite, then [G : H] = [G : NG (H)] · [NG (H) : H]. (The reader should be able to verify this easily in the case that G is finite, which is the most important case for our purposes.) Thus, we have H ⊂ NG (H) ⊂ G. In practice, we can find examples where NG (H) = H, and others where NG (H) strictly contains H and is strictly contained in G. Finally, there is the case where NG (H) = G, which is important enough to merit a name. Definitions 3.7.6. A subgroup H ⊂ G is called normal if NG (H) = G (i.e., xHx−1 = H for all x ∈ G). We write H G to indicate that H is a normal subgroup of G. A subgroup H ⊂ G is called characteristic if f (H) = H for all automorphisms f of G.


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Clearly, a characteristic subgroup of G is normal in G. However, not all normal subgroups are characteristic. For instance, conjugation is trivial in an abelian group, and hence every subgroup of an abelian group is normal. However, as we shall see in the exercises below, the only characteristic subgroups of Z2 × Z2 are e and the group itself. Thus, abelian groups can have subgroups which are not characteristic. We shall consider many more examples of both normal and characteristic subgroups in the exercises. If H G, then we get an action of G on H through automorphisms: Lemma 3.7.7. Let H be a normal subgroup of G. Then for x ∈ G, conjugation by x gives an automorphism, cx , of H: cx (h) = xhx−1 for all h ∈ H. Moreover, there is a homomorphism Γ : G → Aut(H) defined by Γ(x) = cx for all x ∈ G. Proof Since cx gives an automorphism of G, it restricts to a group isomorphism from H to cx (H) = xHx−1 . Since xHx−1 = H, this restriction is precisely an automorphism of H. Now clearly, cx ◦ cy = cxy , and the result follows. Essentially the same proof gives the following lemma. Lemma 3.7.8. Let H be a characteristic subgroup of G. Then every automorphism f : G → G restricts to an automorphism f : H → H. We obtain a restriction homomorphism ρ : Aut(G) → Aut(H), by setting ρ(f ) to be the restriction f : H → H for each f ∈ Aut(G). As we shall begin to see in the next chapter, the study of normal subgroups is one of the most important subjects in group theory. It is also a very complicated subject. For instance, as we shall see in the exercises below, Gertrude Stein’s Theorem fails for normal subgroups: we can find subgroups K ⊂ H ⊂ G such that K H and H G, but K is not normal in G. This failure of Gertrude Stein’s Theorem helps motivate the study of characteristic subgroups: Corollary 3.7.9. Suppose given subgroups K ⊂ H ⊂ G such that H G and K is a characteristic subgroup of H. Then K G. Proof Let x ∈ G. Then conjugation by x restricts to an automorphism of H, and hence preserves K. Recognition of normal subgroups is an important endeavor. Of course, if H is finite, then any injection from H to H is a bijection, so that xHx−1 ⊂ H if and only if xHx−1 = H. However, there are examples of infinite groups H ⊂ G and x ∈ G where xHx−1 is contained in but not equal to H. However, if xHx−1 ⊂ H for all x ∈ G, things work out nicely: Proposition 3.7.10. Let H be a subgroup of G. Then H G if and only if xHx−1 ⊂ H for all x ∈ G. Also, H is characteristic if and only if f (H) ⊂ H for all automorphisms f of G. Proof Suppose that xHx−1 ⊂ H for each x ∈ G. To show that H G, we must show that H ⊂ xHx−1 for each x ∈ G. But x−1 Hx ⊂ H, and hence H = x(x−1 Hx)x−1 ⊂ xHx−1 , as desired. The proof for the case of characteristic subgroups is nearly identical: if f (H) ⊂ H for all f ∈ Aut(G), then f −1 (H) ⊂ H, so H = f (f −1 (H)) ⊂ f (H).


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Notice that the proof of Proposition 3.7.10 actually shows more than was stated. For instance, in the case of normality, it shows that if xHx−1 ⊂ H and x−1 Hx ⊂ H, then xHx−1 = H, and hence x ∈ NG (H). Thus, we have the following proposition. Proposition 3.7.11. Let H and K be subgroups of G. Suppose that xHx−1 ⊂ H for all x ∈ K. Then K ⊂ NG (H). One application of the concept of normality is the following variant of Proposition 2.10.8. Proposition 3.7.12. Let H and K be subgroups of the group G. Then G is the internal direct product of H and K if and only if the following conditions hold. 1. H ∩ K = e. 2. Both H and K are normal in G. 3. G is generated by the elements of H and K. Proof Suppose that G is the internal direct product of H and K. By Proposition 2.10.8, it suffices to show that H and K are normal in G. But by definition of the internal direct product, this will follow if we know that the subgroups H × e and e × K are normal in H × K. The verification of this is left to the reader. Conversely, suppose the three conditions hold. By Proposition 2.10.8, it suffices to show that for each h ∈ H and k ∈ K, the elements h and k commute. By Lemma 3.6.10, it suffices to show that for h ∈ H and k ∈ K, hkh−1 k −1 = e. By condition 1, this says that hkh−1 k −1 ∈ H ∩ K. Now K G, so hkh−1 ∈ K, and hence hkh−1 k −1 ∈ K. Similarly, the normality of H in G gives kh−1 k −1 ∈ H, and hence hkh−1 k −1 ∈ H, so the result follows. Normality is intimately connected with the relationship between left and right cosets. The reader may verify the following lemma. Lemma 3.7.13. Let H be a subgroup of G. Then the following conditions are equivalent. 1. H is normal in G. 2. For each x ∈ G, we have xH = Hx. 3. Every left coset of H in G is equal to a right coset of H in G; i.e., for each x ∈ G there is an element y ∈ G such that xH = Hy. Exercises 3.7.14. 1. Let G be a group and let H ⊂ G × G be given by H = {(g, g)|g ∈ G}. (H is called the diagonal subgroup of G × G.) Show that H G × G if and only if G is abelian. 2. Let G be a group and let y ∈ G. Show that y G if and only if xyx−1 ∈ y for all x ∈ G. Show that y is a characteristic subgroup of G if and only if f (y) ∈ y for each f ∈ Aut(G). 3. Show that the subgroup b ⊂ D2n is normal.


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‡ 4. Show that the subgroup b ⊂ D2n is characteristic for n ≥ 3. 5. Find a non-normal subgroup of D8 . 6. Show that every subgroup of Q8 is normal. 7. Find the unique characteristic subgroup of Q8 which is equal to neither e nor Q8 . ‡ 8. Show that if n ≥ 3, then b ⊂ Q4n is characteristic. ‡ 9. Let H be a subgroup of Zn and let f : Zn → Zn be any homomorphism. Show that f (H) ⊂ H. Deduce that every subgroup of Zn is characteristic. (Hint: Show that H = d = {dk | k ∈ Zn } for some d that divides n.) † 10. Using the fact that An is the subgroup of Sn generated by the 3-cycles, show that An Sn . Thus Lemma 3.7.7 provides a homomorphism Γ : Sn → Aut(An ), where Γ(σ) is induced by conjugation by σ for all σ ∈ Sn . Deduce from Problem 12 of Exercises 3.6.24 that Γ is an injection for n ≥ 4. 11. Show that the only characteristic subgroups of G = Z2 × Z2 are the identity and G itself. † 12. Let H be an index 2 subgroup of the finite group G. Show that H G. 13. Show that any index 2 subgroup of an infinite group is normal. 14. For x ∈ G, show that CG (x) ⊂ NG (x). 15. If x ∈ G has order 2, show that CG (x) = NG (x). Deduce that x G if and only if x ∈ Z(G). 16. Give an example of a group G and an x ∈ G with CG (x) = NG (x). 17. List the distinct subgroups of D2n that are conjugate to a. 18. What is the normalizer of a ⊂ Q16 ? 19. List the distinct subgroups of Q16 that are conjugate to a. 20. What is the normalizer of a ⊂ Q4n ? List the distinct subgroups of Q4n that are conjugate to a. (Hint: The answer depends in a crucial way on the value of n mod 2.) 21. Show that characteristic subgroups satisfy Gertrude Stein’s Theorem: If H is a characteristic subgroup of G, and K is a characteristic subgroup of H, show that K is a characteristic subgroup of G. 22. Show that S∞ S(Z+ ). 23. Show that S({2, 3, . . . }) and S({3, 4, . . . }) are not conjugate as subgroups of S(Z+ ). 24. Let H be a subgroup of G. Show that H is a normal subgroup of NG (H). Show that NG (H) is the largest subgroup of G that contains H as a normal subgroup; i.e., if H ⊂ K ⊂ G and H K, then K ⊂ NG (H). † 25. Let H and K be subgroups of G. Show that NG (H) ∩ NG (K) ⊂ NG (H ∩ K).


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26. Let H be a subgroup of G and let x ∈ G. Show that NG (xHx−1 ) = xNG (H)x−1 . 27. Show that Gertrude Stein’s Theorem fails for normal subgroups: i.e., find a group G with a normal subgroup K G and a subgroup H of K such that H K, but H is not normal in G. (Hint: Find a group G with a subgroup H such that NG (H) is normal in G but not equal to G.) 28. Let H be a subgroup of the finite group G. Show that the number of elements in −1 is less than or equal to [G : NG (H)] · (|H| − 1) + 1. Deduce that no x∈G xHx proper subgroup of G contains an element in every conjugacy class of G. 29. Let X be a G-set and let x ∈ X. Show that the isotropy subgroup of gx is gGx g −1 . 30. (See Problem 18 of Exercises 3.3.23.) What are the elements of the H-fixed-point set (G/H)H ? For which elements yH ∈ (G/H)H is the G-map from G/H to itself which takes eH to yH a G-isomorphism (i.e., which elements of (G/H)H have isotropy subgroup exactly equal to H)? ‡ 31. Let H be a subgroup of G and consider the standard action of G on G/H. Let Λ : G → S(G/H) be the homomorphism induced by this action. Show that ker Λ = xHx−1 . x∈G

32. Let H be a subgroup of G and let X be the set of all subgroups conjugate to H: X = {xHx−1 | x ∈ G}. Thus, X is the orbit of H under the action of G on Sub(G) (the set of subgroups of G) by conjugation. Since X is a G-set, we get an induced homomorphism Λ : G → S(X). Show that ker Λ = xNG (H)x−1 . x∈G

Chapter 4

Normality and Factor Groups The closing section of Chapter 3 gives the definition of a normal subgroup: A subgroup H ⊂ G is normal if xHx−1 = H for all x ∈ G. We write H G to denote that H is a normal subgroup of G. The normal subgroups of a given group are absolutely vital in understanding its structure. Some groups G have no normal subgroups other than the trivial subgroup e and the group G itself. Such groups are called simple. We shall see, via the technique of composition series below, that all finite groups are built up out of the finite simple groups. The finite simple groups are all known. In one of the triumphs of mathematical history, the finite simple groups have been classified, according to a plan of attack developed by Daniel Gorenstein that played out over a sequence of papers by several authors. One of the final steps was the construction, by Robert Griess, of a group that was then known as the “monster group,” as its order is nearly 1054 . In current parlance, the monster group is the sporadic simple group F1 .1 We shall not need the classification by Gorenstein et. al., as we are concentrating here on the groups of relatively low order. As will become apparent in the exercises from the next two chapters, the only simple groups whose order is ≤ 60 are the cyclic groups Zp , where p is a prime < 60, together with the alternating group A5 . In any case, the study of the ways in which finite groups can be put together out of the simple groups is the most important remaining problem for the classification of finite groups. It is known as the Extension Problem. Thus, the Extension Problem is, in the opinion of this author at least, one of the important unsolved problems in mathematics. The study of normal subgroups and factor groups is at its core. We do not yet have the language with which to give a precise statement of the Extension Problem, but we shall do so within this chapter. In fact, we shall develop techniques to study certain special cases of the Extension Problem. These special cases will turn out to be enough, together with the Sylow theorems and the material on G-sets from Chapter 3, to classify the groups in a fair number of the lower orders. 1 See Finite Simple Groups, An Introduction to Their Classification, by Daniel Gorenstein, Plenum Press, 1982.

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CHAPTER 4. NORMALITY AND FACTOR GROUPS

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The most important aspect of normal subgroups is that if H is normal in G, then there is a group structure on the set of left cosets G/H such that the canonical map π : G → G/H is a homomorphism. If G is finite and H is nontrivial, then G/H has a lower order than G, and hence is, in general, easier to analyze. If G has enough normal subgroups (e.g. if G is abelian or solvable), then induction arguments on the order of G may be used to deduce facts about G from facts about the lower order factor groups G/H. In Section 4.1, we give the Noether Isomorphism Theorems, which study the properties of the factor groups G/H when H is normal in G. These are truly fundamental in the understanding of groups, and will open the door to some new classification results. As an application, we shall show that if G is finite and p is the smallest prime dividing the order of G, then any subgroup of G of index p is normal in G. In Section 4.2, we give the technique of composition series, which shows that finite groups are built up out of simple groups. We also show that the alternating groups An are simple for n ≥ 5. Then, in Section 4.3, we give the Jordan–H¨ older Theorem, which shows that the simple groups out of which a given group is built are unique, though the way in which they are put together may not be. Then, by an induction on order, using the fact that every subgroup of an abelian group is normal, we give a complete classification of the finite abelian groups, the Fundamental Theorem of Finite Abelian Groups. In Section 4.5, we calculate the automorphism group of a cyclic group. This calculation will be of essential value in constructing some new groups by the technique of semidirect products and in classifying the extensions of a given cyclic group. Section 4.6 develops the technique of semidirect products, by which new groups may be constructed from old. Unlike the direct product, many of the semidirect products of two abelian groups are nonabelian. We shall construct some interesting new groups in this manner. Section 4.7 defines and studies the extensions of one group by another. Extensions describe some of the ways that groups are built up out of smaller groups. We shall state the Extension Problem here and shall develop some classification techniques, particularly for groups that are semidirect products. We shall make some significant progress in obtaining actual classification results.

4.1

The Noether Isomorphism Theorems

Here, we study normal subgroups H G and the group structure that’s induced on the set of left cosets G/H. One convenient source of normal subgroups (which will also turn out to give all of them) is from the kernels of homomorphisms: Lemma 4.1.1. Let f : G → G be a homomorphism. Then ker f is a normal subgroup of G. Proof Let K = ker f . By Proposition 3.7.10, it suffices to show that xKx−1 ⊂ K for each x ∈ G, i.e., that xkx−1 ∈ K for all x ∈ G and all k ∈ K. But if k ∈ K, then f (xkx−1 ) = f (x) · f (k) · f (x)−1 = f (x) · e · f (x)−1 = e, and hence xkx−1 ∈ K. We shall see that im f need not be normal in G . We next explore the relationship between normality of a subgroup H ⊂ G and group structures on the set of cosets G/H:


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Theorem 4.1.2. Let H be a subgroup of G. Then H is normal in G if and only if there is a group structure on the set G/H of left cosets of H with the property that the canonical map π : G → G/H is a homomorphism. If H G, then the group structure on G/H which makes π a homomorphism is unique: we must have gH · g H = gg H for all g, g ∈ G. Moreover, the kernel of π : G → G/H is H. Thus, a subgroup of G is normal if and only if it is the kernel of a homomorphism out of G. Proof Suppose that there is a group structure on G/H such that π is a homomorphism. Then the identity element of G/H is π(e) = eH. So g ∈ ker π if and only if gH = eH. By Lemma 3.2.3, this is true if and only if g ∈ H. Thus, H is the kernel of π, and hence is normal by Lemma 4.1.1. Clearly, there is at most one group structure on G/H which makes π a homomorphism, as, since π(g) = gH, we must have gH · g H = gg H for all g, g ∈ G. Suppose then that H is normal in G. We wish to show that gH · g H = gg H defines a group structure on G/H. We first show that it gives a well defined binary operation. Thus, if gH = xH and g H = x H, we wish to show that gg H = xx H. But Lemma 3.2.3 shows that x = gh and x = g h for some h, h ∈ H. Thus, xx H = ghg h H = ghg H. By Lemma 3.2.3, this is equal to gg H provided that (gg )−1 ghg ∈ H. But (gg )−1 ghg = (g )−1 g −1 ghg = (g )−1 hg . And (g )−1 hg ∈ H because H G. Thus, the operation is well-defined. That this operation gives a group structure now follows immediately from the group laws in G: associativity in G gives associativity in G/H, and π(e) = eH is an identity element for G/H. Finally, the inverse of gH is g −1 H, so G/H is indeed a group. Definition 4.1.3. When H G, we shall simply write G/H for the group structure on G/H which makes π a homomorphism (i.e., with the multiplication gH · g H = gg H), and call it the factor group, or quotient group, of G modulo H. Example 4.1.4. We’ve already seen examples of factor groups, as the groups Zn are precisely the factor groups Z/n. The point is that for k ∈ Z, the congruence class of k modulo n is precisely the coset k + n. (To see this, note that l ≡ k mod n if and only if l − k ∈ n, i.e., l ∈ k + n.) The canonical map π : Z → Zn that we’ve been using coincides with the canonical map above, as it takes each element in Z to its coset modulo n. And, indeed, the group operation in Zn is defined precisely as in the more general case above. Another way of seeing that Z/n ∼ = Zn will be given using the Noether Isomorphism Theorems, below. Since |G/H| = [G : H], Lagrange’s Theorem gives the following formula. Lemma 4.1.5. Let H G with G finite. Then |G| = |H| · |G/H|.

Notation 4.1.6. Let H G. For brevity, we shall generally write g in place of gH ∈ G/H. Thus, every element of G/H has the form g = π(g) for some g ∈ G, and have g · g = gg . The identity element is e = e = h for any h ∈ H.


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Note that |G/H| = [G : H] is less than |G| whenever G is finite and H = e. In this case, the factor group G/H is less complicated than G, and we may obtain information about G from an understanding of the groups G/H as H ranges over the normal subgroups of G. We can characterize the homomorphisms out of G/H in terms of the homomorphisms out of G. Proposition 4.1.7. Let H G. Then for any group G and any homomorphism f : G → G whose kernel contains H, there is a unique homomorphism f : G/H → G that makes the following diagram commute. / G GC < CC z CC zz z C zz π CC ! zz f G/H f

Conversely, if f : G → G factors by a commutative diagram as above, then H must be contained in ker f . Thus, the passage from a homomorphism f : G/H → G to the composite (f ◦π) : G → G gives a one-to-one correspondence between the homomorphisms from G/H to G and those homomorphisms from G to G whose kernels contain H. Proof Suppose given f : G → G . If a homomorphism f : G/H → G exists with f ◦ π = f , then for each h ∈ H, f (h) = f (h) = f (e) = e, and hence H ⊂ ker f . Moreover, f (g) = f ◦ π(g) = f (g), so f is unique if it exists. It suffices to show that such an f exists whenever H ⊂ ker f . Thus, suppose H ⊂ ker f . As we’ve just seen, f , if it exists, must be given by f (g) = f (g). The key is to show this is a well-defined function from G/H to G . Thus, suppose that g = g in G/H. By Lemma 3.2.3, this means g = gh for some h ∈ H. Thus, f (g ) = f (g)f (h) = f (g), since h ∈ ker f . So f is well-defined. But f (g)f (g ) = f (g)f (g ) = f (gg ) = f (gg ) = f (gg ), and hence f is a homomorphism. We call f the homomorphism induced by f . In the counting argument below, we again make use of the fact that |G/H| = [G : H]. Theorem 4.1.8. (First Noether Isomorphism Theorem) Let f : G → G be a homomorphism. Then f induces an isomorphism from G/ ker f to im f . Thus, if G is finite, |im f | = [G : ker f ], so Lagrange’s Theorem gives |G| = |ker f | · |im f |. Proof Let f : G/ ker f → im f be the homomorphism induced by f . Since f (g) = f (g), f is onto. Thus, it suffices (Lemma 2.5.9) to show that ker f = e. Let g ∈ ker f . Then f (g) = f (g) = e, so g ∈ ker f . But then g = e in G/ ker f . Example 4.1.9. Recall the rotation matrices Rθ ∈ Gl2 (R). We saw in Section 2.7 that the collection of all rotation matrices, {Rθ | θ ∈ R}, forms a subgroup, SO(2), of Gl2 (R). Moreover, as shown in Proposition 2.7.6, there is a surjective homomorphism exp : R → SO(2) given by exp(θ) = Rθ . The kernel of exp is shown there to be the cyclic subgroup 2π ⊂ R. Thus, by the first Noether theorem, SO(2) is isomorphic to R/2π.


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The Noether Theorem gives us another way to recognize Zn as Z/n. Corollary 4.1.10. Let n be a positive integer. Then Zn is isomorphic to the factor group Z/n. Proof The canonical map π : Z → Zn is a surjection, with kernel n. Recall that there is a homomorphism ε : Sn → {±1} defined by setting ε(σ) equal to the sign of σ for σ ∈ Sn . Since ε is onto, its image has order 2. But An is the kernel of ε, so we may calculate its order: Corollary 4.1.11. An has index 2 in Sn , and hence |An | = n!/2. The next result foreshadows the techniques we shall use in Chapter 5. Recall from Problem 12 of Exercises 3.7.14 that any index 2 subgroup of a finite group is normal. Using the First Noether Isomorphism Theorem, we shall give a generalization of this result. First, note from the example of a ⊂ D6 that a subgroup of prime index in a finite group need not be normal. Of course, 2 is the smallest prime, so the next result gives the desired generalization. Proposition 4.1.12. Let G be a finite group and let p be the smallest prime dividing the order of G. Then any subgroup of G of index p is normal. Proof Let H be a subgroup of index p in G, and consider the usual action of G on the set of left cosets G/H. As in Proposition 3.3.16, we get an induced homomorphism Λ : G → S(G/H) ∼ = Sp by setting Λ(g)(xH) = gxH for all g ∈ G and xH ∈ G/H. As shown in Proposition 3.3.16, the kernel of Λ is the intersection of the isotropy subgroups of the elements of G/H. Since the isotropy subgroup of eH is H, we see that ker Λ ⊂ H. Since |Sp | = p! and since p is the smallest prime dividing |G|, the greatest common divisor of |G| and |Sp | is precisely p. The First Noether Isomorphism Theorem shows that |im Λ| must divide |G|. Since |im Λ| also divides |Sp |, the order of im Λ must be either 1 or p. Since ker Λ ⊂ H, Λ cannot be the trivial homomorphism, so we must have |im Λ| = p. By the First Noether Isomorphism Theorem, we see that |ker Λ| = |G|/p = |H|. Since ker Λ ⊂ H, this forces ker Λ = H. Since the kernel of a homomorphism is always normal, this gives H G, as desired. The remaining isomorphism theorems are concerned with a deeper analysis of the canonical maps. The reader may easily verify the following lemma. Lemma 4.1.13. Let f : G → G be a homomorphism and let H ⊂ G. Let f : H → G be the restriction of f to H. Then ker f = H ∩ ker f . Recall from Problem 8 of Exercises 2.5.21 that if f : G → G is a surjective homomorphism, then there is a one-to-one correspondence between the subgroups of G and those subgroups of G that contain ker f . Under this correspondence, the subgroup K of G corresponds to f −1 (K ) ⊂ G, while a subgroup K ⊂ G containing ker f corresponds to f (K) ⊂ G . Let us reconsider this correspondence in the case where f is the canonical map π : G → G/H.


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Lemma 4.1.14. Let H G. Then the subgroups of G/H all have the form K/H, where K is a subgroup of G that contains H. Proof The one-to-one correspondence between the subgroups of G/H and those subgroups of G that contain H is based on the following facts (which we assume were verified by the reader). 1. If K is a subgroup of G/H, then π(π −1 (K )) = K . 2. If K is a subgroup of G which contains H, then π −1 (π(K)) = K. Given K ⊂ G/H, let K = π −1 (K ). Then K contains H, and K = π(K). Now π restricts to give a homomorphism, which we shall call π , from K onto π(K). Now, the kernel of π : K → π(K) is K ∩ ker π = H (Lemma 4.1.13). The first Noether theorem now allows us to identify π(K) with K/H. Example 4.1.15. Let n and k be positive numbers such that k divides n. Then the unique subgroup of Zn of order n/k is k. Let π : Z → Zn be the canonical map. Clearly, k = π(k). But since k divides n, we have ker π = n ⊂ k. Thus, k = π −1 k, and the above gives an isomorphism from k/n to k ∼ = Zn/k . We wish to generalize Lemma 4.1.14 to an analysis of the subgroups π(K) where K ⊂ G does not contain H. First consider this: Definition 4.1.16. Let H and K be subgroups of G. Then there is a subset KH of G defined by KH = {kh | k ∈ K, h ∈ H}. (Of course, if G is abelian, we write K + H = {k + h | k ∈ K, h ∈ H} for this subset.) Examples 4.1.17. Since any element of D2n can be written as either bk or as abk for an appropriate value of k, we see that D2n = ab. Similarly, Q4n = ab. The fact that some elements may be written in more than one way as a product ar bs with 0 ≤ r ≤ 3 and 0 ≤ s ≤ 2n doesn’t change this: the set {ar bs | r, s ∈ Z} is ab, and it includes every element in Q4n . Lemma 4.1.18. Let H be a subgroup of G and let K be a subgroup of NG (H). Then KH is a subgroup of G, and H ⊂ KH ⊂ NG (H). In addition, in this case, KH is equal to the subset HK of G, and is the smallest subgroup of G that contains both H and K. Proof Since K ⊂ NG (H), for any k ∈ K and h ∈ H, we have k −1 hk = h ∈ H, and hence hk = kh . This is what we need to show that KH is closed under multiplication: for k1 , k2 ∈ K and h1 , h2 ∈ H, we have k1 h1 k2 h2 = k1 k2 h1 h2 for some h1 ∈ H. Similarly, (kh)−1 = h−1 k −1 = k −1 h for some h ∈ H, and hence KH a subgroup of G. Since h = eh and k = ke, both H and K are contained in KH. And any subgroup containing both H and K must contain each product kh, so KH is indeed the smallest subgroup containing H and K. Since K ⊂ NG (H), this implies HK ⊂ NG (H). Finally, if h ∈ H and k ∈ K, we have khk −1 = h ∈ H, so that kh = h k ∈ HK, and hence KH ⊂ HK. Similarly, HK ⊂ KH.


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Example 4.1.19. Let K be any subgroup of Sn that is not contained in An (e.g., K = (1 2)). Since An Sn , KAn is a subgroup of Sn . Moreover, since K is not contained in An , [KAn : An ] > 1. But [KAn : An ] divides [Sn : An ]. Since [Sn : An ] = 2, we see that KAn = Sn . Lemma 4.1.18 sets up the second Noether Theorem. Theorem 4.1.20. (Second Noether Isomorphism Theorem) Let H and K be subgroups of G with H G. Write π : K → G/H for the restriction to K of the canonical map π : G → G/H. Then the image of π is KH/H, while the kernel of π is K ∩ H. Thus, the First Noether Isomorphism Theorem provides an isomorphism π : K/(K ∩ H)

∼ =

/ KH/H.

Proof By Lemma 4.1.13, the kernel of π is K ∩ ker π = K ∩ H. Thus, it suffices to show that the image of π is KH/H. Of course, the image of π is just π(K). Now for k ∈ K and h ∈ H, kh = k ∈ G/H, and hence π(kh) = π(k). But this says that π(KH) = π(K). Since H ⊂ KH, π(KH) = KH/H by Lemma 4.1.14, and the result follows. Corollary 4.1.21. Let H and K be subgroups of G with H G. Then [K : H ∩ K] divides [G : H]. Proof [K : H ∩ K] is the order of K/(H ∩ K), which is thus equal to the order of the subgroup KH/H of G/H. By Lagrange’s Theorem, this divides [G : H], the order of G/H. Here is a sample application. Corollary 4.1.22. Let K be a subgroup of Sn and suppose that K is not contained in An . Then [K : K ∩ An ] = 2. Proof Corollary 4.1.21 shows that [K : K ∩ An ] divides [Sn : An ], which is 2. But [K : K ∩ An ] = 1, as K is not contained in An . The Second Noether Theorem may be used for yet other calculations of indices. Corollary 4.1.23. Let f : G → G be a surjective homomorphism between finite groups and let K be a subgroup of G. Then [G : f (K)] divides [G : K]. Proof Let H be the kernel of f . By the First Noether Isomorphism Theorem, there is an isomorphism f : G/H → G such that f ◦ π = f , where π : G → G/H is the canonical map. But then f induces an isomorphism from π(K) to f (K). Thus, we see that [G : f (K)] = [G/H : π(K)]. By the Second Noether Isomorphism Theorem, π(K) ∼ = K/(H ∩ K). Thus, we obtain [G : f (K)]

= = =

|G|/|H| |K|/|H ∩ K| |G|/|K| |H|/|H ∩ K| [G : K] . [H : H ∩ K]

Since these indices are all integers, the result follows.


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We now give a slight refinement of the Second Noether Theorem. Corollary 4.1.24. Let H and K be subgroups of G, with K ⊂ NG (H). Then H ∩K K and K/(H ∩ K) ∼ = KH/H. Proof Just replace G by NG (H) and apply the second Noether Theorem. Now, we give the final Noether Isomorphism Theorem. Theorem 4.1.25. (Third Noether Isomorphism Theorem) Suppose given two subgroups, H and K, of G, which are both normal in G, with H ⊂ K. Then K/H G/H, and the factor group (G/H)/(K/H) is isomorphic to G/K. Proof Write πH : G → G/H and πK : G → G/K for the two canonical maps. Then H ⊂ K = ker πK . By Proposition 4.1.7, there is a homomorphism f : G/H → G/K with f ◦ πH = πK . Clearly, f is onto. It suffices to show that ker f = K/H. Let g = gH ∈ G/H. Since f (g) = πK (g) ∈ G/K, we see that g ∈ ker f if and only if g ∈ K = ker πK . Thus, ker f is precisely K/H. Exercises 4.1.26. 1. Give an example of a homomorphism f : G → G such that im f is not normal in G . † 2. Show that every factor group of a cyclic group is cyclic. (Suggestion: Give at least two different proofs.) 3. Recall that b D8 . For each subgroup H ⊂ b, list all the subgroups K ⊂ D8 with K ∩ b = H. (Hint: What is the order of K/H = K/(K ∩ b)?) 4. List all the normal subgroups of D8 . What is the resulting factor group in each case? 5. List all the groups (up to isomorphism) that appear as subgroups of D8 . 6. Recall that b D2n . Let H be a subgroup of b. How many subgroups of D2n are there whose intersection with b is H? (Hint: The answer depends on [b : H].) What are these subgroups? 7. List all the normal subgroups of D2n . What is the resulting factor group in each case? 8. List all the groups (up to isomorphism) that appear as subgroups of D2n . 9. Recall that b Q4n . Let H be a subgroup of b. How many subgroups of Q4n are there whose intersection with b is H? (Hint: The answer depends on [b : H].) What are these subgroups? 10. List all the normal subgroups of Q4n . What is the resulting factor group in each case? 11. List all the groups (up to isomorphism) that appear as subgroups of Q4n . † 12. Show that every element of Q/Z has finite order. Show also that Q/Z has elements of order n for every n ≥ 1.


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† 13. Show that there is an isomorphism between R/Z and R/x for any 0 = x ∈ R. Show also that the elements of finite order in R/Z are precisely those lying in the subgroup Q/Z. † 14. Let Hi Gi for 1 ≤ i ≤ k. Show that H1 × · · · × Hk G1 × · · · × Gk and that (G1 × · · · × Gk )/(H1 × · · · × Hk ) ∼ = (G1 /H1 ) × · · · × (Gk /Hk ). 15. What are all the subgroups of Z2 × Z4 ? What is the resulting factor group in each case? 16. Suppose given an index three subgroup H ⊂ G such that H is not normal in G. Show that there is a surjective homomorphism from G to S3 . † 17. Let H and K be subgroups of G with K ⊂ NG (H). Show that NG (K) ∩ NG (H) ⊂ NG (KH). 18. Generalize Corollary 4.1.23 as follows: Show that if f : G → G is a surjective homomorphism and if [G : K] is finite, then [G : f (K)] is finite and divides [G : K]. (Hint: Show that [G : f (K)] = [G : KH], where H = ker f .) 19. Let H be a subgroup of G. Show that the factor group NG (H)/H acts on the right of the set of left cosets G/H via gH · x = gxH for all g ∈ G and x ∈ NG (H). Show also that the action of each x ∈ NG (H)/H gives a G-isomorphism from G/H to itself. Finally, show that every G-isomorphism from G/H to itself arises in this manner, and deduce that the group of G-isomorphisms from G/H to itself is isomorphic to the factor group NG (H)/H. 20. Let X be any G-set and let H be a subgroup of G. Show that there is an induced action of the factor group NG (H)/H on the H-fixed point set X H .

4.2

Simple Groups

The finite simple groups, which we define in this section, are the building blocks out of which finite groups are made. We show this here via the technique of composition series. We then show that the alternating groups An are simple for n ≥ 5. Definition 4.2.1. A group G is simple if it has no normal subgroups other than e and G. We can characterize the abelian simple groups immediately. We shall give more complicated examples below. Lemma 4.2.2. An abelian group is simple if and only if it is a cyclic group of prime order. Proof Since every subgroup of an abelian group is normal, we see that an abelian group is simple if and only if it has no subgroups other than e and itself. Thus, an abelian group G is simple if and only if every non-identity element in it generates G. In particular, G must be cyclic, and, since Z has nontrivial proper subgroups, it must be finite. But our calculation of the orders of the elements in Zn (Problem 5 of Exercises 2.5.21) shows that Zn is simple if and only if n is prime.


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The choice of the word “simple” for the simple groups is perhaps unfortunate, as the nonabelian simple groups can be much more complicated than, say, solvable ones (see below). Nevertheless, all finite groups are built up out of simple groups, hence the name. Lemma 4.2.3. Let G = e be a finite group. Then G has a factor group which is simple and is not equal to the trivial group. Proof By the first Noether theorem, the statement is equivalent to saying that each group G = e admits a surjective homomorphism onto a nontrivial simple group. We argue by induction on |G|. If |G| = 2, then G itself is simple. Thus, suppose |G| > 2. If G is simple, we’re done, as G = G/e is a factor group of itself. Otherwise, G has a normal subgroup H, with H not equal to either e or G. But then 1 < |G/H| < |G|, so by induction, there is a surjective homomorphism f : G/H → G , where G is a nontrivial simple group. But then the composite f ◦ π : G → G gives the desired surjection. The way in which finite groups are built up from simple groups is as follows. Definition 4.2.4. Let G be a finite group. A composition series for G is a sequence e = H0 ⊂ H1 ⊂ · · · ⊂ Hk = G, such that Hi−1 Hi , and Hi /Hi−1 is a nontrivial simple group for 1 ≤ i ≤ k. Recall that this does not imply that the groups Hi are normal in G. Proposition 4.2.5. Every nontrivial finite group has a composition series. Proof Again, we argue by induction on the order of G. Once again, if G is simple, then there is nothing to prove. Otherwise, Lemma 4.2.3 provides a subgroup H G such that G/H is a nontrivial simple group. But then |H| < |G|, and hence H has a composition series e = H0 ⊂ · · · ⊂ Hk−1 = H. But then tacking on Hk = G gives a composition series for G. The groups Hi /Hi−1 appearing in a composition series are subquotient groups of G: Definition 4.2.6. A subquotient of a group G is a group of the form H/K, where K H ⊂ G. Composition series are far from unique in many cases. We shall discuss their uniqueness properties in Section 4.3, and shall utilize them here simply to motivate the discussion of simple groups. Our remaining goal for this section is to prove the following theorem, which produces our first examples of nonabelian simple groups. Theorem 4.2.7. The alternating groups An are simple for all n ≥ 5. In the process of proving this, we shall derive some useful information about symmetric groups. We shall start by identifying the conjugacy classes in Sn . First consider this: Definition 4.2.8. We say that the permutations σ, τ ∈ Sn have the same cycle structure if when you write each of them as a product of disjoint cycles, both have the same number of k-cycles for each k ≥ 2. Recall that Lemma 3.6.19 computes the conjugates of a k-cycle. Proposition 4.2.9. Two permutations in Sn are conjugate if and only if they have the same cycle structure.


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Proof Write σ ∈ Sn as a product of disjoint cycles: σ = σ1 . . . σk . Then for τ ∈ Sn , τ στ −1 = τ σ1 τ −1 . . . τ σk τ −1 . By Lemma 3.6.19, each τ σi τ −1 is a cycle of the same length as σi . Thus, σ and its conjugate τ στ −1 have the same cycle structure, provided we show that for two disjoint cycles σ1 and σ2 , and any permutation τ , the cycles τ σ1 τ −1 and τ σ2 τ −1 are disjoint. But that follows from the explicit formula of Lemma 3.6.19. Thus, it suffices to show that any two permutations with the same cycle structure are conjugate. Let σ and τ be permutations with the same cycle structure. Thus, we can write σ and τ as products of disjoint cycles, σ = σ1 . . . σk , and τ = τ1 . . . τk , in such a way that σj and τj have the same length, length rj , for 1 ≤ j ≤ k. Say σj = (ij1 . . . ijrj ), and τj = (ij1 . . . ijrj ), for 1 ≤ j ≤ k. Suppose we can find a permutation μ with μ(ijs ) = ijs for 1 ≤ j ≤ k and 1 ≤ s ≤ rj . Then μσj μ−1 = τj by Lemma 3.6.19 for all j. Thus, μσμ−1 = τ , so it will suffice to find such a μ. Since the permutations σj are disjoint for 1 ≤ j ≤ k, an element of {1, . . . , n} may occur in at most one of the sets {iji , . . . , ijrj }. Thus, setting μ(ijs all of the ) = ijs for indices js gives a well-defined function μ : 1≤j≤k {ij1 , . . . , ijrj } → 1≤j≤k {ij1 , . . . , ijrj }. Moreover, the disjointness ofthe τj shows that this function is bijective. Now the complement of 1≤j≤k {ij1 , . . . , ijrj } in {1, . . . , n} may be placed in one-toone correspondence with the complement of 1≤j≤k {ij1 , . . . , ijrj } in {1, . . . , n}. But any such one-to-one correspondence extends μ to a permutation of {1, . . . , n}, which then gives the desired conjugation. It would make our lives easier in the arguments below if we knew that any two elements of An with the same cycle structure were conjugates in An (i.e., if the conjugating permutation μ could be chosen to lie in An ). Unfortunately, this is not always true, so some care will be required. Lemma 4.2.10. Let σ, σ ∈ An be conjugate in Sn . Suppose that σ commutes with an element of Sn which is not in An . Then σ and σ are conjugate in An . Proof Since σ and σ are conjugate in Sn , there is a permutation τ with τ στ −1 = σ . If τ ∈ An , there’s nothing to prove. Otherwise, the hypothesis provides a τ ∈ Sn , with τ not in An (hence ε(τ ) = −1), which commutes with σ. Now ε(τ τ ) = ε(τ )ε(τ ) = +1, −1 so τ τ ∈ An . And (τ τ )σ(τ τ )−1 = τ (τ στ )τ −1 = τ στ −1 = σ . And here are some situations where the hypothesis is satisfied. Lemma 4.2.11. Let σ ∈ An . Suppose that either of the following conditions hold. 1. There are at least two elements of {1, . . . , n} which are left fixed by σ. 2. At least one of the cycles in the cycle structure of σ has even length. Then σ commutes with an element of Sn which is not in An . Proof Suppose the first conditions holds, and, say, σ fixes i and j. Then σ is disjoint from the transposition (i j), and hence commutes with it. Suppose now that the second condition holds. Then σ commutes with the even length cycle in question. (Said cycle is disjoint from all of the other cycles in the cycle decomposition of σ, and hence commutes with them, but it also commutes with itself.) But even length cycles have sign −1.


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Our strategy for proving Theorem 4.2.7 may now be laid out. Recall from Corollary 3.5.11 that An is generated by the 3-cycles. Since n ≥ 5, any 3-cycle commutes with a transposition in Sn . Thus, by Lemma 4.2.10, any two 3-cycles are conjugate in An . In particular, if H An and if H contains a 3-cycle, it must contain all of its conjugates in An , and hence must contain every 3-cycle. Since the 3-cycles generate An , we must have H = An . Thus, we have verified the following lemma. Lemma 4.2.12. Let H An for n ≥ 5. If H contains a 3-cycle, then H = An .2 Thus, it suffices to show that if n ≥ 5, then any normal subgroup of An which is not the identity must contain a 3-cycle. Given a subgroup e = H G, the only information we have for free is that there is a non-identity element in H. We shall consider various possibilities for the cycle structure of such an element, and verify, in the end, that the presence of any non-identity element in H implies the existence of a 3-cycle in H. We may now generalize Lemma 4.2.12. Lemma 4.2.13. Let H An for n ≥ 5. Suppose that H contains an element of the form στ such that 1. σ is a cycle of length ≥ 3. 2. σ and τ are disjoint. (Note that τ may be the identity here.) 3. στ commutes with an element of Sn which is not in An . Then H = An . Proof Let σ = (i1 . . . ik ). Since the inverse of an r-cycle is an r-cycle, τ −1 has the same cycle structure as τ . Thus, στ is conjugate to σ τ −1 , where σ = (ik−1 ik ik−2 . . . i1 ). But then σ τ −1 στ is in H. Since σ and τ are disjoint, they commute, so this is just σ σ, which is easily seen to be equal to the 3-cycle (ik−2 ik ik−1 ). So H = An by Lemma 4.2.12. We would like to be able to eliminate the condition in the preceding lemma which requires that στ commutes with an element of Sn which is not in An . We shall first consider another special case, which is actually already covered by Lemma 4.2.13 except in the cases of n = 5 or 6. Lemma 4.2.14. Let H An , and suppose that H contains a 5-cycle. Then H = An . Proof Let σ = (a b c d e) be a 5-cycle in H. Then μ = (a b)(c d) is in An . Since H is normal in An , μσμ−1 ∈ H, and hence σ · μσμ−1 is also in H. By Lemma 3.6.19, σ · μσμ−1

= (a b c d e) · (b a d c e) = (a e c),

a 3-cycle. So H = An by Lemma 4.2.12. We can now improve on Lemma 4.2.13, provided the cycle σ in the statement there has length > 3. Lemma 4.2.15. Let H An . Suppose that H contains an element of the form στ , where σ is a cycle of length > 3 and σ and τ are disjoint. Then H = An . 2 This result is still true for n < 5, but with different argumentation. We shall treat these cases in the exercises below.


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Proof First note that the hypothesis forces n ≥ 5, as if σ has length 4, then τ cannot be the identity element, as 4-cycles are not in An . Note also that if σ is a 4-cycle, then στ commutes with an element (σ) which is in Sn but not An , and hence the hypothesis of Lemma 4.2.13 is satisfied. Thus, we may assume that σ has length ≥ 5. Write σ = (i1 . . . ik ), and let μ = (i1 i2 )(i3 i4 ). Then μ ∈ An . Moreover, since the support of μ is contained in the support of σ and since σ and τ are disjoint, μ commutes with τ . Since μ ∈ An and H An , στ · μ(στ )−1 μ−1 is in H. We shall show that στ · μ(στ )−1 μ−1 is a 5-cycle. The result will then follow from Lemma 4.2.14. Expanding (στ )−1 and associating, we get στ · μ(στ )−1 μ−1 = σ(τ μτ −1 )σ −1 μ−1 . But τ commutes with μ, so that τ μτ −1 = μ. Simplifying, we see that στ · μ(στ )−1 μ−1 = σ · μσ −1 μ−1 = (i1 . . . ik )(ik . . . i5 i3 i4 i1 i2 ) = (i5 i4 i2 i1 i3 ). Here, the second equality follows from Lemma 3.6.19, and the last by direct calculation. Thus, there is a 5-cycle in H as claimed, and the result follows. We can now go one step farther and give a version of Lemma 4.2.13 that eliminates the third condition. Proposition 4.2.16. Let H An for n ≥ 5. Suppose there is an element σ ∈ H such that the decomposition of σ as a product of disjoint cycles contains a cycle of length ≥ 3. Then H = An . Proof Suppose that the decomposition of σ as a product of disjoint cycles contains a cycle of length > 3. Then the hypothesis of Lemma 4.2.15 is satisfied, and H = An as desired. Thus, we may assume that σ is the product of disjoint cycles whose lengths are all ≤ 3. But if any of them is a 2-cycle, then σ commutes with an element of Sn which is not in An , and hence the hypothesis of Lemma 4.2.13 is satisfied. Thus, we may assume that σ is a product of disjoint 3-cycles. If σ consists of a single 3-cycle, then H = An by Lemma 4.2.12. Thus, we may assume there are at least two 3-cycles in the decomposition of σ as a product of disjoint cycles. Write σ = (a b c)(d e f )τ , where τ is disjoint from the stated 3-cycles. But an immediate application of Lemma 3.6.19 shows that σ commutes with the permutation (a d)(b e)(c f ), which lies in Sn but not in An . Thus, σ satisfies the hypothesis of Lemma 4.2.13, and hence H = An . Thus, the only remaining possibility for an H An with H = An is for H to consist only of products of disjoint transpositions (2-cycles). Indeed, A4 has such a normal subgroup, as we shall see in the exercises below. So n ≥ 5 will be essential in what is to follow. We begin with an important special case. Lemma 4.2.17. Let H An for n ≥ 5. Suppose that H contains a product of two disjoint transpositions. Then H = An .


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Proof Let σ = (a b)(c d) ∈ H. Since n ≥ 5, there is a fifth letter, e, to play with. Since σ commutes with a 2-cycle, it is conjugate in An to any element of the same cycle structure, by Lemma 4.2.10. In particular, it is conjugate to (b c)(d e). Thus, (a b)(c d)(b c)(d e) = (a b d e c) is in H. The result now follows from Lemma 4.2.14. We may now complete the proof of the theorem. Proof of Theorem 4.2.7 Let e = H An for n ≥ 5. Let σ be any non-identity element of H. If the decomposition of σ as a product of disjoint cycles contains a cycle of length ≥ 3, then H = An by Proposition 4.2.16. Thus, we may assume that σ is a product of disjoint transpositions. Since transpositions have sign −1, there must be an even number of disjoint transpositions in the decomposition of σ. But if σ is the product of two disjoint transpositions, then H = An by Lemma 4.2.17. Thus, we may assume that σ is the product of four or more disjoint transpositions. Thus, we can write σ = (a b)(c d)(e f )τ , where τ is disjoint from the displayed transpositions. Once again, since the decomposition of σ contains a cycle of even length, it is conjugate in An to any permutation with the same cycle structure. In particular, σ is conjugate to (b c)(d e)(f a)τ −1 . Thus, (a b)(c d)(e f )(b c)(d e)(f a)τ τ −1 = (a e c)(b d f ) lies in H. The result now follows from Proposition 4.2.16. Exercises 4.2.18. † 1. Show that A4 has exactly four elements whose order is a power of 2. Show that these elements form a normal subgroup of A4 . † 2. Show that A4 has no subgroup of order 6. Deduce that A4 is the only subgroup of S4 which has order 12. (Hint: One way to proceed is as follows. Use the Noether Theorems to show that if H ⊂ A4 has order 6, then H must have a normal subgroup, K H of order 2. Show that every element of H outside of K must have order 3 or 6, and that K ⊂ Z(H). Deduce that H must have an element of order 6 and derive a contradiction.) ‡ 3. Show that An is a characteristic subgroup of Sn for n ≥ 3. (Hint: Use the simplicity of An for n ≥ 5. For n = 3 or 4, give a separate argument.) 4. Show that any normal subgroup of A4 containing a 3-cycle must be all of A4 . 5. Show that any normal subgroup of A3 containing a 3-cycle must be all of A3 . 6. Let σ ∈ An . Show that the conjugates of σ in Sn are all conjugate to σ in An if and only if the centralizer of σ in Sn is not contained in An . (Hint: Recall that the centralizer, CAn (σ), of σ in An satisfies CAn (σ) = An ∩ CSn (σ). What are the possible values for [CSn (σ) : CAn (σ)]? What does this say about [Sn : CSn (σ)]/[An : CAn (σ)]?)


4.3

96

The Jordan–H¨ older Theorem

Recall that a composition series for a group G is a sequence e = H0 ⊂ · · · ⊂ Hk = G such that Hi−1 Hi for i = 1, . . . , k and the factor groups Hi /Hi−1 are all nontrivial simple groups. Proposition 4.2.5 shows that every nontrivial finite group has a composition series. But composition series are not unique. However, the Jordan–H¨ older Theorem shows that the simple subquotient groups Hi /Hi−1 that appear in a composition series for G are independent of the choice of composition series, though they may occur in a different order in different composition series: Theorem 4.3.1. (Jordan–H¨ older Theorem) Suppose given two different composition series, e = H0 ⊂ · · · ⊂ H k = G

and

e = G0 ⊂ · · · ⊂ Gl = G of the finite group G. Then k = l and there is a permutation σ ∈ Sk such that Hi /Hi−1 is isomorphic to Gσ(i) /Gσ(i)−1 for 1 ≤ i ≤ k. Example 4.3.2. We have two different composition series for Z6 : e ⊂ 2 ⊂ Z6

and

e ⊂ 3 ⊂ Z6 . The successive subquotient groups are Z3 followed by Z2 in the first case, and Z2 followed by Z3 in the second. Rather than present the proof from the bottom up, which would lose the intuition in this case, we shall argue conceptually, leaving the proof (and statement) of the key lemma for the end. We first generalize the notion of composition series: Definition 4.3.3. A subnormal series for a group G is a sequence e = H0 ⊂ · · · ⊂ Hk = G such that Hi−1 Hi for 1 ≤ i ≤ k. Some treatments would call this a normal series, but we prefer subnormal since it is not required that Hi G. Note that a composition series is a subnormal series for which each of the successive factor groups Hi /Hi−1 is a nontrivial simple group. Note also that it is possible in a subnormal series (but not a composition series) to have Hi−1 = Hi . Definition 4.3.4. A refinement of a subnormal series is a subnormal series obtained by inserting more subgroups into the sequence. Thus, every subnormal series is a refinement of the trivial series e ⊂ G.


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Definition 4.3.5. Two subnormal series e = H0 ⊂ · · · ⊂ Hk = G

and

e = G0 ⊂ · · · ⊂ G l = G are equivalent if k = l and there is a permutation σ ∈ Sk such that Hi /Hi−1 is isomorphic to Gσ(i) /Gσ(i)−1 for 1 ≤ i ≤ k. Thus, the Jordan–H¨ older Theorem simply states that any two composition series are equivalent. It will follow rather quickly from the next result. Theorem 4.3.6. (Schreier’s Theorem) Suppose given two subnormal series for the group G. Then we may refine the two series in such a way that the refined series are equivalent. This immediately implies the Jordan–H¨ older Theorem because refining a composition series does nothing other than adding trivial groups to the list of factor groups. So two composition series equivalent after refinement must have been equivalent in the first place. Let us now consider the proof of Schreier’s Theorem. We are given subnormal series e = H0 ⊂ · · · ⊂ H k = G

and

e = G0 ⊂ · · · ⊂ G l = G of G. We shall insert between each pair Hi−1 ⊂ Hi a collection of inclusions derived from the G0 ⊂ · · · ⊂ Gl in such a way that the resulting sequence of k · l inclusions is a subnormal series. Similarly, we shall insert groups between Gi−1 and Gi . Since Hi−1 Hi , the group Hi ∩ Gj normalizes Hi−1 for each j. Thus, the product Hi−1 (Hi ∩ Gj ) is a subgroup of G, and we have inclusions Hi−1 = Hi−1 (Hi ∩ G0 ) ⊂ · · · ⊂ Hi−1 (Hi ∩ Gl ) = Hi . Similarly, we have inclusions of subgroups Gj−1 = Gj−1 (H0 ∩ Gj ) ⊂ · · · ⊂ Gj−1 (Hk ∩ Gj ) = Gj . Schreier’s Theorem now follows by setting Gj−1 = B, Hi−1 = A, Gj = K, and Hi = H in the next lemma. Lemma 4.3.7. (Zassenhaus’ Butterfly Lemma) Let H and K be subgroups of G, and suppose given normal subgroups A H and B K of H and K. Then A(H ∩ B) A(H ∩ K), B(A ∩ K) B(H ∩ K) and there is an isomorphism A(H ∩ K)/A(H ∩ B) ∼ = B(H ∩ K)/B(A ∩ K). Proof We shall show that both of the quotient groups are isomorphic to a third quotient: H ∩K/(A∩K)(H ∩B). (We’ll justify presently that this is a group.) The diagram below,


98

which gives this result its name, illustrates the relations between the various subgroups we shall consider. All arrows are the natural inclusions. A(H ∩ K) O iRRRR RRRR R

H ∩O K

l5 llll llll

B(H ∩ K) O

A(H ∩ B) B(A ∩ K) iRRRR iRRR l5 l ll5 l RRR l R l l RRR RRR lll lll l l RR l ll (A ∩ K)(H ∩ B) A iRRR ll5 B RRR iRRRR lll ll5 RRR l l l R l RRRR RRR ll ll lll lll A∩K H ∩B We shall give the proof that H ∩ K/(A ∩ K)(H ∩ B) ∼ = A(H ∩ K)/A(H ∩ B). The argument that H ∩ K/(A ∩ K)(H ∩ B) ∼ = B(H ∩ K)/B(A ∩ K) is analogous. First note that since H ∩ B ⊂ H ∩ K ⊂ H ⊂ NG (A), the subsets A(H ∩ K) and A(H ∩ B) are subgroups of G. We argue by a straightforward application of the Second Noether Isomorphism Theorem, which we restate as follows (using the form given in Corollary 4.1.24) to avoid conflicting notation: If X and Y are subgroups of G with X ⊂ NG (Y ), then X ∩ Y X, and X/(X ∩ Y ) ∼ = XY /Y . We apply this with X = H ∩ K and Y = A(H ∩ B). We must first verify that for this choice of X and Y that X ⊂ NG (Y ), and then must show that the quotients stated in the Noether theorem are the ones that concern us. First recall from Problem 25 of Exercises 3.7.14 that for subgroups H1 , H2 ⊂ G, we have NG (H1 ) ∩ NG (H2 ) ⊂ NG (H1 ∩ H2 ). Also, Problem 17 of Exercises 4.1.26 shows that if H1 ⊂ NG (H2 ), then NG (H1 ) ∩ NG (H2 ) ⊂ NG (H1 H2 ). Thus, since Y = A(H ∩ B), the inclusion of X in NG (Y ) will follow if we show that X ⊂ NG (A) ∩ NG (H) ∩ NG (B). But X ⊂ H, which is contained in both NG (A) and NG (H), and X ⊂ K, which is contained in NG (B), so X normalizes Y as desired. It suffices to show that X ∩ Y = (A ∩ K)(H ∩ B) and that XY = A(H ∩ K). For the former equality, we have X ∩ Y = [H ∩ K] ∩ [A(H ∩ B)]. Clearly, this contains (A ∩ K)(H ∩ B), so it suffices to show the opposite inclusion, i.e., that [H ∩ K] ∩ [A(H ∩ B)] ⊂ (A ∩ K)(H ∩ B). So let a ∈ A and let b ∈ H ∩ B, and suppose that ab ∈ H ∩ K. It suffices to show that a ∈ K. But since ab ∈ H ∩ K ⊂ K, and since b ∈ B ⊂ K, we must have a ∈ K. Finally, we have XY = [H ∩ K][A(H ∩ B)]. This group clearly contains A(H ∩ K), so it suffices to show the opposite inclusion. Let a ∈ A, b ∈ H ∩ B, and c ∈ H ∩ K. Since H ∩ K normalizes A, cab = a cb with a ∈ A. But cb ∈ H ∩ K and the result follows.

4.4

Abelian Groups: the Fundamental Theorem

Here, we classify all finite abelian groups. Recall that an element x of a group G has exponent n if xn = e. Recall also that if x has exponent n, then the order of x divides n.


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Definitions 4.4.1. We say that a group G has exponent n if every element of G has exponent n. Given an arbitrary element x ∈ G, we say that x is a torsion element if x has finite order. If every element of G has finite order, we call G a torsion group. At the other extreme, if no element of G other than the identity has finite order, we call G torsion-free. Finally, let p be a prime number. We say that x ∈ G is a p-torsion element if the order of x is a power of p. If every element of G is a p-torsion element, we call G a p-group. Recall that an element has finite order if and only if it has exponent n for some n > 0. Since the order then divides all exponents for x, we see that x is a p-torsion element if and only if it has exponent pr for some r ≥ 0. Examples 4.4.2. 1. Let G be a finite group. By the corollary to Lagrange’s Theorem (Corollary 3.2.10), the order of any element of G divides |G|. Thus, the group G has exponent |G|. 2. S∞ (the ascending union of the symmetric groups Sn for all n ≥ 1) has been shown to be an infinite torsion group. But it does not have an exponent, as it contains elements of every finite order. 3. As shown in Problem 12 of Exercises 4.1.26, Q/Z is an example of an infinite abelian torsion group which does not have an exponent. 4. ! The reader acquainted with infinite products should have no trouble verifying that ∞ i=1 Z2 , the product of infinitely many copies of Z2 , has exponent 2. Thus, in addition to having an exponent, it is an infinite abelian 2-group. Let us now specialize to the study of abelian groups. To emphasize this specialization, we shall use additive notation for the group operation in the remainder of this section. Lemma 4.4.3. Let G be abelian and let x, y ∈ G be torsion elements. Then the least common multiple of the orders of x and y is an exponent for x + y. Proof Let n be the least common multiple of the orders of x and y. Then n is an exponent for both x and y: nx = ny = 0. But since G is abelian, n(x + y) = nx + ny = 0. It is immediate from Lemma 4.4.3 and the fact that the inverse of an element of finite order also has finite order (Problem 6 of Exercises 2.5.21) that the torsion elements of an abelian group form a subgroup of it: Definition 4.4.4. Let G be an abelian group. Write Tors(G) ⊂ G for the set of all torsion elements of G. We call it the torsion subgroup of G. Example 4.4.5. As shown in Problem 13 of Exercises 4.1.26, the torsion subgroup of R/Z is Q/Z. Surprisingly, the set of torsion elements in a nonabelian group is not necessarily a subgroup. For instance, Problem 2 of Exercises 10.8.13 shows that Sl2 (Z), the group of 2 × 2 matrices with integer coefficients that have determinant 1, is generated by torsion elements, but the matrix 1 1 0 1 is easily seen to have infinite order.


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Definition 4.4.6. Let G be an abelian group and let p be prime. The p-torsion subgroup, Gp , of G is the set of p-torsion elements of G. The justification is, in this case, immediate from Lemma 4.4.3: Corollary 4.4.7. Let G be an abelian group and let p be prime. Then the p-torsion subgroup Gp is, in fact, a subgroup of G. The analogue for nonabelian groups of the preceding corollary is false even for finite nonabelian groups: Example 4.4.8. Recall from Corollary 3.5.11 that for n ≥ 3, An is generated by 3cycles. But while 3-cycles are 3-torsion elements, if n ≥ 4 there are elements of An (e.g., (1 2)(3 4)) that are not 3-torsion elements. Thus, the 3-torsion elements of An do not form a subgroup if n ≥ 4. Let us now specialize to the study of finite abelian groups. Proposition 4.4.9. Let G be a finite abelian group. Let p1 , . . . , pk be the prime numbers that divide |G|. Then G is the internal direct product of Gp1 , . . . , Gpk , i.e., the function μ : Gp1 × · · · × Gpk → G defined by μ(g1 , . . . , gk ) = g1 + · · · + gk is an isomorphism of groups. Proof Of course, μ restricts on each Gpi to its inclusion into G. Since G is abelian, the images of these inclusions commute with each other, so that μ is a homomorphism by Proposition 2.10.10. Suppose that g = (g1 , . . . , gk ) lies in the kernel of μ. If each gi = 0, then g is the identity element of Gp1 × · · · × Gpk . Otherwise, let i be the smallest index with gi = 0. Then gi + (gi+1 + · · · + gk ) = 0, and we must have i < k, as otherwise gi = μ(g) = 0. By inductive application of Lemma 4.4.3, the element gi+1 + · · · + gk has an exponent which is a product of powers of pi+1 , . . . , pk . But gi+1 + · · · + gk is the inverse of gi , and hence has order equal to a power of pi . This forces the order of −gi to be 1 = p0i , and hence gi = 0. So g had to be 0 in the first place, and hence μ is injective. It suffices to show μ is surjective. Let g ∈ G, and let n = o(g). We argue by induction on the number of primes that divide n. If there is only one prime dividing n, then n is a power of pi for some i, and hence g ∈ Gpi ⊂ im μ. Otherwise, n = piri m for some i, where ri ≥ 1, and (pi , m) = 1. Thus, we can find integers s and t with spri i + tm = 1. But then g = (spri i + tm)g = s(pri i g) + t(mg). By Problem 5 of Exercises 2.5.21, mg has order pri i , and hence lies in Gpi ⊂ im μ, and pri i g has order m. Since m has one less prime divisor than n, pri i g ∈ im μ by induction. Thus, g is the sum of two elements of im μ, and hence lies in im μ as claimed. Note that if p doesn’t divide the order of G, then Gp = 0 by Lagrange’s Theorem. Since the cartesian product of any group H with the trivial group is canonically isomorphic to H, we obtain the following immediate extension of Proposition 4.4.9, which is useful when considering homomorphisms between abelian groups whose orders have differing prime divisors.


101

Corollary 4.4.10. Let G be a finite abelian group. Let p1 , . . . , pk be a collection of primes including all the prime divisors of |G|. Let μG : Gp1 × · · · × Gpk → G be defined by μG (g1 , . . . , gk ) = g1 + · · · + gk . Then μG is an isomorphism. The proof of Proposition 4.4.9 is an elaboration of the proof of the next result, which could have been used inductively to obtain it. We include both proofs as it may be instructive to the beginner. Proposition 4.4.11. Let G be a finite group of order n. Suppose that n = mk, where m and k are relatively prime. Let H ⊂ G be the set of all elements of exponent m and let K ⊂ G be the set of all elements of exponent k. Then H and K are subgroups of G, and G is the internal direct product of H and K. Proof That H and K are subgroups is a consequence of Lemma 4.4.3. For instance, if x, y ∈ H, then the prime divisors of o(x + y) must all divide m. Since o(x + y) must divide |G|, this forces o(x + y) to divide m. By Proposition 2.10.8, since G is abelian, it suffices to show that H ∩ K = 0, and that H and K generate G. Since m and k are relatively prime, any element which has both m and k as an exponent must have order 1 (Corollary 2.5.13), so that H ∩ K is indeed the trivial subgroup. But if g ∈ G with o(g) = n , then we may write n = m k , where m divides m and k divides k. But then m and k are relatively prime, so that sm + tk = 1 for some integers s and t. But now g = s(m g) + t(k g). Since m g and k g have orders k and m , respectively, g is in H + K as desired. We shall now give a special case of Cauchy’s Theorem, showing that the groups Gp are nontrivial for all primes p dividing the order of G. Cauchy’s Theorem has already been given as Problem 24 of Exercises 3.3.23, and will be given again by a different argument, using the case we’re about to give, as Theorem 5.1.1. Both proofs are instructive. Proposition 4.4.12. (Cauchy’s Theorem for Abelian Groups) Let G be a finite abelian group and let p be a prime number dividing the order of G. Then G has an element of order p. Proof We argue by induction on |G|, with the case |G| = p a triviality. Let 0 = g ∈ G. If o(g) is divisible by p, say o(g) = pq, then qg has order p. But if o(g) is not divisible by p, then |G/g| is divisible by p. Since |G/g| < |G|, the induction hypothesis gives an element x ∈ G/g of order p. But since x is the image of x ∈ G under the canonical homomorphism, Corollary 2.5.19 shows that p must divide the order of x. This now serves to clarify the situation of Proposition 4.4.11: Corollary 4.4.13. Let G be a finite abelian group of order n = mk, with (m, k) = 1. Let H ⊂ G be the set of elements in G of exponent m and let K ⊂ G be the set of elements in G of exponent k. Then H has order m and K has order k. Proof By Proposition 4.4.12, the order of H cannot be divisible by any prime that doesn’t divide m, as then H would have an element whose order does not divide m. But then |H| must divide m by Lagrange’s Theorem. Similarly, |K| divides k. But since G ∼ = H × K, |G| = |H| · |K|. Since n = mk, the result must follow.


102

As a special case, Corollary 4.4.13 gives a calculation of the order of the p-torsion subgroup of a finite abelian group for any prime p dividing the order of the group. Corollary 4.4.14. Let G be a finite abelian group of order n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes. Then the pi -torsion subgroup Gpi has order pri i for 1 ≤ i ≤ k. The analogue of the preceding statement for nonabelian groups is the second Sylow Theorem. Its proof is a little deeper than that of the abelian case. We now give another useful consequence of Corollary 4.4.13. Corollary 4.4.15. Let G be a finite abelian group and let H be a subgroup. Suppose that the orders of H and G/H are relatively prime. Then H is the subgroup of G consisting of all elements of exponent |H|, and G is isomorphic to H × G/H. Proof Write n, m and k for the orders of G, H, and G/H, respectively. Then n = mk. Let H ⊂ G be elements of exponent m and let K ⊂ G be the elements of exponent k. We shall show that H = H and that K is isomorphic to G/H. To see the former, let i : H ⊂ G be the inclusion. Since the elements of H have exponent m, we must have i(H) ⊂ H . But then i : H → H is an injective homomorphism. Since H and H both have order m, it must be an isomorphism. Let π : G → G/H be the canonical map. Then ker π = H. Since |K| and |H| are relatively prime, we have H ∩ K = 0, and hence π restricts to an injection from K to G/H. But since K and G/H have the same order, π must induce an isomorphism from K to G/H. Next, we show that for any prime p that any finite abelian p-group is a direct product of cyclic groups. Together with Proposition 4.4.9, this will give the existence part of the Fundamental Theorem: Theorem 4.4.16. (Fundamental Theorem of Finite Abelian Groups) Every finite abelian group is a direct product of cyclic groups of prime power order. We shall restate it with a uniqueness statement below. Lemma 4.4.17. Let G be an abelian group of exponent m. Suppose that g ∈ G has order m. Let x ∈ G/g. Then there exists y ∈ G with y = x, such that the order of y in G is equal to the order of x in G/g. Proof Let k = o(x) and let n = o(x). Because x is the image of x under the canonical map π : G → G/g k divides n (Corollary 2.5.19). So n = kl for l ∈ Z. Note then that kx has order l. But since x has order k, kx = kx = 0, and hence kx ∈ ker π = g. Say kx = sg. Since g has order m, the order of sg is m/(m, s). Since kx = sg, this says l = m/(m, s), and hence l · (m, s) = m. Since n = kl is the order of x, it must divide the exponent of G, which is m. Thus, k divides (m, s). But then s = kt, and hence y = x − tg has order k. For the next result, we only use the special case of Lemma 4.4.17 in which the exponent m is a prime power. The reader is invited to re-examine the proof in this simpler case. Proposition 4.4.18. Let p be a prime. Then any finite abelian p-group is a direct product of cyclic groups.


103

Proof By Proposition 4.4.12, |G| is a power of p. We argue by induction on |G|. The induction begins with |G| = p, in which case G itself is cyclic. By Lagrange’s Theorem, |G| is an exponent for G. Let pj be the smallest power of p which is an exponent for G. Then there must be an element g ∈ G of order pj . Let H = G/g, with π : G → H the canonical map. Then |H| < |G|, so our induction hypothesis gives an isomorphism ν : Zpr1 × · · · × Zprk → H. Let xi = ν ◦ ιi (1), where ιi : Zpri → Zpr1 × · · · × Zprk is the inclusion of the i-th factor of Zpr1 × · · · × Zprk . Then xi has order pri . By Lemma 4.4.17, there are elements yi ∈ G with π(yi ) = xi , such that yi has order pri as well. Define ν : Zpr1 × · · · × Zprk × Zpj → G by ν(m1 , . . . , mk+1 ) = m1 y1 + · · · + mk yk + mk+1 g. Since the orders of the yi and of g are as stated, ν ◦ ιi is a homomorphism for 1 ≤ i ≤ k + 1 by Proposition 2.5.17. Thus, ν is a homomorphism by Proposition 2.10.10. We claim that ν is an isomorphism. Suppose that α = (m1 , . . . , mk+1 ) is in the kernel of ν. Then 0

= π ◦ ν(m1 , . . . , mk+1 ) = ν(m1 , . . . , mk ).

Since ν is an isomorphism, m1 = · · · = mk = 0. But then 0 = ν(α) = mk+1 g. Since o(g) = pj , mk+1 = 0, and hence α = 0. Thus, ν is injective. Let x ∈ G, and suppose that ν −1 ◦ π(x) = (m1 , . . . , mk ). Then π(x − (m1 y1 + · · · + mk yk )) = 0. Since ker π = g, x − (m1 y1 + · · · + mk yk ) = mk+1 g for some integer mk+1 . But then x = ν(m1 , . . . , mk+1 ), and hence ν is onto. This completes the proof of the existence part of the Fundamental Theorem. We shall now discuss uniqueness. First consider this. Lemma 4.4.19. Let f : G → H be a homomorphism of abelian groups. Then for each prime p, f restricts to a homomorphism fp : Gp → Hp . Proof The order of f (g) must divide the order of g. If g ∈ Gp , then the order of g is a power of p, so the same must be true of f (g). Thus, f (g) ∈ Hp . For those who understand the language of category theory, this implies that passage to the p-torsion subgroup provides a functor from abelian groups to abelian p-groups. Even better, the next result shows that the isomorphisms μG of Corollary 4.4.10 are natural transformations. The proof is obtained by chasing elements around the diagram. Lemma 4.4.20. Let G and H be finite abelian groups and let p1 , . . . , pk be the collection of all distinct primes that divide either |G| or |H|. Let f : G → H be a homomorphism. Then the following diagram commutes. Gp1 × · · · × Gpk

μG ∼ =

fp1 ×···×fpk

Hp1

× · · · × Hpk

/ G f

μH ∼ =

/ H

Here, μG and μH are the isomorphisms of Corollary 4.4.10. Note that if p divides |G| but not |H|, then Hp = 0, so that fp is the constant homomorphism to 0. Similarly, if p divides |H| but not |G|, then Gp = 0, and fp is


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the unique homomorphism from 0 to Hp . Note that since μG and μH are isomorphisms, the above diagram shows that f is an isomorphism if and only if fp1 × · · · × fpk is an isomorphism. We obtain an immediate corollary. Corollary 4.4.21. Let f : G → H be a homomorphism of finite abelian groups. Then f is an isomorphism if and only if fp : Gp → Hp is an isomorphism for each prime p dividing either |G| or |H|. Thus, a uniqueness statement for the Fundamental Theorem will follow if we can prove the uniqueness of the decomposition (from Proposition 4.4.18) of an abelian p-group as a product of cyclic groups. Fix a prime p. Then, by ordering the exponents, we can write any finite abelian p-group as a product Zpr1 × · · · × Zprk , where 1 ≤ r1 ≤ · · · ≤ rk . Proposition 4.4.22. Suppose that ν : Zpr1 × · · · × Zprk

∼ =

/ Zps1 × · · · × Zpsl ,

where 1 ≤ r1 ≤ · · · ≤ rk , and 1 ≤ s1 ≤ · · · ≤ sl . Then k = l and ri = si for 1 ≤ i ≤ k. Proof We argue by induction on the order of the group. There is only one way in which we can get a group of order p, so the induction starts there. Write Zp ⊂ Zps for the cyclic subgroup ps−1 . Note that the elements of Zp are the only ones in Zps of exponent p. By Problem 2 of Exercises 2.10.12, the elements of Zpr1 × · · · × Zprk which have exponent p are precisely those in the product Zp × · · · × Zp of the subgroups of order p in each factor. Thus, Zpr1 × · · · × Zprk has pk elements of exponent p, all but one of which has order p. Similarly, Zps1 ×· · ·×Zpsl has pl elements of exponent p. Since isomorphisms preserve order, k = l. Moreover, ν restricts to an isomorphism ν : Zp × · · · × Zp → Zp × · · · × Zp . Passing to factor groups, we obtain an isomorphism ν : (Zpr1 × · · · × Zprk )/(Zp × · · · × Zp ) → (Zps1 × · · · × Zpsk )/(Zp × · · · × Zp ). By Problem 2 of Exercises 4.1.26, Zps /Zp ∼ = Zps−1 . Thus, Problem 14 of the same exercises shows that we may identify ν with an isomorphism ν : Zpr1 −1 × · · · × Zprk −1

∼ =

/ Zps1 −1 × · · · × Zpsk −1 .

The result now follows from induction on order. The full statement of the Fundamental Theorem is now immediate: Theorem 4.4.23. (Fundamental Theorem, with Uniqueness) Any finite abelian group may be written uniquely as a direct product of cyclic groups of prime power order. Uniqueness means that given any two such decompositions of a given group, the number of factors of a given order in the two decompositions must be the same. Exercises 4.4.24. 1. Let G be an abelian group. Show that G/Tors(G) is torsion-free. ‡ 2. Show that a finite abelian group G is cyclic if and only if the p-torsion subgroup Gp is cyclic for each prime p dividing the order of G. (Hint: This does not require the Fundamental Theorem.)


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3. Show that a finite abelian group is cyclic if and only if it has no subgroup of the form Zp × Zp for any prime p. (Hint: Apply the Fundamental Theorem.) † 4. Use the Fundamental Theorem to show that a finite abelian group is cyclic if and only if it has at most d elements of exponent d for each d dividing the order of G. 5. Give a direct proof (not using the Fundamental Theorem) of the preceding problem. (Hint: One direction has already been done in Problem 11 of Exercises 2.5.21. For the other direction, first show via Problem 2 above that we may assume that the group has order pr for some prime p and some r ≥ 1. Then show that there must be a generator.) 6. Show that the full converse to Lagrange’s Theorem holds for finite abelian groups G: if k divides the order of G, then G has a subgroup of order k. 7. Show that any finite abelian group may be written as a product Zn1 × · · · × Znk , where ni divides ni−1 for 1 < i ≤ k. 8. Show that the decomposition of the preceding problem is unique.

4.5

The Automorphisms of a Cyclic Group

We calculate the automorphism group of a cyclic group. Recall that for any group G and for any g ∈ G whose order divides n, there is a unique homomorphism hg : Zn → G with hg (1) = g. Lemma 4.5.1. For n > 1 and m ∈ Z, the homomorphism hm : Zn → Zn is an automorphism if and only if m and n are relatively prime. Proof The image of hm is the cyclic subgroup generated by m. Since the order of m is n/(m, n), hm is onto if and only if (m, n) = 1. But any surjection from a finite set to itself is an injection. Thus, we’ve identified which homomorphisms from Zn to Zn are automorphisms, but we have yet to identify the group structure on Aut(Zn ). Recall that there are two operations defined on Zn , addition and multiplication, and that multiplication gives Zn the structure of a monoid, with unit element 1. Recall that the collection of invertible elements in a monoid M forms a group (which we have denoted Inv(M )) under the monoid operation of M . Definition 4.5.2. For n > 1 we write Z× n for the group of invertible elements in the multiplicative monoid structure on Zn . Thus, m ∈ Z× n if and only if there is a k ∈ Zn with km = 1 ∈ Zn . Z× n is an example of a general construction in rings. The set of elements in a ring A which have multiplicative inverses forms a group under the operation of multiplication. It is called the group of units of A, and may be denoted A× . We wish to identify the elements of Z× n . The answer turns out to be quite relevant to our discussion of automorphisms. Lemma 4.5.3. Let n > 1 and let m ∈ Z. Then m ∈ Z× n if and only if m and n are relatively prime.


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Proof Suppose that m is invertible in Zn , with inverse k. Since km is congruent to 1 mod n, we obtain that km − 1 = sn for some integer s. But then km − sn = 1, and hence (m, n) = 1. Conversely, if rm + sn = 1, then r is the multiplicative inverse for m in Zn . Thus, there is a one-to-one correspondence between the elements of Aut(Zn ) and Z× n. We next show that this correspondence is an isomorphism of groups. Proposition 4.5.4. Define ν : Z× n → Aut(Zn ) by ν(m) = hm , where hm : Zn → Zn is the homomorphism carrying 1 to m. Then ν is an isomorphism of groups. Proof It suffices to show that hm ◦ hm = hmm . But it’s easy to see (e.g., by Corollary 2.5.5 and Proposition 2.5.17) that hm (k) = mk, and the result follows. Since multiplication in Zn is commutative, we obtain the following corollary. Corollary 4.5.5. For any positive integer n, Aut(Zn ) ∼ = Z× is an abelian group. n

In contrast, most other automorphism groups, even of abelian groups, are nonabelian. It will require a bit more work to display the group structures on these groups of units Z× n . It will be easiest to do so in the case that n only has one prime divisor. We will then be able to put this information together, via methods we are about to discuss. First recall from Section 4.4 that if G is abelian and if Gp is its p-torsion subgroup, then any homomorphism f : G → G restricts to a homomorphism fp : Gp → Gp (i.e., fp (g) = f (g) for all g ∈ Gp ). Moreover, Corollary 4.4.21 shows that f : G → G is an automorphism if and only if fp is an automorphism for each p dividing |G|. The next result follows easily. Lemma 4.5.6. Let G be an abelian group and let p1 , . . . pk be the primes dividing the order of G. Let Gpi be the pi -torsion subgroup of G for 1 ≤ i ≤ k. Then there is an isomorphism ∼ =

ϕ : Aut(G)

/ Aut(Gp1 ) × · · · × Aut(Gpk )

defined by ϕ(f ) = (fp1 , . . . , fpk ). Let n = pr11 . . . prkk with p1 , . . . , pk distinct primes. Then the pi -torsion subgroup of Zn is isomorphic to Zpri . We obtain the following corollary, which may also be proven i using the Chinese Remainder Theorem, below. Corollary 4.5.7. With n as above, there is an isomorphism ∼ =

ρ : Z× n

/ Z×r1 × · · · × Z×rk p p 1

k

given by ρ(m) = (m, . . . , m). Proof As the reader may easily check, there is a commutative diagram Z× n ν ∼ = Aut(Zn )

ρ

ϕ ∼ =

/ Z×r1 × · · · × Z×rk p1 pk ∼ = ν×···×ν / Aut(Zpr1 ) × · · · × Aut(Zprk ) 1 k


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Thus, the calculation of the automorphism groups of cyclic groups is reduced to the × calculation of Z× pr for p a prime and r ≥ 1. We begin by calculating the order of Zpr . Lemma 4.5.8. For any prime p and any integer r > 0, the group of units Z× pr has order pr−1 (p − 1). Proof The elements of Zpr that do not lie in Z× pr are the elements m such that (p, m) = 1. But since p is prime, this means that p divides m, and hence m ∈ p ⊂ Zpr . But p r r−1 = pr−1 (p − 1). has order pr−1 , and hence |Z× pr | = p − p Corollary 4.5.7 now allows us to calculate |Z× n |. Corollary 4.5.9. Let n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes and the exponents ri are all positive. Then r1 −1 |Aut(Zn )| = |Z× . . . (pk − 1)prkk −1 . n | = (p1 − 1)p1

In particular, Z× p has order p − 1. In Corollary 7.3.15, we shall show, using some elementary field theory, that Z× p is always a cyclic group. However, the proof is theoretical, and doesn’t provide any hint as to what elements generate this group. And indeed there is no known algorithm to obtain a generator other than trial and error. (Some of the deeper aspects of number theory are lurking in the background.) So for the practical purpose of identifying the orders of individual elements of Z× p , direct calculation is necessary. The reader with some computer expertise may find it interesting to write a program for this purpose. On the other hand, when r > 1, the orders of the elements of Z× p will be the only open question in our understanding of Z× . We need to understand what happens to Z× r p pr s r as r varies. Thus, let 1 ≤ s < r. Since p divides p , there is a homomorphism, which we shall denote by π : Zpr → Zps , with π(1) = 1. In fact, π(m) = m for all m ∈ Z. We call π the canonical homomorphism. We have m ∈ Z× pr if and only if (m, p) = 1, which × is true if and only if m ∈ Zps . The next lemma is now immediate. Lemma 4.5.10. For 1 ≤ s < r and for p a prime, the canonical homomorphism π : Zpr → Zps induces a surjective homomorphism of multiplicative groups × π : Z× pr → Zps .

At this point, the cases p > 2 and p = 2 diverge. One difference that is immediately × apparent is that Z× 2 = e, while Zp is nontrivial for p > 2. Thus, for p > 2 we shall define × K(p, r) to be the kernel of π : Z× pr → Zp , while K(2, r) is defined to be the kernel of × × π : Z2r → Z4 (and hence K(2, r) is only defined for r ≥ 2). Note that Z× 4 = {±1} has order 2. Lemma 4.5.10 and the First Noether Theorem now give us the order of K(p, r). Corollary 4.5.11. For p > 2 and r ≥ 1, K(p, r) has order pr−1 . Also, for r ≥ 2, K(2, r) has order 2r−2 . For p > 2, the orders of K(p, r) and Z× p are relatively prime, so Corollary 4.4.15 gives a splitting of Z× . r p


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Corollary 4.5.12. For p > 2 and r ≥ 1, K(p, r) is the p-torsion subgroup of Z× pr , and we have an isomorphism × ∼ Z× pr = K(p, r) × Zp .

The prime 2 is a little different, in that K(2, r) and Z× 4 are both 2-groups. But we get a product formula regardless. Lemma 4.5.13. For r ≥ 2, we have an isomorphism × ∼ Z× 2r = K(2, r) × Z4 . × Proof Let H = {±1} ⊂ Z× 2r . Then H is a subgroup of Z2r , and is carried isomorphically × × onto Z4 by π. We shall show that Z2r is the internal direct product of K(2, r) and H, i.e., the map μ : H × K(2, r) → Z× 2r given by μ(h, k) = hk is an isomorphism. The key is that since −1 is not congruent to 1 mod 4, −1 is not in K(2, r), and hence H ∩ K(2, r) = 1. (The unit element here is denoted 1.) This easily implies that μ is injective. But since m · k = mk in both Z2r and Z4 , either m or −m is in K(2, r) for each m ∈ Z× 2r . Thus, μ is onto.

We shall show that K(p, r) is cyclic for all primes p and all r ≥ 2. (Unlike the case of Z× p , we shall provide a criterion to determine the order of each element.) The key is to understand the order of the elements 1 + ps . As the group operation is multiplication modulo pr , we can get a handle on this if we can calculate the powers (x + y)n of the sum of two integers x and y. The formula for (x + y)n is given by what’s called the Binomial Theorem, which in turn depends on an understanding of the binomial coefficients. Definition 4.5.14. Let 0 ≤ k ≤ n ∈ Z. We define the binomial coefficient nk by n n! = . k!(n − k)! k Here, the factorial numbers are defined as usual: 0! = 1, 1! = 1, and n! is defined inductively by n! = (n − 1)! · n for n > 1. The next lemma is useful for manipulating the binomial coefficients. Lemma 4.5.15. The binomial coefficients are all integers. Moreover, if 1 ≤ k < n we have n−1 n−1 n + = . k−1 k k Proof An easy inspection of the definition shows that n0 = nn = 1 for all n. But

the only other binomial coefficients are the nk with 1 ≤ k < n. Thus, if the displayed equation is true, then the binomial coefficients are all integers by induction on n, starting with the case n = 1, which we’ve just verified by inspection. The left-hand side of the displayed equation expands to (n − 1)! (n − 1)! (n − 1)![k + (n − k)] + = , (k − 1)!(n − k)! k!(n − k − 1)! k!(n − k)! and the result follows.


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Our next result is the Binomial Theorem, which is true, with the same proof, not only in the integers, but in any commutative ring. Theorem 4.5.16. (Binomial Theorem) Let x and y be integers, and n ≥ 1 be another. Then n n k n−k (x + y)n = x y k k=0 n 2 n−2 = y n + nxy n−1 + x y + · · · + nxn−1 y + xn 2 Proof The second line just

gives n an expansion of the preceding expression, using the = n, in addition to the already known calculations (easy) calculations that n1 = n−1 n n of 0 and n . We prove the theorem by induction on n. For n = 1, it just says that x + y = x + y. Thus, assume that n > 1 and the theorem is true for n − 1. Then # "n−1 n − 1 xk y n−k−1 (x + y). (x + y)n = (x + y)n−1 (x + y) = k k=0

Applying the distributive law to the last expression and collecting terms, we get $ % n−1 n−1 n−1 + . xn + y n + xk y n−k k−1 k k=1

But this is exactly the desired sum, by Lemma 4.5.15. We wish to calculate the order of 1 + ps in K(p, r). By the definitions, the order will be the smallest integer m such that (1 + ps )m is congruent to 1 mod pr . Since the order of K(p, r) is a power of p, m must be also. Thus, we shall study the binomial expansion k for (1 + ps )p . For this purpose, we shall need to know the largest power of p that divides k the binomial coefficients pj . Lemma 4.5.17. Let p be prime and let k ≥ 1. Let 1 ≤ j ≤ pk and write j = pa b, with (p, b) = 1. Then k p = pk−a c, j with (p, c) = 1. Proof After making appropriate cancellations, we have k p pk (pk − 1) . . . (pk − j + 1) = . j 1 · 2 . . . (pa b) Consider the right-hand side as a rational number and rearrange the terms slightly. We get the product of pk /pa b with the product k k k p −2 p − (j − 1) p −1 · ... 1 2 j−1 But for 1 ≤ i < pk , pk − i and i have the same p-divisibility, so that the only contribution to the p-divisibility of our binomial coefficient comes from pk /pa b. But this gives the desired result.


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We shall now calculate the order of the generic element of K(p, r). Note that if p > 2, then x ∈ K(p, r) if and only if x = 1 + y for some y divisible by p. Similarly, x ∈ K(2, r) if and only if x = 1 + y for some y divisible by 4. Proposition 4.5.18. Let p be a prime and let x = 1 + ps a, where (p, a) = 1. If p = 2, assume that s > 1. Then x has order pr−s in K(p, r). Proof As discussed above, the order of x in K(p, r) is the smallest power, pk , of p for k which xp is congruent to 1 mod pr . So we want to find the smallest value of k such k k that (ps a + 1)p − 1 is divisible by pr . Consider the binomial expansion for (ps a + 1)p . k k We have (ps a + 1)p = 1 + pk ps a plus terms of the form pj psj aj with j > 1. But then k k (ps a + 1)p − 1 is equal to ps+k a plus the terms pj psj aj with j > 1. k It suffices to show that if j > 1, then pj psj aj is divisible by ps+k+1 . k By Lemma 4.5.17, if j = pb c, with (p, c) = 1, then pj is divisible by pk−b . Thus, pk sj j b sj+k−b = psp c+k−b , so it suffices to show that spb c+k−b > s+k. j p a is divisible by p But this is equivalent to showing that s[pb c − 1] > b. If b = 0, then c = j > 1, and the result holds. Otherwise, b > 0, and we may and shall assume that c = 1. For p > 2, we must show that s[pb − 1] > b for all s ≥ 1. But the general case will follow from that of s = 1, where we show that pb − 1 > b for all b > 0. For p = 2, our hypothesis is that s > 1, so it suffices to show that 2b − 1 ≥ b for all b > 0. For any p, pb − 1 = [(p − 1) + 1]b − 1, so we may use the Binomial Theorem again. We see that pb − 1 is equal to b(p − 1) plus some other non-negative terms. But this is greater than b if p > 2, and is ≥ b if p = 2, precisely as desired. Corollary 4.5.19. For p > 2 and r ≥ 1, K(p, r) is the cyclic group (of order pr−1 ) generated by 1 + p. For p = 2 and r ≥ 2, K(2, r) is the cyclic group (of order 2r−2 ) generated by 5. Proof The orders of the groups in question were calculated in Corollary 4.5.11. By Proposition 4.5.18, the 1 + p has the same order as K(p, r) for p > 2, while 5 has the same order as K(2, r). Lemma 4.5.13 now gives an explicit calculation of Z× 2r . Corollary 4.5.20. For r ≥ 2, Z× 2r is isomorphic to Z2r−2 × Z2 . In particular, Z× 2r is not cyclic for r ≥ 3. On the other hand, when we have established Corollary 7.3.15, we will know that Z× p is a cyclic group of order p − 1 for all primes p. We shall state the consequence here, though the reader is advised to read ahead for the proof. This time we make use of Corollary 4.5.12: Corollary 4.5.21. Let p be an odd prime and let r ≥ 1. Then Z× pr is a cyclic group of order (p − 1)pr−1 . Exercises 4.5.22. 1. Find generators for Z× p for all primes p ≤ 17. 2. Find a generator for Z× 3r for any r. 3. Find a generator for Z× 5r for any r.


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4. Find a generator for Z× 49 . 5. Find all elements of order 2 in Z× 2r . × × 6. Find all elements of order 2 in Z× 15 , Z30 , and Z60 .

7. Write a computer program to calculate the order of an element in Z× pr , where the input is p, r, and the element in question. Be sure it checks that the element is a unit. 8. Write a computer program to calculate the order of an element in Z× n . Be sure it checks that the element is relatively prime to n.

4.6

Semidirect Products

The simplest way to construct new groups from old is to take direct products. But there are ways to construct more interesting groups if we can calculate the automorphisms of one of the groups in question. Suppose given an action of K on H through automorphisms. We may represent this by a homomorphism α : K → Aut(H), so that the action of k ∈ K takes h ∈ H to α(k)(h). We shall define a group multiplication on the set H × K, to obtain a new group, which we denote H α K, called the semidirect product of H and K given by α: Definition 4.6.1. Let α : K → Aut(H) be a homomorphism. By the semidirect product of H and K with respect to α, written H α K, we mean the set H × K with the binary operation given by setting (h1 , k1 ) · (h2 , k2 ) = (h1 · α(k1 )(h2 ), k1 k2 ). Proposition 4.6.2. H α K, with the above product, is a group. There are homomorphisms π : H α K → K and ι : H → H α K, given by π(h, k) = k, and ι(h) = (h, e). The homomorphism ι is an injection onto the kernel of π. Finally, there is a homomorphism s : K → H α K given by s(k) = (e, k). This homomorphism has the property that π ◦ s is the identity map of K. Proof We first show associativity. ((h1 , k1 ) · (h2 , k2 )) · (h3 , k3 )

= (h1 · α(k1 )(h2 ), k1 k2 ) · (h3 , k3 ) = (h1 · α(k1 )(h2 ) · α(k1 k2 )(h3 ), k1 k2 k3 ) = (h1 · α(k1 )(h2 ) · α(k1 )(α(k2 )(h3 )), k1 k2 k3 ).

Here, the last equality holds because α is a homomorphism, so that α(k1 k2 ) = α(k1 ) ◦ α(k2 ). Since α(k1 ) is an automorphism of H, we have α(k1 )(h2 ) · α(k1 )(α(k2 )(h3 )) = α(k1 )(h2 · α(k2 )(h3 )), and hence ((h1 , k1 ) · (h2 , k2 )) · (h3 , k3 ) = (h1 · α(k1 )(h2 · α(k2 )(h3 )), k1 k2 k3 ). But this is precisely what one gets if one expands out (h1 , k1 ) · ((h2 , k2 ) · (h3 , k3 )).


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The pair (e, e) is clearly an identity element for this product, since α(e) is the identity homomorphism of H. We claim that the inverse of (h, k) is (α(k −1 )(h−1 ), k −1 ). The reader is welcome to verify this fact, as well as the statements about π, ι, and s. Note that if α(k) is the identity automorphism for all k ∈ K (i.e., α : K → Aut(H) is the trivial homomorphism), then H α K is just the direct product H × K. Notation 4.6.3. In the semidirect product H α K, we shall customarily identify H with ι(H) and identify K with s(K), so that the element (h, k) is written as hk. Thus, H α K = {hk | h ∈ H, k ∈ K}. Lemma 4.6.4. With respect to the above notations, the product in H α K is given by h1 k1 · h2 k2 = h1 α(k1 )(h2 )k1 k2 for h1 , h2 ∈ H and k1 , k2 ∈ K. In particular, this gives khk −1 = α(k)(h). Thus, H α K is abelian if and only if H and K are both abelian and α : K → Aut(H) is the trivial homomorphism. Proof The multiplication formula is clear. But this gives kh = α(k)(h)k, which in turn implies the stated conjugation formula. Thus, when α is nontrivial, the new group H α K may have some interesting features. Indeed, this method of constructing groups will give us a wide range of new examples. But it also gives new ways of constructing several of the groups we’ve already seen. For instance, as we shall see below, the dihedral group D2n is the semidirect product obtained from an action of Z2 on Zn , and A4 is the semidirect product obtained from an action of Z3 on Z2 × Z2 . We wish to analyze the semidirect products obtained from actions of cyclic groups on other cyclic groups. Since these are, in general, nonabelian, we shall use multiplicative notation for our cyclic groups. Thus, we shall begin with a discussion of notation. × Recall first the isomorphism ν : Z× n → Aut(Zn ) of Proposition 4.5.4. Here, Zn is the group of invertible elements with respect to the operation of multiplication in Zn , and, for m ∈ Z× n , ν(m) is the automorphism defined by ν(m)(l) = ml. In multiplicative notation, we write Zn = b = {e, b, . . . , bn−1 }. Thus, the element of Zn written as l in additive notation becomes bl . In other words, in this notation, we have an isomorphism ν : Z× n → Aut(b), where ν(m) is the automorphism defined by ν(m)(bl ) = (bml ). Now write Zk = a in multiplicative notation, and recall from Proposition 2.5.17 that there is a homomorphism hg : Zk → G with hg (a) = g ∈ G if and only if g has exponent k in G. Moreover, if g has exponent k, there is a unique such homomorphism, and it is given by the formula hg (aj ) = g j . Thus, suppose given a homomorphism α : Zk = a → Z× n , with α(a) = m. Then m × k has exponent k in Zn , meaning that m is congruent to 1 mod n. Moreover, α(aj ) = mj . By abuse of notation, we shall write α : Zk → Aut(Zn ) for the composite of this α with the isomorphism ν above. Thus, as an automorphism of Zn = b, α(aj ) is given j by α(aj )(bi ) = bm i . × Corollary 4.6.5. Let m have exponent k in Z× n , and let α : Zk → Zn be the homomorphism that takes the generator to m. Then writing Zn = b, Zk = a, and identifying Z× n with Aut(Zn ), we obtain

Zn α Zk = {bi aj | 0 ≤ i < n, 0 ≤ j < k},


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where b has order n, a has order k, and the multiplication is given by

j

bi aj bi aj = bi+m i aj+j . Moreover, the nk elements {bi aj | 0 ≤ i < n, 0 ≤ j < k} are all distinct. Proof The multiplication formula comes via Lemma 4.6.4 from the formula α(aj )(bi ) = j bm i . Examples 4.6.6. 1. Let α : Z2 → Z× n be the homomorphism that carries the generator to −1. Then in Zn α Z2 , with the notations above, we see that b has order n, a has order 2, and aba−1 = b−1 . Thus, Problem 6 of Exercises 2.8.5 provides a homomorphism f : D2n → Zn α Z2 with f (a) = a and f (b) = b. Since a and b generate Zn α Z2 , f is onto, and hence an isomorphism, as the two groups have the same order. Thus, semidirect products provide a construction of the dihedral groups that does not require trigonometry. 2. Let α : Z2 → Z× 8 be the homomorphism that carries the generator to 3. We obtain a group of order 16, Z8 α Z2 = {bi aj | 0 ≤ i < 8, 0 ≤ j < 2},

where a has order 2, b has order 8, and the multiplication is given by bi aj bi aj = j bi+3 i aj+j . 3. Let β : Z2 → Z× 8 be the homomorphism that carries the generator to 5. We obtain a group of order 16, Z8 β Z2 = {bi aj | 0 ≤ i < 8, 0 ≤ j < 2},

where a has order 2, b has order 8, and the multiplication is given by bi aj bi aj = j bi+5 i aj+j . 4. Let α : Z4 → Z× 5 be the homomorphism that carries the generator to 2. We obtain a group of order 20, Z5 α Z4 = {bi aj | 0 ≤ i < 5, 0 ≤ j < 4},

where a has order 4, b has order 5, and the multiplication is given by bi aj bi aj = j bi+2 i aj+j . 5. Let α : Z3 → Z× 7 be the homomorphism that carries the generator to 2. We obtain a group of order 21, Z7 α Z3 = {bi aj | 0 ≤ i < 7, 0 ≤ j < 3},

where a has order 3, b has order 7, and the multiplication is given by bi aj bi aj = j bi+2 i aj+j . 6. Let p be an odd prime. Recall from Proposition 4.5.18 that p + 1 has order p in × Z× p2 . Thus, there is a homomorphism α : Zp → Zp2 that carries the generator of Zp to p + 1. We obtain a group of order p3 , Zp2 α Zp = {bi aj | 0 ≤ i < p2 , 0 ≤ j < p},

where a has order p, b has order p2 , and the multiplication is given by bi aj bi aj = j bi+(p+1) i aj+j .


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Other than the first example, these groups are all new to us. Indeed, the groups of order 16 in Examples 2 and 3 turn out to be the lowest-order groups that we had not constructed previously. The group of Example 5 turns out to be the smallest odd order group that is nonabelian, while for each odd prime p, the group of Example 6 represents one of the two isomorphism classes of nonabelian groups of order p3 . We may also construct examples of semidirect products where at least one of the two groups is not cyclic. For instance, since the groups Z× n are not generally cyclic, we can obtain a number of groups of the form Zn α K, where K is not cyclic. We are also interested in the case where group “H” in the construction of semidirect products is not cyclic. The simplest case is where H = Zn × Zn . As above, write Zn × Zn = {bi cj | 0 ≤ i, j < n}, where b and c have order n and commute with each other. Since every element of Zn × Zn has exponent n, Propositions 2.10.6 and 2.5.17 show that there is a homomorphism f : Zn × Zn → Zn × Zn with the property that f (b) = bc and f (c) = c: explicitly, f (bi cj ) = bi ci+j . Lemma 4.6.7. The homomorphism f just described is an automorphism of Zn × Zn . It has order n in the group of automorphisms of Zn × Zn . Proof For k ≥ 1 the k-fold composite of f with itself is given by f k (c) = c and f k (b) = f k−1 (bc) = f k−1 (b)f k−1 (c) = (f k−1 (b))c. An easy induction now shows that f k (b) = bck . Thus, f k is the identity if and only if k is congruent to 0 mod n. Since f n is the identity, f is indeed an automorphism, with inverse f n−1 . Thus, if we write Zn = a, there is a homomorphism α : Zn → Aut(Zn × Zn ) with α(a) = f . We obtain a construction which is of special interest when n is prime: Corollary 4.6.8. There is a group G of order n3 given by G = {bi cj ak | 0 ≤ i, j, k < n}, where a, b, and c all have order n, and b commutes with c, a commutes with c, and aba−1 = bc. Thus,

bi cj ak · bi cj ak = bi+i cj+j

+ki k+k

a

.

There are, of course, at least as many examples of semidirect products as there are groups with nontrivial automorphism groups. Thus, we shall content ourselves for now with these examples. In our studies of the dihedral and quaternionic groups, we have made serious use of our knowledge of the homomorphisms out of these groups. There is a similar result for the generic semidirect product. First, we need some discussion. Let ι : H → H α K and s : K → H α K be the structure maps: in the ordered pairs notation, ι(h) = (h, e) and s(k) = (e, k). Then every element of H α K may be written uniquely as ι(h) · s(k) with h ∈ H and k ∈ K. This is, indeed, the basis for the HK notation. In particular, H α K is generated by the elements of ι(H) and s(K), so that any homomorphism f : H α K → G is uniquely determined by its restriction to the subgroups ι(H) and s(K). Since ι and s are homomorphisms, this says that f is uniquely determined by the homomorphisms f ◦ ι and f ◦ s.


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Thus, we will understand the homomorphisms out of H α K completely if we can give necessary and sufficient conditions for a pair of homomorphisms f1 : H → G and f2 : K → G to be obtained as f1 = f ◦ ι and f2 = f ◦ s for a homomorphism f : H α K → G. Proposition 4.6.9. Let α : K → Aut(H) be a homomorphism. Suppose given homomorphisms f1 : H → G and f2 : K → G for a group G. Then there is a homomorphism f : H α K → G with f1 = f ◦ ι and f2 = f ◦ s if and only if f2 (k)f1 (h)(f2 (k))−1 = f1 (α(k)(h)) for all h ∈ H and k ∈ K. Such an f , if it exists, is unique, and is given in the HK notation by f (hk) = f1 (h)f2 (k). Proof Suppose first that f exists with f ◦ ι = f1 and f ◦ s = f2 . In the HK notation we have that hk = ι(h)s(k), so that f (hk) = (f ◦ ι)(h) · (f ◦ s)(k) = f1 (h)f2 (k) for all h ∈ H and k ∈ K. As shown in Lemma 4.6.4, khk −1 = α(k)(h) in H α K, for all h ∈ H and k ∈ K. Applying f to both sides, we obtain f2 (k)f1 (h)(f2 (k))−1 = f1 (α(k)(h)), as claimed. Conversely, suppose that f2 (k)f1 (h)(f2 (k))−1 = f1 (α(k)(h)) for all h ∈ H and k ∈ K. We define f : H α K → G by f (hk) = f1 (h)f2 (k). It suffices to show that f is a homomorphism. But f (h1 k1 )f (h2 k2 )

= f1 (h1 )f2 (k1 )f1 (h2 )f2 (k2 ) = f1 (h1 )f1 (α(k1 )(h2 ))f2 (k1 )f2 (k2 ) = f1 (h1 α(k1 )(h2 ))f2 (k1 k2 ) = f (h1 k1 h2 k2 ),

for all h1 , h2 ∈ H and k1 , k2 ∈ K, where the second equality follows from the formula f2 (k1 )f1 (h2 )(f2 (k1 ))−1 = f1 (α(k1 )(h2 )). This permits a slight simplification when H and K are cyclic. Corollary 4.6.10. Let α : Zk → Z× n be a homomorphism, with α(a) = m, where Zk = a in multiplicative notation. Write Zn α Zk in multiplicative notation, with Zn = b. Then for any group G, there is a homomorphism f : Zn α Zk → G with f (a) = x and f (b) = y if and only if the elements x, y ∈ G satisfy the following conditions: 1. x has exponent k 2. y has exponent n 3. xyx−1 = y m . Proof The conditions are clearly necessary, as a has order k, b has order n, and aba−1 = bm in Zn α Zk . Conversely, if the three conditions hold, there are homomorphisms f1 : b → G and f2 : a → G given by f1 (bi ) = y i and f2 (aj ) = xj . Thus, by Proposition 4.6.9, it suffices j to show that xj y i x−j = y m i . j Since conjugation by x is a homomorphism, xj y i x−j = (xj yx−j )i , so it suffices by the j law of exponents to show that xj yx−j = y m . But since xj yx−j = x(xj−1 yx−(j−1) )x−1 , the desired result follows from the third condition by an easy induction argument. The above is an important ingredient in calculating the automorphism groups of the semidirect products of cyclic groups. But it does not give good criteria for showing


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when the homomorphism f is an isomorphism. For this it will be convenient to use the language and concepts of extensions, as developed in Section 4.7. We shall now begin to consider the question of the uniqueness of the semidirect product: When are two different semidirect products isomorphic? We will not answer this question completely, but will study some aspects of it. First, what happens if we switch to a different homomorphism K → Aut(H)? Recall from Section 3.6 that G acts on the set of homomorphisms from K to G as follows: if f : K → G is a homomorphism and g ∈ G, then g · f is the homomorphism defined by (g · f )(k) = gf (k)g −1 for all k ∈ K. By abuse of notation, we shall also write gf g −1 for g · f . We say that the homomorphisms f and gf g −1 are conjugate homomorphisms. So it makes sense to ask whether H α K and H α K are isomorphic when α and α are conjugate homomorphisms of K into Aut(H). Another fairly minor way in which we could alter a homomorphism α : K → Aut(H) is by replacing it with α ◦ g, where g ∈ Aut(K). We wish to study the effect of this sort of change as well. Since we’re considering two different semidirect products of the same groups, we shall use different notations for their structure maps. Thus, we write ι : H → H α K, π : H α K → K, and s : K → H α K for the structure maps of H α K, and write ι : H → H α K, π : H α K → K, and s : K → H α K for the structure maps of H α K. Proposition 4.6.11. Let α and α be homomorphisms of K into Aut(H). Then the following conditions are equivalent: 1. α is conjugate to α ◦ g, with g ∈ Aut(K). 2. There is an isomorphism ϕ : H α K → H α K such that im(ϕ ◦ ι) = im ι and im(ϕ ◦ s) = im s . 3. There is a commutative diagram H f H

ι

ι

/ H α K o ϕ / H α K o

s

s

K g K

in which the vertical maps are all isomorphisms. Proof Suppose that the second condition holds. Then ϕ restricts to an isomorphism from ι(H) to ι (H), and hence there exists an isomorphism f : H → H making the left-hand square of the diagram commute. Similarly, ϕ restricts to an isomorphism from s(K) to s (K) and there exists an automorphism g of K such that the right-hand square of the diagram commutes. Thus, the second condition implies the third. Suppose then that the third condition holds. Then identifying H and K with their images under ι and s , we see that ϕ ◦ ι = f and ϕ ◦ s = g. By Proposition 4.6.9, this says that g(k)f (h)(g(k))−1 = f (α(k)(h)) for all h ∈ H and k ∈ K. But since f (h) ∈ H and g(k) ∈ K, we have g(k)f (h)(g(k))−1 = α (g(k))(f (h)) by the conjugation formula that holds in H α K.


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Putting these two statements together, we see that α (g(k))(f (h)) = (f ◦ α(k))(h), and hence α (g(k)) ◦ f = f ◦ α(k) as automorphisms of H. But then as maps of K into Aut(H), this just says that α ◦ g is the conjugate homomorphism to α obtained by conjugating by f ∈ Aut(H). Thus, the third condition implies the first. Now suppose that the first condition holds, and we’re given g ∈ Aut(K) such that α ◦g is conjugate to α. Thus, there is an f ∈ Aut(H) such that (α ◦ g)(k) = f ◦ α(k) ◦ f −1 for all k ∈ K. Reversing the steps above, this implies that g(k)f (h)(g(k))−1 = f (α(k)(h)) for all h ∈ H and k ∈ K. Thus, by Proposition 4.6.9, There is a homomorphism ϕ : H α K → H α K given by ϕ(hk) = f (h)g(k). Clearly, ϕ restricts to an isomorphism, f , from ι(H) to ι (H) and restricts to an isomorphism, g, from s(K) to s (K), so it suffices to show that ϕ is an isomorphism. Since the elements of H α K may be written uniquely as an element of H times an element of K, f (h)g(k) = e if and only if f (h) = e and g(k) = e. Since f and g are both injective, so is ϕ. Finally, since f and g are both surjective, every element of H α K may be written as a product f (h)g(k), and hence ϕ is onto. Thus, the first condition implies the second. This is far from giving all the ways in which semidirect products can be isomorphic, but we shall find it useful nevertheless, particularly when H is abelian. As a sample application, consider this. Corollary 4.6.12. Let α, α : Z5 → Z× 11 be homomorphisms such that neither α nor α is the trivial homomorphism. Then the semidirect products Z11 α Z5 and Z11 α Z5 are isomorphic.

Proof By Corollary 7.3.15, Z× 11 is a cyclic group of order 10. Alternatively, we can verify it by hand: 2k is not congruent to 1 mod 11 for k ≤ 5. Since the order of 2 in Z× 11 divides 10, we see that 2 has order 10, and hence generates Z× 11 . Thus, Z× 11 has a unique subgroup H of order 5. Since the images of α and α must have order 5, they induce isomorphisms of Z5 onto H. Thus, the composite Z5

α

/ H

α−1

/ Z5

gives an isomorphism, which we’ll denote by g : Z5 → Z5 . But then α = α ◦ g, and hence Z11 α Z5 ∼ = Z11 α Z5 by Proposition 4.6.11. Using the Sylow Theorems in Chapter 5, we’ll be able to show that every group of order 55 is a semidirect product of the form Z11 α Z5 . Together with the above corollary, this shows that there are exactly two isomorphism classes of groups of order 55: the product Z11 ×Z5 (which is cyclic), and the semidirect product Z11 α Z5 obtained from the homomorphism α : Z5 → Z× 11 that sends the generator of Z5 to 4. Exercises 4.6.13. 1. Show that the subgroup s(K) ⊂ H α K is normal if and only if α : K → Aut(H) is the trivial homomorphism. 2. Consider the groups of Examples 4.6.6. For the groups of Examples 2, 3, 4, and 5, find the orders of all the elements not contained in b. Deduce that the groups of Examples 2 and 3 cannot be isomorphic. 3. Show that the group of Example 4 of Examples 4.6.6 contains a subgroup isomorphic to D10 .


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4. Let α : Z4 → Z× 3 be the homomorphism that carries the generator of Z4 to the unique nontrivial element of Z× 3 . Construct an isomorphism f : Q12 → Z3 α Z4 . 5. Consider the group G = Z8 β Z2 of Example 3 of Examples 4.6.6. Show that the subgroup b is not characteristic in G, by constructing an automorphism of G which carries b outside of itself. (Of course, b is normal in G, as it is the group being acted on in the semidirect product.) 6. Show that if m divides the order of an element x ∈ G, then o(x) = m · o(xm ). Deduce that if f : G → G is a homomorphism and if x ∈ G, then xo(f (x)) ∈ ker f and o(x) = o(f (x)) · o(xo(f (x)) ). Deduce that if hk ∈ H α K, then (hk)o(k) ∈ H, and o(hk) = o(k) · o((hk)o(k) ). Thus, the determination of the order of hk depends only on calculating the element (hk)o(k) ∈ H and on understanding the orders of the elements in H. † 7. Let x, y ∈ G, such that xyx−1 = y m . Show that (yx)k = y 1+m+···+m k ≥ 2.

k−1

xk for all

† 8. Let m > 1 and k > 1 be integers. Show that (1 + m + · · · + mk−1 )(m − 1) = mk − 1, and hence 1 + m + · · · + mk−1 = (mk − 1)/(m − 1). 9. Find the orders of the elements in the group of Example 6 of Examples 4.6.6. Show also that the subgroup b is not characteristic. (Hint: Take a look at the proof of Proposition 4.5.18.) ‡ 10. Show that the group constructed in Corollary 4.6.8 for n = 2 is isomorphic to D8 . 11. Consider the groups constructed in Corollary 4.6.8 for n a prime number p. When p > 2, show that every element in this group has exponent p. (Why is this different from the case n = 2?) Recall that in Problem 3 of Exercises 2.1.16, it was shown that any group of exponent 2 must be abelian. As we see here, this property is not shared by primes p > 2. 12. Now consider the groups of Corollary 4.6.8 for general values of n. For which values of n does the group in question have exponent n? If the exponent is not equal to n, what is it? 13. In this problem we assume a familiarity with the group of invertible 3 × 3 matrices over the ring Zn . Let ⎫ ⎧⎛ ⎞ ⎬ ⎨ 1 x y G = ⎝ 0 1 z ⎠ x, y, z ∈ Zn . ⎭ ⎩ 0 0 1 Show that G forms a group under matrix multiplication. Show, using Proposition 4.6.9, that G is isomorphic to the group constructed in Corollary 4.6.8. † 14. There are three non-identity elements in Z2 × Z2 , which must be permuted by any automorphism. Show that the automorphism group of Z2 × Z2 is isomorphic to the group of permutations on these elements, so that Aut(Z2 × Z2 ) ∼ = S3 . (Hint: Take a look at the proof of Corollary 3.2.11.) 15. Show that any two nontrivial homomorphisms from Z2 to S3 are conjugate. Deduce from Problem 14 that if α, α : Z2 → Aut(Z2 × Z2 ) are nontrivial, then (Z2 ×Z2 )α Z2 ∼ = (Z2 ×Z2 )α Z2 . Now deduce from Problem 10 that if α : Z2 → Aut(Z2 × Z2 ) is nontrivial, then (Z2 × Z2 ) α Z2 ∼ = D8 .


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16. Show that Aut(Z) ∼ = {±1} and write α : Z2 → Aut(Z) for the nontrivial homomorphism. The semidirect product Z α Z2 is called the infinite dihedral group. Show that the dihedral groups D2n are all factor groups of the infinite dihedral group. 17. Let α : Z → Aut(Z) be the unique nontrivial homomorphism. The semidirect product Z α Z is called the Klein bottle group, as it turns out to be isomorphic to the fundamental group of the Klein bottle. Show that the infinite dihedral group is a factor group of the Klein bottle group. 18. Let Z[ 12 ] ⊂ Q be the set of rational numbers that may be written as 2mk with m ∈ Z and k ≥ 0. In other words, if placed in lowest terms, these elements have denominators that are powers of 2. Show that Z[ 12 ] is a subgroup of the additive group of rational numbers and that multiplication by 2 induces an automorphism of Z[ 12 ]. Deduce the existence of a homomorphism μ2 : Z → Aut(Z[ 12 ]) such that μ2 (k) is multiplication by 2k for all k ∈ Z. ‡ 19. Let G = Z[ 12 ] μ2 Z. Let H ⊂ Z[ 12 ] be the standard copy of the integers in Q. Find an element g ∈ G such that gHg −1 is contained in H but not equal to H.3

4.7

Extensions

Here, loosely, we study all groups G that contain a given group H as a normal subgroup, such that the factor group G/H is isomorphic to a particular group K. This is the first step in trying to classify composition series. Definitions 4.7.1. Let H and K be groups. An extension of H by K consists of a group G containing H as a normal subgroup, together with a surjective homomorphism f : G → K with kernel H.4 Examples 4.7.2. 1. Let f : D2n → Z2 be induced by f (a) = 1 and f (b) = 0. Then f is an extension of Zn = b by Z2 . 2. The projection map π : Zn × Z2 → Z2 is an extension of Zn by Z2 . 3. The unique nontrivial homomorphism Z2n → Z2 is yet another extension of Zn = 2 by Z2 . 4. Let f : Q4n → Z2 be induced by f (a) = 1 and f (b) = 0. Then f is an extension of Z2n = b by Z2 . 5. Let α : K → Aut(H) be a homomorphism and let π : H α K → K be the projection. Then π is an extension of H by K. 3 We could have used the rational numbers Q here in place of Z[ 1 ], obtaining a construction that 2 would be slightly simpler for the beginner. However, by using Z[ 12 ], we obtain a group that is finitely generated (say, by the generator of H and the generator of the Z that acts), whereas we would not have if we’d used Q. 4 It is sometimes useful to think of an extension of H by K as a group G, together with an embedding ι : H → G and a surjective homomorphism f : G → K with kernel ι(H). In this language, the entire f ι sequence H −→ G −→ K constitutes the extension in question. Note that given such data, we can form an extension in the sense of Definition 4.7.1 by replacing G by the group G obtained by appropriately renaming the elements in im ι.


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Clearly, for appropriate choices of H and K, there may be a number of nonisomorphic extensions of H by K. Moreover, an understanding of the extensions of one group by another is a key to understanding groups in general. Recall from Proposition 4.2.5 that every nontrivial finite group G has a composition series, which is a sequence e = H0 ⊂ H1 ⊂ · · · ⊂ Hk = G, such that Hi−1 Hi , and Hi /Hi−1 is a nontrivial simple group for 1 ≤ i ≤ k. But this says that Hi is an extension of Hi−1 by a nontrivial simple group. In other words, every finite group may be obtained by a sequence of extensions of lower order groups by simple groups. But the finite simple groups are all known. Thus, the next step in a systematic study of finite group theory is to classify all possible extensions of one finite group by another. The search for such a classification is known as the Extension Problem, and constitutes one of the major unsolved problems in mathematics. Notice that there are various ways in which we could desire to classify the extensions of H by K. For our purposes here, we’d be most interested in simply determining the isomorphism classes of groups G that occur as extensions of H by K. Such a classification takes no account of the precise embedding of H in G, nor of the map f : G → K that determines the extension. Thus, such a classification forgets some of the data that’s part of the definition of an extension. A classification that would forget less data would be to classify the extensions of H by K up to the equivalence relation obtained by identifying extensions f : G → K and f : G → K if there is a commutative diagram H

⊂

/ K

f

g2

g1

H

/ G

⊂

/ G

g3 f

/ K

in which the vertical maps are all isomorphisms. An even more rigid equivalence relation is the one given by such diagrams where the maps g1 and g3 are required to be the identity maps of H and K, respectively. All three equivalence relations are relevant for particular questions. We shall discuss these issues further. In any case, we shall not be able to solve all the extension questions that might interest us, but we can make considerable progress in special cases. These cases will be sufficient to classify all groups of order ≤ 63. Here is a useful result for obtaining equivalences between extensions. Lemma 4.7.3. (Five Lemma for Extensions) Let f : G → K be an extension of H by K and let f : G → K be an extension of H by K . Suppose given a commutative diagram of group homomorphisms H

⊂

g1

H

/ G

f

g2 ⊂

/ G

/ K g3

f

/ K

where g1 and g3 are isomorphisms. Then g2 is also an isomorphism.


121

Proof Let x ∈ ker g2 . Then (f ◦ g2 )(x) = e, and hence g3 (f (x)) = e by commutativity of the diagram. Since g3 is injective, f (x) = e, and hence x ∈ ker f = H. But on H ⊂ G, g2 acts as g1 : H → H , and hence x ∈ ker g1 . But g1 is injective, so we must have x = e. Thus, ker g2 = e, and hence g2 is injective. Thus, it suffices to show that each y ∈ G lies in the image of g2 . Since g3 is surjective, f (y) = g3 (z ) for some z ∈ K. But f is also surjective, so z = f (z) for some z ∈ G. Thus, f (g2 (z)) = g3 (f (z)) = g3 (z ) = f (y), so that g2 (z) and y have the same image under f . But then y(g2 (z))−1 lies in the kernel of f , which is H . Thus, since g1 is surjective, there is an x ∈ H such that g1 (x) = y(g2 (z))−1 . Thus, y = g1 (x)g2 (z) = g2 (xz), and hence y ∈ im g2 as desired. The Five Lemma for Extensions is a special case of a more general result which is given for abelian groups as the Five Lemma in Lemma 7.7.48. The statement of the general case involves five vertical maps, rather than the three displayed here, hence the name. In both cases, the technique of proof is called a diagram chase. As noted above, if α : K → Aut(H) is a homomorphism, then the projection map π : H α K → K is an extension of H by K. As a general rule, the semidirect products are the easiest extensions to analyze, so it is useful to have a criterion for an extension to be one. Definition 4.7.4. A section, or splitting, for f : G → K is a homomorphism s : K → G, such that f ◦ s is the identity map of K. A homomorphism f : G → K that admits a section is said to be a split surjection. An extension of H by K is called a split extension if f : G → K admits a section. The section s is said to split f . The homomorphism s : K → H α K is clearly a section for π. We also have a converse. Proposition 4.7.5. Split extensions are semidirect products. Specifically, let f : G → K be a split extension of H by K, with splitting map s : K → G. Then s induces a homomorphism α : K → Aut(H) via α(k)(h) = s(k)h(s(k))−1 . Moreover, there is an isomorphism ϕ : H α K → G such that the following diagram commutes. H H

⊂ ι

/ H α K o ϕ ⊂ / G o

s1

s

K K

Here, s1 is the standard section for the semidirect product. Explicitly, ϕ is defined by ϕ(hk) = hs(k). Proof Recall from Lemma 3.7.7 that, since H G, we have a homomorphism Γ : G → Aut(H), given by setting Γ(g)(h) = ghg −1 for all g ∈ G and h ∈ H. The above specification for α is just the composite Γ ◦ s : K → Aut(H), and hence is a homomorphism as claimed. Recall from Proposition 4.6.9 that if f1 : H → G and f2 : K → G are homomorphisms such that f2 (k)f1 (h)(f2 (k))−1 = f1 (α(k)(h)) for all h ∈ H and k ∈ K, then there


122

is a unique homomorphism f : H α K → G such that f ◦ ι = f1 and f ◦ s1 = f2 : specifically, f (hk) = f1 (h)f2 (k). We apply this with f1 equal to the inclusion of H in G and with f2 equal to the section s : K → G. But then the condition we need to verify to construct our homomorphism is that s(k)h(s(k))−1 = α(k)(h), which holds by the definition of α. Thus, there is a homomorphism ϕ : H α K → G with ϕ◦ι equal to the inclusion of H in G, and ϕ◦s1 = s: specifically, ϕ(hk) = hs(k). But this just says that the stated diagram commutes, so it suffices to show that ϕ is an isomorphism. Since ϕ(hk) = hs(k), we obtain that (f ◦ ϕ)(hk) = (f ◦ s)(k) = k. But this says that f ◦ ϕ = π, where π : H α K → K is the projection map. Thus, we have an additional commutative diagram. H H

⊂ ι

/ H α K ϕ ⊂ / G

π

f

/ K / K

But this shows ϕ to be an isomorphism by the Five Lemma for Extensions. Of course, if α : K → Aut(H) is the trivial homomorphism, then the semidirect product H α K is simply H × K. Thus, the corollary below is immediate from the way that α depends on s in the statement of Proposition 4.7.5. Corollary 4.7.6. Let f : G → K be a split extension of H by K, with splitting map s : K → G. Suppose that s(k) commutes with each element of H for all k ∈ K. Then G∼ = H × K. In particular, if f : G → K is a split surjection and if G is abelian, then G ∼ = ker f × K. Semidirect products are the easiest extensions to construct, and hence may be viewed as the most basic of the extensions of one group by another. For this reason, it is useful to be able to recognize whether an extension splits. The easiest case is when the two groups have orders that are relatively prime. Proposition 4.7.7. Let H and K be groups whose orders are relatively prime and let f : G → K be an extension of H by K. Then the extension splits if and only if G has a subgroup of order |K|. Proof If s : K → G is a section of f , then s is injective, so s(K) has the same order as K. Conversely, suppose that K ⊂ G has order |K| and write f1 : K → K for the restriction of f to K . Then ker f1 = K ∩ H. Since the orders of K and H are relatively prime, K ∩ H = e, by Lagrange’s Theorem. Thus, f1 is injective. Since K and K have f −1

1 the same order, f1 is an isomorphism, and the composite K −− → K ⊂ G gives a section for f .

In fact, under the hypotheses above, G always has a subgroup of order |K|. Theorem (Schur–Zassenhaus Lemma) Let f : G → K be an extension of H by K, where H and K have relatively prime order. Then G has a subgroup of order |K|.


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The proof requires the Sylow Theorems and could be given in Chapter 5, but we shall not do so. In practice, when considering a particular extension, one can generally construct the section by hand. Let us now consider the issue of classifying the split extensions of one group by another. Note that the definition of the homomorphism α : K → Aut(H) constructed in Proposition 4.7.5 makes use of the specific section s. Thus, we might wonder whether a different homomorphism α would have resulted if we’d started with a different section. Indeed, this can happen when the group H is nonabelian. Example 4.7.8. Let ε : Sn → {±1} take each permutation to its sign. Then ε is an extension of An by Z2 . Moreover, setting s(−1) = τ ∈ Sn determines a section of ε if and only if τ has order 2 and is an odd permutation. Since the order of a product of disjoint cycles is the least common multiple of the orders of the individual cycles, we see that τ determines a section of ε if and only if it is the product of an odd number of disjoint 2-cycles. For such a τ , we write sτ for the section of ε with sτ (−1) = τ . Given the variety of such τ , we see that there are many possible choices of section for ε. Write ατ : Z2 → Aut(An ) for the homomorphism induced by sτ . By definition, ατ = Γ ◦ sτ , where Γ : Sn → An is the homomorphism induced by conjugation. Thus, ατ (−1) = Γ(τ ). But Problem 10 in Exercises 3.7.14 shows that Γ is injective for n ≥ 4. Thus, if n ≥ 4, then each different choice for a section of ε induces a different homomorphism from Z2 to Aut(An ). If τ and τ have the same cycle structure, then they are conjugate in Sn , and hence sτ and sτ are conjugate as homomorphisms into Sn . But then ατ and ατ are conjugate as homomorphisms into Aut(An ). However, as we shall see in the exercises below, if τ is a 2-cycle and τ is a product of three disjoint 2-cycles, then ατ and ατ are not conjugate as homomorphisms into Aut(An ) if n > 6. Thus, in this case, Proposition 4.6.11 would not have predicted that An ατ Z2 and An ατ Z2 would be isomorphic; nevertheless they are isomorphic, as each one of them admits an isomorphism to Sn . However, the criterion in Proposition 4.6.11 is the most general sufficient condition we shall give for the isomorphism of two semidirect products. Thus, the classification of even the split extensions of a nonabelian group offers substantial difficulty. But we shall be mainly concerned here with the extensions of an abelian group, a problem for which we can develop some reasonable tools. Lemma 4.7.9. Let f : G → K be an extension of H by K, where H is abelian. For x ∈ G, write cx : H → H for the automorphism obtained from conjugation by x. Suppose that f (x) = f (y). Then cx = cy as automorphisms of H. Moreover, there is a homomorphism αf : K → Aut(H) defined by setting αf (k) equal to cx for any choice of x with f (x) = k. In particular, if f happens to be a split extension and if s : K → G is a section of f , then the homomorphism from K to Aut(H) induced by s coincides with the homomorphism αf above. Thus, if H is abelian, the homomorphism from K to Aut(H) induced by a section of f is independent of the choice of section. Proof If f (x) = f (y), then y −1 x ∈ ker f = H, so x = yh for some h ∈ H. But then cx = cy ◦ ch . Since H is abelian, conjugation of H by h is the identity map, so cx = cy as claimed.


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Thus, there is a well defined function αf : K → Aut(H) obtained by setting αf (k) equal to cx for any x with f (x) = k. But if f (x) = k and f (y) = k , then f (xy) = kk , so that αf (kk ) = cxy = cx ◦ cy = αf (k)αf (k ), so αf is a homomorphism. If s : K → G is a section of f , then the homomorphism α : K → Aut(H) induced by this section is obtained by setting α(k) = cs(k) : H → H. Since f (s(k)) = k, this just says that α = αf . Thus, we may make the following definition. Definition 4.7.10. Let f : G → K be an extension of H by K, where H is abelian. Then the action of K on H induced by f is the homomorphism αf : K → Aut(H) obtained by setting αf (k) equal to the automorphism obtained by conjugating H by any given element of f −1 (k). The existence of the homomorphisms αf gives a valuable method of studying the extensions of an abelian group: fix a homomorphism α : K → Aut(H) and classify the extensions whose induced action is α. Do this for all possible homomorphism α, and look to see if a given group can be written as an extension in more than one way. We shall illustrate this process by classifying the groups of order 8. Lemma 4.7.11. Let f : G → Z2 be an extension of Z4 by Z2 . Suppose that the induced action αf : Z2 → Z× 4 carries the generator of Z2 to −1. Then G is isomorphic to either D8 or Q8 . Proof Let Z4 = b and Z2 = a. For any x ∈ G with f (x) = a, we have xbk x−1 = αf (a)(bk ) = b−k for all k. Also, f (x2 ) = a2 = e, so x2 ∈ ker f = b. Thus, x2 = bk for some k. Note that x fails to commute with b and b−1 , so neither of these can equal x2 . Thus, either x2 = e or x2 = b2 . If x2 = e, then x has order 2, and hence there is a section of f , specified by s(a) = x. In this case, G ∼ = Z4 αf Z2 , which, for this value of αf , is isomorphic to the dihedral group D8 by the first example of Examples 4.6.6. The remaining case gives x2 = b2 . But then there is a homomorphism h : Q8 → G with h(a) = x and h(b) = b by Problem 5 of Exercises 2.9.7. As the image of h contains more than half the elements of G, h must be onto, and hence an isomorphism. (Alternatively, we could use the Five Lemma for Extensions to show that h is an isomorphism.) Q8 , of course, is not a split extension of b by Z2 , for two reasons. First, we know that every element of Q8 which is not in b has order 4, and hence it is impossible to define a section from Q8 /b to Q8 . Second, if it were a split extension, then it would be isomorphic to D8 , which we know is not so. We have classified the extensions of Z4 by Z2 corresponding to the homomorphism × from Z2 to Z× 4 that takes the generator of Z2 to −1. Since Z4 is isomorphic to Z2 , there × is only one other homomorphism from Z2 to Z4 : the trivial homomorphism. Let us begin with some general observations about extensions whose induced action is trivial. Lemma 4.7.12. Let f : G → K be an extension of H by K, where H is abelian. Then the induced homomorphism αf : K → Aut(H) is the trivial homomorphism if and only if H ⊂ Z(G).


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Proof Suppose that αf is the trivial homomorphism. Then for any x ∈ G, let y = f (x). By the definition of αf , we have xhx−1 = αf (y)(h) for all h ∈ H. But αf is the trivial homomorphism, so that αf (y) is the identity map of H. Thus xhx−1 = h for all h ∈ H, and hence each h ∈ H commutes with every element of G. Thus, H ⊂ Z(G). But the converse is immediate: if H ⊂ Z(G), then conjugating H by any element of G gives the trivial homomorphism, and hence αf (y) is the identity element of Aut(H) for every y ∈ K. The extensions of H containing H in their centers are important enough to merit a name. Definition 4.7.13. An extension f : G → K of H by K is a central extension if H ⊂ Z(G). Example 4.7.14. As one may easily see, the center of D8 is b2 . Moreover, D8 /b2 is isomorphic to Z2 × Z2 , with generators a and b. Thus, D8 is a central extension of Z2 by Z2 × Z2 . In particular, a central extension of a cyclic group by an abelian group is not necessarily abelian. But a central extension of an abelian group by a cyclic group is abelian: Lemma 4.7.15. Let f : G → Zk be a central extension of the abelian group H by the cyclic group Zk . Then G is abelian. Proof Write Zk = a, and let x ∈ G with f (x) = a. Let g ∈ G. Then f (g) = ai = f (xi ) for some i, and hence gx−i ∈ ker f = H. But then g = hxi for some h ∈ H. If g is any other element of G, we may write g = h xj for h ∈ H and j ∈ Z. But then gg = hxi h xj = hh xi+j , as h ∈ Z(G). But multiplying out g g gives the same result, and hence g and g commute. We now return to the groups of order 8. Lemma 4.7.16. Let f : G → Z2 be a central extension of Z4 by Z2 . Then G is isomorphic to either Z8 or Z4 × Z2 . Proof Since Z2 is cyclic, Lemma 4.7.15 shows that G is abelian. So we could just appeal to the Fundamental Theorem of Finite Abelian Groups, which tells us that Z8 , Z4 × Z2 , and Z2 × Z2 × Z2 are the only abelian groups of order 8. Clearly, the first two are the only ones that are extensions of Z4 . Alternatively, we can argue directly. Write Z4 = b and Z2 = a and let f (x) = a. Then f (x2 ) = e, and hence x2 ∈ b. Now x has even order, so that o(x) = 2o(x2 ). Thus, if x2 = b or b−1 , then x has order 8, and hence G is cyclic. If x has order 2, then s(a) = x gives a splitting of f , and hence G is the semidirect product obtained from the action of Z2 on Z4 . But in this case, the action is trivial, so that the semidirect product thus obtained is just the direct product. If x has order 4, then x2 = b2 . But then b−1 x has order 2, and setting s(a) = b−1 x gives a section of f , and hence G ∼ = Z4 × Z2 as above. Once again, we have a nonsplit extension: the unique nontrivial homomorphism f : Z8 → Z2 is a nonsplit extension of Z4 by Z2 . We now classify the groups of order 8. Proposition 4.7.17. Every group of order 8 is isomorphic to exactly one of the following groups: Z8 , D8 , Q8 , Z4 × Z2 , and Z2 × Z2 × Z2 .


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Proof If G has an element of order 4, say o(b) = 4, then b has index two in G, and hence is normal in G by Proposition 4.1.12 (or by the simpler argument of Problem 12 of Exercises 3.7.14). But then G is an extension of Z4 by Z2 . As Z× 4 = {±1}, there are , the homomorphism that carries the only two different homomorphisms from Z2 to Z× 4 generator of Z2 to −1 and the trivial homomorphism. Thus, if G is an extension of Z4 by Z2 , then G is isomorphic to one of Z8 , D8 , Q8 , and Z4 × Z2 by Lemmas 4.7.11 and 4.7.16. If G does not have an element of order 4, then every element of G has exponent 2. But then G is abelian by Problem 3 of Exercises 2.1.16. But the fundamental theorem of finite abelian groups shows Z2 × Z2 × Z2 to be the only abelian group of order 8 and exponent 2. The reader may easily check that no two of the listed groups are isomorphic. We shall classify all groups of order p3 for odd primes p as well, but this will require theoretical results to be given in the next chapter. We now return to the general theory of extensions. Lemma 4.7.18. Let f : G → K and f : G → K be extensions of H by K, where H is abelian. Suppose given a commutative diagram H

⊂

f

⊂

/ G

/ K g3

g2

g1

H

/ G

f

/ K,

where g1 , g2 , and g3 are isomorphisms. Then the actions αf , αf : K → Aut(H) induced by f and f are related by g1 ◦ αf (k) ◦ g1−1 = αf (g3 (k)), so that αf is conjugate to αf ◦ g3 as a homomorphism from K to Aut(H). Proof For k ∈ K, let x ∈ G with f (x) = k. Then f (g2 (x)) = g3 (k), so αf (g3 (k))(g1 (h)) = g2 (x)g1 (h)g2 (x)−1 = g2 (xhx−1 ) = g1 (αf (k)(h)) for all h ∈ H. Thus, αf (g3 (k)) ◦ g1 = g1 ◦ αf (k), and the result follows. Remarks 4.7.19. There is a general method that may be used to classify the extensions of an abelian group H by a group K up to the following equivalence relation: The extensions f : G → K and f : G → K are equivalent if there is a commutative diagram H

⊂

/ G

f

/ K

f

/ K.

g

H

⊂

/ G

Here, g is forced to be an isomorphism, by the Five Lemma for Extensions. Notice that Lemma 4.7.18 implies that if f and f are equivalent, then the homomorphisms αf and αf are equal. Thus, the method under discussion will give a separate classification for each different homomorphism α : K → Aut(H). Thus, fixing such an α, we shall write Hα to denote H with this particular action of K. The method of classification is given using the cohomology of groups, a subtopic of homological algebra. The classification theorem gives a one-to-one correspondence


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between the equivalence classes of extensions corresponding to the homomorphism α and the elements of the cohomology group H 2 (K; Hα ).5 Nevertheless, a couple of deficiencies in this approach should be pointed out. First, given an element in H 2 (K; Hα ), it is quite tedious to construct the associated extension. And it comes in a form that is difficult to analyze. Second, the classification is too rigid for many of the questions of interest: there are extensions whose associated actions are both α which are not equivalent in the above sense, but are equivalent under weaker hypotheses. Thus, there may be fewer isomorphism classes of groups G that may be written as extensions of H by K with action α than there are elements of the cohomology group H 2 (K; Hα ). Nevertheless, the homological approach can be useful. It is sometimes useful to be able to go back and forth between the existence of a commutative diagram such as that in Lemma 4.7.18 and the existence of a suitable isomorphism from G to G . Lemma 4.7.20. Let f : G → K and f : G → K be extensions of H by K, and write ι : H → G and ι : H → G for the inclusions. Then the following conditions are equivalent. 1. There is a commutative diagram H

⊂ ι

f

⊂ ι

/ G

/ K g3

g2

g1

H

/ G

f

/ K,

where g1 , g2 , and g3 are isomorphisms. 2. There is an isomorphism g : G → G with g(im ι) = im ι . Proof If the first condition holds, then the second one does also, as we may take g = g2 . Thus, suppose that the second condition holds, and we are given an isomorphism g : G → G with g(im ι) = im ι . Then g induces an isomorphism from im ι to im ι . Since ι and ι are embeddings, there is an isomorphism g1 : H → H such that ι ◦ g1 = g ◦ ι. Thus, if we set g2 = g, then the left square of the displayed diagram commutes. We must now construct g3 , which comes from a couple of applications of the First Noether Theorem. First, we have surjective maps f : G → K and f : G → K, with kernels given by ker f = H and ker f = H. Thus, the First Noether Theorem gives ∼ ∼ = = isomorphisms f : G/H −→ K and f : G /H −→ K such that the following diagrams commute. /K GD DD z< z DD zz D zzf π DD z ! z G/H f

f

/K G E y< EE y EE yy E yyf π EE y y " G /H

5 See, for instance, A Course in Homological Algebra, by P.J. Hilton and U. Stammbach, SpringerVerlag, 1971, for an exposition of this classification.

CHAPTER 4. NORMALITY AND FACTOR GROUPS g

128 π

→ G − → G /H is precisely H, so Since g(im ι) = im ι , the kernel of the composite G − ∼ = the First Noether Theorem provides an isomorphism k : G/H −→ G /H that makes the following diagram commute. H g1 ∼ = H

Now set g3 = f ◦ k ◦ f

−1

⊂ ι

⊂ ι

/ G ∼ g = / G

π

π

/ G/H k ∼ = / G /H

f ∼ =

f ∼ =

/ K

/ K

.

Note that the group H of Lemma 4.7.20 need not be abelian. We now wish to explore the question of when two different homomorphisms, α, α : K → Aut(H) produce isomorphic semidirect products. In this regard, the above lemmas permit an improvement of the result in Proposition 4.6.11 in the case that H is abelian. Proposition 4.7.21. Let H be an abelian group and let α and α be homomorphisms of K into Aut(H). Then α is conjugate to a composite α ◦ g with g ∈ Aut(K) if and only if there is an isomorphism ϕ : H α K → H α K such that ϕ(im ι) = im ι , where ι and ι are the canonical inclusions of H in H α K and H α K, respectively. Proof If α is conjugate to a composite α ◦ g with g ∈ Aut(K), then Proposition 4.6.11 provides an isomorphism ϕ : H α K → H α K with ϕ(im ι) = im ι . Thus, it suffices to show the converse. So suppose given an isomorphism ϕ : H α K → H α K with ϕ(im ι) = im ι . Then Lemma 4.7.20 provides a commutative diagram of the sort appearing in the hypothesis of Lemma 4.7.18. By Lemma 4.7.9, the action map απ : K → Aut(H) induced by π : H α K → K is precisely α. A similar result holds for α . Thus, the output of Lemma 4.7.18 is precisely that α is conjugate to α ◦ g for some g ∈ Aut(K). We would like to eliminate the hypothesis that ϕ(im ι) = im ι from the above, and obtain a necessary and sufficient condition for the semidirect products H α K and H α K to be isomorphic. We can do this if H and K are finite groups whose orders are relatively prime. Corollary 4.7.22. Let H and K be finite groups whose orders are relatively prime, and suppose that H is abelian. Let α and α be two homomorphisms from K to Aut(H). Then the semidirect products H α K and H α K are isomorphic if and only if α is conjugate to α ◦ g for some automorphism g of K. Proof It suffices to show that any homomorphism from H α K to H α K carries ι(H) into ι (H), and that the analogous statement holds for homomorphisms in the opposite direction. But this will follow if we show that in either one of the semidirect products, the only elements of exponent |H| are those in H itself. If |H| is an exponent for x, then it is also an exponent for f (x) for any homomorphism f . Since any element whose order divides both |H| and |K| must be the identity, we see that any element of either semidirect product which has exponent |H| must be carried to the identity element by the projection map (π or π ) into K. Since H is the kernel of the projection map, the result follows.


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We now give a generalization of Corollary 4.7.22. Corollary 4.7.23. Let H be an abelian group and let α, α : K → Aut(H). Suppose that α has the property that any embedding f : H → H α K must have image equal to ι(H). Then H α K is isomorphic to H α K if and only if α is conjugate to α ◦ g for some g ∈ Aut(K). Proof Again, it suffices to show that ϕ(im ι) = im ι , or, equivalently, that im ι = im(ϕ−1 ◦ ι ), for any isomorphism ϕ : H α K → H α K. But since ϕ−1 ◦ ι is an embedding, this follows from the hypothesis. Corollary 4.7.23 may be applied, for instance, if ι(H) is the only subgroup of H α K which is isomorphic to H. An example where this occurs is for H = b ⊂ D2n for n ≥ 3. Here, we make use of the fact that every element of D2n outside of b has order 2. Since b has order n > 2, every embedding of b in D2n must have image in b. Since b is finite, this forces the image to be equal to b. Finally, let us consider the extensions of a nonabelian group H. As shown in Example 4.7.8, if f : G → K is even a split extension of H by K, different choices of splittings may define different homomorphisms from K into Aut(H). In particular, we do not have a homomorphism from K to Aut(H) which is associated to the extension itself. But we can define something a little weaker. As usual, we write ch : H → H for the automorphism induced by conjugating by h ∈ H: ch (h ) = hh h−1 for all h ∈ H. Recall that an automorphism f ∈ Aut(H) is an inner automorphism if f = ch for some h ∈ H. The inner automorphisms form a subgroup, Inn(H) ⊂ Aut(H). Lemma 4.7.24. The group of inner automorphisms of a group G is a normal subgroup of Aut(G). Proof Let x ∈ G and write cx for conjugation by x. Let f ∈ Aut(G). We wish to show that f ◦ cx ◦ f −1 is an inner automorphism. But f ◦ cx ◦ f −1 (g)

= f (xf −1 (g)x−1 ) = f (x)f (f −1 (g))f (x)−1 ,

and hence f ◦ cx ◦ f −1 = cf (x) . Thus, we may make the following definition. Definition 4.7.25. We write Out(G) for the quotient group Aut(G)/Inn(G) and call it the outer automorphism group of G. Proposition 4.7.26. Let f : G → K be an extension of H by K. Then f determines a homomorphism α : K → Out(H). Here, for k ∈ K, α(k) is represented by the automorphism of H given by conjugation by g, for any g ∈ G with f (g) = k. In particular, if f is a split extension and s is a splitting of f , then α = π ◦ αs , where αs : K → Aut(H) is the homomorphism induced by s and π is the canonical map from Aut(H) to Out(H) = Aut(H)/Inn(H). Thus, if s and s are two different sections of f , then the homomorphisms αs and αs agree in Out(H).


130

Proof If f (g) = f (g ) = k, then g = gh for some h ∈ H. But then conjugating by g is the same as conjugating first by h and then by g. But conjugation by h is an inner automorphism of H, and hence represents the identity in Out(H). Thus, conjugation by g represents the same element of Out(H) as conjugation by g, and hence the prescription above gives a well-defined function α : K → Out(H). It suffices to show that α is a homomorphism. But for k, k ∈ K, if f (g) = k and f (g ) = k , then f (gg ) = kk . But conjugation by gg is the composite of the conjugations by g and by g . We should also note that the above is not always very useful. For instance, for n = 6, Out(Sn ) is the trivial group. Exercises 4.7.27. † 1. Problem 1 of Exercises 4.2.18 provides a normal subgroup H ⊂ A4 of order 4. Show that H ∼ = Z2 × Z2 and that A4 is a split extension of H by Z3 . 2. Show that any nonabelian group G of order 12 which contains a normal subgroup of order 4 must be isomorphic to A4 . (Hint: Find a G-set with 4 elements and analyze the induced homomorphism from G to S4 . Alternatively, show that G is a split extension of a group of order 4 by Z3 , and, using the calculation of Aut(Z2 × Z2 ) given in Problem 14 of Exercises 4.6.13, show that there is only one such split extension which is nonabelian.) † 3. Consider the subgroup H ⊂ An of Problem 1. Show that H is normal in S4 and that S4 is isomorphic to the semidirect product (Z2 × Z2 ) α S3 , where α : S3 → Aut(Z2 × Z2 ) is an isomorphism. (Recall the calculation of Aut(Z2 × Z2 ) given in Problem 14 of Exercises 4.6.13. In the situation at hand, you may enjoy giving a new proof of this calculation, based on showing that the map α above is injective.) 4. Show that if k is odd, then Q4k is isomorphic to Zk α Z4 for some α : Z4 → Aut(Zk ). Calculate α explicitly. 5. Find all the central extensions of Z2 by Z2 × Z2 . 6. Let f : G → Z2 be an extension of Z8 by Z2 and suppose that the induced homomorphism αf : Z2 → Z× 8 carries the generator of Z2 to 5. Show that f is a split extension, and hence that G is isomorphic to the group Z8 β Z2 of Example 3 in Examples 4.6.6. (Hint: Write Z8 = b and Z2 = a and let x ∈ G with f (x) = a. Note that x2 = bk for some k, and use this to calculate the order of each of the elements bi x.) 7. Let f : G → Z2 be an extension of Z8 by Z2 and suppose that the induced homomorphism αf : Z2 → Z× 8 carries the generator of Z2 to 3. Show that f is a split extension, and hence that G is isomorphic to the group Z8 α Z2 of Example 2 in Examples 4.6.6. (Hint: Argue as in the preceding problem. Here, if x2 = bk , note that the fact that x must commute with x2 places restrictions on which values of k can occur.) 8. Let f : G → Z2 be an extension of Z8 by Z2 and suppose that the induced homomorphism αf : Z2 → Z× 8 carries the generator of Z2 to −1. Show that f is isomorphic to either D16 or Q16 . 9. Classify those groups of order 16 that contain an element of order 8.


131

10. Let H be any group and let f : G → Z be an extension of H by Z. Show that f splits. 11. We show that the inner automorphisms of Sn induced by different sections of the sign homomorphism are not always conjugate in Aut(Sn ). (a) Let f ∈ Aut(G) and let Gf = {g ∈ G | f (g) = g}, the fixed points of G under f . Show that Gf is a subgroup of G. Show also that if f and f are conjugate in Aut(G), then Gf and Gf are isomorphic as groups. (b) Let τ ∈ Sn be a 2-cycle and write An τ for the fixed points of An under the automorphism obtained from conjugating by τ . Show that An τ is isomorphic to Sn−2 . (c) Let H ⊂ S6 be the subgroup consisting of those elements of S6 that commute with (1 2)(3 4)(5 6). Show that H is a split extension of Z2 × Z2 × Z2 by S3 . (d) Let τ ∈ Sn be a product of three disjoint 2-cycles. Show that the subgroup An τ fixed by τ , is isomorphic to an index 2 subgroup of H × Sn−6 , where H ⊂ S6 is the group of the preceding part. Deduce that if n > 6, then the automorphism of An obtained from conjugation by a 2-cycle cannot be conjugate in Aut(An ) to the automorphism obtained from conjugation by the product of three disjoint 2-cycles. ‡ 12. Let α : K → Aut(H) and β : K → Aut(H ) be homomorphisms. Let f : H → H be a group homomorphism such that f (α(k)(h)) = β(k)(f (h)) for all k ∈ K and h ∈ H (i.e., f is a K-map between the K-sets H and H ). Define f∗ : H α K → H β K by f∗ (hk) = f (h)k. Show that f∗ is a homomorphism. 13. Let α : K → Aut(H) be a homomorphism. Suppose given f ∈ Aut(H) that commutes with every element of the image of α. Show that the homomorphism f∗ : H α K → H α K of Problem 12 is defined and gives an automorphism of H α K. ‡ 14. In this problem, we calculate the automorphisms of a dihedral group. The case of D4 is exceptional, and indeed has been treated in Problem 14 in Exercises 4.6.13. Thus, we shall consider D2n for n ≥ 3. A key fact is given by either Problem 6 of Exercises 2.8.5 or by Proposition 4.6.10: The homomorphisms from D2n to a group G are in one-to-one correspondence with the choices of x = f (a) and y = f (b) such that x has exponent 2, y has exponent n, and xyx−1 = y −1 . The reader should first recall from Problem 4 of Exercises 3.7.14 that since n ≥ 3, the subgroup b ⊂ D2n is characteristic. Thus, Lemma 3.7.8 provides a restriction homomorphism ρ : Aut(D2n ) → Aut(b), given by setting ρ(f ) equal to its restriction f : b → b. As usual, by abuse of notation, we shall identify Aut(b) with × Z× n , and write ρ : Aut(D2n ) → Zn . (a) Show that any homomorphism f : D2n → D2n whose restriction to b gives an automorphism of b must have been an automorphism of D2n in the first place. (b) Show that the restriction homomorphism ρ : Aut(D2n ) → Z× n is a split sur× jection (i.e., ρ : Aut(D2n ) → Z× n is a split extension of ker ρ by Zn ).


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(c) Show that an automorphism f is in the kernel of ρ if and only if f (b) = b and f (a) = bi a for some i. Deduce that the kernel of ρ is isomorphic to Zn . (d) Show that the action of Z× n on Zn = ker ρ induced by the split extension is the same as the usual action of Z× n = Aut(Zn ). Deduce that Aut(D2n ) is isomorphic to the semidirect product Zn id Aut(Zn ), where id is the identity map of Aut(Zn ). 15. Show that every automorphism of S3 is inner. 16. Recall from Problem 8 of Exercises 3.7.14 that if n ≥ 3, the subgroup b ⊂ Q4n is characteristic. Using this as a starting point, give a calculation of Aut(Q4n ), n ≥ 3, by methods analogous to those of Problem 14 above. Note that the lack of a semidirect product structure on Q4n necessitates greater care in the construction of a splitting for ρ than was required in the dihedral case. (We shall treat the exceptional case of Aut(Q8 ) in Problem 9 of Exercises 5.4.9.) ‡ 17. Recall from Problem 3 of Exercises 4.2.18 that An is a characteristic subgroup of Sn for n ≥ 3. Thus, Lemma 3.7.8 provides a restriction homomorphism ρ : Aut(Sn ) → Aut(An ). Let Γ : Sn → Aut(An ) be the homomorphism induced by conjugation. By Problem 10 in Exercises 3.7.14, Γ is injective for n ≥ 4. Use this fact, together with Proposition 4.6.9, to show that ρ is injective for n ≥ 4. 18. We compute the automorphism groups of A4 and S4 . (a) Let H ⊂ A4 be the subgroup of order 4. Show that H is characteristic in A4 . Thus, Lemma 3.7.8 provides a restriction homomorphism ρ : Aut(A4 ) → Aut(H). (b) Show that the kernel of ρ is isomorphic to H, and hence to Z2 × Z2 . (c) Show that ρ is a split surjection. Deduce that the homomorphism Γ : S4 → Aut(A4 ) induced by conjugation is an isomorphism. (d) Deduce from Problem 17 that the homomorphism Γ : S4 → Aut(S4 ) induced by conjugation is an isomorphism. ‡ 19. We consider the automorphism groups of the semidirect products G = Zn α Zk for selected homomorphisms α : Zk → Aut(Zn ). As usual, we write ι : Zn → G and s : Zk → G for the standard inclusions and write π : G → Zk for the standard projection. The one assumption we shall insist on is that ι(Zn ) be a characteristic subgroup of G. As the reader may easily verify, this is always the case if n and k are relatively prime. But we have seen other examples, e.g., D2n = Zn α Z2 or Example 2 of Examples 4.6.6, where Zn is characteristic in G, and our arguments will apply to these cases as well. In particular, the material here generalizes that in Problem 14. As usual, we use multiplicative notation, with Zn = b and Zk = a. As discussed j in Section 4.6, we have α(aj )(bi ) = bm i for an integer m with the property that mk ≡ 1 mod n. The elements of G may be written uniquely as bi aj with 0 ≤ i < n j and 0 ≤ j < k, and bi aj bi aj = bi+m i aj+j . Since b is assumed characteristic in G, we have a restriction homomorphism ρ : Aut(G) → Aut(b) ∼ = Z× n.


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(a) Let f ∈ Aut(G). Show that there is a unique element f ∈ Aut(a) such that the following diagram commutes: G

f

π

a

/ G π

f

/ a

Show that the passage from f to f gives a homomorphism η : Aut(G) → Aut(a) ∼ = Z× k. (b) Show, via Corollary 4.6.10, that the image of η consists of those f ∈ Aut(Zk ) such that the following diagram commutes: Zk

α

/ Z× n

α

/ Z× . n

f

Zk

Show that such automorphisms f ∈ Aut(Zk ) correspond to elements j ∈ Z× k such that mj ≡ m mod n. Deduce that if α is injective (i.e., if m has order k in Z× n ), then η is the trivial homomorphism. (c) Consider the map (ρ, η) : Aut(G) → Aut(b) × Aut(a) given by (ρ, η)(f ) = (ρ(f ), η(f )). Show that (ρ, η) induces a split surjection of Aut(G) onto Aut(b)× (im η). (d) Show that an automorphism f is in the kernel of (ρ, η) if and only if f (b) = b and f (a) = bi a, where i is chosen so that bi a has order k in G. Show, via Problems 7 and 8 of Exercises 4.6.13, that bi a has order k if and only if i · (mk − 1)/(m − 1) is divisible by n. Show also that passage from f to bi induces an isomorphism from the kernel of (ρ, η) onto a subgroup H ⊂ b. (e) Let β : Aut(Zn ) × (im η) → Aut(H) be the action induced by the extension (ρ, η). Show that β restricts on Aut(Zn ) to the restriction map r : Aut(Zn ) → Aut(H) induced by the fact that H is characteristic in Zn . (Recall from Problem 9 of Exercises 3.7.14 that every subgroup of a cyclic group is characteristic.) Deduce that if α is injective, then Aut(G) ∼ = H r Aut(Zn ). (f) Let β : Aut(Zn ) × (im η) → Aut(H) be as above. Compute the restriction of β to im η. (g) Show that if n is prime, then H = Zn . (h) Give an example of α, n, and k where H is not all of Zn .

Chapter 5

Sylow Theory, Solvability, and Classification

The main theorems of this chapter, e.g., Cauchy’s Theorem, the Sylow Theorems, and the characterization of solvable and nilpotent groups, could have been given much earlier in the book. In fact, a proof of Cauchy’s Theorem has appeared as Problem 24 in Exercises 3.3.23. Indeed, these theorems depend on very little other than the theory of G-sets (which it might be advisable to review at this time) and the Noether Isomorphism Theorems. Our main reason for delaying the presentation of these theorems until now was so that we could develop enough theory to support their applications. Given our analysis of abelian groups, extensions, and some automorphism groups, we will be able to use the results in this chapter to classify the groups of most orders less than 64, and, hopefully, to bring the study of finite groups into a useful perspective. Lagrange’s Theorem tells us that if the group G is finite, then the order of every subgroup must divide the order of G. But there may be divisors of |G| that do not occur in this manner. For instance, Problem 2 of Exercises 4.2.18 shows that A4 has no subgroup of order 6. In this chapter, we shall give some existence theorems for subgroups of particular orders in a given finite group. The first and most basic such result is Cauchy’s Theorem, which says that if G is a finite group whose order is divisible by a prime p, then G has an element (hence a subgroup) of order p. We give this in Section 5.1, along with applications to classification. In Section 5.2, we study finite p-groups, which, because of Cauchy’s Theorem, are precisely the finite groups whose order is a power of the prime p. By a proof similar to the one given for Cauchy’s Theorem, we show that any p-group has a nontrivial center. We give some applications, including most of the pieces of a classification of the groups of order p3 for p odd. The last step will be given via the Sylow Theorems in Section 5.3. Section 5.3 is devoted to the Sylow Theorems, which study the p-subgroups of a finite group. Extending the result of Cauchy’s Theorem, the Sylow Theorems show that if p is prime and if |G| = pj m with (p, m) = 1, then G must have a subgroup of order pj . Such a subgroup is the highest order p-group that could occur as a subgroup of G, and is known as a p-Sylow subgroup of G (though that is not the initial definition we shall give for a p-Sylow subgroup). 134

CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION

135

The Sylow Theorems are actually a good bit stronger than a simple existence proof for subgroups whose order is the highest power of p dividing the order of G. They also show that any two p-Sylow subgroups of G are conjugate and that the number of p-Sylow subgroups of G is congruent to 1 mod p. This leads to some useful theoretical results regarding p-Sylow subgroups which are normal, as well as to some valuable counting arguments for the classification of groups of particular orders. In particular, Sylow theory can be useful for finding normal subgroups in a particular group. The Sylow Theorems give a useful point of view for considering finite groups. Finite p-groups are more tractable than general finite groups, and a useful first step in understanding a given group is to understand its Sylow subgroups and their normalizers. For many low order groups, an understanding of the entire structure of the group will follow from this analysis. Given the Sylow Theorems and the material on semidirect products, extensions, and the automorphisms of cyclic groups, we are able to classify the groups of many of the orders less than 64. We give numerous classifications in the text and exercises for Section 5.3. Section 5.4 studies the commutator subgroup and determines the largest abelian factor group that a group can have. Then, in Section 5.5, and using the commutator subgroups, we study solvable groups. Solvable groups are a generalization of abelian groups. In essence, in the finite case, they are the groups with enough normal subgroups to permit good inductions on order. One characterization we shall give of the finite solvable groups is that they are the finite groups whose composition series have no nonabelian simple groups in the list of successive simple subquotients. As expected, the induction arguments available for solvable groups provide for stronger results than we get for arbitrary groups. In Section 5.6, we prove Hall’s Theorem, which shows that if G is a finite solvable group and if |G| = mn with (m, n) = 1, then G has a subgroup of order m. This fact turns out to characterize the solvable groups, though we won’t give the converse here. The nilpotent groups form a class of groups that sits between the abelian groups and the solvable groups. They are defined by postulating properties that always hold in p-groups, as we showed in Section 5.2. By the end of Section 5.7, we show that the finite nilpotent groups are nothing other than direct products of finite p-groups over varying primes p. In the process we will learn some more about the p-groups themselves. We shall also see that the full converse of Lagrange’s Theorem holds for nilpotent groups: A nilpotent group G has subgroups of every order that divides the order of G. We close the chapter with a discussion of matrix groups over commutative rings. Here, we assume an understanding of some of the material in Chapters 7 and 10. The reader who has not yet studied this material should defer reading Section 5.8 until having done so. One of the reasons for studying matrix groups is that the automorphism group of a direct product, Zkn , of k copies of Zn is isomorphic to Glk (Zn ), so by studying the latter, we can obtain classification results for groups containing normal subgroups isomorphic to Zkn . This situation arises for some low order groups. We also calculate the order of Gln (Fq ), where Fq is the finite field with q = pr elements with p prime. We describe some important subquotient groups, PSln (Fq ) of Gln (Fq ). Most of these projective special linear groups turn out to be simple, but we shall not show it here.


5.1

136

Cauchy’s Theorem

We now prove Cauchy’s Theorem. An alternative proof was given as Problem 24 of Exercises 3.3.23. Theorem 5.1.1. (Cauchy’s Theorem) Let G be a finite group and let p be a prime dividing the order of G. Then G has an element of order p. Proof Note that we’ve already verified the theorem in the case that G is abelian: Cauchy’s Theorem for abelian groups was given as Proposition 4.4.12. We argue by induction on the order of G. The induction starts with |G| = p, in which case G ∼ = Zp , and the result is known to be true. Recall that the nontrivial conjugacy classes in G are those with more than one element and that the conjugacy class of x is nontrivial if and only if x ∈ Z(G). Recall also that a set of class representatives for the nontrivial conjugacy classes in G is a set X ⊂ G such that 1. X ∩ Z(G) = ∅. 2. If y ∈ G is not in Z(G), then y is conjugate to some x ∈ X. 3. If x, x ∈ X with x = x , then x and x are not conjugate in G. Then the class formula (Corollary 3.6.18), shows that if X ⊂ G is a set of class representatives for the nontrivial conjugacy classes in G, then [G : CG (x)], |G| = |Z(G)| + x∈X

where CG (x) is the centralizer of x in G. As stated, we already know Cauchy’s Theorem to be true for abelian groups. Thus, if p divides the order of Z(G), then Z(G) has an element of order p, and hence G does also. Thus, we may assume that p does not divide the order of Z(G). Since p does divide the order of G, the class formula tells us that there must be some x ∈ Z(G) such that p does not divide [G : CG (x)]. Since |G| = [G : CG (x)] · |CG (x)|, this says that p must divide the order of CG (x). Since x ∈ Z(G), [G : CG (x)], which is equal to the number of conjugates of x in G, cannot equal 1. Thus, |CG (x)| < |G|, so the induction hypothesis provides an element of order p in CG (x), and hence also in G. As a sample application, we classify the groups of order 6. Corollary 5.1.2. Let G be a group of order 6. Then G is isomorphic to either Z6 or D6 . Proof By Cauchy’s Theorem, G has an element, b, of order 3. But then b has index 2 in G, and hence b G by Proposition 4.1.12 (or by the simpler argument of Problem 12 of Exercises 3.7.14). Let π : G → G/b be the canonical map. A second application of Cauchy’s Theorem given an element a ∈ G of order 2. But then a ∈ b, and hence π(a) = a is not the identity element of G/b. Thus, a generates G/b. Since a and a both have order 2, setting s(a) = a gives a section of π.


137

∼ Z× = {±1}, there are Thus, G is a split extension of Z3 by Z2 . Since Aut(Z3 ) = 3 exactly two homomorphisms from Z2 to Aut(Z3 ): the trivial homomorphism and the homomorphism that sends the generator of Z2 to −1. The semidirect product given by the trivial homomorphism is Z3 × Z2 , which is Z6 , because 2 and 3 are relatively prime (e.g., by Problem 4 of Exercises 2.10.12). The semidirect product induced by the nontrivial homomorphism from Z2 to Aut(Z3 ) was shown to be isomorphic to D6 in the first example of Examples 4.6.6. Exercises 5.1.3. 1. Show that any group of order 10 is isomorphic to either Z10 or D10 . 2. Assuming that Z× p is cyclic for any odd prime p (as shown in Corollary 7.3.15), show that any group of order 2p is isomorphic to either Z2p or D2p . 3. Recall that Cauchy’s Theorem for abelian groups was available prior to this chapter. Give a proof of the preceding problem that doesn’t use Cauchy’s Theorem for nonabelian groups. (Hint: If G is a group of order 2p, then all of its elements must have order 1, 2, p, or 2p. If all nonidentity elements have order 2, then G must be abelian, and hence Cauchy’s Theorem for abelian groups gives a contradiction. So G must have an element, b, of order p. Now study the canonical map G → G/b, and show that G must have an element of order 2.)

5.2

p-Groups

Recall that a p-group is a group in which the order of every element is a power of p. Lagrange’s Theorem shows that any group whose order is a power of p must be a pgroup. But by Cauchy’s Theorem, if |G| has a prime divisor q = p, then it must have an element of order q, and hence it is not a p-group. We obtain the following corollary. Corollary 5.2.1. A finite group is a p-group if and only if its order is a power of p. The proof used above for Cauchy’s Theorem may be used, with suitable modifications, to prove some other useful facts. For instance: Proposition 5.2.2. A finite p-group has a nontrivial center (i.e., Z(G) = e). Proof Let X be a set of class representatives for the nontrivial conjugacy classes in G and consider the class formula: |G| = |Z(G)| + [G : CG (x)]. x∈X

If x ∈ Z(G), then [G : CG (x)], which is equal to the number of conjugates of x in G, cannot equal 1. Thus, since [G : CG (x)] divides |G|, we must have [G : CG (g)] = pr for some r > 0, and hence p divides [G : CG (g)]. Thus, p must divide every term in the class formula other than |Z(G)|, so it must divide |Z(G)| as well. In particular, |Z(G)| = 1, and hence Z(G) = e. We first give an immediate consequence of Proposition 5.2.2 to classification. Corollary 5.2.3. Every group of order p2 is abelian.


138

Proof We argue by contradiction. If G is not abelian, then Z(G) = G. But then if |G| = p2 , we must have |Z(G)| = p. But conjugation acts trivially on Z(G), so that Z(G) G. Thus, G/Z(G) has order p, and hence is cyclic. But now G must be abelian by Lemma 4.7.15, which gives the desired contradiction. We now consider some more general corollaries to Proposition 5.2.2. Corollary 5.2.4. The only finite p-group which is simple is Zp . Proof Let G be a finite simple p-group. The center of any group is a normal subgroup. Since the center of G is nontrivial and G is simple, G must equal its center. Thus, G is abelian. But the only abelian simple groups are the cyclic groups of prime order, by Lemma 4.2.2. Recall that a composition series for a group G is a sequence e = H0 ⊂ H1 ⊂ · · · ⊂ H j = G such that Hi−1 Hi and Hi /Hi−1 is a nontrivial simple group for 1 ≤ i ≤ j. By Proposition 4.2.5, every nontrivial finite group has a composition series. Since a quotient of a subgroup of a p-group is again a p-group, the simple groups Hi /Hi−1 associated to the composition series for a nontrivial finite p-group must all be Zp . But a direct proof of this fact will give us a stronger statement: We can construct a composition series for a p-group G such that all the subgroups in the series are normal in G itself, and not just normal in the next group in the series. Corollary 5.2.5. Let G be a finite group of order pj , with p prime and j > 0. Then there is a sequence e = H0 ⊂ H1 ⊂ · · · ⊂ Hj = G, with Hi G and Hi /Hi−1 ∼ = Zp for 1 ≤ i ≤ j. Proof We argue by induction on |G|, with the result being trivial for |G| = p. By Proposition 5.2.2, Z(G) = e, so Cauchy’s Theorem provides a subgroup H1 ⊂ Z(G) with |H1 | = p. But any subgroup of Z(G) is normal in G, so we can form the factor group G/H1 . Since |G/H1 | < |G|, the induction hypothesis provides a sequence e = K1 ⊂ · · · ⊂ Kj = G/H1 , such that Ki G/H1 and Ki /Ki−1 ∼ = Zp for 2 ≤ i ≤ j. Let π : G → G/H1 be the canonical map and set Hi = π −1 (Ki ) for 2 ≤ i ≤ j. Since the pre-image under a homomorphism of a normal subgroup is always normal, we have Hi G for all i. Also, since Ki = Hi /H1 for i ≥ 1, the Third Noether Theorem shows that Hi /Hi−1 ∼ = Ki /Ki−1 ∼ = Zp for 2 ≤ i ≤ j, and the result follows. Since the subgroup Hi above has order pi , we see that the full converse of Lagrange’s Theorem holds in p-groups. Corollary 5.2.6. Let G be a finite p group. Then every divisor of |G| is the order of a subgroup of G.


139

We shall now begin to try to classify the groups of order p3 for primes p > 2. We shall not yet succeed entirely, because not all the groups we must consider are p-groups. We must also understand the automorphism groups of the groups of order p2 , which in the case of Zp × Zp will require the Sylow theorems. As in the case of p = 2, the abelian groups of order p3 are Zp3 , Zp2 × Zp , and Zp × Zp × Zp . For the nonabelian groups of order p3 , the whole story will depend on whether there exists an element of order p2 . Proposition 5.2.7. Let G be a nonabelian group of order p3 where p is an odd prime. Suppose that G contains an element of order p2 . Then G is isomorphic to the semidirect 1 product Zp2 α Zp , where α : Zp → Z× p2 is induced by α(1) = p + 1. Proof Let b ∈ G have order p2 . Then b has index p in G, so b G by Proposition 4.1.12. Thus, G is an extension of Zp2 by Zp , via f : G → Zp . Since G is not abelian, 2 Lemma 4.7.15 shows that the induced homomorphism α : Zp → Z× p2 = Aut(Zp ) is nontrivial. We shall use strongly the fact that Z× p2 is abelian. Because of this, the p-torsion elements form a subgroup, which, according to Corollary 4.5.12, is K(p, 2), the subgroup consisting of those units congruent to 1 mod p. Corollary 7.3.15 then shows that K(p, 2) is a cyclic group of order p, generated by 1 + p. × Since K(p, 2) is the p-torsion subgroup of Z× p2 , the image of α : Zp → Zp2 must lie in K(p, 2). Since α is nontrivial and Zp is simple, α must be an isomorphism from Zp to K(p, 2). But then p + 1 = α(x) for some generator x of Zp . But we may change our identification of Zp to make x the canonical generator of Zp . Thus, we may assume that α carries the canonical generator of Zp to 1 + p. Thus, it suffices to show that f is a split extension, as then G ∼ = Zp2 α Zp , with α as claimed. Let us continue to write x for our preferred generator of Zp , so that α(x) = p + 1. Since α is the action induced by the extension f : G → Zp , Lemma 4.7.9 shows that ghg −1 = h1+p for any h ∈ ker f and any g with f (g) = x. To show that f splits, it suffices to find g ∈ G of order p such that f (g) = x: The splitting is then defined by setting s(x) = g. Let a ∈ G with f (a) = x. Then a cannot have order p3 , as G would then be cyclic. Alternatively, if a has order p, then a itself gives the desired splitting of f . Thus, we shall assume that a has order p2 . For simplicity of notation, let m = p + 1, so that aha−1 = hm for any h ∈ ker f . j Moreover, since α(aj ) = mj , aj ha−j = hm for h ∈ ker f . Thus, as shown in Problem 7 of Exercises 4.6.13, an easy induction on k shows that k−1

(ha)k = h1+m+···+m

ak

for k ≥ 1. Thus, taking k = p, Problem 8 of Exercises 4.6.13 gives (ha)p = h

mp −1 m−1

ap .

Recall that m = p + 1. Then m − 1 = p, while the proof of Proposition 4.5.18 shows that mp is congruent to 1 + p2 mod p3 . But then (mp − 1)/(m − 1) is congruent to p mod p2 , so that (ha)p = hp ap . Now ap has order p and lies in b = ker f . Thus, ap = bjp for some j relatively prime to p. But then if we take h = b−j , then the element ha has order p, and gives the desired splitting for f . 1 Thus,

G is the group of Example 6 of Examples 4.6.6.


140

Thus, there is no analogue at odd primes of Q8 : Every nonabelian extension of Zp2 by Zp is split when p > 2. Lemma 5.2.8. Let p be an odd prime and let G be a nonabelian group of order p3 that does not contain an element of order p2 . Then G is a split extension of Zp × Zp by Zp . Proof By Corollary 5.2.5, G has a normal subgroup H of order p2 . Since groups of order p2 are abelian, and since G contains no element of order p2 , H must be isomorphic to Zp × Zp . Since H has index p, G is an extension of H by Zp . But all elements of G have exponent p, so any lift of the generator of G/H to G provides a splitting for this extension. We have constructed an example of a nonabelian group of order p3 in which every element has exponent p, in Corollary 4.6.8. We would like to show that every nonabelian split extension of Zp × Zp by Zp is isomorphic to this one. To show this we first need to understand the automorphism group of Zp × Zp . We will then need to understand the conjugacy classes of homomorphisms from Zp into this group of automorphisms. But the automorphism group in question is nonabelian, so that a complete picture of the conjugacy classes of its subgroups of order p will depend on the Sylow Theorems of Section 5.3. We will content ourselves here with a calculation of the order of Aut(Zp × Zp ). Together with the Sylow theorems, this will be sufficient to complete our classification. Given a basic knowledge of linear algebra, the following result becomes a little more transparent, and may be extended to a calculation of the order of the automorphism group of a product of n copies of Zp . Lemma 5.2.9. The automorphism group of Zp × Zp has order (p2 − 1)(p2 − p). In particular, this group has order jp, where (j, p) = 1. Proof Write Zp × Zp = {ai bj | 0 ≤ i, j ≤ p}. Of course, a and b commute and each has order p. Since every element of Zp × Zp has exponent p, the homomorphisms, f , from Zp × Zp to itself are in one-to-one correspondence with all possible choices of the elements f (a) and f (b). Since there are p2 choices for f (a) and p2 choices for f (b), this gives p2 · p2 = p4 different homomorphisms from Zp × Zp to itself. We’d like to know which of these are isomorphisms. Let f : Zp × Zp → Zp × Zp be a homomorphism. Then the image of f has order 1, p or p2 . It is an isomorphism if and only if the image has order p2 . So one way to argue at this point is to count up the homomorphisms whose images have order 1 or p, and subtract this number from p4 . Since Zp × Zp has exponent p, every e = x ∈ Zp × Zp is contained in exactly one subgroup of order p. That subgroup is x. Thus, the number of subgroups of order p is the number of nonidentity elements in Zp × Zp divided by the number of nonidentity elements of Zp : we have (p2 − 1)/(p − 1) = p + 1, subgroups of order p in Zp × Zp . But if H ⊂ Zp × Zp has order p, then the homomorphisms from Zp × Zp to H are in one-to-one correspondence with the choices of pairs of elements of H, so there are p2 different homomorphisms from Zp × Zp to H. Only one of these is the trivial homomorphism, and hence there are p2 − 1 homomorphisms from Zp × Zp onto H. Multiplying this by the number of subgroups of order p, we see that there are p3 + 2 p − p − 1 homomorphisms from Zp × Zp to itself with an image of order p. Since the trivial homomorphism is the only homomorphism whose image has order 1, this gives exactly p3 + p2 − p homomorphisms from Zp × Zp to itself that are not automorphisms. But then there are exactly p4 − p3 − p2 + p automorphisms of Zp × Zp . But this is easily seen to factor as (p2 − 1)(p2 − p), as desired.


141

A perhaps simpler argument is as follows. Choose f (a). Any nonidentity choice will do, so there are p2 − 1 choices for f (a). Now consider the choice of f (b). As long as f (b) is not in f (a), f will be an automorphism. Thus, there are p2 − p possible choices for f (b) once f (a) has been chosen. In particular, there are (p2 − 1)(p2 − p) ways to choose an automorphism f . Exercises 5.2.10. 1. Show that every nonabelian group of order p3 is a central extension of Zp by Zp ×Zp . 2. Construct a group of order p4 whose center is isomorphic to Zp2 . 3. Find a group of order 16 whose center has order 2. 4. Construct a group of order 81 whose center has order 3. 5. Let p be an odd prime. Construct a group of order p4 whose center has order p. 6. Show that every group of order pn may be generated by n elements.

5.3

Sylow Subgroups

Cauchy’s Theorem shows that if p divides the order of a finite group G, then G has a subgroup of order p. But what if |G| is divisible by p2 ? Can we then find a subgroup of order p2 ? We cannot argue here by Cauchy’s Theorem and induction, as there may not be a normal subgroup of order p. (E.g., the group G may be simple.) So we need a new approach to the question. Note that there are examples of groups G and numbers n such that |G| is divisible by n, but G has no subgroup of order n. For instance, we’ve seen in Problem 2 of Exercises 4.2.18 that A4 has no subgroup of order 6. Definitions 5.3.1. Let G be a finite group and let p be a prime. A p-subgroup of G is a subgroup which is a p-group. A p-Sylow subgroup P of G is a maximal p-subgroup of G, i.e., P is a p-subgroup of G, and any p-subgroup of G containing P must be equal to P. Examples 5.3.2. In D6 , b has order 3. Since no higher power of 3 divides the order of D6 , b cannot be properly contained in a 3-subgroup of D6 . Thus, b is a 3-Sylow subgroup of D6 . Similarly, a, ab, and ab2 are all 2-subgroups of D6 . Since 2 itself is the highest 2-power that divides |D6 |, each of these is a 2-Sylow subgroup of D6 . Lemma 5.3.3. Let H be a p-subgroup of a finite group G. Then H is contained in a p-Sylow subgroup of G. Proof Let |G| = pi m with p relatively prime to m, and let |H| = pj . If i = j, then H is a p-Sylow subgroup of G by Lagrange’s Theorem. Otherwise, we argue by induction on i − j. From what we know at this point, H itself might be a p-Sylow subgroup of G even if j < i. If this is the case, we are done. Otherwise, there is a p-subgroup K of G that strictly contains H. But then the index of K in G is smaller than that of H, so by induction, K is contained in a p-Sylow subgroup.


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Lemma 5.3.4. Let G be a finite group and let p be a prime dividing the order of G. Let P be a p-Sylow subgroup of G. Then the index of P in its normalizer is relatively prime to p. Proof Recall that P NG (P ). Thus, the order of NG (P )/P is the index of P in NG (P ). If this order is divisible by p, then Cauchy’s Theorem gives a subgroup H ⊂ NG (P )/P with |H| = p. But then if π : NG (P ) → NG (P )/P is the canonical map, then π −1 H is a p-subgroup of G that strictly contains P . Corollary 5.3.5. Let G be a finite group and let p be a prime dividing the order of G. Let P be a p-Sylow subgroup of G. Then any p-subgroup of NG (P ) is contained in P . Proof Let K ⊂ NG (P ) be a p-group. Then K/(K ∩ P ) is isomorphic to a subgroup of NG (P )/P . Since K is a p-group, so is K/(K ∩ P ). Since |NG (P )/P | is prime to p, K/(K ∩ P ) = e, and hence K ⊂ P . Lemma 5.3.6. Let P be a p-Sylow subgroup of the finite group G. Then any conjugate of P is also a p-Sylow subgroup of G. Proof Since conjugation gives an automorphism of G, any conjugate of a p-subgroup of G is again a p-subgroup of G. Now for x ∈ G, suppose that xP x−1 not a p-Sylow subgroup of G. Then xP x−1 must be properly contained in a p-subgroup P of G. But then P is properly contained in x−1 P x, contradicting the statement that P is a p-Sylow subgroup of G. We now give yet another variant of the proof of Cauchy’s Theorem. Theorem 5.3.7. (First Sylow Theorem) Let G be a finite group and let p be a prime dividing the order of G. Then any two p-Sylow subgroups of G are conjugate, and the number of p-Sylow subgroups of G is congruent to 1 mod p. Proof Let P be a p-Sylow subgroup of G and let X ⊂ Sub(G) be the orbit of P under the G-action given by conjugation. (Recall that Sub(G) is the set of subgroups of G.) By Lemma 5.3.6, each element of X is a p-Sylow subgroup of G. The G-set X may be considered as a P -set by restricting the action. For P ∈ X, the isotropy subgroup of P under the action of P on X is given by PP

= {x ∈ P | xP x−1 = P } = P ∩ NG (P ).

In particular, P is fixed by P if and only if P ⊂ NG (P ). By Corollary 5.3.5, this can happen if and only if P = P , as both P and P are p-Sylow subgroups of G. Thus, the action of P on X has exactly one trivial orbit: {P }. (Here, an orbit is trivial if it has only one element.) The number of elements in the orbit of P ∈ X under the P -action on X is [P : P ∩ NG (P )], which divides |P |. Thus, since P is a p-group, the number of elements in any nontrivial orbit is divisible by p (hence congruent to 0 mod p). In particular, if Y ⊂ X is a set of orbit representatives for the nontrivial orbits of the action of P on X (i.e., the intersection of each nontrivial P -orbit in X with Y consists of exactly one element), then the G-set Counting Formula (Corollary 3.3.15) gives |X| = |{P }| + [P : P ∩ NG (P )]. P ∈Y


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Since each [P : P ∩ NG (P )] is congruent to 0 mod p and since |{P }| = 1, we see that the number of elements in X is congruent to 1 mod p. It remains to show that X contains every p-Sylow subgroup of G. Let P be an arbitrary p-Sylow subgroup of G, and consider the action of P on X by conjugation. If P is not in X, then every orbit of this P -action must be nontrivial. But that would make the number of elements in X congruent to 0 mod p, rather than 1 mod p. So we must have P ∈ X. Theorem 5.3.8. (Second Sylow Theorem) Let G be a finite group and let p be a prime dividing the order of G. Let P be a p-Sylow subgroup of G. Then the index of P in G is prime to p. Thus, if |G| = pi m, with p and m relatively prime, then |P | = pi . Moreover, the number of p-Sylow subgroups in G divides m in addition to being congruent to 1 mod p. Proof The index, m, of P in G is the product of [NG (P ) : P ] and [G : NG (P )]. The former is prime to p by Corollary 5.3.5. The latter is equal to the number of conjugates of P , and hence is 1 mod p by the First Sylow Theorem. Recall from Corollary 5.2.6 that if p is prime and if P is a finite group of order pi , then P has a subgroup of every order dividing the order of P . Applying this to a p-Sylow subgroup of a group G, the Second Sylow Theorem produces the following corollary. Corollary 5.3.9. Let G be a finite group and let q be any prime power that divides the order of G. Then G has a subgroup of order q. We now give the final Sylow Theorem. Theorem 5.3.10. (Third Sylow Theorem) Let G be a finite group and let p be a prime dividing the order of G. Let P be a p-Sylow subgroup of G. Then the normalizer of P is its own normalizer: NG (NG (P )) = NG (P ). Proof We claim that P is a characteristic subgroup of NG (P ). The point here is that if x ∈ P and if f is any automorphism of NG (P ), then f (x) must be a p-torsion element. But by Corollary 5.3.5, every p-torsion element of NG (P ) must be contained in P . Now suppose that NG (P ) H. Then conjugation by an element of H gives an automorphism of NG (P ). But this must preserve P , so P is normal in H. But then H ⊂ NG (P ). We now summarize some important facts implicit in the above discussion. Corollary 5.3.11. Let G be a finite group and let p be a prime dividing the order of G. Suppose there is a p-Sylow subgroup P of G which is normal in G. Then P is the only p-Sylow subgroup of G, and P is the set of all p-torsion elements in G. Indeed, P is a characteristic subgroup of G. Conversely, if there is only one p-Sylow subgroup of G, then it is normal in G. Proof The p-Sylow subgroups of G are all conjugate. But if P G, then it has no conjugates other than itself. Moreover, since NG (P ) = G, P contains all the p-torsion elements of G by Corollary 5.3.5. And since P is a p-group, all its elements are p-torsion, so that P is, in fact, the set of p-torsion elements of G. But then P is characteristic, since any automorphism preserves the p-torsion elements. If there is only one p-Sylow subgroup, P , of G, then P has no conjugates other than itself. Thus, [G : NG (P )] = 1, and hence P G.


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In many cases a normal p-Sylow subgroup is sufficient to unfold the entire structure of a group. The point is that if P is a normal p-Sylow subgroup of G, then |P | and |G/P | are relatively prime. Thus, the argument of Proposition 4.7.7 gives the following corollary. Corollary 5.3.12. Suppose that G has a normal p-Sylow subgroup, P , and that G has a subgroup K of order [G : P ]. Then G is the semidirect product P α K, where α : K → Aut(P ) is induced by conjugation in G. According to the Schur–Zassenhaus Lemma, which we will not prove, such a subgroup K always exists. However, if [G : P ] is a prime power, the existence of such a K is automatic from the second Sylow Theorem. Corollary 5.3.13. Let G be a group of order pr q s where p and q are distinct primes. Suppose that the p-Sylow subgroup, P , of G is normal. Let Q be a q-Sylow subgroup of G. Then G is isomorphic to the semidirect product P α Q, where α : Q → Aut(P ) is induced by conjugation in G. The Sylow Theorems are extremely powerful for classifying low-order finite groups. We shall give a few of their applications now. First, let us complete the classification of groups of order p3 . Proposition 5.3.14. Let p be an odd prime and let G be a nonabelian group of order p3 that contains no element of order p2 . Then G is isomorphic to the group constructed in Corollary 4.6.8 for n = p. Proof By Lemma 5.2.8, G is a semidirect product (Zp × Zp ) α Zp for some α : Zp → Aut(Zp × Zp ). Moreover, α cannot be the trivial homomorphism, as then G would be abelian by Lemma 4.7.15. But by Lemma 5.2.9, the p-Sylow subgroups of Aut(Zp × Zp ) have order p, so α must be an isomorphism onto one of these p-Sylow subgroups. But the p-Sylow subgroups are all conjugate, and hence an isomorphism onto one of them is conjugate as a homomorphism into Aut(Zp × Zp ) to an isomorphism onto any of the others. Moreover, if α and α are isomorphisms onto the same p-Sylow subgroup of Aut(Zp × Zp ), then because they are isomorphisms, we have α = α ◦ g for an automorphism g of Zp . Thus, any two nonabelian semidirect products (Zp ×Zp )α Zp and (Zp × Zp ) α Zp are isomorphic by Proposition 4.6.11. Let us return to the theme of discovering normal Sylow subgroups and using them to split the group as a semidirect product. Corollary 5.3.15. Let G be a group of order pr m, where m < p, with p prime. Then the p-Sylow subgroup of G is normal. Proof The number of p-Sylow subgroups of G is congruent to 1 mod p and divides m. Since m < p, there must be only one p-Sylow subgroup, which is thus normal. When used in conjunction with Corollary 5.3.13, we can get some concrete results: Corollary 5.3.16. Let G be a group of order pr q s , where p and q are distinct primes. Suppose that q s < p. Then G is a semidirect product P α Q for some α : Q → Aut(P ), where P and Q are the p-Sylow and q-Sylow subgroups of G, respectively. This becomes more powerful when used in conjunction with what we know about automorphisms:


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Corollary 5.3.17. Every group of order 15 is isomorphic to Z15 . Proof Let |G| = 15. By Corollary 5.3.16, G is a semidirect product Z5 α Z3 for some × homomorphism α : Z3 → Z× 5 . But Z5 has order 4, which is relatively prime to 3, and hence α must be the trivial homomorphism. But then G ∼ = Z5 × Z3 , which is isomorphic to Z15 . Now let us consider a deeper sort of application of the Sylow Theorems. Sometimes, under the assumption that a given Sylow subgroup is nonnormal, we can use the count of the number of its conjugates to find the number of elements of p-power order. This will sometimes permit us to find other subgroups that are normal. Corollary 5.3.18. Let G be a group of order 12 whose 3-Sylow subgroups are not normal. Then G is isomorphic to A4 . Proof We begin by showing that the 2-Sylow subgroup of G is normal. Let P be a 3-Sylow subgroup of G. Then the number of conjugates of P is 1 mod 3 and divides 4. Thus, P has either 1 or 4 conjugates. But we’ve assumed that P is nonnormal, so it must have 4 conjugates. Note that if P and P are distinct 3-Sylow subgroups, then P ∩ P = e, since P and P both have order 3. But since each 3-Sylow subgroup has exactly two elements of order 3, there must be 8 elements of order 3 in G. The Second Sylow Theorem, on the other hand, shows that G has a subgroup, H, of order 4. Since there are 8 elements of order 3 in G, every element outside of H must have order 3. Since the elements of H are all 2-torsion, none of their conjugates may lie outside of H, and hence H must be normal. By Corollary 5.3.16, G ∼ = H α Z3 for some α : Z3 → Aut(H). Since there are no nontrivial homomorphisms of Z3 into Z× 4 = {±1} and since G is nonabelian, H must be isomorphic to Z2 × Z2 . As shown in either Problem 14 of Exercises 4.6.13 or Problem 3 of Exercises 4.7.27, the automorphism group of Z2 × Z2 is isomorphic to S3 . It’s easy to see that there are only two nontrivial homomorphisms from Z3 into S3 ∼ = D6 , and that one is obtained from the other by precomposing with an automorphism of Z3 . Thus, Proposition 4.6.11 shows that there is exactly one isomorphism class of semidirect products (Z2 × Z2 ) α Z3 which is nonabelian. As shown in Problem 1 of Exercises 4.7.27, A4 is a semidirect product (Z2 ×Z2 )α Z3 , and the result follows. Thus, the classification of groups of order 12 depends only on classifying the split extensions of Z3 by the groups of order 4. We shall leave this to the exercises. Here is a slightly more complicated application of the Sylow Theorems. Lemma 5.3.19. Let G be a group of order 30. Then the 5-Sylow subgroup of G is normal. Proof We argue by contradiction. Let P5 be a 5-Sylow subgroup of G. Then the number of conjugates of P5 is congruent to 1 mod 5 and divides 6. Thus, there must be six conjugates of P5 . Since the number of conjugates is the index of the normalizer, we see that NG (P5 ) = P5 . Since the 5-Sylow subgroups of G have order 5, any two of them intersect in the identity element only. Thus, there are 6 · 4 = 24 elements in G of order 5. This leaves


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6 elements whose order does not equal 5. We claim now that the 3-Sylow subgroup, P3 , must be normal in G. The number of conjugates of P3 is congruent to 1 mod 3 and divides 10. Thus, if P3 is not normal, it must have 10 conjugates. But this would give 20 elements of order 3, when there cannot be more than 6 elements of order unequal to 5, so that P3 must indeed be normal. But then P5 normalizes P3 , and hence P5 P3 is a subgroup of G. Moreover, the Second Noether Theorem gives (P5 P3 )/P3 ∼ = P5 /(P5 ∩ P3 ). But since |P5 | and |P3 | are relatively prime, P5 ∩ P3 = e, and hence P5 P3 must have order 15. Thus, P5 P3 ∼ = Z15 , by Corollary 5.3.17. But then P5 P3 normalizes P5 , which contradicts the earlier statement that NG (P5 ) has order 5. Corollary 5.3.20. Every group of order 30 is a semidirect product Z15 α Z2 for some α : Z2 → Z× 15 . Proof Since the 5-Sylow subgroup P5 is normal, it is normalized by any 3-Sylow subgroup P3 , and hence the product P3 P5 is a subgroup of order 15, and hence index 2. Since index 2 subgroups are normal, G is an extension of a group of order 15 by Z2 . By Cauchy’s Theorem, this extension must be split. Since there is only one group of order 15, the result follows. We shall classify these semidirect products in Exercises 5.3.22. Often in mathematics the statement of a theorem is insufficient, by itself, to derive all the applications of its method of proof. This is, in general, a good thing. It means that the methods have enough substance to be applied in more varied settings. (The alternative would be to continually learn or write proofs that one never needed to see again, which, in the end, would become boring.) Here is an example of a proof based on the method of proof of the Sylow Theorems. Proposition 5.3.21. Let G be a group of order 24 that is not isomorphic to S4 . Then one of its Sylow subgroups is normal. Proof Suppose that the 3-Sylow subgroups are not normal. The number of 3-Sylow subgroups is 1 mod 3 and divides 8. Thus, if there is more than one 3-Sylow subgroup, there must be four of them. Let X be the set of 3-Sylow subgroups of G. Then G acts on X by conjugation, so we get a homomorphism f : G → S(X) ∼ = S4 . As we’ve seen in the discussion on G-sets, the kernel of f is the intersection of the isotropy subgroups of the elements of X. Moreover, since the action is that given by conjugation, the isotropy subgroup of H ∈ X is NG (H). Thus, ker f = NG (H). H∈X

For H ∈ X, the index of NG (H) is 4, the number of conjugates of H. Thus, the order of NG (H) is 6. Suppose that K is a different element of X. We claim that the order of NG (H) ∩ NG (K) divides 2. To see this, note that the order of NG (H) ∩ NG (K) cannot be divisible by 3. This is because any p-group contained in the normalizer of a p-Sylow subgroup must be contained


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in the p-Sylow subgroup itself (Corollary 5.3.5). Since the 3-Sylow subgroups have prime order here, they cannot intersect unless they are equal. But if the order of NG (H) ∩ NG (K) divides 6 and is not divisible by 3, it must divide 2. In consequence, we see that the order of the kernel of f divides 2. If the kernel has order 1, then f is an isomorphism, since G and S4 have the same number of elements. Thus, we shall assume that ker f has order 2. In this case, the image of f has order 12. But by Problem 2 of Exercises 4.2.18, A4 is the only subgroup of S4 of order 12, so we must have im f = A4 . By Problem 1 of Exercises 4.2.18, the 2-Sylow subgroup, P2 , of A4 is normal. But since ker f has order 2, f −1 P2 has order 8, and must be a 2-Sylow subgroup of G. As the pre-image of a normal subgroup, it must be normal, and we’re done. The results we have at this point are sufficient to classify the groups of most orders less than 64. Almost all of them are semidirect products. We shall treat many examples in the exercises below. But for some of the orders less than 64, we must understand more about the subnormal series of p-groups. We shall develop the material to do this in the next three sections. Then, we must make a deeper analysis of some automorphism groups. For this, we will need some information from our analysis of matrix groups over finite fields below. Exercises 5.3.22. 1. Show that the 2-Sylow subgroup of S4 is isomorphic to D8 . Is it normal in S4 ? 2. What is the 2-Sylow subgroup of S6 ? 3. What is the 2-Sylow subgroup of S8 ? 4. Let H be any group and write H n for the product H × · · · × H of n copies of H. Recall from Problem 21 of Exercises 3.6.24 that Sn acts on H n through automorphisms via σ · (h1 , . . . , hn ) = (hσ−1 (1) , . . . , hσ−1 (n) ). For any subgroup K ⊂ Sn we get an induced semidirect product H n α K, called the wreath product of H and K, and denoted H K. Write Pr for the 2-Sylow subgroup of S2r . (a) Show that Pr ∼ = Pr−1 S2 for r ≥ 2. (b) Suppose that 2r−1 < n < 2r . Show that the 2-Sylow subgroup of Sn is isomorphic to the product of Pr−1 with the 2-Sylow subgroup of Sn−2r−1 . (c) Express the 2-Sylow subgroup of Sn in terms of the binary expansion of n. 5. Prove the analogue of the preceding problem for primes p > 2. 6. Let p and q be primes, with q < p. Assume, via Corollary 7.3.15, that Z× p is cyclic. Show that there are at most two groups of order pq. When are there two groups, and when only one? 7. Let p and q be primes. Show that every group of order p2 q has a normal Sylow subgroup. 8. Give a simpler, direct argument for Corollary 5.3.18 by studying the homomorphism Λ : G → S(G/P3 ) obtained via Proposition 3.3.16 from the standard action of G on the left cosets of a 3-Sylow subgroup of G. 9. Classify the groups of order 12. How does Q12 fit into this classification?


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10. Classify the groups of order 20. 11. Classify the groups of order 24 whose 2-Sylow subgroup is either cyclic or dihedral. 12. Classify the groups of order 30. 13. We show here that any group G of order 36 without a normal 3-Sylow subgroup must have a normal 2-Sylow subgroup. Thus, assume that G has a non-normal 3-Sylow subgroup P3 . Then the action of G on the set of left cosets G/P3 induces a homomorphism Λ : G → S(G/P3 ) ∼ = S4 . (a) Show, using Problem 31 of Exercises 3.7.14, that the kernel of Λ must have order 3. (b) Deduce from Problem 2 of Exercises 4.2.18 that the image of Λ is A4 . Write V ⊂ A4 for the 2-Sylow subgroup of A4 and let H = Λ−1 V ⊂ G. (c) Show that H, as an extension of Z3 by Z2 × Z2 , must be isomorphic to either D12 or to Z3 × Z2 × Z2 . (d) Show that every 2-Sylow subgroup of G must be contained in H. Deduce that if H ∼ = Z3 × Z2 × Z2 , then the 2-Sylow subgroup of G is normal. (e) Show that if H ∼ = D12 , then G has exactly three 2-Sylow subgroups. Deduce that if P2 is a 2-Sylow subgroup of G, then |NG (P2 )| = 12. Deduce that Λ induces an isomorphism from NG (P2 ) to A4 . (f) Show that the only split extension of Z3 by A4 is the trivial extension Z3 ×A4 . Deduce that H must have been isomorphic to Z3 × Z2 × Z2 . 14. Classify those groups of order 40 whose 2-Sylow subgroup is isomorphic to one of the following: (a) Z8 (b) D8 (c) Z4 × Z2 . 15. Show that any group of order 42 is a split extension of Z7 by a group of order 6. Classify all such groups. 16. Alternatively, show that any group of order 42 is a split extension of a group of order 21 by Z2 . Classify these, and relate this classification to the preceding one. (To do this, you should calculate the automorphism group of the nonabelian group of order 21. See Problem 19 of Exercises 4.7.27.) 17. Classify the groups of order 45. 18. Show that any group G of order 60 whose 5-Sylow subgroups are nonnormal must be simple. (Hint: First calculate the orders of the normalizers of the 5-Sylow and the 3-Sylow subgroups of G and then go through all possible orders for the subgroups of G and show that no subgroup of that order can be normal. In some cases, it is possible to show that no subgroup of the stated order can exist at all. The classifications above of the groups whose order properly divides 60 will be useful.)


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19. Show that any simple group G of order 60 is isomorphic to A5 . (Hint: You definitely want to use the information derived in the solution of the previous exercise. The key is finding a subgroup of index 5. It may help to calculate the order of the centralizer of an element of order 2 in G.) 20. Classify the groups of order 60. 21. Show that every group G of order 90 has a normal 5-Sylow subgroup. Deduce that G has a subgroup of index two. 22. Classify the groups of order 90 whose 3-Sylow subgroup is cyclic.

5.4

Commutator Subgroups and Abelianization

The commutator of two group elements, a and b, may be thought of as a measurement of how far a and b are from commuting with each other. The commutators of all pairs of elements in a group generate a useful normal subgroup called the commutator subgroup of G, and denoted [G, G]. We shall see that G/[G, G] is the largest abelian quotient group of G. Definition 5.4.1. Let a and b be elements of the group G. The commutator of a and b, written [a, b], is defined by [a, b] = aba−1 b−1 . We’ve already used the next lemma repeatedly. Lemma 5.4.2. The elements a and b commute if and only if [a, b] = e. The next lemma may be seen by inspection. Lemma 5.4.3. For a, b ∈ G, [a, b]−1 = [b, a]. Definition 5.4.4. The commutator subgroup of G, written [G, G], is the subgroup generated by all the commutators of the elements of G. Thus, an element of G is in [G, G] if and only if it is a product of commutators. Note, however, that not every element of [G, G] is actually a commutator itself. Lemma 5.4.5. Let G be any group. Then the commutator subgroup is a characteristic subgroup of G, and hence is normal. Proof Let f be an automorphism of G. Then clearly, f ([a, b]) = [f (a), f (b)]. Thus, since f is a homomorphism, it must take a product of commutators to a product of commutators. Proposition 5.4.6. The quotient group G/[G, G] is abelian. In addition, any homomorphism f : G → H with H abelian factors through the canonical map π : G → G/[G, G]. Proof In G/[G, G], we have ab(a)−1 (b)−1 = aba−1 b−1 = e, so that each pair of elements a and b of G/[G, G] commutes. Let f : G → H with H abelian. To see that f factors through the canonical map to G/[G, G], it suffices to show that the commutator subgroup [G, G] is contained in the kernel of f . But f ([a, b]) = [f (a), f (b)], and the latter is the identity element of H, since f (a) and f (b) commute. But then the generators of [G, G] are contained in the kernel of f , and hence [G, G] must be also.


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Definition 5.4.7. We write Gab = G/[G, G], and call it the abelianization of G. Corollary 5.4.8. Every subgroup of G that contains [G, G] is normal. Moreover, if H G, then H contains [G, G] if and only if G/H is abelian. Proof If [G, G] ⊂ H, then H = π −1 π(H), where π is the canonical map from G to Gab . But every subgroup of the abelian group Gab is normal, and hence H is normal as the pre-image of a normal subgroup. If G/H is abelian, then the canonical map from G to G/H factors through Gab . But this happens if and only if [G, G] ⊂ H. Conversely, if [G, G] ⊂ H, then G/H is a quotient group of Gab by the third Noether Theorem, and hence is abelian. Exercises 5.4.9. † 1. Show that a group G is abelian, if and only if [G, G] = e. † 2. Show that if G is nonabelian and simple, then [G, G] = G, and Gab = e. 3. Show that for nonabelian groups of order p3 , p prime, the commutator subgroup and center coincide. 4. What is the commutator subgroup of D2n ? What is D2n ab ? 5. What is the commutator subgroup of Q4n ? What is Q4n ab ? 6. What is the commutator subgroup of A4 ? What is A4 ab ? 7. What is the commutator subgroup of S4 ? What is S4 ab ? 8. What is the commutator subgroup of Sn for n ≥ 5? What is Sn ab ? ‡ 9. We calculate the automorphism group of Q8 . (a) Show that any automorphism f of Q8 induces an automorphism f of Q8 ab . Show that passage from f to f induces a surjective homomorphism η : Aut(Q8 ) → Aut(Q8 ab ). (b) Show η is a split extension whose kernel is isomorphic to Z2 × Z2 . (c) Deduce that Aut(Q8 ) is isomorphic to S4 . 10. Classify the groups of order 24 whose 2-Sylow subgroups are quaternionic. Show that there is a unique such group whose 3-Sylow subgroups are not normal. We shall meet this group again later as the matrix group Sl2 (Z3 ).

5.5

Solvable Groups

The most tractable groups to analyze are the abelian groups. Every subgroup is normal, and hence lots of induction arguments are available. This is amply illustrated by the classification of finite abelian groups given in Section 4.4. At the other extreme, we have the nonabelian simple groups. With no nontrivial proper normal subgroups available, it is difficult to get our hands on the structure of the group. The solvable groups give a generalization of abelian groups that share some of the nice properties found in the abelian setting. In particular, knowing that a group is solvable will open up a collection of tools for revealing its structure.


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Solvability has long been known to be an important idea. For instance, solvability plays a key role in Galois’ famous proof that there does not exist a formula based on the operations of arithmetic and n-th roots for finding solutions of polynomials of degree 5 or higher. We shall give an argument for this in Section 11.11. Thus, it is of interest to determine whether a given group is solvable. Of particular importance in this regard is the famous Feit–Thompson Theorem: Theorem (Feit–Thompson Theorem) Every group of odd order is solvable. The Feit–Thompson Theorem is considered to be the first major breakthrough along the way to the classification of the finite simple groups. Its proof is quite lengthy and is well beyond the scope of this book.2 But the fact that odd order groups are solvable is an important one to keep in mind for intuition’s sake. We shall need a more compact notation for commutator subgroups, as we shall be discussing commutator subgroups of commutator subgroups, etc. Definition 5.5.1. For i ≥ 0, define G(i) inductively by G(0) = G and G(i) = [G(i−1) , G(i−1) ] for i ≥ 1. We call G(i) the i-th derived subgroup of G. Thus, G(1) = [G, G]. Given this, let us make the following definition. Definition 5.5.2. A group G is solvable if G(k) = e for some k ≥ 0. The virtue of this definition is that it is easy to work with. We shall give an equivalent definition below that may be more intuitive. In the meantime, we can give a quick workup of some basic properties. The next lemma is immediate from Problems 1 and 2 of Exercises 5.4.9. Lemma 5.5.3. Every abelian group is solvable. A simple group, on the other hand, is solvable if and only if it is abelian. Lemma 5.5.4. Let f : G → H be a surjective homomorphism. Then f (G(i) ) = H (i) for all i ≥ 0. Proof We shall show that f ([G, G]) = [H, H]. This then shows that f : [G, G] → [H, H] is surjective, and the result follows by induction on k. But since f ([a, b]) = [f (a), f (b)], f ([G, G]) ⊂ [H, H]. And since f is surjective, the generators of [H, H] must lie in f ([G, G]). Corollary 5.5.5. A quotient group of a solvable group is solvable. Proof If G(k) = e, so is H (k) for any quotient group H of G. Lemma 5.5.6. Let H be a subgroup of G. Then H (i) ⊂ G(i) for all i ≥ 0. Proof Once again, by induction, it suffices to treat the case of i = 1. But since H ⊂ G, if h, k ∈ H, then [h, k] ∈ [G, G]. Of course, H (i) need not equal G(i) ∩ H. Corollary 5.5.7. A subgroup of a solvable group is solvable. 2 The proof takes up the entirety of a 255-page journal issue. See W. Feit and J. Thompson, Solvability of groups of odd order, Pacific J. Math. 13 (1963), 775–1029.


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Recall that a subnormal series for G consists of a sequence of subgroups e = Gk Gk−1 · · · G0 = G. Here, each Gi+1 is normal in Gi , but need not be normal in G if i > 1. (Note that we’ve reversed the order of the indices from that used previously.) Definition 5.5.8. A subnormal series is abelian if the quotient groups Gi /Gi+1 are abelian for 0 ≤ i < k. Proposition 5.5.9. A group G is solvable if and only if it has an abelian subnormal series. Proof Suppose that G is solvable, with G(k) = e. Consider the sequence e = G(k) G(k−1) · · · G(0) . ab

Since G(i+1) = [G(i) , G(i) ], the quotient group Gi /Gi+1 is just (G(i) ) , which is certainly abelian, so that the sequence above is an abelian subnormal series. Conversely, given an abelian subnormal series e = Gk Gk−1 · · · G0 = G, we claim that G ⊂ Gi for all i, which then forces G(k) = e. Clearly, G(0) = G0 , so assume inductively that G(i−1) ⊂ Gi−1 . But then G(i) ⊂ [Gi−1 , Gi−1 ] by Lemma 5.5.6, so it suffices to see that [Gi−1 , Gi−1 ] ⊂ Gi . But this follows immediately from Corollary 5.4.8, since Gi−1 /Gi is abelian. (i)

The above characterization of solvable groups can be quite useful. Corollary 5.5.10. Let H be a normal subgroup of G. Then G is solvable if and only if both H and G/H are solvable. Proof Suppose that G is solvable. Then H is solvable by Corollary 5.5.7, and G/H is solvable by Corollary 5.5.5. Conversely, suppose that H and G/H are solvable. Let e = Hk · · · H0 = H be an abelian subnormal series for H and let e = Kl · · · K0 = G/H be an abelian subnormal series for G/H. Then if π : G → G/H is the canonical map, we have an abelian subnormal series e = Hk · · · H0 = π −1 Kl π −1 Kl−1 · · · π −1 K0 = G for G. Also, the following corollary is immediate from Corollary 5.2.5. Corollary 5.5.11. Let p be a prime. Then any finite p-group is solvable. Note that other than this last result, we have made no use of finiteness in the discussion of solvability. Recall that a composition series for a group G is a subnormal series e = Gk Gk−1 · · · G0 = G where the subquotient groups Gi /Gi+1 of G are all nontrivial simple groups. By Proposition 4.2.5, every nontrivial finite group has a composition series. The Jordan–H¨ older Theorem shows that any two composition series for a group G have the same length, and give the same collection of subquotient groups Gk−1 /Gk , . . . , G/G1 , though possibly in a different order. However, for economy of means, we shall not use it in the following proof.


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Proposition 5.5.12. A finite group is solvable if and only if it has a composition series in which the simple subquotient groups are all cyclic of prime order. If G is solvable, then every composition series for G has that form. Proof A composition series whose simple subquotient groups are cyclic is certainly an abelian subnormal series, so any group that has such a composition series is solvable by Proposition 5.5.9. Conversely, suppose that G has a composition series in which one of the quotients Gi /Gi+1 is not cyclic of prime order. Since the only abelian simple groups have prime order (Lemma 4.2.2), Gi /Gi+1 must be a nonabelian simple group. But if G were solvable, the subgroup Gi would be also. And this would force the quotient Gi /Gi+1 to be solvable as well, contradicting Lemma 5.5.3. Exercises 5.5.13. 1. Show that Sn is not solvable if n ≥ 5. 2. Show that A5 is the only group of order ≤ 60 that is not solvable. 3. Show that the Feit–Thompson Theorem is equivalent to the statement that every nonabelian simple group has even order.

5.6

Hall’s Theorem

There is a generalization of Sylow theory that works in solvable groups. It is due to Philip Hall. Definition 5.6.1. Let G be a finite group. A subgroup H ⊂ G is a Hall subgroup if |H| and [G : H] are relatively prime. Our purpose in this section is to prove the following theorem, which shows that a finite solvable group G has Hall subgroups of all possible orders n dividing |G| such that (n, |G|/n) = 1. Theorem 5.6.2. (Hall’s Theorem) Let G be a finite solvable group, and let |G| = nk with (n, k) = 1. Then G has a subgroup of order n. Moreover, any two subgroups of order n are conjugate. In fact, Hall also showed a converse to this. We shall state it here, but its proof is beyond the scope of this book. Theorem (The Hall Converse) Let G be a finite group with |G| = pr11 . . . prkk with p1 , . . . , pk distinct primes. Suppose that G has a Hall subgroup of order |G|/pri i for each i = 1, . . . , k. Then G is solvable. Thus, the existence of Hall subgroups of every order n dividing G with (n, |G|/n) = 1 is equivalent to solvability. There is an interesting special case of the Hall Converse. Suppose that |G| = pr q s with p and q prime. Then a p-Sylow subgroup of G is a Hall subgroup of order |G|/q s and a q-Sylow subgroup of G is a Hall subgroup of order |G|/pr . The Hall Converse produces an immediate corollary. Corollary (Burnside’s Theorem) Let G be a finite group whose order has only two prime divisors. Then G is solvable.


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In fact, Burnside’s Theorem is one of the ingredients in the proof of the Hall Converse, and, like the converse itself, is beyond the scope of this book. We shall now concentrate on proving Hall’s Theorem. Lemma 5.6.3. Let G be a finite group and let H be a Hall subgroup of G of order n. Let K G with |K| = m. Then H ∩ K is a Hall subgroup of K of order (m, n) and HK/K is a Hall subgroup of G/K of order n/(m, n). Proof Write |G| = nk. Since H is a Hall subgroup, this gives (n, k) = 1. Write m = n k with n dividing n and k dividing k. Thus, n = (m, n). Of course, |H ∩ K| divides (m, n) = n , so the order of H/(H ∩ K) is divisible by n/(m, n) = n/n . The Noether Isomorphism Theorem gives H/(H ∩ K) ∼ = HK/K, so |H/(H ∩ K)| divides |G/K| = (n/n ) · (k/k ). But also, |H/(H ∩ K)| divides |H| = n. Since n and k are relatively prime, |H/(H ∩ K)| must divide n/n . Thus, |H/(H ∩ K)| = n/(m, n), which forces |H ∩ K| = (m, n). Recall that a “minimal subgroup” of G was defined to be a nontrivial subgroup H ⊂ G such that H has no nontrivial proper subgroups. Thus, H is minimal among the nontrivial subgroups of G. We define minimal normal subgroups by analogy. Definition 5.6.4. We say that H is a minimal normal subgroup of G if e = H G and if no nontrivial proper subgroup of H is normal in G. If G is finite, then the existence of minimal normal subgroups in G is easy. Just choose the nontrivial normal subgroup of the lowest order. Lemma 5.6.5. Let G be a finite solvable group and let H be a minimal normal subgroup of G. Then H is an abelian group of exponent p for some prime p. Proof The commutator subgroup [H, H] is a characteristic subgroup of H. Since H G, this gives [H, H] G by Corollary 3.7.9. Thus, since H is a minimal normal subgroup of G, either [H, H] = H or [H, H] = e. Suppose that [H, H] = H. Then the derived subgroups H (i) = H for all i > 0. But solvable groups are by definition those for which H (i) = e for some i. Since H is nontrivial, it thus cannot be solvable, contradicting the fact that a subgroup of a solvable group is solvable (Corollary 5.5.7). Thus, [H, H] = e, and hence H is abelian. Let p be a prime dividing |H| and let Hp be the p-torsion subgroup of G. Then Hp is a characteristic subgroup of H (Lemma 4.4.19), and hence Hp G. Since Hp is nontrivial by Cauchy’s Theorem, we see that H must be an abelian p-group for some prime p. But then the elements of exponent p form a nontrivial characteristic subgroup of H, so H must have exponent p as well. We are now ready to prove Hall’s Theorem. Proof of Theorem 5.6.2 We wish to show that for any divisor n of |G| such that (n, |G|/n) = 1 there are Hall subgroups of G of order n and that any two such subgroups are conjugate. We argue by induction on |G|. The proof breaks up into two parts. First, we assume that G contains a nontrivial normal subgroup, K, whose order is not divisible by k. Write |K| = n k with n dividing n and k dividing k. The induction hypothesis gives a Hall subgroup H/K of G/K of order n/n . Thus, H ⊂ G has order nk . Since


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k < k, the induction hypothesis shows that there’s a Hall subgroup H ⊂ H of order n. This gives the desired subgroup of G. Suppose now that H and H are two subgroups of G of order n. By Lemma 5.6.3, HK/K and H K/K are Hall subgroups of order n/n in G/K, and hence are conjugate by the induction hypothesis. But if x conjugates HK/K to H K/K, then x conjugates HK to H K. But then x conjugates H to a Hall subgroup of H K of order n. And induction shows that xHx−1 is conjugate to H in H K. Thus, we are left with the case where the order of every nontrivial normal subgroup of G is divisible by k. Let P G be a minimal normal subgroup. Then Lemma 5.6.5, shows that P is an abelian group of exponent p for a prime number, p. Since |P | = pr is divisible by k and since (n, k) = 1, we must have k = pr . Thus, P is a p-Sylow subgroup of G. Since P G, there is only one subgroup of G of order k = pr , and hence P is the unique minimal normal subgroup of G. The quotient group G/P is solvable, so it has a minimal normal subgroup Q/P , whose order is q s for some prime number q. Let Q be a q-Sylow subgroup of Q. Then the canonical map π : G → G/P restricts to an isomorphism Q ∼ = Q/P (Corollary 5.3.12), so Q = QP . We claim that NG (Q) is the desired subgroup of G of order n. To show this, it suffices to show that [G : NG (Q)] = pr . To see this, first note that, since Q is normal in G, every conjugate of Q in G must lie in Q. Thus, every conjugate of Q in G is a q-Sylow subgroup of Q, and is conjugate to Q in Q by the First Sylow Theorem. So Q has the same number of conjugates in Q as in G, and hence [G : NG (Q)] = [Q : NQ (Q)]. We claim now that NQ (Q) = Q. Since [Q : Q] = pr , this will complete the proof that |NG (Q)| = n. Since P is the p-Sylow subgroup of Q and is normal there, it contains every p-torsion element of Q. Since Q ⊂ NQ (Q), the only way the two could differ is if |NQ (Q)| is divisible by p. Thus, it suffices to show that NQ (Q) ∩ P = e. Note first that if x ∈ P normalizes Q and if y ∈ Q, then xyx−1 y −1 ∈ P ∩ Q = e. Thus, x commutes with every element of Q. Since P is abelian and Q = QP , we see that NQ (Q) ∩ P ⊂ Z(Q), so it suffices to show that Z(Q) = e. Recall that Q/P G/P , and hence Q G. Since the center of a group is a characteristic subgroup, we see that Z(Q) G. Suppose that Z(Q) is nontrivial. Since P is the unique minimal normal subgroup of G, this implies that P ⊂ Z(Q). But then P normalizes Q. Since Q = QP , we must have NQ (Q) = Q, so that Q Q. But Q is the q-Sylow subgroup of Q, so this implies that Q is characteristic in Q and hence normal in G, contradicting the assumption that G has no normal subgroups whose order is not divisible by k. Thus, Z(Q) = e, and hence NG (Q) is a Hall subgroup of G of order n, as desired. It now suffices to show that if H is another subgroup of G of order n, then H is conjugate to NG (Q). Since |H| = |G/P | is prime to |P |, the canonical map induces an isomorphism H → G/P . Thus, H ∩ Q = Q maps isomorphically to Q/P under the canonical map, and hence is a q-Sylow subgroup of Q. Thus Q and Q are conjugate in Q, hence in G. But then NG (Q) and NG (Q ) are conjugate as well. Now Q = (H ∩ Q) H, so H ⊂ NG (Q ). Since the two groups have the same order, the result follows.


5.7

156

Nilpotent Groups

Nilpotent groups form a class of groups that lie between the solvable and the abelian groups. They form a generalization of the properties found among the finite p-groups. The fact that a finite p-group, P , has a nontrivial center is a very useful one, and it demonstrates that finite p-groups are, in fact, better behaved than the general run of solvable groups. And since P/Z(P ) is again a p-group, the center of P/Z(P ) must be nontrivial as well. Thus, induction arguments are possible. This forms the basic idea that we will generalize to define nilpotency. In the end, we shall see that the finite nilpotent groups are nothing more than cartesian products of p-groups. But the abstractions we will develop along the way turn out to be useful for our understanding of the p-groups, if nothing else. Definition 5.7.1. Let G be a group. Define the subgroups Z i (G) inductively as follows. Z 0 (G) = e and Z 1 (G) = Z(G). Suppose given Z i−1 (G), which is assumed to be normal in G, and let π : G → G/Z i−1 (G) be the canonical map. Then Z i (G) = π −1 (Z(G/Z i−1 (G))). Of course, as the pre-image of a normal subgroup of G/Z i−1 (G), Z i (G) G, and the induction continues. Definition 5.7.2. A group G is nilpotent if Z k (G) = G for some k ≥ 0. Just so the abstraction doesn’t get away from us here, we prove the following lemma. Lemma 5.7.3. Finite p-groups are nilpotent. Proof Suppose that Z i−1 (G) = G. Then G/Z i−1 (G) is a nontrivial p-group, and must have a nontrivial center. Thus, Z i (G) is strictly larger than Z i−1 (G). But then if |G| = pj , we must have Z k (G) = G for some k ≤ j. Since the quotient group Z i (G)/Z i−1 (G) is the center of G/Z i−1 (G), and hence abelian, the sequence e = Z 0 (G) · · · Z k (G) = G is an abelian subnormal series when G is nilpotent. Proposition 5.7.4. Nilpotent groups are solvable. The next result gives a useful property of nilpotent groups, which we have not yet verified for p-groups. (We dealt with a special case in Proposition 4.1.12 by a different, but related, argument.) Proposition 5.7.5. Let H be a subgroup of the nilpotent group G, with H = G. Then H cannot be its own normalizer in G. Proof Let H ⊂ G be its own normalizer. We shall show that H must equal G. Since G is nilpotent, it suffices to show that Z i (G) ⊂ H for all i. We show this by induction on i, with the case i = 0 immediate. Suppose that Z i−1 (G) ⊂ H, and let π : G → G/Z i−1 (G) be the canonical map. Since Z i−1 (G) ⊂ H, we have H = π −1 (π(H)). This is easily seen to imply that NG (H) = π −1 (NG/Z i−1 (G) (π(H))), so π(H) must be its own normalizer in G/Z i−1 (G). Since Z i (G) = π −1 (Z(G/Z i−1 (G))), it suffices to show that Z(G/Z i−1 (G)) ⊂ π(H). In other words, it suffices to assume that i = 1, and show that Z(G) ⊂ H. But the elements of Z(G) normalize every subgroup of G, so that Z(G) ⊂ NG (H) = H, and the result follows.


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Corollary 5.7.6. Every subgroup of a finite nilpotent group is part of a subnormal series. Proof Let H be the subgroup and G the group. If H is not already normal in G, then its normalizer at least strictly contains it, so that the index of the normalizer is strictly smaller than the index of H. Thus, the result follows by induction on the index. We can use this in our efforts to classify the low order groups. Corollary 5.7.7. Let G be a nonabelian group of order 16 that does not contain an element of order 8. Then G has an index 2 subgroup H such that H has an element of order 4. Proof If every element of G has order 2, then G is abelian by Problem 3 of Exercises 2.1.16. Thus, G must have an element x of order 4. If x is not normal in G, then we may take H to be its normalizer. Otherwise, take H to be the pre-image of a subgroup of order 2 in G/x. Thus, we shall be able to classify the groups of order 16 without first understanding the structure of Aut(Z2 × Z2 × Z2 ). This is useful, because Aut(Z2 × Z2 × Z2 ) turns out to be a simple group of order 168. Lemma 5.7.8. A product H × K is nilpotent if and only if both H and K are nilpotent. Proof This will follow once we show that Z i (H × K) = Z i (H) × Z i (K) for all i ≥ 0. We show this by induction on i, with the case i = 0 immediate. Suppose Z i−1 (H × K) = Z i−1 (H) × Z i−1 (K). Then the product of canonical maps

π × π : H × K → H/Z i−1 (H) × K/Z i−1 (K) has kernel Z i−1 (H × K). Thus, it suffices to show that

Z( H/Z i−1 (H) × K/Z i−1 (K) ) = Z(H/Z i−1 (H)) × Z(K/Z i−1 (K)). But it is easy to see that for any groups H and K , Z(H × K ) = Z(H ) × Z(K ). But now we can characterize finite nilpotent groups. Proposition 5.7.9. A finite group is nilpotent if and only if it is the internal direct product of its Sylow subgroups, meaning that if p1 , . . . , pk are the primes dividing |G|, then the pi -Sylow subgroup, Gpi , of G is normal for all i, and G is the internal direct product of Gp1 , . . . , Gpk . Proof If G is isomorphic to Gp1 × · · · × Gpk , then G is nilpotent by Lemmas 5.7.3 and 5.7.8. Thus, we prove the converse. Note that for abelian groups, the converse is given as Proposition 4.4.9. Let G be nilpotent and let Gpi be a pi -Sylow subgroup of G. Then the third Sylow Theorem shows that the normalizer of Gpi is its own normalizer in G. But since G is nilpotent, Proposition 5.7.5 shows that the normalizer of Gpi must be G itself, so that Gpi G. We shall show that this is sufficient to obtain the stated product formula. If 1 ≤ i, j ≤ k and if i = j, let x ∈ Gpi and let y ∈ Gpj . Consider the commutator [x, y] = xyx−1 y −1 . Since Gpj is normal in G, xyx−1 ∈ Gpj , and hence the commutator [x, y] is also. But the normality of Gpi shows that yx−1 y −1 and hence also [x, y] are in Gpi as well. But because the identity is the only element that has prime power order for more than one prime, Gpi ∩ Gpj = e, and hence [x, y] = e.


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Thus, the elements of Gpi commute with the elements of Gpj whenever i = j, and we have a homomorphism μ : Gp1 × · · · × Gpk → G given by μ(g1 , . . . , gk ) = g1 · · · gk . We claim that μ is an isomorphism. By the second Sylow Theorem, Gp1 × · · · × Gpk has the same order as G, so it suffices to show that μ is injective. Suppose that μ(g1 , . . . , gk ) = e. We wish to show that gi = e for all i. Assume by contradiction that i is the smallest index for which gi = e. We cannot have i = k, as then gi = e as well. Thus, gi−1 = gi+1 · · · gk . But the order of (e, . . . , e, gi+1 , · · · , gk ) in Gp1 × · · · × Gpk is the least common multiple of the orders of gi+1 , . . . , gk (by Problem 2 of Exercises 2.10.12), and hence the order of its image, under μ is divisible only by the primes pi+1 , . . . , pk . But the image in question happens to be gi−1 , which is a pi -torsion element, and we have our desired contradiction. We obtain a converse to Lagrange’s Theorem for nilpotent groups. Corollary 5.7.10. Let G be a finite nilpotent group. Then every divisor of the order of G is the order of a subgroup of G. Proof Write |G| = pr11 . . . prkk where p1 , . . . , pk are distinct primes. Then any divisor of |G| has the form m = ps11 . . . pskk with 0 ≤ si ≤ ri for i = 1, . . . , k. Let Gpi be the pi -Sylow subgroup of G. Then there is a subgroup Hi of Gpi of order psi i by Corollary 5.2.6. By Proposition 5.7.9, G has a subgroup isomorphic to H1 × · · · × H k . Exercises 5.7.11. 1. Give an example of a solvable group that’s not nilpotent. 2. Classify the groups of order 16 containing a copy of D8 . 3. Classify the groups of order 16 containing a copy of Q8 . 4. Classify the groups of order 16 containing a copy of Z4 × Z2 . 5. Which groups appear in more than one of the three preceding problems? Which of them contain an element of order 8? 6. Classify the groups of order 24 whose 2-Sylow subgroups are isomorphic to Z4 ×Z2 .

5.8

Matrix Groups

Perhaps the most important missing piece at this point in being able to classify low order groups is an understanding of the automorphism groups of the Sylow subgroups that arise in them. So far, we know how to calculate the automorphisms only of cyclic groups, and groups such as dihedral and quaternionic groups, whose automorphisms are calculated in Exercises 4.7.27 and 5.4.9. Here, we shall investigate the automorphisms of groups of the form Zn × · · · × Zn , with special attention paid to the case where n is prime. There are a number of groups of order < 64 which have Sylow subgroups of this form. Unlike the preceding material, we shall need to make use of some theory we have not yet developed: basic ring theory and linear algebra. The reader who has not yet been exposed to this material should come back to this section after reading the appropriate chapters.


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Of course, for any ring A, the cartesian product A × · · · × A of n copies of A is the underlying abelian group of the free A-module of rank n. Consonant with this, we shall write A × · · · × A = An . n times

Linear algebra gives techniques of studying the A-module homomorphisms from An to itself. Recall that if N is an A-module, then EndA (N ) denotes the set of A-module homomorphisms from N to itself, while AutA (N ) denotes the group (under the operation of composition of homomorphisms) of A-module isomorphisms from N to itself. Recall also that an A-module homomorphism is an isomorphism if and only if it is bijective. Since every A-module isomorphism is an isomorphism of abelian groups, forgetting the A-module structure gives an injective group homomorphism ϕ : AutA (N ) → Aut(N ). For A = Zn , Corollary 7.7.16 shows that every group homomorphism between Zn modules is a Zn -module homomorphism. Thus, the next lemma is immediate. Lemma 5.8.1. For all k ≥ 1, forgetting the Zn -module structure gives an isomorphism ∼ = ϕ : AutZn (Zkn ) −→ Aut(Zkn ). Recall that the group of invertible n × n matrices over a ring A is denoted by Gln (A). Here, the group structure comes from matrix multiplication. Corollary 7.10.9 shows that allowing matrices to act on the left on column vectors gives a group isomorphism from Gln (A) to AutA (An ).3 Thus, for any A, allowing matrices to act on the left on column vectors gives an injective ring homomorphism ιn : Gln (A) → Aut(An ). For A = Zn , Lemma 5.8.1 allows us to do even better. Corollary 5.8.2. The standard left action of matrices on column vectors induces a group isomorphism ιk : Glk (Zn ) → Aut(Zkn ).

In particular, the study of matrix groups will give us calculations of automorphism groups of groups that we care about. Recall that the transpose, M t , of an n × n matrix M = (aij ) is the matrix whose ij-th entry is aji . Lemma 5.8.3. For any commutative ring A, there is a group automorphism α : Gln (A) → Gln (A) defined by α(M ) = (M t )−1 . Proof It is easy to check that for any pair of matrices M1 , M2 ∈ Mn (A), we have (M1 M2 )t = M2t M1t . This is enough to show that the transpose of an invertible matrix is invertible, with (M t )−1 = (M −1 )t . Since inversion of matrices also reverses order, the result follows. 3 Here,

if A is not commutative, then this applies only to the standard right module structure on An .


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For finite rings A, the set of n × n matrices is finite, and hence so is Gln (A). (If A is infinite, it is not hard to show that Gln (A) is infinite for n ≥ 2.) The next result will assist in calculating the order of Gln (A) when A is finite. Proposition 5.8.4. Let A be a commutative ring. Consider the standard action of Gln (A) on columns, and let en be the n-th canonical basis vector. Then the isotropy subgroup of en under this action is isomorphic to An−1 ιn−1 Gln−1 (A). Proof The isotropy subgroup, Gln (A)en , of en consists of those invertible matrices whose lastcolumn is en . Thus, we are concerned with the invertible matrices of the form M 0 , where M is (n−1)×(n−1), X is a 1×(n−1) row matrix, 0 is the (n−1)×1 X 1 column matrix whose coordinates are all 0, and the 1 in the lower corner is the 1 × 1 matrix 1. Note that M must lie in Gln−1 (A), as otherwise the first n − 1 rows of the full matrix would be linearly dependent. if M is any element of Gln−1 (A) and X Conversely, is any 1 × (n − 1) row matrix, M 0 M −1 0 is invertible, with inverse , and hence lies in the isotropy X 1 −XM −1 1 group Gln (A)en . Thus, if α is the automorphism of Gln (A) replacing a matrix withthe inverseof its M X transpose, then H = α(Gln (A)en ) is the set of all matrices of the form with 0 1 M ∈ Gln−1 (A) and X ∈ An−1 , where we identify An−1 with the set of (n − 1) × 1 column matrices. Of course, Gln (A)en is isomorphic to H, so we may study the latter. Note that the elements of H multiply according to the formula M X M Y MM X + MY = . 0 1 0 1 0 1 M X Thus, there is an isomorphism f : An−1 ιn−1 Gln−1 (A) → H, via f (X, M ) = . 0 1 We can apply this to calculate the order of the general linear group of a finite field. It is shown in Section 11.3 that the order of any finite field must have the form q = pr for p prime, and that for any such q, there is a unique field, which we shall denote by Fq , of order q. Of course, Fp = Zp . Corollary 5.8.5. Let Fq be the field of order q = pr . Then for n ≥ 1, |Gln (Fq )| = (q n − 1)(q n − q) . . . (q n − q n−1 ). Proof We argue by induction on n. For n = 1, Gl1 (Fq ) ∼ = F× q has order q − 1, and the assertion holds. Suppose the result is true for n − 1. Then the formula for the size of an orbit gives |Gln (Fq )| = |Gln (Fq ) · en | · |Fn−1 ιn−1 Gln−1 (Fq )| q = |Gln (Fq ) · en | · q n−1 · (q n−1 − 1) . . . (q n−1 − q n−2 ) = |Gln (Fq ) · en | · (q n − q) . . . (q n − q n−1 ). Thus, it suffices to show that the orbit of en under the action of Gln (Fq ) has q n − 1 elements, and hence that for any nonzero element v ∈ Fnq , there is an M ∈ Gln (Fq ) with M en = v.


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But this follows immediately from elementary linear algebra: v is linearly independent over Fq , so there is a basis of Fnq that contains v. Matrix groups over finite fields have another important use in the theory of finite groups. They are a source of finite simple groups. Recall that if A is a commutative ring, then the determinant induces a surjective group homomorphism det : Gln (A) → A× for n ≥ 1. The special linear group Sln (A) is defined to be the kernel of det. Since a matrix is invertible if and only if its determinant is a unit, we see that Sln (A) consists of all n × n matrices over A of determinant 1. Recall that aIn is the matrix whose diagonal entries are all equal to a and whose off-diagonal entries are all 0. It’s an easy exercise to show that det(aIn ) = an . Definition 5.8.6. Let A be a commutative ring and define H ⊂ Sln (A) by H = {aIn | a ∈ A

with an = 1}.

Then the projective special linear group is defined as PSln (A) = Sln (A)/H. It turns out that the subgroup H above is the center of Sln (A). Theorem Let Fq be the field with q = pr elements. The projective special linear group PSln (Fq ) is simple for all q if n ≥ 3 and is simple for q > 3 if n = 2. The proof is accessible to the techniques of this volume, but we shall not give it. It is somewhat analogous to the proof that An is simple for n ≥ 5. There, the cycle structure gave us a handle on the conjugacy classes. Here, the conjugacy classes are determined by the rational canonical form. Exercises 5.8.7. 1. Give a new proof that Aut(Z22 ) = Gl2 (Z2 ) is isomorphic to S3 . 2. Show that Gl3 (Z2 ) has a subgroup of index 7 that is isomorphic to S4 . 3. What are the normalizers of all the Sylow subgroups of Gl3 (Z2 )? 4. Show that Gl3 (Z2 ) is simple. 5. Complete the classification of the groups of order 24. 6. Let q = pr with p prime. Show that one of the p-Sylow subgroups of Gln (Fq ) is the set of matrices with 1’s on the diagonal and 0’s below it (i.e., the upper triangular matrices with 1’s on the diagonal). 7. Show that the p-Sylow subgroups of Gl3 (Zp ) are isomorphic to the group constructed in Corollary 4.6.8 for n = p. 8. Write Tn ⊂ Gln (Fq ) for the subgroup of upper triangular matrices with 1’s on the diagonal. Then Tn acts on Fnq through automorphisms by an action ιn obtained by restricting the action of the full general linear group. Show that Tn is isomorphic to Fn−1 ιn−1 Tn−1 . q 9. Show that D8 embeds in Gl2 (A) whenever A has characteristic = 2.


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10. Show that any element of order 4 in Z24 is part of a Z4 -basis. Calculate the order of Gl2 (Z4 ). 11. Classify the groups of order 48 whose 2-Sylow subgroup is Z4 × Z4 . 12. What are the orders of Sln (Fq ) and PSln (Fq )? 13. Show that the 2-Sylow subgroup of Sl2 (Z3 ) is isomorphic to Q8 and is normal in Sl2 (Z3 ). (Hint: Look at the elements of Sl2 (Z3 ) of trace 0.) 14. Deduce that the 2-Sylow subgroup of Gl2 (Z3 ) is a group of order 16 containing both D8 and Q8 . 15. Classify the groups of order 18. 16. What is the 2-Sylow subgroup of Sl2 (Z5 )? 17. What is the 2-Sylow subgroup of Sl2 (Z4 )? 18. Show that the subgroup H of Definition 5.8.6 is the center of Sln (A).

Chapter 6

Categories in Group Theory Category theory, a.k.a. abstract nonsense, is a branch of mathematics that attempts to identify phenomena that crop up in many different types of mathematical structures. One may think of it as the study of those things that happen for reasons which are purely formal. For this reason, the facts that hold for categorical reasons are often not the deepest ones in a given field. But categorical patterns repeat in so many contexts that they form a collection of truths that may be instantly recognized and used. What category theory gives us is a language by which to describe patterns common to different areas of mathematics. Verifying that these patterns hold in a given context can take some work. For instance, we will give some concepts regarding commonly occurring universal properties. In each category of interest, one must make a construction, often unique to that category, to show that objects satisfying the universal property actually exist. Once that construction has been made, one can revert to formal arguments in using it. Categorical language is well suited to describing the overview of the relationships between objects in a given context. For instance, classification results may be expressed in categorical terms. For reasons like this, some of the more important mathematical questions can be phrased in terms of categorical language. Formal study of category theory will not generally be useful in solving them, but categorical thinking can sometimes suggest a viable approach. In the first two sections, we give the most general categorical concepts: categories and functors. The next three sections introduce some important kinds of universal properties: initial and terminal objects, products, coproducts, pushouts, and pullbacks. We give constructions realizing these universal properties in various categories of interest, particularly in the categories of sets, groups, and abelian groups. The constructions realizing these universal properties in these three categories can be quite different. Thus, the reader can see that a general categorical notion can behave quite differently in different categories. The insight coming from the categorical concept tells us nothing about whether or how that concept might be realized in a particular category. In Section 6.6, we define the notion of a free object on a set in a given category and construct the free groups and the free abelian groups. Then in Section 6.7, we give the technique of describing a group by generators and relations. This technique can be valuable in studying infinite groups, and arises frequently in topology, as the fundamental group of a CW complex with one vertex is expressed naturally in this form. We then define direct and inverse limits, giving constructions for them in various categories. Direct limits will be used in the construction of algebraic closures of fields.

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CHAPTER 6. CATEGORIES IN GROUP THEORY

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p. Inverse limits are used to construct the ring of p-adic integers, Z Then, generalizing free functors, we define adjoint functors, giving examples and showing their utility for deducing general results. We then close with a description of colimits and limits indexed on a partially ordered set or a small category. This material has applications to sheaves and to various of the constructions in algebraic geometry and homotopy theory, and generalizes many of the universal properties described earlier in the chapter. Because of the number of concepts and constructions here that may be new to the reader, Sections 6.2, 6.3, 6.4, 6.5, and 6.8 have been divided into subsections, each with its own set of exercises.

6.1

Categories

The notion of a category is very abstract and formal, and is best understood by going back and forth between the definition and some examples. The point is that the word “object” has to be taken as a black box at first, as there are categories with many different sorts of objects. The objects in a category are the things you want to study. So the category of sets has sets as objects, the category of groups has groups as objects, etc. The morphisms in a category determine the ways in which the objects are related. For instance, when studying sets, one generally takes the functions between two sets as the morphisms. With this in mind, here is the abstraction. Definition 6.1.1. A category, C, consists of a collection of objects, ob(C), together with morphisms between them. Specifically, if X and Y are objects of C (written X, Y ∈ ob(C)) there is a set of morphisms from X to Y , denoted MorC (X, Y ). Morphisms may be composed, just as functions or homomorphisms may be composed: If Z is another object of C, then for any f ∈ MorC (Y, Z) and any g ∈ MorC (X, Y ) there is a composite morphism f ◦ g ∈ MorC (X, Z). This composition rule is required to satisfy two properties. First, there’s an associativity law: if h ∈ MorC (W, X), then for f and g as above, we have f ◦ (g ◦ h) = (f ◦ g) ◦ h as elements of MorC (W, Z). Second, each object has an identity morphism. We write 1X ∈ MorC (X, X) for the identity of X. The defining property of the identity morphisms is that g ◦ 1X = 1Y ◦ g = g for all g ∈ MorC (X, Y ). Example 6.1.2. One of the most ubiquitous categories is the category of sets and functions, which we denote by Sets. The objects of Sets consist of all possible sets. The morphisms between X and Y in Sets are all the functions from X to Y . The identity function is the usual one, and the properties of composition of functions are easily seen to satisfy the requirements for this to be a category. A very large number of the categories we will look at can be thought of as subcategories of the category of sets and functions: Definition 6.1.3. A category D is a subcategory of C if ob(D) is a subcollection of ob(C), MorD (X, Y ) is a subset of MorC (X, Y ) for all objects X and Y of D, and the laws of composition for the two categories coincide, as do the identity morphisms. It may come as a surprise that not all categories are equivalent (in a formal sense we shall give below, but which includes the intuitive notion of two categories being


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isomorphic) to subcategories of Sets.1 But the structure of Sets does provide a very good intuition for what a category is. Of course, one can define categories from now until doomsday, and very few of the ones chosen randomly will be mathematically interesting. In particular, if one wants to study something like groups, one should not take the morphisms between two groups to be all functions between their underlying sets. The morphisms in a given category should convey information about the basic structure of the objects being considered. Without that, a study of the category will not shed light on the mathematics one wants to do. Examples 6.1.4. 1. The category in which we’ve been working for the most part here is the category Gp, of groups and homomorphisms. Thus, the objects are all groups and the morphisms between two groups are the group homomorphisms between them. 2. An interesting subcategory of Gp is the category of abelian groups and homomorphisms, which we denote by Ab. Thus, the objects of Ab are all abelian groups, while the morphisms between two abelian groups are the group homomorphisms between them. The inclusion of Ab in Gp is an example of what’s called a full subcategory: Definition 6.1.5. A subcategory D of C is a full subcategory if for any objects X, Y ∈ ob(D), the inclusion MorD (X, Y ) ⊂ MorC (X, Y ) is bijective. It is sometimes useful to restrict the morphisms in a category, rather than the objects. Thus, we can form a subcategory of Gp whose objects are all groups, but whose morphisms are the injective homomorphisms. Note that in many cases here, the set of morphisms between two objects is the null set. But that doesn’t cause any problems as far as the definition of a category is concerned. It is sometimes useful to make new categories out of old ones. Definition 6.1.6. Let C and D be categories. The product category C × D is defined as follows. The objects of C × D are the ordered pairs (X, Y ) with X ∈ ob(C) and Y ∈ ob(D). The morphisms of C × D are given by MorC×D ((X, Y ), (X , Y )) = MorC (X, X ) × MorD (Y, Y ). Here, the latter product is the cartesian product of sets. We write f ×g : (X, Y ) → (X , Y ) for the morphism corresponding to f ∈ MorC (X, X ) and g ∈ MorD (Y, Y ), and define composition by (f × g) ◦ (f × g ) = (f ◦ f ) × (g ◦ g ) for f × g : (X , Y ) → (X , Y ) in C × D. One of the most important ideas about the relationships between objects in a category is that of isomorphism: Definition 6.1.7. An isomorphism in a category C is a morphism f : X → Y in C which admits an inverse. Here, an inverse for f is a morphism g : Y → X such that f ◦ g = 1Y and g ◦ f = 1X . 1 One example of a category not equivalent to a subcategory of Sets is the homotopy category encountered in topology.


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Clearly, a morphism in Sets is an isomorphism in Sets if and only if it is a bijection. Also, a morphism in Gp (or Ab) is an isomorphism in Gp (resp. Ab) if and only if it is a group isomorphism in the sense we have been studying. As in the case of finite group theory, one of the most important problems in studying a given category is being able to classify the objects. Definitions 6.1.8. The isomorphism class containing a given object consists of the collection of all objects isomorphic to it. Clearly, any two isomorphism classes having an object in common are equal. A classification of the objects that have a given property consists of a determination of all the individual isomorphism classes with that property. One example of a classification is the determination of a set of representatives of the isomorphism classes with a given property. This means a set S = {Xi | i ∈ I} of objects such that every isomorphism class in C having the given property contains exactly one of the objects in S. Not every collection of isomorphism classes has a set of representatives. For instance, the isomorphism classes of torsion-free abelian groups do not form a set. Even if there is a classification of the objects with a given property, it can be difficult to determine which isomorphism class contains a given object. In particular, given two objects in a given category, there may not be any algorithmic procedure to determine whether the objects are isomorphic. For instance, think about the classifications we’ve seen of the groups in different orders. At first glance, it can be quite difficult to decide whether two groups are isomorphic. And the classifications for different orders can involve very different criteria. The development of criteria to determine whether two objects are isomorphic is one of the most important problems in studying a given category. It should be noted that general category theoretic techniques are rarely useful for solving this problem. However, the notion of functors, to be studied in Section 6.2, can be useful for showing that two objects are not isomorphic. There are partial forms of invertibility which can be useful. Definitions 6.1.9. A left inverse for a morphism f : X → Y in C is a morphism g : Y → X in C such that g ◦ f = 1X . A left inverse is often called a retraction, and X is often called a retract of Y in C. Similarly, a right inverse for f : X → Y in C is a morphism g : Y → X in C such that f ◦ g = 1Y . A right inverse for f is often called a section of f . We shall also consider categories of monoids. We write Mon for the category whose objects are all monoids and whose morphisms are the homomorphisms of monoids. We write Amon for the full subcategory whose objects are the abelian monoids. Exercises 6.1.10. 1. Show that if f : X → Y is an isomorphism in C, then its inverse is unique. 2. Show that every injection in Sets admits a left inverse, and that every surjection in Sets admits a right inverse. Are these inverses unique? 3. Does every injective group homomorphism have a left inverse in Gp? Does every surjective group homomorphism have a right inverse in Gp? 4. Let g be a left inverse for f : X → Y in C. Suppose that g admits a left inverse also. Show then that f is an isomorphism with inverse g.


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5. Suppose that f : G → K is a morphism in Gp that admits a right inverse. Show then that G is a split extension of ker f by K. 6. Suppose that i : H → G is a morphism in Gp that admits a left inverse. Show that i must be injective. Show that if i(H) G, then G ∼ = H × G/i(H). What can you say if i(H) is not normal in G? 7. Let G be a group and let G-sets be the category whose objects are all G-sets and whose morphisms are all G-maps. Show that the isomorphisms in G-sets are precisely the G-isomorphisms. 8. Let C be any category and let X ∈ ob(C). Show that MorC (X, X) is a monoid under composition of morphisms. We call it the monoid of endomorphisms of X, and write MorC (X, X) = EndC (X). Show that this monoid is a group if and only if every morphism from X to itself is an isomorphism in C. 9. Let M be a monoid. Show that there is a category which has one object, O, such that the monoid of endomorphisms of O is M . Thus, monoids are essentially equivalent to categories with one object.

6.2

Functors

Just as we can learn about a group by studying the homomorphisms out of it, we can learn about a category by studying its relationship to other categories. But to get information back, we shall need to define ways to pass from one category to another that preserve structure. We study covariant functors here, and study contravariant functors in a subsection below. Definition 6.2.1. Let C and D be categories. A functor, F : C → D assigns to each object X ∈ ob(C) an object F (X) ∈ ob(D) and assigns to each morphism f ∈ MorC (X, Y ) a morphism F (f ) ∈ MorD (F (X), F (Y )), such that F (1X ) = 1F (X) for all X ∈ ob(C), and F (f ) ◦ F (g) = F (f ◦ g) for each pair of morphisms f and g in C for which the composite f ◦ g is defined. Functors are often used in mathematics to detect when objects of a given category are not isomorphic. Lemma 6.2.2. Let F : C → D be a functor and suppose that f : X → Y is an isomorphism in C. Then F (f ) : F (X) → F (Y ) is an isomorphism in D. Thus, if Z and W are objects of C, and if F (Z) and F (W ) are not isomorphic in D, then Z and W must not be isomorphic in C. Proof Let g be the inverse of f . Then F (g) ◦ F (f ) = F (g ◦ f ) = F (1X ) = 1F (X) . Similarly, F (f ) ◦ F (g) = 1F (Y ) . Thus, F (g) is an inverse for F (f ) in D.


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Indeed, in a lot of ways, the preceding lemma is the main reason for looking at functors, to the point where the definition of functor is tailored to make it true. The utility of this approach lies in the fact that we can often find functors to categories that are much more manageable than the one we started in. Examples 6.2.3. 1. Let p be a prime. Then for an abelian group G, the set Gp of p-torsion elements of G is a subgroup of G. As shown in Lemma 4.4.19, if f : G → H is a homomorphism with H abelian, then f restricts to a homomorphism fp : Gp → Hp . Clearly, gp ◦ fp = (g ◦ f )p for any homomorphism g : H → K with K abelian, and (1G )p = 1(Gp ) , so there is a functor Fp : Ab → Ab given by setting Fp (G) = Gp for all abelian groups G, and setting Fp (f ) = fp for f : G → H a homomorphism of abelian groups. Notice that while Fp (f ) may be an isomorphism even if f is not, Corollary 4.4.21 shows that if f : G → H is a homomorphism between finite abelian groups, then f is an isomorphism if and only if Fp (f ) is an isomorphism for each prime p dividing either |G| or |H|. Thus, on the full subcategory of Ab whose objects are the finite abelian groups, the collection of functors Fp , for p prime, together detect whether a morphism is an isomorphism. 2. Let C be any category and let X be an object of C. We define a functor F : C → Sets as follows. For an object Y of C, we set F (Y ) = MorC (X, Y ). If f : Y → Z is a morphism from Y to Z in C, we define F (f ) : MorC (X, Y ) → MorC (X, Z) by setting F (f )(g) = f ◦ g for all g ∈ MorC (X, Y ). (It is often customary to write f∗ instead of F (f ).) The reader may easily check that F is a functor. 3. Define φ : Gp → Sets by setting φ(G) = G, the underlying set of G. For a homomorphism f : G → H, we set φ(f ) = f . We call φ the forgetful functor that forgets the group structure on G. More generally, any functor that simply forgets part of the structure of a given object is called a forgetful functor. Thus, there is also a forgetful functor from Gp to Mon. Definition 6.2.4. A functor C × C → D is often called a bifunctor on C, and may be thought of as a functor of two variables. Example 6.2.5. There is a bifunctor F : Sets × Sets → Sets obtained by setting F (X, Y ) = X × Y and setting F (f × g) = f × g, for f ∈ MorSets (X, X ) and g ∈ MorSets (Y, Y ). Here, the right-hand function f ×g is the usual mapping f ×g : X ×Y → X × Y given by (f × g)(x, y) = (f (x), g(y)). Exercises 6.2.6. 1. Show that functors preserve left inverses and right inverses. 2. Show that there is a functor F : Gp → Gp such that F (G) is the commutator subgroup of G for every group G. Here, if f : G → G is a group homomorphism, then F (f ) is obtained by restricting f to the commutator subgroup. 3. Give an example of groups H and G such that H is a retract of G, but Z(H) is not a retract of Z(G). Deduce that there is no functor on Gp that takes each group to its center.


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4. Show that there is a functor F : Gp → Ab such that F (G) = Gab for all groups G, and if f : G → G is a group homomorphism, then F (f ) = f ab , the unique homomorphism such that the following diagram commutes. G πG Gab

f

f ab

/ G πG / ab G

Here, for any group H, we write πH for the canonical map from H to H ab = H/[H, H]. 5. Show that the inclusion of a subcategory is a functor. In particular the inclusion of a category in itself is called the identity functor. 6. Let φ : Gp → Sets be the forgetful functor. Suppose that φ(f ) is an isomorphism. Is f then an isomorphism? Alternatively, suppose that φ(G) and φ(G ) are isomorphic in Sets. Are G and G then isomorphic in Gp? 7. Show that passage from a monoid M to its group of invertible elements Inv(M ) gives a functor from Mon to Gp (or from Amon to Ab). 8. Let G be a group and let G be the category with one object, O, such that the monoid of endomorphisms of O is G. Show that there is a one-to-one correspondence between G-sets and the functors from G into Sets. 9. Let M and M be monoids and let M and M be categories with one object whose monoids of endomorphisms are M and M , respectively. Show that there is a oneto-one correspondence between the monoid homomorphisms from M to M and the functors from M to M . 10. Let C and D be categories. (a) Show that there are functors πC : C × D → C and πD : C × D → D obtained by setting πC (X, Y ) = X, πD (X, Y ) = Y , πC (f × g) = f and πD (f × g) = g. (b) For an object Y of D, show that there is a functor ιY : C → C × D obtained by setting ιY (X) = X × Y and ιY (f ) = f × 1Y .

Contravariant Functors Now we come to the fact that there’s more than one kind of functor. The functors we’ve considered so far are called covariant functors. The other kind of functors are called contravariant. They behave very much like covariant ones, except that they reverse the direction of morphisms. Definition 6.2.7. A contravariant functor F : C → D associates to each X ∈ ob(C) an object F (X) ∈ ob(D) and to each morphism f ∈ MorC (X, Y ) a morphism F (f ) ∈ MorD (F (Y ), F (X)), such that F (1X ) = 1F (X)


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for all X ∈ ob(C), and F (f ) ◦ F (g) = F (g ◦ f ) for each pair of morphisms f and g in C for which the composite g ◦ f is defined. Contravariant functors may be viewed as covariant functors out of a different category. Definition 6.2.8. Let C be a category. The opposite category, C op , is defined by setting ob(C op ) = ob(C)

and setting

MorC op (X, Y ) = MorC (Y, X)

for all X, Y ∈ ob(C). Here, if f : X → Y is a morphism in C, we write f op : Y → X for the associated morphism in C op . We define composition in C op by setting f op ◦ g op = op op (g ◦ f ) , and set the identity morphism of X ∈ ob(C op ) equal to (1X ) . The point here is that any contravariant functor on C may be viewed as a covariant functor on C op . The reader may supply the details: Proposition 6.2.9. For any category C, there is a contravariant functor I : C → C op defined by setting I(X) = X for X ∈ ob(C) and setting I(f ) = f op for f ∈ MorC (X, Y ). If F : C → D is a contravariant functor, then there is a covariant functor F op : op C → D defined by setting F op (X) = F (X) for X ∈ ob(C) = ob(C op ) and setting F op (f op ) = F (f ) for f ∈ MorC (X, Y ). Thus, F itself factors as the composite I

F op

C− → C op −−→ D.

Exercises 6.2.10. 1. Let C be any category and let Y be an object of C. For X ∈ ob(C), let F (X) = MorC (X, Y ). For f ∈ MorC (X, X ), define F (f ) : MorC (X , Y ) → MorC (X, Y ) by F (f )(g) = g ◦ f . (It is often customary to write f ∗ instead of F (f ).) Show that F defines a contravariant functor from C to Sets. 2. Let C be a category. Show that there is a functor Hom : C op × C → Sets, given by setting Hom(X, Y ) = MorC (X, Y ) for each pair of objects X, Y of C. Here, if f : X → X and g : Y → Y are morphisms in C, we define Hom(f op , g) : MorC (X, Y ) → MorC (X , Y ) by Hom(f op , g)(h) = g ◦ h ◦ f for all h ∈ MorC (X, Y ). 3. Show that a contravariant functor takes isomorphisms to isomorphisms. 4. Show that a contravariant functor takes morphisms that admit left inverses to morphisms that admit right inverses and vice versa. 5. Let G be a group and let G be the category with one object, O, such that the monoid of endomorphisms of O is G. Show that there is a one-to-one correspondence between right G-sets and the contravariant functors from G into Sets.


6.3

171

Universal Mapping Properties: Products and Coproducts

An object, or the relationship between a family of objects connected by morphisms, is sometimes uniquely determined by what morphisms exist either in or out of the object or family. This phenomenon is known as a universal mapping property, and is perhaps best understood through examples. We treat initial and terminal objects here, and then give subsections for products, coproducts, and free products. The simplest examples of universal mapping properties are initial and terminal objects for a category: Definitions 6.3.1. Suppose that X ∈ ob(C) has the property that MorC (X, Y ) has exactly one element for each Y ∈ ob(C). Then X is called an initial object for C. Alternatively, if MorC (Y, X) has exactly one element for each Y ∈ ob(C), then X is called a terminal object for C. If X is both an initial object and a terminal object for C, we call X a zero object for C. One example that may not be immediately obvious is that the null set is an initial object for Sets. The point here is that a function is determined by where it sends the elements of its domain. Since the null set has no elements, there is exactly one way to specify where its elements go. Exercises 6.3.2. 1. Show that any two initial objects for C must be isomorphic. 2. Show that any two terminal objects for C must be isomorphic. 3. Show that a set with one element is a terminal object for Sets. 4. Show that the trivial group e is a zero object in both Gp and in Ab. 5. Let G be a group. Show that the G-set with one element is a terminal object in G-sets. 6. Let G be any nontrivial group. Show that G-sets does not have an initial object.

Products Other universal mapping properties express relationships between more than one object. These properties often recur in many different categories, and are given names appropriate to that fact. The one we’ve seen the most of is products: Definition 6.3.3. Let X, Y ∈ ob(C). A product for X and Y in C consists of an object, Z ∈ ob(C), together with two morphisms, πX : Z → X and πY : Z → Y , called projections. The universal property that makes these a product is as follows: For any object W ∈ ob(C) and any two morphisms f : W → X and g : W → Y , there is a unique


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morphism (f, g) : W → Z such that the following diagram commutes: W/ ?OO // ??OOOO // ??? h OOgOO OOO // ?? OO' // Z πY / Y f // // // πX / X, Here, h = (f, g). Thus, the projection maps are part of the definition of a product, and not just the object Z itself. Examples 6.3.4. 1. The cartesian product, together with the usual projections, is the product in Sets: a function into the product is determined by its component functions, which are precisely the composites of the original function with the projection maps. Indeed, a complete argument showing that the cartesian product is the product in the category of sets is given in Proposition 1.6.5. 2. In Gp, the direct product has been shown in Lemma 2.10.5 to be the categorical product, and this remains true in Ab. In Gp and Ab, the product has some extra features (like the inclusion maps) that do not occur in Sets or more general categories. The universal mapping property that defines the product has at most one solution (up to isomorphism) for a given pair of objects X and Y . This is because of the uniqueness of the map from W to Z that makes the diagram commute in the definition of the product: Proposition 6.3.5. Suppose given objects and morphisms π

π

X Y X ←− − Z −−→ Y

π

π

X Y − Z −−→ Y and X ←−

in C such that both (Z, πX , πY ) and (Z , πX , πY ) satisfy the universal mapping property for the product of X and Y in C. Let f = (πX , πY ) be the unique morphism from Z to Z which makes the following diagram commute. (Here, we use the universal mapping π

π

X Y property of X ←− Y .) − Z −−→

Z / ?OO // ??OOOO // ??? fOOOπOY OOO // ?? OOO // '/ / Y Z πX / πY // // // πX X Then f is an isomorphism.

CHAPTER 6. CATEGORIES IN GROUP THEORY π

173

π

X Y Proof By the universal property of X ←− − Z −−→ Y , there is a unique morphism g : Z → Z such that πX ◦ g = πX and πY ◦ g = πY . Now g◦f : Z → Z is the unique morphism such that πX ◦g◦f = πX and πY ◦g◦f = πY . Thus, g ◦ f = 1Z . Similarly, f ◦ g = 1Z , and the result follows.

Exercises 6.3.6. 1. What is the product in G-sets? Is the product of two orbits an orbit? What are its isotropy groups? 2. Suppose that for each pair (X, Y ) of objects of C we are given a product for X and Y . Show that passage from a pair of objects to their product gives a functor from C × C to C.

Coproducts It is frequent in categorical mathematics to ask what happens to a definition if you turn all the arrows around. This is called dualization. The dual of a product is called a coproduct. Definition 6.3.7. Let X, Y ∈ ob(C). A coproduct for X and Y in C is an object Z together with morphisms ιX : X → Z and ιY : Y → Z, called the natural or canonical inclusions, with the following universal property: For any pair of morphisms f : X → W and g : Y → W , there is a unique morphism f, g : Z → W such that the following diagram commutes: Y/ // // ιY // /g ιX / X OO Z ? /// OOO OOO ??h? // OOO ?? // OOO?? / f O' W, Here, h = f, g. Examples 6.3.8. 1. Proposition 2.10.6, when applied strictly to abelian groups and their homomorphisms, shows that the direct product, together with the canonical inclusions, ι

ι

1 2 G × H ←− H, G −→

is the coproduct in the category Ab of abelian groups. 2. The coproduct in Sets is called the disjoint union. Here, the disjoint union of X and Y is denoted X Y . Its elements are the union of those in X and those in Y , where the elements from X are not allowed to coincide with any of the elements of Y . Formally, this may be accomplished as follows. Set theory tells us that there’s some set, S, containing both X and Y . The disjoint union of X and Y is defined to be the subset of S × {1, 2} consisting of the elements (x, 1) with x ∈ X and the elements (y, 2) with y ∈ Y . We then have ιX (x) = (x, 1) and ιY (y) = (y, 2).


174

Now if f : X → W and g : Y → W are functions, we define f, g : X Y → W by setting f, g(x, 1) = f (x) and setting f, g(y, 2) = g(y). Since ιX and ιY are disjoint and have X Y as their union, this is well defined and satisfies the universal property for the coproduct. Thus, unlike the case of the product, the forgetful functor from Ab to Sets does not preserve coproducts. As we shall see presently, the forgetful functors Ab → Gp and Gp → Sets also fail to preserve coproducts. Indeed, the coproducts in these three categories are given by completely distinct constructions. Exercises 6.3.9. 1. What is the coproduct in G-sets? 2. Show that the direct product is not the coproduct in Gp, even if the two groups are abelian. 3. Prove the analogue of Proposition 6.3.5 for coproducts, showing that any two coproducts for a pair of objects are isomorphic via a morphism that makes the appropriate diagram commute.

Free Products The coproduct in Gp is called the free product. We shall give the construction now. Let H and K be groups, and write X = H K for the disjoint union of the sets H and K. By a word in X we mean a formal product x1 x2 · · · xk , with xi ∈ X for all i. (We write for the formal product operation to distinguish it from the multiplications in H and K.) The empty word is the word with no elements. The elements in the free product H ∗ K will not be the words themselves, but rather equivalence classes of words. The equivalence relation we shall use is derived from a process called reduction of words. Words may be reduced as follows. First, if xi and xi+1 are both in the image of ιH : H → X, then x1 · · · xk reduces to x1 · · · (xi xi+1 ) · · · xk , where xi xi+1 is the product of xi and xi+1 in H. There is a similar reduction if xi and xi+1 are both in the image of ιK : K → X. Finally, if xi is the identity element of either H or K, then x1 · · ·xk reduces to x1 · · ·xi−1 xi+1 · · ·xk . In particular, for either e ∈ H or e ∈ K, then the singleton word e reduces to the empty word. e We shall refer to the reductions above as elementary reductions, and write w w or e w w if the word w is obtained from the word w by one of these elementary reductions. e e e w1 ... wn of elementary reductions, we say More generally, given a sequence w0 that w0 reduces to wn , and write w0 wn (or wn w0 ). We shall also declare that w w, so that is reflexive. The equivalence relation we shall use is the equivalence relation generated by the process of reduction, in the sense of Section 1.5. This means that the words w and w are equivalent, written w ∼ w , if there is a sequence w = w0 , . . . , wn = w of words such that for i = 0, . . . , n − 1, either wi wi+1 or wi wi+1 . Note that some words (e.g., h h eK k, where h, h ∈ H, k ∈ K, and eK is the identity of K) can be reduced in more than one way. (Also, as in the case of words containing an adjacent pair such as eK k above, two different elementary reduction operations can produce exactly the same word as a result. However, both cases are e represented by the same symbols, w w .) An examination of words which admit more than one reduction shows the following important fact.


175

Lemma 6.3.10. Suppose given two different elementary reduction operations on a word w, giving w e w1 and w e w2 . Then either w1 = w2 or there is a word w3 with e e w1 w3 w2 . It is convenient to be able to represent an element of the free product by what’s known as a reduced word: Definition 6.3.11. We say that a word x1 · · · xk is reduced if xi = e for 1 ≤ i ≤ k and if no adjacent pair xi , xi+1 comes from the same group (H or K). We also declare the empty word to be reduced. The next lemma is fundamental for the understanding of free products. Lemma 6.3.12. Let w be a word in X = H K. Then there is a unique reduced word, w such that w w . Proof We argue by induction on the number of letters in w. If w is reduced, then it cannot be reduced further, and we take w = w. Otherwise, there is an elementary reduction from w to a word v. By induction, there is a unique reduced word w with v w . But then w w . For the uniqueness e e e w1 ... w . statement, suppose that w is reduced and that w w , say, via w If w1 = v, then w = w by the induction hypothesis. Otherwise, by Lemma 6.3.10, there e e is a word w ) with v w ) w1 . But if w ) w with w reduced, then the uniqueness statement of the induction hypothesis gives w = w = w . Recall that ∼ is the equivalence relation generated by reduction. Proposition 6.3.13. Every equivalence class under ∼ contains exactly one reduced word. If w is a reduced word, then every word in the equivalence class of w under ∼ reduces to w. Proof Suppose that w ∼ w . Then there is a sequence w = w0 , . . . , wn = w such that for each i = 0, . . . , n − 1, either wi wi+1 or wi wi+1 . In either case, Lemma 6.3.12 tells us that the unique reduced word w to which wi reduces is equal to the unique reduced word to which wi+1 reduces. In particular, each wi reduces to w , and there can be at most one reduced word in such a sequence. We define the elements of the free product H ∗ K to be the equivalence classes of words under ∼. Thus, the elements of H ∗ K are in one-to-one correspondence with the reduced words. We define a binary operation on the words themselves by concatenation: (x1 · · · xk ) · (y1 · · · yl ) = x1 · · · xk y1 · · · yl . In word notation, we write w · w for the product in this sense of the words w and w . But clearly, if w reduces to w1 and w reduces to w1 , then w · w reduces to w1 · w1 . Thus, concatenation induces a well defined binary operation on the equivalence classes of words. Lemma 6.3.14. Concatenation of words induces a group structure on the free product H ∗ K. Proof Concatenation is clearly an associative operation on the set of words themselves, with the empty word as an identity element. Since it preserves equivalence classes, the free product inherits a monoid structure from the set of words themselves. But the −1 inverse of x1 · · · xk in H ∗ K is given by x−1 k · · · x1 .


176

The inclusions of H and K in X induce group homomorphisms from H and K to H ∗ K, which we’ll denote by ιH and ιK . Proposition 6.3.15. The free product H ∗ K, together with the homomorphisms ιH and ιK is the coproduct of H and K in the category of groups. Proof Given group homomorphisms f : H → G and g : K → G, we define f, g on X to be f on H ⊂ X and to be g on K ⊂ X. Given a word x1 · · · xk in X, we may then define f, g(x1 · · · xk ) to be (f, g(x1 )) . . . (f, g(xk )). Stipulating that f, g carries the empty word to the identity element of G, we obtain a monoid homomorphism from the monoid of words in X into G. Since two words differing by an elementary reduction are carried to the same element of G (since f and g are homomorphisms), passage to equivalence classes of words gives a group homomorphism from H ∗ K into G. Clearly, f, g ◦ ιH = f and f, g ◦ ιK = g. Since the images of ιH and ιK generate H ∗ K, f, g is the unique homomorphism with this property. Free products can be somewhat intractable, especially in comparison to the groups we have considered previously. But they are important in understanding groups in general. For instance, finitely generated free groups, which we shall study in Section 6.6, are iterated free products of copies of the integers. We shall see that any finitely generated group is a factor group of such a free group, and this opens up a new way of looking at groups, called generators and relations, which we shall study in Section 6.7. A concrete example of free products “in nature” is PSl2 (Z), which turns out to be isomorphic to Z2 ∗ Z3 . It should be noted that not all categories have products, coproducts, or other of the constructions we shall give. Exercises 6.3.16. 1. Show that if H and K are both nontrivial, then H ∗ K is infinite. 2. Show that if H and K are both nontrivial, then H ∗ K is nonabelian even if H and K are both abelian. 3. Show that the homomorphisms ιH : H → H ∗ K and ιK : K → H ∗ K are injective. 4. What is H ∗ e? 5. Show that the same constructions that give the product and coproduct in Gp give a product and coproduct in Mon. 6. Show that the same constructions that give the product and coproduct in Ab give a product and coproduct in Amon.

6.4

Pushouts and Pullbacks

We now give a generalization of products and coproducts. These are universal mapping properties not just of pairs of objects, but of particular collections of objects and morphisms. We treat pullbacks now, and treat pushouts in a separate subsection. In further subsections, we will construct pushouts in Gp, Ab, and Sets.


177

Definition 6.4.1. Suppose given a pair of morphisms into Z, f : X → Z and g : Y → Z. This gives us the following diagram: X .

f g

Y

/ Z

A pullback for this diagram consists of the following data: We have an object W and morphisms g : W → X and f : W → Y such that 1. W f

g

/ X f / Z

Y

g

W

g / X

commutes. 2. For every commutative diagram

f Y

f g

/ Z

there is a unique morphism h : W → W such that the following diagram commutes. W / ?OO // ?? OOOO // ??? OOOgO OOO // ??h? OOO // '/ / W X f / g // // f // f g / Z Y In other words, the pullback is a universal completion of the original 3-object diagram to a commutative square. When the context is totally unambiguous, one often refers to the object W as the pullback, but the proper usage is that “pullback” refers to the whole square. Exercises 6.4.2. 1. Show that pullbacks in Sets, Gp, and Ab may all be constructed as follows. Given a diagram X f Y

g

/ Z

,


178

let Y ×Z X ⊂ Y × X be given by Y ×Z X = {(y, x) | g(y) = f (x)}. Then the pullback of the original diagram is given by Y ×Z X f g Y

g / X f , / Z

where f (y, x) = y and g(y, x) = x. The construction Y ×Z X is sometimes called the fibre product of Y and X over Z. 2. Show that pullbacks, when they exist, are unique. 3. Show that if C has a terminal object, ∗, then the pullback of the unique diagram X / ∗

Y

is the product of X and Y (i.e., show that the pullback of this diagram satisfies the same universal property as the product). Thus, pullbacks do generalize products. 4. Suppose given a pullback diagram W f

g

/ X f

Y

g

/ Z

in either Gp or Ab. Show that ker f ∼ = ker f and that ker g ∼ = ker g. 5. In either Gp or Ab, show that the pullback of X f

e

/ Y

is ker f .

Pushouts Pushout is the notion dual to pullback. In Sets and Gp, pushouts are more delicate and harder to construct than pullbacks, but they can be very important.


179

Definition 6.4.3. The pushout of a diagram X

f

/ Y

f

/ Y

g Z is a commutative diagram X g

g

Z

f

/ W

which is universal in the sense that if f

X

/ Y g

g Z

f / W

is another commutative diagram, then there is a unique morphism h : W → W such that the following diagram commutes. / Y // // g g // // f /g / Z OO W ? // OOO OOO ??h? /// OOO ?? / OOO ??// f OO' W X

f

As in the case of pullbacks, it is customary to refer to the object W as the pushout as well. Exercises 6.4.4. 1. Show that pushouts, when they exist, are unique. 2. Show that if C has an initial object, ∗, then the pushout of the unique diagram ∗ Z is the coproduct of Y and Z.

/ Y


180

3. Suppose given a pushout diagram f

X

/ Y

g

g

Z

f

/ W

in C, where f is an isomorphism. Deduce that f is also an isomorphism. 4. Show, using the universal mapping property, that in Sets, Gp, or Ab, the pushout of a product diagram πY / X ×Y Y πX X is the terminal object. (In Sets, we must assume that neither X nor Y is the null set.)

Pushouts in Gp and Ab Pushouts are closely related to coproducts, and are often constructed using them. We first give the construction of pushouts in Ab. Proposition 6.4.5. The pushout in Ab of a diagram f / A B g C is the diagram f

A g C

/ B ιB

ιC / (C × B)/D,

where D = im(g, −f ) = {(g(a), −f (a)) | a ∈ A} ⊂ C × B, and where the maps ι are the composites of the standard inclusions of B and C in the product with the canonical map from C × B to its factor group. Proof First, note that factoring out by D makes the square commute. Now given a commutative square f / A B g

g C

f

/ G


181

in Ab, there is a unique homomorphism h from the coproduct C × B into G such that h ◦ ιB = g and h ◦ ιC = f . But then h((g(a), −f (a)) = f g(a) − g f (a) = 0 by commutativity of the diagram. Thus, h factors through the factor group of C × B by D. But factorizations of homomorphisms through factor groups are unique. In order to understand pushouts in the category of groups, we will need the following concept. Definition 6.4.6. Let G be a group and let S be any subset of G. We write S ⊂ G for the subgroup generated by T = {gxg −1 | g ∈ G, x ∈ S}, the set of all conjugates of the elements of S. We call S the normal subgroup generated by S. As usual, if S = {g1 , . . . , gn }, we write S = g1 , . . . , gn . From a different viewpoint, if S is contained in the subgroup H G and if the subgroup S = T above equals H, we say that H is normally generated by S in G. Lemma 6.4.7. Let S be a subset of the group G. Then S is the smallest normal subgroup of G that contains S. Proof Clearly, any normal subgroup containing S must contain T , and hence must contain the subgroup generated by T . Thus, it suffices to show that S is normal in G. A generic element of S is a product of powers of elements of T . Since any conjugate of an element of T is also in T , the result follows. The pushout in Gp is called the free product with amalgamation. The construction proceeds precisely as in the abelian case. We give it as an exercise. We shall see below that every group may be presented in terms of “generators and relations,” which is an example of a free product with amalgamation. Exercises 6.4.8. 1. Show that the pushout in Gp of the diagram f

G

/ H

g K is the diagram G g K

f

/ H ιH

ιK / (K ∗ H)/S,

where S = {g(x) · f (x)−1 | x ∈ G}, and where the maps ι are the composites of the standard inclusions of H and K in the free product with the canonical map from K ∗ H to its factor group.


182

2. What is the pushout in Gp of a diagram of the following form? f / H G e What would the pushout be in Ab if the diagram were also in Ab? 3. Suppose given a pushout diagram A

f

g C

/ B g

f

/ D

in Ab. (a) If f is injective, show that f is also. (b) Show that f is surjective if and only if f is. 4. Suppose the diagram of the previous problem is a pushout square in Gp rather than Ab. (a) If f is surjective, show that f is also. (b) Show that f may be surjective when f is not.

Pushouts in Sets In Sets, the pushout is again a quotient of the coproduct. This time we proceed by placing an equivalence relation on the coproduct and passing to the set of equivalence classes. Thus, suppose given a diagram f / X Y g Z in Sets. Our equivalence relation on the coproduct Y Z will be the equivalence relation, ∼, generated by the following notion of basic equivalence: There is a basic equivalence between ιY (y) with ιZ (z) in Y Z if there is an element x ∈ X with f (x) = y and g(x) = z. In this case, we write either ιY (y) ιZ (z) or ιZ (z) ιY (y), as the relation in question is symmetric. Since ∼ is the equivalence relation generated by , we see that two elements v, v ∈ Y Z are equivalent under ∼ if and only if there is a sequence v = v0 v1 . . . vn = v .

We define the pushout object Pto be the set of equivalence classes in Y Z under ∼. We then have a function π : Y Z → P that takes each element to its equivalence class.


183

Exercises 6.4.9. 1. Show that the pushout in sets of the above diagram is f / X Y g Z

ιY

ιZ / P,

where the functions ι are the composites of the standard inclusions into the coproduct with the quotient function π. 2. If f is injective, show that ιZ is also. (In this case, P is often denoted Z ∪g Y .) 3. If f is surjective, show that ιZ is also.

6.5

Infinite Products and Coproducts

We’ve discussed products and coproducts of pairs of objects in a category. This is sufficient, by induction, to provide products or coproducts for any finite collection of objects. But in some cases, as we shall see in our discussion of rings and modules, it is necessary to make use of products or coproducts of infinite families of elements. We define infinite products and coproducts here and construct the products. We give constructions of infinite coproducts in a separate subsection. We shall consider sets of objects {Xi | i ∈ I}. In other words, I is a set, called the indexing set, and Xi is an object of our category for each i ∈ I. Definitions 6.5.1. The product of a set {Xi | i ∈ I} of objects of C is an object X, together with morphisms πi : X → Xi , called projections, with the following universal property. If fi : Y → Xi is a morphism in C for each i ∈ I, then there is a unique morphism f : Y → X such that πi ◦ f = fi for all i ∈ I. The coproduct of a set {Xi | i ∈ I} of objects of C is an object X, together with morphisms ιi : Xi → X for each i ∈ I, called the natural inclusions, with the following universal property. If fi : Xi → Y is a morphism in C for each i ∈ I, then there is a unique morphism f : X → Y such that f ◦ ιi = fi for all i ∈ I. Once again, the products in Sets, Gp, and Ab come from the same construction. We shall see in our discussion of adjoint functors in Section 6.9 that this is no coincidence. ! The construction of products in Sets is as follows. The elements of the product i∈I Xi are collections (xi | i ∈ I) such that xi ∈ Xi for all i ∈ I. In other words, the element (xi | i ∈ I) consists of a choice, for each i ∈ I, of an element of Xi . We call (xi | i ∈ I) a tuple or an I-tuple. Another way of thinking about the elements of a product is as follows. Let S be a set that contains each Xi . (The ! existence of such a set is one of the standard axioms of set theory.) Then an I-tuple in i∈I Xi is a function from I to S whose value on i ∈ I must lie in Xi . From this viewpoint, we ! can see that the collection of all such I-tuples, ! i∈I Xi , is a set. The projection πi : i∈I Xi → Xi takes (xi | i ∈ I) to xi . The act of choosing elements xi ∈ Xi for each i ∈ I suggests that the Axiom of Choice is involved here. This is indeed the case. Axiom of Choice (second form) Let {Xi | i ∈ I} be a family of nonempty sets. Then ! the product i∈I Xi is nonempty.


184

We shall discuss the relationship between the two given forms of the Axiom of Choice in Exercises 6.5.11. The following argument does not make ! use of the Axiom of Choice. The output, if one does not accept that axiom, is that if i∈I Xi = ∅, then ∅ is the product of {Xi | i ∈ I} in Sets. In particular, any infinite family of sets has a product in Sets. ! Proposition 6.5.2. The product i∈I Xi , together with its projection maps, is the product of {Xi | i ∈ I} in Sets. Proof Suppose given a collection of functions fi : Y → Xi for each i ∈ I. Define ! f : Y → i∈I Xi by f (y) = (fi (y) | i ∈ I). Then πi ◦ f = fi for all i ∈ I. But two elements of the product are equal if their coordinates are all equal, so f is the unique function with this property. Exercises 6.5.3. 1. Show that the category of finite groups and homomorphisms does not have infinite products. ! 2. Show that if i∈I Xi is the product of {Xi | i ∈ I} in an arbitrary category!C, then for each ! Y ∈ ob(C) there is an isomorphism of sets between MorC (Y, i∈I Xi ) and i∈I MorC (Y, Xi ), where the latter product is that in Sets. (In fact, this isomorphism is natural, in the sense of natural transformations, Section 6.9.) 3. Suppose given objects {Gi | i ∈ I} in either Gp or Ab. Define a!binary operation ! on i∈I Gi by (xi | i ∈ I) · (yi | i ∈ I) = (xi yi | i ∈ I). Show that i∈I Gi , with this multiplication, is the product of {Gi | i ∈ I} in the category in question.

Constructions of Infinite Coproducts The infinite coproduct in Sets is the infinite disjoint union.

Definition 6.5.4. Let {Xi | i ∈ I} be a family of sets. We define i∈I Xi as follows. Let S be a set that contains each Xi . Then i∈I Xi is the subset of S × I consisting of the pairs (x, i) such that x ∈ Xi . The inclusion ιi : Xi → i∈I Xi takes x ∈ Xi to the pair (x, i). Proposition 6.5.5. The disjoint union, together with the structure maps ιi , is the coproduct in the category of sets. Thus, given a family {Xi | i ∈ I} ofsets, together with functions fi : Xi → Y for all i ∈ I, there is a unique function f : i∈I Xi → Y such that f ◦ ιi = fi for all i. Proof Given the functions fi : Xi → Y , define f : i∈I Xi → Y by f (x, i) = fi (x) for each pair (x, i) ∈ i∈I Xi . Since (x, i) = ιi (x), f satisfies the requirement that f ◦ ιi = fi for all i, and is the unique function that does so. The infinite coproduct in Gp is called the infinite free product. It is constructed by precise analogy to the free product of two groups. Thus, given a collection {Gi | i ∈ I} of groups, we set X = i∈I Gi , the disjoint union of the sets Gi . The infinite free product i∈I Gi is obtained by a passage to equivalence classes from an equivalence relation, ∼, on the set of all finite words in X. Here, as in the case of the free product of two groups, the equivalence relation ∼ is the equivalence relation generated by a notion of elementary reductions. The elementary reductions are defined as follows.

∗


Definition 6.5.6. Let X =

i∈I

185

Gi and let x1 · · · xn be a word in X.

1. If xi is the identity in the group Gj that contains it, then e x1 · · · xn x1 · · · xi−1 xi+1 · · · xn .

2. If xi and xi+1 both lie in the same group Gj ⊂ X, then e x1 · · · xn x1 · · · (xi xi+1 ) · · · xn .

∗

As in the case of the free product of two groups, i∈I Gi is a group under the operation induced by concatenation of words. The canonical map ιi : Gi → i∈I Gi takes g ∈ Gi to the equivalence class of the singleton word g. The proof of the following proposition proceeds precisely as did that of Proposition 6.3.15.

∗

Proposition 6.5.7. Let {Gi | i ∈ I} be any family of groups. Then the infinite free product i∈I Gi , together with the canonical maps ιi : Gi → i∈I Gi , is the coproduct of {Gi | i ∈ I} in the category of groups.

∗

∗

Intuition might now suggest that the infinite coproduct in Ab would coincide with the infinite product. However, this is not the case (mainly because we cannot add an infinite number of group elements). Rather, the coproduct is a subgroup of the infinite product, called the direct sum. Definition 6.5.8. Let {Ai | i ∈ I}!be a family of abelian groups. We define the direct * sum i∈I Ai to be the subset of i∈I Ai consisting of * the tuples (ai | i ∈ I) for which ai = 0 for all but finitely many i. We define ιi : Ai → i∈I Ai as follows: For a ∈ Ai , the coordinates of ιi (a) are all 0except for the i-th coordinate, which *is a. We shall customarily write a for the I-tuple (a | i ∈ I) ∈ i i∈I i i∈I Ai . Note that the sum i∈I ai is actually finite, as all but finitely many of the ai are 0. * ! i∈I Ai is a subgroup of the product i∈I Ai , and the maps ιi : Ai → * Clearly, i∈I Ai are homomorphisms. The next lemma is immediate from the definitions. Lemma 6.5.9. Let {Ai | i ∈ I} be a family of abelian groups. Then in the direct sum * A , we have i i∈I ai = ιi (ai ), i∈I

i∈I

where the sum on the right is the sum, with respect to the group operation in of the finitely many terms ιi (ai ) for which ai = 0.

* i∈I

Ai ,

Proposition 6.5.10. Let {Ai | i ∈ I} be a family of abelian * * groups. Then the direct sum, A , together with the structure maps ι : A → i i i∈I i i∈I Ai , is the coproduct in the category of abelian groups. In other words, if B is an abelian group and if fi * : Ai → B is a homomorphism for each i ∈ I, then there is a unique homomorphism f : i∈I Ai → B such that f ◦ ιi = fi for all i ∈ I. Explicitly, the homomorphism f is given by " # f ai = fi (ai ), i∈I

i∈I

where the right-hand side is the sum in B of the finite number of terms fi (ai ) for which ai = 0.


186

Proof That f must satisfy the stated formula is immediate from Lemma 6.5.9 and the requirement that f ◦ ιi = fi for all i ∈ I. Moreover, the definition of ιi shows that if * f : i∈I → B is defined by the stated formula, = fi for all i, as required. then f ◦ ιi = a Thus, it suffices to show that setting f i∈I i i∈I fi (ai ) defines a homo* A to B. But that is immediate from the fact that each fi is a morphism from i∈I i homomorphism. Coproducts are so important in the study of abelian groups that it is often customary to refer to the finite products as direct sums. Thus, A1 ⊕· · ·⊕Ak is an alternative notation for A1 × · · · × Ak . Exercises 6.5.11. 1. Suppose given a family {Xi | i ∈ I} of sets. Define π : i∈I Xi → I by setting π ◦ ιi equal to the constant map from Xi to i, for each i ∈ I. Show that the ! elements of X are in one-to-one correspondence with the functions s : I → i i∈I i∈I Xi such that π ◦ s = 1I . Deduce that the two forms of the Axiom of Choice are equivalent. * ! 2. Show that the direct sum i∈I Ai is the subgroup of the product i∈I Ai generated by i∈I ιi (Ai ). 3. What are the infinite coproducts in Mon and Amon? 4. For any category C, show that if X is the coproduct of {Xi | i ∈ I}, ! then for any object Y , there is an isomorphism of sets between MorC (X, Y ) and i∈I MorC (Xi , Y ).

6.6

Free Functors

Free functors are actually a special case of the adjoint functors discussed in Section 6.9. For simplicity, and because they constitute a very important special case, we give a separate treatment of them. We give an illustration first. Let X be a set. Since abelian groups are sets with additional structure, we can consider the functions f : X → A, where A is an abelian group. The free abelian group on X is an abelian group, F A(X), together with a function ηX : X → F A(X) which is universal in the sense that if f : X → A is any function from X to an abelian group, then there is a unique group homomorphism f : F A(X) → A such that the following diagram commutes. f /A XG ; GG xx x GG x GG x xx ηX GG xx f # F A(X)

As usual, we shall refer to F A(X) as the free abelian group on X, even though that name more properly applies to the function ηX . We shall give a construction of free abelian groups below. The following proposition shows how useful they can be. Proposition 6.6.1. Let f : A → F A(X) be an extension of the abelian group B = ker f by the free abelian group F A(X). Suppose that A is also abelian. Then the extension splits, and A is isomorphic to B × F A(X).


187

Proof We shall see below that the function ηX : X → F A(X) is always injective. For simplicity, we shall make use of this fact, and treat X as a subset of F A(X). Group extensions are always onto, so that X is contained in the image of f . Thus, for each x ∈ X there is an element g(x) ∈ A such that f (g(x)) = x. But this defines a function g : X → A. Since A is abelian, the universal property of free abelian groups provides a homomorphism g : F A(X) → A such that g(x) = g(x) for all x ∈ X. We claim that g is a section of f . Thus, we wish to show that f ◦ g is the identity map of F A(X). But the universal property shows that homomorphisms out of F A(X) are determined by their restriction to X. And f ◦ g(x) = f (g(x)) = x. Thus, f ◦ g has the same effect on X as the identity map, so that f ◦ g is the identity as desired. Thus, the extension splits. Since A is abelian, F A(X) acts trivially on B = ker f , and hence A is the product of B and F A(X). We define free functors in general as follows. (More general definitions are possible.) Definition 6.6.2. Suppose given a category C with a forgetful functor to Sets. (In other words, the objects of C are sets with additional structure, and passage from the objects to the underlying sets and from morphisms to the underlying functions provides a functor φ : C → Sets.) Then a free C-object on a set X consists of an object F (X) ∈ ob(C) together with a function ηX : X → F (X) from X to the underlying set of F (X) with the universal property that if f : X → A is any function from X to a C-object, then there is a unique C-morphism f : F (X) → A such that the following diagram commutes. /A XE z< EE z EE zz E zzf ηX EE z " z F (X) f

The free objects we’re most interested in will be very easy to construct. The key is the following lemma. Lemma 6.6.3. Let C be a category with a forgetful functor to Sets. Suppose given a family of sets {Xi | i ∈ I} such that 1. Each set Xi has a free C-object ηi : Xi → F (Xi ). 2. The collection {F (Xi ) | i ∈ I} has a coproduct in C. * Write i∈I F (Xi ) for the coproduct in C of the F (Xi ) and write η : i∈I Xi → * F (X ) for the unique function that makes the following diagram commute for each i i∈I i. ηi / F (Xi ) Xi ιi ιi , + η / Xi F (Xi ) i∈I

i∈I

Here, the maps ιi are the canonical maps for the coproducts, in Sets on the left and in C on the right. Then η is the free C-object on i∈I Xi .


Proof Suppose given a function f : write fi : Xi → A for the composite

i∈I

ι

i Xi −→

188

Xi → A, with A and object of C. For i ∈ I,

+

f

Xi − → A.

i∈I

Then since ηi is the free C-object on Xi , there is a unique C-morphism f i : F (Xi ) → A such that f i ◦ ηi = fi . Now apply * the universal property of the coproduct in C, obtaining the unique morphism f from i∈I F (Xi ) to A such that f ◦ ιi = f i . An easy diagram chase shows that f ◦ η = f , and uniqueness follows by composition of the uniqueness property of the coproduct in C and the uniqueness properties of the free C-objects ηi . For any set X, we have X = x∈X {x}. Thus, if C has arbitrary coproducts, it will have arbitrary free objects, provided that a one-point set has a free object. But we have seen free objects for the one-point set already in Gp and Ab: if G is a group and if g ∈ G, there is a unique homomorphism f : Z → G with f (1) = g. Thus, if η : {x} → Z takes x to 1, then η is the free object on {x} in both Gp and Ab. (One-point sets are the only sets for which free groups are abelian, and therefore are the only sets for which the free groups and the free abelian groups coincide.) Proposition 6.6.4. A group is a free abelian group if and only if it is a direct sum of copies of Z. A group is a free group if and only if it is a free product of copies of Z. * Proof By the preceding lemma, x∈X Z is the free abelian group on X, while x∈X Z is the free group on X. But given a direct sum or free product of copies of Z, just let X be the indexing set for the sum or free product, and this says that our sum or free product is free in the appropriate category.

∗

We can turn free objects into free functors as follows. The proof follows from the universal property and is left to the reader. Proposition 6.6.5. Let C be a category with a forgetful functor to Sets. Suppose that for every set X we are given a free C-object ηX : X → F (X). Then for any function f : X → Y there is a unique C-morphism F (f ) : F (X) → F (Y ) such that F (f ) ◦ ιX = ιY ◦ f . The correspondence that takes X to F (X) and takes f to F (f ) is a functor F : Sets → C. In practice, we identify X with its image under ιX in any of the categories under discussion. Thus, the free group on X is given by x∈X x, where each x ∈ X has infinite order, and the generic element may be written xn1 1 . . . xnk k where x1 , . . . , xk ∈ X are not necessarily distinct, and n1 , . . . , nk ∈ Z. There is a similar additive notation for the free abelian group on X. Its elements may be written in the form n1 x1 + · · · + nk xk , where x1 , . . . , xk ∈ X and n1 , . . . , nk ∈ Z. In this case, because the group is abelian, we may collect terms, and assume that x1 , . . . , xk are distinct.

∗

Exercises 6.6.6.


189

1. Show that if f : G → F (X) is an extension of a group H by the free group on X, then f is a split extension. 2. Show that there exist group extensions f : G → F A(X) which cannot be split, where F A(X) is a free abelian group. 3. Let f : F (X) → G be a homomorphism, with F (X) the free group on X. Show that f is onto if and only if G is generated by {f (x) | x ∈ X}. 4. Let X be a set and let F (X) and F A(X) be the free group and the free abelian group on X, respectively. Let f : F (X) → F A(X) be the unique homomorphism such that the following diagram commutes. X? ?? η η ?? ?? ? f / F (X) F A(X) Here, the maps η are the canonical inclusions of X in the free group and free abelian group, respectively. ab Show that f factors uniquely through an isomorphism f : F (X) ∼ = F A(X) from the abelianization of F (X) onto F A(X). 5. Let f : F A(X) → A be a homomorphism, with A an abelian group and with F A(X) the free abelian group on X. Show that f is onto if and only if G is generated by {f (x) | x ∈ X}. 6. What is the free G-set on a set X? 7. What is the free monoid on a set X? 8. What is the free abelian monoid on a set X?

6.7

Generators and Relations

Let G be a group and suppose that the set X ⊂ G generates G. Then there is a unique homomorphism f : F (X) → G from the free group on X whose restriction to X is its inclusion in G. By Problem 3 of Exercises 6.6.6, f is surjective. In this way, we see that every group is a factor group of a free group. But that, by itself, isn’t very useful information. We need to be able to say something about the kernel, K, of f : F (X) → G before we can get any useful information out of it. Free groups are very complicated objects, and we shall not attempt to say anything about the case where X is infinite. Thus, we shall assume that G is finitely generated, by the set X = {x1 , . . . , xk }. We shall write F (X) = F (x1 , . . . , xk ). It happens that subgroups of free groups are always free groups.2 But even if the free group is finitely generated, if it is free on more than one generator, then it turns out to have many subgroups that are not finitely generated. But some infinitely generated subgroups of a free group turn out to be normally generated by a finite set. 2 Our favorite proof of this uses the topology of covering spaces of graphs. See Section 3.8 of E.H. Spanier’s Algebraic Topology, McGraw–Hill, 1966.


190

Definitions 6.7.1. A finite presentation for a group G consists of a finite subset X = {x1 , . . . , xk } ⊂ G which generates G, together with a finite subset R = {R1 , . . . , Rl } of the free group F (x1 , . . . , xk ) on x1 , . . . , xk , such that the kernel of the homomorphism f : F (x1 , . . . , xk ) → G that’s induced by {x1 , . . . , xn } ⊂ G is normally generated by R in F (x1 , . . . , xk ). Given the above presentation for G, we say that G has generators X and relations R. In general, given variables x1 , . . . , xn and elements R1 , . . . , Rm of the free group F (x1 , . . . , xn ), we write x1 , . . . , xn | R1 , . . . , Rm for the group with generators x1 , . . . , xn and relations R1 , . . . , Rm . Thus, x1 , . . . , xn | R1 , . . . , Rm = F (x1 , . . . , xn )/R1 , . . . , Rm . If a group G admits a finite presentation, then we say that G is finitely presented. Note that a finite presentation of G is the construction of a pushout diagram F (r1 , . . . , rm )

/ F (x1 , . . . , xn )

e

/ G.

It turns out that not every finitely generated group is finitely presented. Thus, subgroups of finitely generated free groups can be very complicated indeed. One can show, via the theory of covering spaces, that the commutator subgroup of the free group, F (x, y), on two letters is infinitely generated. But recall that the ab abelianization, F (x, y) , is naturally isomorphic to F A(x, y), the free abelian group of x and y. Thus, the next example shows that the commutator subgroup of F (x, y) is normally generated by a single element, xyx−1 y −1 . Example 6.7.2. We show here that the free abelian group on two letters has the presentation F A(x, y) = x, y | xyx−1 y −1 . First note that since F A(x, y) is abelian, xyx−1 y −1 is in the kernel of the map π : F (x, y) → F A(x, y) which takes x to x and y to y. Thus, there is an induced map π : G → F A(x, y), where G = x, y | xyx−1 y −1 = F (x, y)/xyx−1 y −1 . Note that x and y commute in G. Thus, since x and y generate G, G is abelian. Thus, the universal property of the free abelian group provides a homomorphism ϕ : F A(x, y) → G with ϕ(x) = x and ϕ(y) = y. Now notice that the composites ϕ ◦ π and π ◦ ϕ restrict to the identity map on generators for the two groups. Thus, ϕ and π are inverse isomorphisms. The hard part in showing that a given group H is given by a particular presentation is generally in producing a homomorphism from H to the stated quotient of the free group on the generators. In the above example, the homomorphism is obtained via the universal property for the free abelian group. Exercises 6.7.3. 1. Give a finite presentation for the free abelian group on n generators.


191

2. Show that Zn = x | xn . 3. Show that D2n = a, b | a2 , bn , aba−1 b. 4. Show that Q4n = a, b | b2n , a2 bn , aba−1 b. 5. Find a presentation for A4 . 6. Show that x, y | x2 , y 2 , (xy)3 is a presentation for S3 . 7. Show that x, y | x2 , y 2 , (xy)n is a presentation for D2n for each n ≥ 2. 8. Show that x, y, z | x2 , y 2 , z 2 , (xy)3 , (yz)3 , xzx−1 z −1 is a presentation for S4 , via setting x = (1 2), y = (2 3), and z = (3 4). 9. Let G = x, y | xn , y m , xyx−1 y k for integers n, m, k with m, n > 0. Show that G is finite and that y G. Do x and y necessarily have order n and m, respectively? 10. Let G = x, y | x4 , y 4 , (xy)4 . Show that Q8 and Z4 × Z4 are both factor groups of G. What extra relations, when added to those of G, will give Q8 ? What extra relations, when added to those of G, will give Z4 × Z4 ? Do any of the relations originally in G become redundant when we do this? 11. Show that x, y | xyx−1 y −2 is a presentation for the group Z[ 12 ]μ2 Z of Problem 19 of Exercises 4.6.13. 12. Given finite presentations for H and K and a homomorphism α : K → Aut(H), derive a finite presentation for H α K.

6.8

Direct and Inverse Limits

We consider yet another pair of examples of universal properties that replace one diagram with another. We treat inverse limits here and treat direct limits in a separate subsection. Inverse limits are related to pullbacks. We suppose given objects Xi for i ≥ 0 and morphisms pi : Xi → Xi−1 for i ≥ 1. (The indices i here are integers.) We call this data an inverse system. Pictorially, this gives / X 3 p3 / X 2 p2 / X 1 p1 / X 0 . ··· The inverse limit of this system can be thought of as a universal object to tack on at −∞. What this means is that the inverse limit is a universal diagram of the following sort. X JTTTT JJ TTTT JJ TTT TTTT JJ $ TT* / ··· X1 p / X0 . 1 Definition 6.8.1. Suppose given an inverse system, consisting of objects Xi for i ≥ 0 and morphisms pi : Xi → Xi−1 for i ≥ 1. An inverse limit for this system consists of an object X = ← lim −Xi , together with morphisms πi : X → Xi for i ≥ 0 such that pi ◦ πi = πi−1 for i ≥ 1, and the following universal property is satisfied: For each object Y and each collection of morphisms fi : Y → Xi such that pi ◦ fi = fi−1 for all i ≥ 1, there is a unique morphism f : Y → X such that πi ◦ f = fi for all i ≥ 0.


192

We give the construction of inverse limits in Sets as follows. Suppose given an inverse system of sets: ···

/ X3

p3 / X2

p2 / X1

p1 / X0 .

! Let X be the subset of the infinite product i≥0 Xi consisting of those tuples (xi | i ≥ 0) with the property that pi (xi ) = xi−1 for all i ≥ 1, and let πi : X → Xi be the restriction to X of the i-th projection map of the product (which we shall also write as ! πi : i≥0 Xi → Xi ). Proposition 6.8.2. Using the notations just given, the set X, together with the maps πi : X → Xi , constitutes the inverse limit of the stated inverse system in Sets. Proof Suppose given functions fi : Y → Xi , i ≥ 0, such that p!i ◦ fi = fi−1 for i ≥ 1. Then the universal property of products gives a function f : Y → i≥0 Xi with πi ◦f = fi for all i. As tuples, this says f (y) = (fi (y) | i ≥ 0) for all y ∈ Y . But since pi ◦ fi = fi−1 , we have f (y) ∈ lim ←− Xi for all y ∈ Y . In other words, f takes values in the inverse limit. By construction, πi ◦ f = fi . Moreover, f is the unique map with this property because X is a subset of the product, and a map into the product is determined by its coordinate functions. We now give a useful way to interpret the construction ! of inverse limits. We shall write (. . . , xi , . . . , x1 , x0 ) for an element of the product i≥0 Xi , reversing the usual order. ! ! Then there is a “shift” map s : i≥0 Xi → i≥0 Xi given by s(. . . , xi , . . . , x1 , x0 ) = (. . . , pi+1 (xi+1 ), . . . , p2 (x2 ), p1 (x1 )). It is easy to see that the inverse limit lim ←− Xi is the ! set of all x ∈ i≥0 Xi such that s(x) = x. In other words, the inverse limit is the set of ! fixed points of the shift map on i≥0 Xi . As was the case for products and pullbacks, the construction of inverse limit that works in Sets also produces the inverse limit in most of the other categories we’ve been considering. Proposition 6.8.3. Suppose given an inverse system ···

/ G3

h3 / G2

h2 / G1

h1 / G0

! in the category of groups, and let G ⊂ i≥0 Gi be the inverse limit of the induced inverse system of sets (i.e., G = {(gi | i ≥ 0) | hi (gi ) = gi−1 for all i ≥ 0}). Then G is a subgroup of the direct product of the Gi , and ! with this multiplication, together with the restrictions to G of the projection maps πi : i≥0 Gi → Gi , forms the inverse limit of this system in the category of groups. In addition, if the groups Gi are abelian, so is G, and in this case, G, together with the projection maps is the inverse limit of the given system in the category of abelian groups. Proof That G is a subgroup of the direct product follows from the fact that the hi are group homomorphisms. Since the direct product is the product in the category of groups, G, together with the projection maps, satisfies the universal property for the inverse limit by exactly the same proof as that given for the category of sets. A similar argument works for abelian groups.


193

We shall see in Chapter 7 that the same construction gives the inverse limit in the p , in some category of rings. We shall study one such example, the p-adic integers, Z depth. Zp is the inverse limit of the unique system of ring homomorphisms ···

/ Zp3

/ Zp2

/ Zp .

Exercises 6.8.4. 1. Show that if each of the functions pi : Xi → Xi−1 in an inverse system of sets is lim surjective, then so is each function πi : ← − Xi → Xi . 2. Show that if each of the functions pi : Xi → Xi−1 in an inverse system of sets is bijective, then so is each function πi : ← lim − Xi → Xi . 3. Show that the construction given for Sets, Gp, and Ab also produces the inverse limit in G-sets, Mon, and Amon. 4. Show that if X = ← lim − Xi is the inverse limit for an inverse system consisting of objects Xi , i ≥ 0, and morphisms pi : Xi → Xi−1 in a category C, then for any object Y of C, MorC (Y, X) is the inverse limit in Sets of the MorC (Y, Xi ).

Direct Limits Direct limit is the concept dual to inverse limit: Definition 6.8.5. A direct system in C consists of objects Xi for i ≥ 0, together with morphisms ji : Xi → Xi+1 for i ≥ 0. Pictorially, we get X0

j0 / X1

j1 / X2

j2 / X3

/ ···

A direct limit for this system consists of an object X = − lim → Xi , together with morphisms ιi : Xi → X for i ≥ 0, such that ιi+1 ◦ ji = ιi for i ≥ 0 and such that the following universal property is satisfied. Given an object Y and morphisms fi : Xi → Y such that lim fi+1 ◦ji = fi for all i ≥ 0, there is a unique morphism f : − → Xi → Y such that f ◦ιi = fi for all i ≥ 0. Surprisingly (since the direct limit is related to coproducts and pushouts in the same way that the inverse limit is related to products and pullbacks), the construction of the direct limit is the same in all the categories we’ve been discussing. The intuition is that if the morphisms ji are all injective, then the direct limit is the union of the Xi . (See Problem 6 of Exercises 6.8.8.) If the ji are not injective, we need to work a little harder to construct the direct limit. Assume that we’re working in Sets. We shall construct − lim by passage to equiva→X i lence classes from an equivalence relation on the disjoint union i≥0 Xi . Here, we use the equivalence relation generated by the following notion of elementary equivalence: if x ∈ Xi , we set x equivalent to ji (x) ∈ Xi+1 . We write x ∼e ji (x). We then declare two elements x, x ∈ i≥0 Xi to be equivalent, written x ∼ x , if there is a sequence x = x0 , . . . , xn = x such that for i = 0, . . . , n − 1, either xi ∼e xi+1 or xi+1 ∼e xi . We define − lim → Xi to be the set of equivalence classes under ∼, and write π : i≥0 Xi → lim −→ Xi are defined to be the −→ Xi for the quotient map. The structure maps ιi : Xi → lim


194

composites Xi where ιi : Xi →

i≥0

ιi /

+

Xi

i≥0

π / lim X , −→ i

Xi is the natural inclusion of Xi in the coproduct.

Proposition 6.8.6. The above construction gives the direct limit in Sets. Proof Given functions fi : Xi → Y with fi+1 ◦ ji = fi for all i ≥ 0, there is a unique function f from the coproduct i≥0 Xi to Y such that f ◦ ιi = fi for all i, where ιi is the natural inclusion of Xi in the coproduct. Write ≈ for the equivalence relation on i≥0 Xi given by setting x ≈ x if f (x) = f (x ). Since fi+1 ◦ ji = fi for all i ≥ 0, we see that x ∼e x implies that x ≈ x . By Proposition 1.5.3, if x ∼ x , then f (x) must equal f (x ). Thus, f factors (necessarily uniquely, since π is onto) through a function f : − lim → Xi → Y . But now f clearly has the desired properties. Example 6.8.7. Let p be a prime. We shall construct a group by taking direct limits of cyclic p-groups. Specifically, note the equivalence class of p mod pi has order pi−1 in the additive group Zpi . Thus, there is a group homomorphism ji−1 : Zpi−1 → Zpi which takes 1 to p: explicitly, ji−1 (n) = pn. Since p has order pi−1 , ji−1 is injective. Clearly, the maps {ji | i ≥ 1} form a direct system in Ab. We write Tp for the direct limit of this system, which by Problem 4 below is an abelian group. Exercises 6.8.8. 1. For any direct system of sets, show that any element of the direct limit lies in the image of ιi for some i. 2. In Sets, show that if x, y ∈ Xi are equivalent in − lim → Xi , then there exists k ≥ 0 such that ji+k ◦ · · · ◦ ji (x) = ji+k ◦ · · · ◦ ji (y). 3. Suppose given a direct system in Gp. Show that the direct limit of the underlying sets has a unique group structure that makes it the direct limit of this system in Gp. † 4. Repeat the preceding problem in Ab, G-sets, Mon, and Amon. 5. In Sets, suppose there is a set U such that Xi ⊂ U for all i ≥ 0. Suppose also that each composite ji

Xi −→ Xi+1 ⊂ U coincides with the inclusion Xi ⊂ U . Show that we may identify the direct limit lim X with the union X i i ⊂ U. i≥0 −→ † 6. More generally, suppose given a direct system of sets such that each function ji : Xi → Xi+1 is injective. Show that the structure maps ιi : Xi → lim −→ Xi are also injective. In addition, show that − lim X is the union of the images of the Xi . → i 7. Let Tp , as defined above, be the direct limit of the cyclic groups Zpi and let ιi : Zpi → Tp be the structure map for the direct limit. By Problem 6, ιi is an injection. Show that Tp is a p-torsion group and that the image of ιi the subgroup consisting of those elements of exponent pi .


195

8. Let p be a prime and let Z[ p1 ] ⊂ Q be the subgroup of the rational numbers consisting of all quotients of the form pmi , where m ∈ Z and i ≥ 0. Show that Z[ p1 ] is isomorphic to the direct limit of the system p

p

Z → Z → Z → ... where each group in the direct system is Z and each map is multiplication by p. 9. Show that the torsion group Tp above is isomorphic to the factor group Z[ p1 ]/Z. 10. Show that any countable abelian torsion group is a direct limit of finite abelian groups. 11. Show that if G is finite, then a countable G-set is a direct limit of finite G-sets. 12. Show that if X = − lim → Xi is the direct limit for a direct system consisting of objects Xi , i ≥ 0, and morphisms ji : Xi → Xi+1 in a category C, then for any object Y of C, MorC (X, Y ) is the inverse limit in Sets of the MorC (Xi , Y ).

6.9

Natural Transformations and Adjoints

In our study of semidirect products, we found that it was interesting to be able to determine when two homomorphisms are related in certain ways (e.g., conjugation). Similarly, it is often of interest to relate the effect of two different functors. Definition 6.9.1. Let C and D be categories, and let F and G be two functors from C to D. A natural transformation η : F → G consists of a D-morphism ηX : F (X) → G(X) for each X ∈ ob(C) such that for each f ∈ MorC (X, Y ), the following diagram commutes. F (f ) / F (Y ) F (X) ηX ηY G(f ) / G(Y ) G(X) We say that η is a natural isomorphism if ηX is an isomorphism in D for each object X of C. With straightforward modifications, we obtain the definitions of natural transformations or isomorphisms between two contravariant functors. The value of naturality is not immediately apparent. However, it turns out to be a strong tool in certain kinds of mathematical argument. On the theoretical level, naturality permits us to define equivalence of categories. Definition 6.9.2. An equivalence of categories between C and D consists of functors F : C → D and G : D → C, together with natural isomorphisms between G ◦ F and the identity functor of C, and between F ◦ G and the identity functor of D. One place where equivalence of categories turns out to be useful is in the notion of skeletal subcategories.


196

Definition 6.9.3. Let D be a subcategory of C, and suppose that the inclusion functor i : D → C is an equivalence of categories. Then we say that D is a skeleton (or skeletal subcategory) of C. We shall illustrate a nice feature of skeletal subcategories presently. First, we have another definition. Definition 6.9.4. A small category is a category whose objects form a set. For instance, consider the category of finite groups. It is not small. In fact, the collection of all objects that are isomorphic to the trivial group is too big to be a set. But we can recover most of the benefits of smallness if we can construct a skeleton which is small. Recall from Corollary 2.6.6 that there are countably many isomorphism classes of finite groups. In particular, the isomorphism classes of finite groups form a set. (This does not happen in Gp, as there is no bound on the cardinality of the underlying sets of the objects in Gp.) Proposition 6.9.5. Let D be a full subcategory of C, and suppose that we’re given, for each X ∈ ob(C) an isomorphism ηX : X ∼ = F (X) for some object F (X) of D. Then D is a skeleton for C. Proof We claim that there is a unique way to make F into a functor from C to D in such a way that η is a natural transformation. But if this is the case, we’re done, since η then provides a natural isomorphism from the identity functor of C to i ◦ F , as well as from the identity functor of D to F ◦ i, where i is the inclusion of D in C. Because the morphisms ηX are all isomorphisms, it is easy to give the functoriality −1 of F : For f : X → Y in C, we take F (f ) = ηY ◦ f ◦ ηX . This immediately gives the naturality diagram for η, and the functoriality of F follows by cancelling inverses. However, the fact that we have to choose an isomorphism ηX : X ∼ = F (X) for each object X of C can be problematical from a set-theoretic point of view, as it requires an extension from sets to classes of the Axiom of Choice. Thus, it is not clear that the category of finite groups has a small skeleton. For any group G, we could use Cayley’s Theorem to identify G with a subgroup of S(G), the group of symmetries on G, but there’s no natural ordering of the elements of G, and hence no natural identification of S(G) with S|G| . However, this need not be a serious issue. Any real mathematical question that we cared about would probably either concern isomorphism classes of objects or concern information relating to particular sorts of diagrams of objects. The latter questions can generally be phrased in terms of functors from a small category into the category one wishes to study. Thus, for the questions we care about, the next result will show that we do not really need a small skeleton. The proof of the next proposition is based on the method of proof of Proposition 6.9.5. It is left to the reader. Proposition 6.9.6. Let D be a full subcategory of C such that D contains an object in every isomorphism class of objects of C. Let E be any small category. Then any functor from E into C is naturally isomorphic to a functor that takes value in D. Also, since D is full in C, any two functors from E into D that are naturally isomorphic as functors into C are also naturally isomorphic as functors into D.


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Thus, the category of finite subgroups of S∞ , which is small, contains all the information of interest with regard to the category of finite groups. Just as group isomorphisms are special examples of homomorphisms, equivalences of categories are special examples of what are called adjoint functors. Definitions 6.9.7. We say that a pair of functors R : C → D and L : D → C is an adjoint pair if for each X ∈ ob(D) and Y ∈ ob(C), there is an isomorphism (of sets) ΦX,Y : MorC (L(X), Y ) ∼ = MorD (X, R(Y )) which is natural in the following sense: given morphisms f1 : X → X in D and f2 : Y → Y in C, then the following diagram commutes. ΦX,Y / MorD (X, R(Y )) MorC (L(X), Y ) L(f1 )∗ ◦ f2 ∗

f1∗ ◦ R(f2 )∗

ΦX ,Y / MorD (X , R(Y )) MorC (L(X ), Y )

Here, g∗ (h) = g ◦ h = h∗ (g). If R and L form an adjoint pair, we call R a right adjoint for L and call L a left adjoint for R. The pair itself can be called an adjunction. Another way of expressing the naturality property for adjoint functors is as follows. Fixing X ∈ ob(D), then filling in the blanks provides two functors, MorC (L(X), −) and MorD (X, R(−)), from C to Sets, and ΦX,− provides a natural isomorphism between them. At the same time, for each Y ∈ ob(C), we have contravariant functors MorC (L(−), Y ) and MorD (−, R(Y )) from D to Sets, and Φ−,Y provides a natural isomorphism between them. It turns out that we’ve already seen several examples of adjoint functors. For instance, if C comes equipped with a forgetful functor to Sets and if ηX : X → F (X) is a free C-object for each set X, then the universal property of free objects says precisely that ηX ∗ : MorC (F (X), Y ) ∼ = MorSets (X, Y ) for each object Y of C. The right adjoint here is the forgetful functor to Sets, which doesn’t really require notation. Naturality in the variable Y is immediate, while naturality in the first variable is shown in Proposition 6.6.5. We’ve seen some other examples as well. Exercises 6.9.8. 1. Show that abelianization is a left adjoint to the inclusion of Ab in Gp. 2. Let H be a subgroup of the group G. Then there is a forgetful functor from G-sets to H-sets. Show that this forgetful functor has a left adjoint, given by L(X) = G×H X for any H-set X. 3. Let H be a subgroup of the group G and let R : G-sets → Sets be given by R(X) = X H , the H-fixed point set of X. Define L : Sets → G-sets by L(X) = (G/H) × X. Show that these form a pair of adjoint functors. 4. Show that passage from a monoid to its group of invertible elements is a right adjoint to the forgetful functor from groups to monoids.


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5. Suppose given an adjunction ΦX,Y : MorC (L(X), Y ) ∼ = MorD (X, R(Y )). For X ∈ ob(D), let ηX : X → R(L(X)) be given by ηX = ΦX,L(X) (1L(X) ). Show that η is a natural transformation from the identity functor to R ◦ L. It is customarily called the unit of the adjunction. Show also that this unit determines the isomorphism Φ as follows: for f ∈ MorC (L(X), Y ), ΦX,Y (f ) = R(f ) ◦ ηX . 6. Suppose given an adjunction ΦX,Y : MorC (L(X), Y ) ∼ = MorD (X, R(Y )). For Y ∈ ob(C), let εY : L(R(Y )) → Y be given by εY = Φ−1 R(Y ),Y (1R(Y ) ). Show that ε gives a natural transformation from L ◦ R to the identity functor. We call it the counit of the adjunction. Show that it determines the isomorphism Φ as follows: for g ∈ MorD (X, R(Y )), Φ−1 X,Y (g) = εY ◦ L(g). 7. Show that left adjoints preserve coproducts, pushouts, and direct limits. This means, for instance, that if we apply a left adjoint to a pushout diagram, the resulting diagram is also a pushout diagram. 8. Show that right adjoints preserve products, pullbacks, and inverse limits. The last two problems should explain some of the results in the preceding sections, as well as predict some others that we haven’t proven yet. That, indeed is one of the main points in favor of category theory. And in some cases, knowing about adjoints can help explain why a theorem is true. For instance, it may be instructive to reconsider the notion of adjoint when one looks at the Frobenius Reciprocity Theorem in representation theory.

6.10

General Limits and Colimits

Products, coproducts, pushouts, pullbacks, direct and inverse limits are examples of a couple of much more general types of construction. In all these cases, we’re given some kind of diagram in C, and we’re asked to extend it to a larger diagram that satisfies a universal property. The key to generalizing is in what constitutes a diagram. Definition 6.10.1. Let I be a small category and let C be an arbitrary category. By an I-diagram in C, we mean a functor X : I → C. We shall also refer to X as a diagram in C with indexing category I. We’ll show in Exercises 6.10.7 how to interpret the universal constructions we’ve already seen in terms of functors. By analogy with the constructions we’ve seen, the limit of an I-diagram in C will be defined to be a diagram defined on a larger indexing category that satisfies a certain universal property. Definition 6.10.2. For a category I, let I + be the category defined as follows. The objects of I + consist of those in I, together with one new object, denoted ∗. The morphisms in I + between objects that come from I are the same as those in I (i.e., I is a full subcategory of I + ). In addition, we declare that MorI + (∗, i) has a single element for all objects i of I + , while MorI + (i, ∗) = ∅ if i = ∗. There is, of course, a unique composition law which makes this a category. Clearly, ∗ is an initial object for I + . Indeed, I + is the result of adjoining an initial object to I. (If I already had an initial object, that object will no longer be initial in I + .)


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Definition 6.10.3. Let I be a small category and let X : I → C be a functor. A limit + + for X, denoted ← lim − X, is a functor X : I → C whose restriction to I ⊂ I is X, and which is universal in the following sense: if Y : I + → C is another such extension of X to I + , then there is a unique natural transformation η : Y → X such that ηi is the identity map of Xi whenever i is an object of I. Thus, the limit of an I-diagram is an I + -diagram. Colimits are defined in terms of the category obtained by adjoining a terminal object to I. Definition 6.10.4. For a category I, let I+ be the category defined as follows. The objects of I+ consist of those in I, together with one new object, denoted ∗. The morphisms in I+ between objects that come from I are the same as those in I (i.e., I is a full subcategory of I+ ). In addition, we declare that MorI+ (i, ∗) has a single element for all objects i of I+ , while MorI+ (∗, i) = ∅ if i = ∗. Colimits are obtained as follows. Definition 6.10.5. Let I be a small category and let X : I → C be a functor. A colimit for X, denoted − lim → X, is a functor X : I+ → C whose restriction to I ⊂ I+ is X, and which is universal in the following sense: if Y : I+ → C is another such extension of X to I+ , then there is a unique natural transformation η : X → Y such that ηi is the identity map of Xi whenever i is an object of I. Limit and colimit constructions indexed by arbitrary small categories can be quite complicated. It is often desirable to restrict attention to diagrams indexed by the category associated to a partially ordered set. (Recall that a partial ordering on a set I is a relation, generally denoted ≤ by default, that is reflexive, transitive, and antisymmetric.) Definition 6.10.6. Let I be a partially ordered set. The category, I, associated to I is the category whose objects are the elements of I, and whose morphisms are defined as follows: for i, j ∈ I, MorI (i, j) is empty unless i ≤ j, in which case it has one element, which we denote by i ≤ j. I is a category, as transitivity gives the composition of morphisms, while reflexivity gives the existence of identity morphisms. Since I is a set, I is small. Exercises 6.10.7. 1. Show that a small category I comes from a partially ordered set if and only if the following properties hold. (a) For each i, j ∈ ob(I), MorI (i, j) has at most one element. (b) If i = j and if MorI (i, j) = ∅, then MorI (j, i) = ∅. 2. Let I be a set. Then we may consider it to be a partially ordered set in which i ≤ j if and only if i = j. Let I be the induced category and let C be an arbitrary category. Note that an I-diagram, X, in C is simply a family {X(i)|i ∈ I} of Cobjects indexed by I. Show that specifying a limit for X is equivalent to specifying a product for {X(i)|i ∈ I} and that specifying a colimit for X is equivalent to specifying a coproduct for {X(i)|i ∈ I}.


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3. Let I be a partially ordered set with three elements, i, j and k, where i ≤ j, i ≤ k, and there are no other nonidentity morphisms in the induced category I. Show that ordinary diagrams in a category C which have the form •

/ •

• are in one-to-one correspondence with I-diagrams in C. Under this correspondence, show that specifying a colimit for a functor X : I → C is equivalent to specifying a pushout for the associated ordinary diagram in C. 4. Give the dual of the preceding exercise. 5. Let N be the non-negative integers with the usual ordering, and with associated category N . Show that direct systems in a category C are in one-to-one correspondence with functors from N into C. Under this correspondence, show that specifying a colimit for an N -diagram is equivalent to specifying a direct limit for the associated direct system. 6. Give the dual of the preceding exercise. 7. Let I be a small category and let C be any category. Show that there is a category, which we denote C I , whose objects are the I-diagrams in C and whose morphisms are the natural transformations between them. 8. Suppose we’re given a construction that produces a colimit for every I-diagram in C. Show that we get an induced functor from C I to C I+ . 9. Suppose we’re given a construction that produces a limit for every I-diagram in C. + Show that we get an induced functor from C I to C I . 10. Show that left adjoints preserve colimits of I-diagrams. 11. Show that right adjoints preserve limits of I-diagrams.

Chapter 7

Rings and Modules The study of rings and their modules is vastly more complicated than that of groups. In this chapter, long as it is, we shall do little more than set the stage for this study. We shall give some examples that we shall study in greater depth in later chapters, and shall give a number of basic definitions, constructions, and questions for further study. Section 7.1 defines concepts such as zero-divisors, unit groups, and algebras over commutative rings. Examples are given from matrix rings, the rings Zn , the complex numbers, C, and its subrings, the division ring H of quaternions, and the p-adic integers, p. Z Section 7.2 defines left, right, and two-sided ideals. Quotient rings are defined, and their universal property is given. Principal ideals and generators for ideals are discussed. Operations on ideals are considered, and their use is illustrated via the Chinese Remainder Theorem. Section 7.3 develops polynomials in one and several variables. Evaluation maps are studied, with the usual applications to the theory of roots. Finitely generated algebras are discussed. The cyclotomic extensions Z[ζn ] and Q[ζn ] are featured as examples. Section 7.4 studies the symmetric polynomials, showing that they form a polynomial algebra on the elementary symmetric functions. Applications are given to the theory of discriminants. Section 7.5 develops group rings and monoid rings. It is shown that polynomials form an example of the latter. Augmentation ideals are studied, and are generalized to the notion of augmented algebras. The relationship between Z[Zn ] and Z[ζn ] is studied in the exercises. Then, in Section 7.6, the theory of prime and maximal ideals is developed for commutative rings. Krull dimension is introduced. So are the ascending and descending chain conditions for ideals, in the context of the discussion of maximality principles for families of ideals. Section 7.7 introduces modules. Cyclic modules, generating sets, annihilators, and internal and external operations on modules are studied. The theories of free modules and of exact sequences are developed. The section closes with a discussion of rings without identity. Using the techniques developed for studying modules, the ascending and descending chain conditions are studied for modules and for left and right ideals in Section 7.8. The relationship between chain conditions and extensions of modules is studied, as is the relationship between the Noetherian property and finite generation. The section closes with the Hilbert Basis Theorem and its application to finitely generated algebras over a

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CHAPTER 7. RINGS AND MODULES

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Noetherian ring. Then, Section 7.9 classifies the vector spaces over a division ring, and shows that the finitely generated free modules over a commutative ring have a well defined rank. Section 7.10 gives the relationship between matrix rings and the endomorphism rings of finitely generated free modules. We close the chapter with a development of rings and modules of fractions. Local rings and localization are studied. The field of fractions of an integral domain is constructed, showing that a ring is an integral domain if and only if it is a subring of a field.

7.1

Rings

We give some basic definitions and examples of rings. We then treat the complex numbers, the quaternions, and the p-adic integers in subsections. We shall follow the tradition that insists that a ring should have a multiplicative identity. We give a brief discussion of rings without identity in the last subsection of Section 7.7. Definitions 7.1.1. A ring A is a set with two binary operations, addition and multiplication.1 Addition gives A the structure of an abelian group, with identity element 0. Multiplication provides A with a (not necessarily abelian) monoid structure, with identity element 1. The two operations interrelate via the distributive laws: a(b + c) = ab + ac

and

(b + c)a = ba + ca

for all a, b, c ∈ A. If the multiplication does satisfy the commutative law, we call A a commutative ring. Let A and B be rings. Then a ring homomorphism f : A → B from A to B is a function such that f (a + b) = f (a) + f (b) and f (ab) = f (a)f (b) for all a, b ∈ A, and f (1) = 1. As usual, a ring isomorphism is a ring homomorphism that is bijective. A subring of a ring A is a subgroup B of the additive group of A, such that B is closed under multiplication and contains the multiplicative identity element 1. Finally, the invertible elements in the multiplicative monoid of a ring A are called the units of A. They form a group under multiplication, denoted A× . Let f : A → B be a ring homomorphism. Then f gives a group homomorphism between the additive groups of A and B, so we can talk about its kernel. Note that the kernel of f cannot be a subring of A unless 1 = 0 in B, because f (1) = 1. Nevertheless, ker f is more than just a subgroup of A. It is a two-sided ideal, a concept we shall study below. Images are a different story: Lemma 7.1.2. Let f : A → B be a ring homomorphism. Then the image of f is a subring of B. The next example is the exception, rather than the rule. Example 7.1.3. There is a ring with only one element. We denote the element by 0 and set 0 + 0 = 0 and 0 · 0 = 0. In other words, both addition and multiplication are given by the unique binary operation on a one-element set. The element 0 serves as identity for both operations. Thus, with the multiplicative identity set equal to 0, the properties of a ring are satisfied. We call this ring the 0 ring and denote it by 0. 1 The

use of A for a ring derives from anneau, the French word for ring.


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The next lemma shows that the 0 ring is the only ring in which 0 = 1. Lemma 7.1.4. Let A be a ring. Then 0 · a = a · 0 = 0 for all a ∈ A. Proof We have 0a = (0 + 0)a = 0a + 0a. But then subtracting 0a from both sides gives 0 = 0a. The case of a · 0 is similar. The distributive laws now give a useful fact. Corollary 7.1.5. Let A be a ring and let −1 ∈ A be the additive inverse of 1. Then for any a ∈ A, (−1) · a = a · (−1) is the additive inverse of a. The next result is an important generalization of the preceding ones. The proof is left to the reader. Lemma 7.1.6. Let A be a ring. For m ∈ Z, write m for the m-th power of 1 in the additive group of A. (Thus, if m is positive, m is the sum of m copies of 1 in A.) Then for any a ∈ A, m · a = a · m is the m-th power of a in the additive group of A. Examples 7.1.7. The most obvious examples of rings are the integers, rational numbers, and real numbers, Z, Q, and R, with the usual addition and multiplication. The latter two examples turn out to have some interesting subrings, but Z does not: Since 1 generates Z as an abelian group, the only subring of Z is Z itself. We’ve also seen (in Proposition 2.4.7) that multiplication in Zn induces a commutative ring structure on Zn such that the canonical map π : Z → Zn is a ring homomorphism. (Since π is surjective, this is the only ring structure that makes π a ring homomorphism.) The integers play a special role in ring theory: Lemma 7.1.8. Let A be a ring. Then there is a unique ring homomorphism from Z to A. Moreover, the image of each m ∈ Z under this homomorphism commutes2 with every element of A. Proof Ring homomorphisms are required to preserve the multiplicative identity elements. But recall from Proposition 2.5.6 that for any group G, and any element x ∈ G, there is a unique group homomorphism from Z to G that carries 1 to x. Thus, there is at most one ring homomorphism from Z to A: as a homomorphism of additive groups, it must be the homomorphism f : Z → A for which f (1) = 1. It suffices to show that this group homomorphism is also a ring homomorphism for any ring A, and that f (m) commutes with every element of A. Note that since f is a homomorphism of additive groups and since f (1) = 1, f (n) must be the n-th power of 1 in the additive group structure on A, for all n ∈ Z. Thus, Lemma 7.1.6 says that for each n ∈ Z, f (n) commutes with every element of a and that f (n) · a is the n-th power of a in the additive group structure on A. Thus, f (m)f (n) is the m-th power of the n-th power of 1 with respect to the additive group operation in A. So f is a ring homomorphism by the power laws for a group. 2 When we say that two elements of a ring commute, we mean this with respect to the operation of multiplication. The point is that the addition operation is abelian so that two elements always commute additively.


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In categorical language, this implies that Z is an initial object for the category of rings. Clearly, 0 is a terminal object. We now give some examples of unit groups. Examples 7.1.9. 1. It is easy to see, e.g., via the order inequalities for multiplication, that Z× = {±1}. 2. In Lemma 4.5.3, it is shown that m ∈ Z× n if and only if (m, n) = 1. Then, in Corollary 4.5.7, it is shown that if n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes, then × ∼ × Z× n = Zpr1 × · · · × Zprk . 1

k

Finally, the unit groups Z× pr are explicitly calculated in Corollaries 4.5.20 and 4.5.21. We now give some language for describing properties of rings and their elements. Definitions 7.1.10. Suppose we have two elements a, b in a ring A, neither of which is 0, such that ab = 0. Then a and b are said to be zero-divisors in A. More specifically, if A is noncommutative, a is a left zero-divisor and b is a right zero-divisor. We say that an element a ∈ A is nilpotent if an = 0 for some integer n. Note that any nonzero nilpotent element is a zero-divisor. Generalizing the obvious facts about the integers, we say that an integral domain, or domain, for short, is a nonzero commutative ring with no zero-divisors. A nonzero ring is called a division ring if every nonzero element in it is a unit. If the ring is also commutative, we call it a field. The next lemma connects some of these definitions. Lemma 7.1.11. A division ring has no zero-divisors. Thus, every field is an integral domain. Proof Let A be a division ring and let a, b ∈ A such that a = 0 and ab = 0. It suffices to show that b = 0. Because A is a division ring and a = 0, a has a multiplicative inverse, a−1 . But then 0 = a−1 · 0 = a−1 · ab = (a−1 a)b = b. Of course, Z is an example of an integral domain, and Q and R are examples of fields. The following lemma could have been left as an exercise, but we shall make sufficient use of it that a proof has been given. The reader may wish to investigate alternative proofs. Lemma 7.1.12. Let n > 1. Then Zn is an integral domain if and only if n is prime. Moreover, for any prime number p, Zp is a field. Proof Suppose that n is not prime. Then we can write n = kl, where k and l are both greater than 1. But then 1 < k, l < n, so k and l are nonzero in Zn . But k · l = kl = 0, and hence k and l are zero-divisors in Zn . On the other hand, if p is prime, then each nonzero element of Zp is shown to be a unit in Lemma 4.5.3. In a noncommutative ring, it is useful to keep track of the elements that commute with all other elements of the ring.


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Definition 7.1.13. Let A be a ring. Then the center of A is the set {a ∈ A | ab = ba for all b ∈ A} consisting of those elements of a that commute with every element of A. Lemma 7.1.14. Let A be a ring. Then the center of A is a subring of A. Our first examples of noncommutative rings come from matrices. Definitions 7.1.15. Let A be a ring (not necessarily commutative). We write Mn (A) for the collection of n × n matrices with coefficients in A. Given elements aij ∈ A with 1 ≤ i ≤ n and 1 ≤ j ≤ n, we write (aij ) ∈ Mn (A) for the matrix whose ij-th entry is aij . We define addition in Mn (A) so that the ij-th entry of (a ijn) + (bij ) is aij + bij and define multiplication so that the ij-th entry of (aij ) · (bij ) is k=1 aik bkj . Note that the order of the factors in the last expression is important if A is noncommutative. We write 0 for the matrix whose entries are all 0, and write In for the matrix 1 0 0 whose off-diagonal entries are all 0 and whose diagonal entries are all 1, e.g., I3 = 0 1 0 . 001 More generally, for a ∈ A, we write aIn for the matrix whose diagonal entries are all equal to a and whose off-diagonal entries are all 0. We write ι : A → Mn (A) for the map given by ι(a) = aIn for all a ∈ A. We summarize the most basic properties of matrix rings: Lemma 7.1.16. Let A be a ring. Then Mn (A) is a ring under the operations of matrix addition and matrix multiplication, and the map ι : A → Mn (A) is a ring homomorphism. Moreover, the elements ι(a) = aIn behave as follows: if M = (aij ) ∈ Mn (A), then aIn · M is the matrix whose ij-th entry is a · aij for all i, j, while M · aIn is the matrix whose ij-th entry is aij · a for all i, j. Thus, if B is the center of A, then ι(B) is contained in the center of Mn (A). The units in Mn (A) are those matrices M for which there exists an M ∈ Mn (A) such that M M = M M = In . Note that this precisely says that M is invertible in the usual sense from linear algebra. ×

Definitions 7.1.17. Let A be a ring. We write Gln (A) for Mn (A) , the group of invertible n × n matrices. We call it the n-th general linear group of A. The reader should be warned that if A is not commutative, then the usual sort of determinant theory fails for matrices over A, and hence invertibility cannot be detected by the familiar methods. When A is commutative, on the other hand, there is a good theory of determinants, which we shall develop in Chapter 10. We shall show that a matrix in Mn (A) is invertible if and only if its determinant is in A× . Another example relates to a generalization of matrix rings. Definition 7.1.18. Let G be an abelian group and let EndZ (G) be the set of all group homomorphisms from G to itself. We call it the endomorphism ring of G. Here, the ring operations are defined as follows. For f, g ∈ EndZ (G), we let f + g be the homomorphism given by (f + g)(x) = f (x) + g(x), where additive notation is used for the group operation in G. The product, f · g, of f, g ∈ EndZ (G) is simply the composition f ◦ g. As usual, the justification for this definition is left to the reader:


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Lemma 7.1.19. Under these operations, EndZ (G) is a ring, with additive identity the trivial homomorphism, and multiplicative identity the identity map. We shall generalize this when we study A-modules. If M is an A-module, then the A-module homomorphisms from M to itself will be shown to form a subring, EndA (M ), of EndZ (M ). If A is commutative, and if An is the free A-module of rank n, we shall see that EndA (An ) is isomorphic as a ring to Mn (A). Thus, endomorphism rings do generalize matrix rings when A is commutative. A useful concept in ring theory is that of an algebra over a commutative ring A. Definitions 7.1.20. An algebra over a commutative ring A consists of a ring B together with a ring homomorphism ν : A → B such that the image of ν is contained in the center of B. We call ν the structure map for B as an A-algebra. If B is a commutative ring, we call it a commutative A-algebra. If B and C are A-algebras, via ν : A → B and μ : A → C, then an A-algebra homomorphism from B to C is a ring homomorphism f : B → C such that the following diagram commutes. ν ooo7 B ooo f A OOO OOO ' μ C Examples 7.1.21. 1. Let B be any ring and let A be the center of B. Then the inclusion of A in B makes B an A-algebra. 2. Let B be any ring and let A be the center of B. Let ι : B → Mn (B) be the ring homomorphism given by ι(b) = bIn for b ∈ B. Then Lemma 7.1.16 shows that ι(A) is contained in the center of Mn (B).3 Thus, Mn (B) is an A-algebra via ι ◦ i, where i : A ⊂ B is the inclusion, and ι : B → Mn (B) is an A-algebra map. We now show that the notions of rings and of Z-algebras are equivalent. Lemma 7.1.22. Every ring A has a unique structure as a Z-algebra, and every ring homomorphism f : A → B is a Z-algebra homomorphism. Proof The fact that a ring has a unique Z-algebra structure is just a restatement of Lemma 7.1.8: For any ring A, there is a unique ring homomorphism from Z to A, and its image lies in the center of A. Now if ν : Z → A and μ : Z → B are the unique ring homomorphisms and if f : A → B is any ring homomorphism, we must have f ◦ ν = μ. Thus, f is a Z-algebra homomorphism. As in the case of groups, the direct product gives the simplest method of obtaining new rings from old. Definition 7.1.23. Let A and B be rings. Then the direct product A × B is the ring whose additive group is the direct product of the additive groups of A and B and whose multiplication is given coordinate-wise. In other words, the underlying set of A × B is the cartesian product of A and B, and the operations are given by (a, b) + (c, d) = (a + c, b + d) 3 Indeed,

and

(a, b) · (c, d) = (ac, bd).

Problem 10 of Exercises 7.1.27 shows that ι(A) is equal to the center of Mn (B).


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The reader should supply the justification that A × B is a ring: Lemma 7.1.24. Let A and B be rings. Then the direct product A × B is a ring and the projection maps from A × B to A and B are both ring homomorphisms. Moreover, if C is a ring and if f : C → A × B is a map, say f (c) = (f1 (c), f2 (c)), then f is a ring homomorphism if and only if f1 : C → A and f2 : C → B are ring homomorphisms. In category theoretic terms, Lemma 7.1.24 just says that the direct product is the product in the category of rings. The finite direct products A1 × · · · × An are defined analogously, and satisfy a similar universal property. Indeed, the product of an infinite family of rings is a ring via coordinate-wise multiplication: ! Definition 7.1.25. Suppose given a family {Ai | i ∈ I} of rings. We write i∈I Ai for the ring structure on the product of the Ai whose operations are coordinate-wise addition and multiplication. Thus, (ai | i ∈ I) + (bi | i ∈ I) = (ai + bi | i ∈ I)

and

(ai | i ∈ I) · (bi | i ∈ I) = (ai bi | i ∈ I).

Again, the reader should justify the definition: ! Lemma 7.1.26. Let {Ai | i ∈ I} be a family of rings. Then i∈I Ai is a ring under coordinate-wise addition and multiplication. Here 0 is the I-tuple whose coordinates are ! all 0, and 1 is the I-tuple whose coordinates are all 1. The projection maps πj : for all j ∈ I. i∈I Ai → Aj are ring homomorphisms ! If B is a ring and if f : B → i∈I Ai is a map, say f (b) = (fi (b) | i ∈ I) for each b ∈ B, then f is a ring homomorphism if and only if fi is a ring homomorphism for all i ∈ I. Exercises 7.1.27. 1. Give the proof of Lemma 7.1.6. 2. Show that every subring of a field is an integral domain.4 3. We say that n ∈ Z is square free if it is not divisible by the square of any prime number. Show that Zn contains no nonzero nilpotent elements if and only if n is square free. 4. Suppose given a ring structure on the additive group Zn . Show that the multiplicative identity element must generate Zn as an additive group. Deduce that any ring structure on Zn is isomorphic to the usual one. (Hint: Let e be the multiplicative identity for a given ring structure on Zn . Show that e fails to generate Zn additively if and only if there is an integer d > 1 such that d divides n and such that e is the additive d-th power of some element x ∈ Zn . Now show that no such ring structure could exist.) 5. Let f : A → B be a ring homomorphism. Show that f restricts to a group homomorphism f : A× → B × . (Note then that the passage from rings to their unit groups gives a functor from rings to groups.) 6. Give the proof of Lemma 7.1.16. 4 We

shall see in Section 7.11 that a ring is an integral domain if and only if it is a subring of a field.


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7. Let A and B be rings. Show that Mn (A × B) is isomorphic to Mn (A) × Mn (B). 8. Show that Mn (A) contains nonzero nilpotent elements, and hence zero-divisors, for any nonzero ring A, provided that n ≥ 2. 9. Let A and B be subrings of a ring C. Let a c R= a ∈ A, b ∈ B and c ∈ C . 0 b Show that R is a subring of M2 (C). † 10. Let B be the center of A. Show that ι : A → Mn (A) restricts to an isomorphism from B onto the center of Mn (A). 11. Let f : A → B be a ring homomorphism. Show that there is a ring homomorphism f∗ : Mn (A) → Mn (B) defined by setting f∗ ((aij )) equal to the matrix whose ij-th entry is f (aij ). 12. Give the proof of Lemma 7.1.19. 13. Show that the unit group of EndZ (G) is the automorphism group Aut(G) of G. 14. Let G be the direct sum of a countably infinite number of copies of Z. Find an element of EndZ (G) which has a left inverse, but is not a unit. 15. Show that the center of a direct product A × B is the direct product of the centers of A and B. 16. Show that the group of units of A × B is A× × B × . ‡ 17. Suppose given ring homomorphisms f : A → C and g : B → C. Recall that the pullback of f and g is the diagram f / B

A ×C B

g

g A

f

/ C

where A ×C B = {(a, b) ∈ A × B | f (a) = g(b)}, and f and g are defined by f (a, b) = b and g(a, b) = a. (a) Show that A ×C B is a subring of A × B and that f and g are ring homomorphisms. (b) Suppose given a ring R and ring homomorphisms f : R → B and g : R → A such that f ◦ g = g ◦ f . Show that there is a unique ring homomorphism h : R → A ×C B such that f ◦ h = f and g ◦ h = g . Show that h must satisfy h(r) = (g (r), f (r)) for all r ∈ R. (c) Show that the natural map (A ×C B)× → A× ×C × B × is an isomorphism. 18. Show that a direct product A × B of commutative rings is an integral domain if and only if one of A and B is an integral domain and the other is the zero ring.


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19. Show that a direct product A × B of rings is a division ring if and only if one of A and B is a division ring and the other is the zero ring. 20. Give the proof of Lemma 7.1.24. 21. Give the proof of Lemma 7.1.26. 22. Let A be a ring. Show that Mm (A) × Mn (A) to the subring of is isomorphic M 0 Mm+n (A) consisting of matrices of the form , where M is an m × m 0 N matrix, N is an n × n matrix, and the 0’s are zero-matrices of the appropriate size.

The Complex Numbers The complex numbers, C, give yet another example of a field. We review the definition and basic properties of C. Definitions 7.1.28. The elements of the complex numbers are formal sums a + bi, with a, b ∈ R. This presentation is unique in the sense that if a, b, c, d ∈ R, then a+bi = c+di if and only if a = c and b = d. Thus, the elements of C are in one-to-one correspondence with ordered pairs (a, b) of real numbers. So we may identify C with the plane R2 , which gives us a geometric picture of the complex numbers. We identify a ∈ R with the complex number a + 0i. We say that a complex number is in R if it has this form. Similarly, we write i for the complex number 0 + 1i, and refer to the numbers ai = 0 + ai, with a ∈ R as the pure imaginary numbers. If z = a + bi and w = c + di are complex numbers, with a, b, c, d ∈ R, we define their sum by z + w = (a + c) + (b + d)i, and define their product by zw = (ac − bd) + (ad + bc)i. For z = a + bi with a, b ∈ R, we define the complex conjugate, z, of z by z = a − bi. Again, there are verifications left to the reader. Lemma 7.1.29. The above operations of addition and multiplication give C the structure of a commutative ring. Moreover, the standard inclusion R ⊂ C is a ring homomorphism, and hence the additive and multiplicative identities of C are given by 0 = 0 + 0i and 1 = 1 + 0i, respectively. The function C → C that carries each z ∈ C to its complex conjugate, z, is a ring homomorphism. For z = a + bi with a, b ∈ R, we have the equality zz = a2 + b2 ∈ R ⊂ C. In particular, zz ≥ 0 for all z, and zz = 0 if and only if z = 0. We can now find a multiplicative inverse for every nonzero element of C. Corollary 7.1.30. The complex numbers, C, is a field. Explicitly, if 0 = z ∈ C, then z is a unit in C with inverse z −1 =

1 · z. zz

Here, 1/(zz) is the inverse of the nonzero real number zz.


210

Proof Since zz = 0, the result follows from the facts that C is commutative and that R is a subring of C. The notion of planar distance from analytic geometry may be derived from the complex product zz. Definition 7.1.31. We define the absolute value | | : C → R by √ |z| = zz = a2 + b2 for z = a + bi with a, b ∈ R. A very useful tool in studying the complex numbers is the complex exponential function. Definition 7.1.32. Let x ∈ R. We define the complex exponential eix ∈ C by eix = cos x + i sin x. More generally, if z = x + iy ∈ C, with x, y ∈ R, we define ez ∈ C by ez = ex eiy = ex cos y + iex sin y, where ex is the ordinary real exponential function applied to x. We shall be primarily concerned with the exponentials eix with x ∈ R. Lemma 7.1.33. Let x, y ∈ R, then ei(x+y) = eix · eiy for x, y ∈ R, where the product on the right is complex multiplication. Moreover, ei·0 = 1. Thus, there is a group homomorphism exp : R → C× defined by exp(x) = eix . Proof Complex multiplication gives eix · eiy

= (cos x + i sin x) · (cos y + i sin y) = (cos x cos y − sin x sin y) + i(cos x sin y + sin x cos y) = cos(x + y) + i sin(x + y),

by the formulæ for the sine and cosine of a sum. But the last line is precisely ei(x+y) . From the known values of the sine and cosine of 0, we see that ei·0 = 1, as claimed, so that setting exp(x) = eix gives a monoid homomorphism from R to the multiplicative monoid of C. But any monoid homomorphism from a group to a monoid must take value in the group of invertible elements of the monoid, which in this case is C× . The exponential allows us to define some important elements of C, the standard primitive n-th roots of unity. Definition 7.1.34. Let n ≥ 1. The standard primitive n-th root of unity, ζn ∈ C, is defined by 2π 2π ζn = ei·2π/n = cos + i sin . n n The fact that 2π is the smallest positive real number whose sine is 0 and whose cosine is 1 gives us the following lemma.


211

Lemma 7.1.35. Let n ≥ 1. Then ζn has order n in C× . There is a noncommutative division ring, called the quaternions, H, that is given by a construction similar to that of the complex numbers. We shall treat it in the next set of exercises. Both C and H are examples of what are called Clifford algebras over R. We shall treat the general case of Clifford algebras in Chapter 9. We shall not prove it here, but C and H are the only division rings that are algebras over R in such a way that the induced real vector space structure is finite dimensional. Exercises 7.1.36. 1. Write Z2 = T = {1, T }. Show that Z2 acts on C via T · z = z for all z ∈ C. Show that the fixed point set CZ2 is equal to R. 2. Write Zn = T = {1, T, . . . , T n−1 } for n ≥ 2. Show that Zn acts on C via T k · z = ζnk · z for all k, where ζn = ei·2π/n is the standard primitive n-th root of unity in C. Show that 1 + ζn + · · · + ζnn−1 is a fixed point under this action. Deduce that 1 + ζn + · · · + ζnn−1 = 0. 3. Show that |zw| = |z| · |w| for all z, w ∈ C, that |z| ≥ 0 for all z ∈ C, and that |z| = 0 if and only if z = 0. 4. Show that |z + w| ≤ |z| + |w| for all z, w ∈ C. 5. Show that i is a square root of −1 in C. 6. Let S 1 = {z ∈ C | |z| = 1}. Show that S 1 is a subgroup of C× . 7. Show that eix ∈ S 1 for all x ∈ R, and hence exp gives a homomorphism from R to S 1 . What is the kernel of exp? 8. Show that exp : R → S 1 is onto. Deduce that S 1 is isomorphic as a group to R/Z. 9. Show that there is a homomorphism expC : C → C× given by expC (z) = ez . 10. Show that eiπ = −1 ∈ C. 11. Show that ζ1 = 1, ζ2 = −1, and ζ4 = i. 12. Let n = mk with m and k positive integers. Show that ζnk = ζm . 13. Show that if n = 2m with m a positive integer, then ζnk+m = −ζnk . 14. Let m > 0 be an odd integer. Show that the subgroup of C× generated by −ζm coincides with the subgroup generated by ζ2m .

when a, b ∈ R. Show that φ is an 15. Define φ : C → M2 (R) by φ(a + bi) = ab −b a injective ring homomorphism. Show that φ(S 1 ) = SO(2) and that φ(eiθ ) = Rθ , the matrix that rotates R2 through the angle θ, for all θ ∈ R. 16. We construct the ring of quaternions. The quaternionic groups Q4n , which we analyzed in the material on group theory, will be seen to embed naturally in the group of units of the quaternions. An alternative name for the quaternions is the Hamiltonians. Because of this, and the fact that Q is already taken, we shall write H for the quaternions. The elements of H are formal sums a + bi + cj + dk, where a, b, c, and d are real


212

numbers. Here, a + bi + cj + dk = a + b i + c j + d k if and only if a = a , b = b , c = c , and d = d , and hence we may identify H with the set R4 of 4tuples (a, b, c, d) of real numbers. We define the addition in H accordingly, so that (a + bi + cj + dk) + (a + b i + c j + d k) = (a + a ) + (b + b )i + (c + c )j + (d + d )k. We write 1 = 1 + 0i + 0j + 0k, i = 0 + 1i + 0j + 0k, j = 0 + 0i + 1j + 0k and k = 0 + 0i + 0j + 1k. For a ∈ R, we write a = a · 1 = a + 0i + 0j + 0k. Thus, we identify the real multiples of 1 with R. The multiplication in H is defined so that 1 is the identity element, i2 = j 2 = k 2 = −1, ij = −ji = k, jk = −kj = i and ki = −ik = j. Assembling this information, we see that if α = a + bi + cj + dk and α = a + b i + c j + d k, then αα = (aa − bb − cc − dd ) + (ab + ba + cd − dc )i + (ac + ca + db − bd )j + (ad + da + bc − cb )k. Unlike the complex numbers, we see that multiplication in H is noncommutative. (a) Show that H is a ring and that the real numbers R ⊂ H form a subring. (b) Show that the center of H is R, so that H is an R-algebra. (c) Show that if we set x equal to either i, j, or k, then the set of all elements of H of the form a + bx with a, b ∈ R forms a subring of H that is isomorphic to C. (d) Define conjugation in H as follows: for α = a + bi + cj + dk, we set α = a − bi − cj − dk. Show that the following equalities hold for all α, α ∈ H. i. α + α = α + α ii. α · α = α · α iii. 1 = 1. iv. α = α. (According to definitions we shall give below, the first three of the above properties say that conjugation is an antiautomorphism of H, and hence H is a self-opposite ring. This is useful in the study of matrices over H.) (e) Show that α and α commute for all α ∈ H, and that if α = a + bi + cj + dk, then αα is the real number a2 + b2 + c2 + d2 . (f) Define the absolute value | | : H → R by √ |α| = αα = a2 + b2 + c2 + d2 for α = a + bi + cj + dk with a, b, c, d ∈ R. Show that |α · α | = |α| · |α | for all α, α ∈ H, that |α| ≥ 0 for all α ∈ H, and that |α| = 0 if and only if α = 0. (g) Show that every nonzero element α ∈ H is a unit in H (and hence H is a division ring). Here, if α = 0, show that α−1 =

1 α. αα

(h) Show that if we set b = cos(π/n) + i sin(π/n) and set a = j, then the subgroup of H× generated by a and b is isomorphic to Q4n , the quaternionic group of order 4n. (In particular, note that Q8 is isomorphic to the subgroup {±1, ±i, ±j, ±k} of H× . When studying Q8 alone, rather than the quaternionic groups in general, it is customary to use the notation ±1, ±i, ±j, ±k for its elements.)


213

(i) Let S 3 = {α ∈ H | |α| = 1}. Show that S 3 is a subgroup of H × containing the above copies of Q4n . Show also that S 1 embeds as a subgroup of S 3 . 17. Let S = {α = ai + bj + ck ∈ H | a, b, c ∈ R and |α| = 1}.5 Show that α2 = −1 for all α ∈ S. Deduce that H has infinitely many subrings isomorphic to C.

The p-adic Integers p . For k ≥ 2, let εk : Zpk → Zpk−1 be the unique ring We construct the p-adic integers, Z homomorphism (in fact, the unique group homomorphism that carries 1 to 1) from Zpk to Zpk−1 . The p-adic integers are the inverse limit of the maps εk . To avoid a dependence on category theory here, we shall give the construction explic p is a subring of the infinite product ! itly. Z k≥1 Zpk . It is customary to write the tuples in the infinite product in the reverse order from the usual one. Thus, a typical element looks like (. . . , mk , . . . , m2 , m1 ), where each mk lies in Zpk . Under this convention, we have p = {(. . . , mk , . . . , m2 , m1 ) | εk (mk ) = mk−1 for all k ≥ 2}. Z Recall that the ring operations in the infinite product are defined via coordinate-wise addition and multiplication, so the fact that the εk are ring homomorphisms verifies the assertion above: p is a subring of ! Lemma 7.1.37. Z k≥1 Zpk . The additive and multiplicative identities, of course, are 0 = (. . . , 0, . . . , 0, 0) and 1 = (. . . , 1, . . . , 1, 1), respectively. The next lemma is elementary, but important. p → Zpk be induced by the projection onto the k-th factor. Lemma 7.1.38. Let πk : Z Then πk is a ring homomorphism for all k ≥ 1. The p-adic integers are important in number theory. Despite the fact that they are made up from quotient rings of Z, the next lemma will show that the p-adic integers themselves form a ring of characteristic 0. (See Definition 7.2.12.) p is an embedding. Lemma 7.1.39. The unique ring homomorphism ι : Z → Z Proof We have ι(m) = (. . . , m, . . . , m, m), so m ∈ ker ι if and only if m is divisible by pk for all k ≥ 0. p . Recall that a The structure of the unit group is one of the interesting features of Z splitting map, or section, for a homomorphism f : G → K is a homomorphism s : K → G such that f ◦ s is the identity map of K. A map f that admits a section is called a split surjection. Recall from Corollary 4.7.6 that if f : G → K is a split surjection with G abelian, then G ∼ = ker f × K. p is a unit if and only Proposition 7.1.40. An element a = (. . . , mk , . . . , m2 , m1 ) of Z if m1 is a unit in Zp . The map ×

→ Z× π1 : Z p p is a split surjection. 5 Geometrically,

S is a 2-dimensional subsphere of S 3 .


214

p . Since the multiplication in Z p is coordinateProof Let a = (. . . , mk , . . . , m2 , m1 ) ∈ Z wise, if a is a unit in Zp , then each mk must be a unit in Zpk . Conversely, suppose that m1 is a unit. According to Lemma 4.5.3, an element m is a unit in Zpk if and only if (m, p) = 1. In particular, m is a unit in Zpk if and only if εk (m) = m is a unit in Zpk−1 . Thus, if m1 is a unit, then each mk must be a unit in Zpk . Let nk be the inverse of mk in Zpk . Since each εk is a ring homomorphism, we must p have εk (nk ) = nk−1 . Thus, setting b = (. . . , nk , . . . , n2 , n1 ) specifies an element of Z that is easily seen to be a multiplicative inverse for a. × → Z× . For p = 2, there’s nothing It remains to construct a splitting map for π1 : Z p p to show, as Z× is the trivial group. For p > 2, we consider the proof of Corollary 4.5.12. 2 Let ηk : Zpk → Zp be the unique ring homomorphism between these rings. Then the × kernel of the induced map ηk : Z× → Z× p has order relatively prime to |Zp | = p − 1. pk consisting Thus, the proof of Corollary 4.4.15 shows that if K is the subgroup of Z× pk ∼ =

of those elements of exponent p − 1, then ηk : K −→ Z× p . But this says that there is a × unique section sk : Z× → Z of η . k p pk But then εk ◦ sk is a section for ηk−1 : Z× → Z× p , so εk ◦ sk = sk−1 . Thus, setting pk−1 s(m) = (. . . , sk (m), . . . , s2 (m), m) gives a well defined function s : Z× p → Zp , that is easily seen to give a section for × × →Z . π1 : Z p p Exercises 7.1.41. p is an integral domain. (Hint: Suppose that ab = 0 in Z p , with † 1. Show that Z a = (. . . , mk , . . . ) and b = (. . . , nk , . . . ). Suppose that mk = 0 in Zpk . What does this say about the p-divisibility of mk+l and nk+l ?) × → Z× is isomorphic to the 2. Let p be an odd prime. Show that the kernel of π1 : Z p p × additive group of Zp . Deduce that the torsion subgroup of Zp maps isomorphically to Z× p under π1 . × → Z× is isomorphic to the additive group of Z 2. 3. Show that the kernel of π2 : Z 2 4 × maps isomorphically to Z× under π2 . Deduce that the torsion subgroup of Z 2

4

p → Zp . Show that † 4. Let a be in the kernel of the additive homomorphism π1 : Z p . (Hint: Consider the diagram a = pb for some b ∈ Z Zpk εk Zpk−1

⊂ / Zpk+1 εk+1 ⊂ / Zpk

where the inclusion maps send 1 to p.) p . Suppose that mk = 0 in Zpk for all k ≤ r and † 5. Let a = (. . . , mk , . . . , m2 , m1 ) ∈ Z ×. that mr+1 = 0 in Zpr+1 . Show that a = pr u, where u ∈ Z p


215

6. Let fk : Ak → Ak−1 be a ring homomorphism for k ≥ 2. Give an isomorphism × × ∼ lim A = lim ←− k ←−(Ak ). 7. Give Zpk the discrete topology for ! all k. Show that Zp is a closed subspace of the product (Tikhonov) topology on k≥1 Zpk , and hence is compact by the Tikhonov p . For this reason, Z p may Theorem. Show that ι(Z) is dense in this topology on Z p. be thought of as a completion of Z in the topology that Z inherits from Z

7.2

Ideals

The ideals in a ring are a very important part of its structure. We give the most basic results on ideals in this section, and shall continue to study them throughout the material on rings. Definitions 7.2.1. Let A be a ring. A left ideal of A is a subgroup a ⊂ A of the additive group of A with the additional property that if x ∈ a, then for any a ∈ A, we have ax ∈ a. (In other words, a is closed on the operation of multiplying its elements on the left by any element of A.) From the other side, a right ideal of A is a subgroup a ⊂ A of the additive group of A with the additional property that if x ∈ a, then for any a ∈ A, we have xa ∈ a. A two-sided ideal of A is a subgroup of the additive group of A which is simultaneously a left ideal and a right ideal. If A is commutative, all ideals are two-sided, and we refer to them simply as ideals. Examples 7.2.2. 1. Let A be a ring. Then A is a two-sided ideal of itself. 2. Let A be a ring and let 0 ⊂ A be the trivial subgroup. Since a · 0 = 0 · a = 0 for all a ∈ A, 0 is a two-sided ideal of A. Another set of examples comes from the principal ideals. Definition 7.2.3. Let A be a ring and let x ∈ A. The principal left ideal, Ax, generated by x is defined by Ax = {ax | a ∈ A}. Similarly, the principal right ideal generated by x is xA = {xa | a ∈ A}. If A is commutative, then Ax = xA, and we call it simply the principal ideal generated by x, and denote it by (x). Principal ideals play a role in ideal theory similar to the role of cyclic subgroups in group theory: Lemma 7.2.4. Let A be a ring and let x ∈ A. Then Ax and xA are the smallest left and right ideals, respectively, that contain x. Some commutative rings have no ideals other than the principal ones. Definitions 7.2.5. A principal ideal ring is a commutative ring in which every ideal is principal. A principal ideal domain, or P.I.D., is an integral domain in which every ideal is principal.


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Indeed, the integers are better behaved than the general integral domain: Lemma 7.2.6. Every subgroup of Z is a principal ideal. In particular, Z is a P.I.D. Proof We know (Proposition 2.3.2) that the subgroups of Z all have the form n = {nk | k ∈ Z} for some n ∈ Z. But n is precisely (n), the principal ideal generated by n. We shall see below that if K is a field,6 then the polynomial ring K[X] is a principal ideal domain. Principal ideals are useful in recognizing division rings and fields. Lemma 7.2.7. A nonzero ring A is a division ring if and only if the only left ideals of A are 0 and A. Proof Let A be a division ring and let a be a nonzero left ideal of A. Thus, there is an x ∈ a with x = 0. But then x is a unit in A. Now 1 = x−1 x ∈ a, as a is a left ideal. But then a = a · 1 ∈ a for the same reason, for all a ∈ A. Thus, a = A. Conversely, suppose that A is a ring with no left ideals other than 0 and A. Let 0 = x ∈ A. Then Ax is a nonzero left ideal, and hence Ax = A. Thus, there is an element y ∈ A with yx = 1. Thus, every nonzero element in the multiplicative monoid of A has a left inverse. As in Problem 13 of Exercises 2.1.16, it is easy to see that every nonzero element of A is a unit. Two-sided ideals play a role in ring theory somewhat analogous to that played by normal subgroups in the theory of groups. Proposition 7.2.8. Let a be any subgroup of the additive group of a ring A, and let π : A → A/a be the canonical map. Then A/a has a ring structure that makes π a ring homomorphism if and only if a is a two-sided ideal of A. Such a ring structure, if it exists, is unique. Proof Suppose given a ring structure on A/a such that π is a ring homomorphism. Then writing a ∈ A/a for π(a), we must have a · b = ab, so the ring structure is indeed unique. Moreover, for x ∈ a, 1 = 1 + x. Thus, for a ∈ A, a = a · 1 = a · 1 + x = a + ax. In other words, a + ax represents the same element of A/a as a does. But this says a + ax − a ∈ a, and hence ax ∈ a for a ∈ A and x ∈ a. Thus, a has to be a left ideal. But a similar argument, obtained by multiplying 1 and 1 + x on the right by a, shows that a must also be a right ideal, so that a is in fact two-sided. For the converse, suppose that a is a two-sided ideal. We wish to show that the multiplication a · b = ab gives a well defined binary operation on A/a. This will be sufficient to complete the proof, as then π is a homomorphism with respect to this operation, so that the verifications that A/a is a ring under this multiplication will follow from the fact that A is a ring. Thus, for a, b ∈ A and x, y ∈ a, we must show that ab and (a + x)(b + y) represent the same element of A/a, and hence that (a + x)(b + y) − ab ∈ a. But (a + x)(b + y) − ab = ay + xb + xy, which is in a because a is a two-sided ideal. The rings A/a satisfy a universal property similar to that of factor groups in group theory. We call them quotient rings of A. 6 The use of K for a field derives from the German K¨ orper , which means field in the mathematical sense. Amusingly, the German word for a farmer’s field is a cognate of ours: Feld.


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Proposition 7.2.9. Let f : A → B be a ring homomorphism. Then ker f is a two-sided ideal of A. If a is another two-sided ideal of A, then a ⊂ ker f if and only if there is a ring homomorphism f : A/a → B which makes the following diagram commute, where π : A → A/a is the canonical map: /B AB = BB || BB | | B || π BB || f ! A/a f

Moreover, if a ⊂ ker f , then the homomorphism f : A/a → B making the diagram commute is unique. When a = ker f , f induces a ring isomorphism between A/a and the image of f . Proof By Proposition 4.1.7, there is a unique homomorphism of additive groups, f : A/a → B making the diagram commute if and only if a ⊂ ker f . And if a = ker f , this homomorphism f gives a group isomorphism onto the image of f by the First Noether Isomorphism Theorem. Thus, it suffices to show that ker f is a two-sided ideal and that f is a ring homomorphism whenever a is a two-sided ideal contained in ker f . We leave these verifications to the reader. Since π : Z → Zn is a surjective ring homomorphism and has kernel (n), we obtain the following corollary. Corollary 7.2.10. There is an isomorphism of rings between Zn and the quotient ring Z/(n). The next corollary is another consequence of Proposition 7.2.9. Corollary 7.2.11. Let f : D → B be a ring homomorphism, where D is a division ring and B is any nonzero ring. Then f is injective. Proof The kernel of f is a two-sided ideal, and hence must be either 0 or D. As B is nonzero, ker f = D. So ker f = 0, and f is injective. Recall that for any ring B, there is a unique ring homomorphism f : Z → B. We may obtain useful information about B by applying Proposition 7.2.9 with a = ker f . Definition 7.2.12. Let A be a ring and let f : Z → A be the unique ring homomorphism from Z to A. Let ker f = (n), with n ≥ 0. Then we say that A has characteristic n. In particular, A has characteristic 0 if and only if f is injective. Examples 7.2.13. Since the kernel of the canonical map π : Z → Zn is (n), Zn has characteristic n. On the other hand, Z is a subring of Q, R, C, and H, so these have characteristic 0. Proposition 7.2.14. Let n > 0. Then a ring A has characteristic n if and only if it has a subring isomorphic as a ring to Zn . If A does have characteristic n, then the subring in question is the image of the unique ring homomorphism f : Z → A.


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Proof If A has characteristic n with n > 0, then the image of f is isomorphic to Z/(n) by Proposition 7.2.9. Conversely, suppose that A has a subring B that is isomorphic as a ring to Zn . Then if f : Z → B is the unique ring homomorphism, we may identify f with the canonical map π : Z → Z/(n). In particular, the kernel of f is (n). But if i : B → A is the inclusion, then i ◦ f is the unique ring homomorphism from Z to A. Since i is injective, the kernel of i ◦ f , whose generator is by definition the characteristic of A, is precisely the kernel of f . Let K be a field. Then the image of the unique ring homomorphism f : Z → K, as a subring of K, must be an integral domain. Thus, Lemma 7.1.12 shows that if f is not injective (i.e., if the image of f is finite), then f must have image Zp for some prime p. Corollary 7.2.15. Let K be a field of nonzero characteristic. Then the characteristic of K is a prime number. We can generalize the notion of a principal ideal as follows. Definitions 7.2.16. Let A be a ring and let S be any subset of A. Then the left ideal generated by S, written AS, is given as follows: AS = {a1 x1 + · · · + ak xk | k ≥ 1, ai ∈ A and xi ∈ S for 1 ≤ i ≤ k}. Similarly, the right ideal generated by S is given by SA = {x1 a1 + · · · + xk ak | k ≥ 1, ai ∈ A and xi ∈ S for 1 ≤ i ≤ k}. When A is commutative, of course, AS = SA, and we denote it by (S). If S is finite, say S = {x1 , . . . , xn }, we write (S) = (x1 , . . . , xn ). For A noncommutative, the two-sided ideal generated by S is ASA = {a1 x1 b1 + · · · + ak xk bk | k ≥ 1, ai , bi ∈ A and xi ∈ S for 1 ≤ i ≤ k}. The next lemma is immediate from the definitions. Lemma 7.2.17. Let A be a ring and let S be a subset of the center of A. Then the left, right, and two-sided ideals generated by A all coincide, i.e., AS = SA = ASA. In particular, if A is commutative, then ASA = (S). Warning: In the case of left ideals, we may apply the distributive law to reduce the sums a1 x1 + · · · + ak xk until the elements xi are all distinct. Thus, for instance, A{x} = Ax = {ax | a ∈ A}, the principal left ideal generated by x. The analogous result holds for the right ideals. But in the case of two-sided ideals in noncommutative rings, we cannot make this kind of reduction in general. Thus, in A{x}A, the principal two-sided ideal generated by x, we must consider all elements of the form a1 xb1 + · · · + ak xbk with k ≥ 1 and with ai , bi ∈ A for 1 ≤ i ≤ k. Lemma 7.2.18. Let A be a ring and let S ⊂ A. Then AS, SA, and ASA are the smallest left, right, and two-sided ideals, respectively, that contain S. This now gives the following universal property.


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Corollary 7.2.19. Let A be a ring and let S ⊂ A, and let a ⊂ A be the two-sided ideal generated by S. Let f : A → B be a ring homomorphism whose kernel contains S. Then there is a unique homomorphism f : A/a → B such that the following diagram commutes /B, AB = BB || BB | | B || π BB || f ! A/a f

where π : A → A/a is the canonical map. Proof The kernel of f is a two-sided ideal of A containing S, so we must have a ⊂ ker f by Lemma 7.2.18. The result now follows from Proposition 7.2.9. When a ring homomorphism f : B → A is surjective, we can use information about A and about ker f to deduce information about B. An important topic in ideal theory is a discussion of the various ways in which we can obtain new ideals from old. Recall first (from Lemma 4.1.18) that if G is a group and if H and K are subgroups of G with K ⊂ NG (H), then KH = {kh | k ∈ K, h ∈ H} is a subgroup of G. Thus, if G is abelian, written in additive notation, and if H and K are any subgroups of G, we have a subgroup H + K given by H + K = {h + k | h ∈ H, k ∈ K}. The next lemma is elementary. Lemma 7.2.20. Let a and b be left (resp. right) ideals in the ring A. Then the subgroups a + b and a ∩ b are also left (resp. right) ideals of A. We have already seen an important use of sums of ideals. Indeed, the greatest common divisor of two integers, m and n, was defined to be the generator of the ideal (m) + (n). We can generalize the notion of the ideals generated by a set. Definitions 7.2.21. Let A be any ring. Let a be a left ideal of A, let b be a right ideal of A, and let S be any subset of A. We write aS = {a1 x1 + · · · + ak xk | k ≥ 1, ai ∈ a and xi ∈ S for 1 ≤ i ≤ k} Sb = {x1 b1 + · · · + xk bk | k ≥ 1, xi ∈ S and bi ∈ b for 1 ≤ i ≤ k} Thus, under either of the above definitions, ab = {a1 b1 + · · · + ak bk | k ≥ 1, ai ∈ a and bi ∈ b for 1 ≤ i ≤ k}. These constructions behave as follows. Lemma 7.2.22. Let A be any ring. Let a be a left ideal of A, let b be a right ideal of A, and let S be any subset of A. Then aS is a left ideal of A and Sb is a right ideal of A. Combining these two facts, we see that ab is a two-sided ideal of A. Finally, if a and b are both two-sided ideals of A, then we have an inclusion of twosided ideals ab ⊂ a ∩ b.


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Proof We leave the assertions of the first paragraph to the reader. For the final statement, consider an element of the form a1 b1 + · · · + ak bk with ai ∈ a and bi ∈ b for all i. Since b is a left ideal, each term ai bi lies in b. Since a is a right ideal, each term ai bi lies in a. So their sum lies in a ∩ b. We shall explore the effects of taking products and intersections of ideals in Z in the next set of exercises. The following use of sums and intersections can be very useful in the study of rings, as well as for the module theory we shall introduce in Section 7.7. Proposition 7.2.23. (Chinese Remainder Theorem) Suppose given two-sided ideals a1 , . . . , ak of the ring A such that ai + aj = A whenever i = j. Let f : A → A/a1 × · · · × A/ak be given by f (a) = (a, . . . , a). Then f induces a ring isomorphism ∼ =

f : A/b −→ A/a1 × · · · × A/ak , where b =

k i=1

ai .

Proof By the obvious extension of Lemma 7.1.24 to k-fold products, kf is a ring homomorphism. Thus, it suffices to show that f is surjective, with kernel i=1 ai . The latter statement is immediate, as the kernel of a map into a product is always the intersection of the kernels of the associated maps into the factors. Thus, we need only show that f is surjective. But the product A/a1 × · · · × A/ak is generated as an abelian group by the images of the A/ai under the canonical inclusions. Thus, it suffices to show that these images are contained in the image of f . We show this for i = 1. Now, we claim that it suffices to show that (1, 0, . . . , 0) is in the image of f . To see this, suppose that f (x) = (1, 0, . . . , 0). Since f is a ring homomorphism f (ax) = (a, . . . , a) · (1, 0, . . . , 0) = (a, 0, . . . , 0) for all a ∈ A, and hence A/a1 × 0 × · · · × 0 is indeed contained in the image of f . We now make use of the fact that a1 + ai = A for all i = 1. Thus, for 2 ≤ i ≤ k, there are elements ai ∈ a1 and bi ∈ ai , with ai + bi = 1. Then if πi : A → A/ai is the canonical map for 1 ≤ i ≤ k, we have π1 (bi ) = 1 and πi (bi ) = 0 for 2 ≤ i ≤ k. Let b = b2 · · · bk . Since the canonical maps πi are ring homomorphisms, π1 (b) = 1, while πi (b) = 0 for 2 ≤ i ≤ k. But this just says that f (b) = (1, 0, . . . , 0) as desired. Exercises 7.2.24. 1. Let A and B be rings. Show that the left ideals of A × B all have the form a × b, where a is a left ideal of A and b is a left ideal of B. 2. Let A be an integral domain and let a, b ∈ A. Show that (a) = (b) if and only if there is a unit x ∈ A× such that ax = b. 3. Let A be a ring. Show that there is a ring homomorphism from Zn to A if and only if the characteristic of A divides n. Show that this ring homomorphism, if it exists, is unique. 4. Let f : A → B be a ring homomorphism. Show that f −1 (a) is a left (resp. right or two-sided) ideal of A for each left (resp. right or two-sided) ideal a of B.


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† 5. Let f : A → B be a surjective ring homomorphism. Show that f (a) is a left (resp. right or two-sided) ideal of B whenever a is a left (resp. right or two-sided) ideal of A. Deduce that there is a one-to-one correspondence between the ideals of B and the ideals of A containing ker f . 6. Show that every subgroup of Zn is a principal ideal, and hence that Zn is a principal ideal ring. 7. Find a ring homomorphism f : A → B that is not surjective and an ideal a ⊂ A such that f (a) is not an ideal of B. 8. Let a ⊂ A be a left ideal. Show that a = A if and only if a contains an element which has a left inverse. 9. Show that a principal left ideal Ax = A if and only if x has a left inverse. 10. A division ring has been shown to have no two-sided ideals other than 0 and the ring itself. Show that the paucity of two-sided ideals does not characterize division rings: Show that if K is a field and n ≥ 2, then Mn (K) has no two-sided ideals other than 0 and itself. (Hint: The cleanest way to show this uses linear algebra. The reader may wish to come back to this problem after reading Sections 7.9 and 7.10.) † 11. Repeat the preceding problem, replacing K by a division ring D. † 12. Let A be a ring. Define the opposite ring Aop as follows. As an abelian group, Aop is just A. For a ∈ A, we write ) a for the associated element of Aop . We define the ) Show that Aop is a ring. Show that the left multiplication in Aop by ) a · )b = ba. ideals of A are in one-to-one correspondence with the right ideals of Aop , and that the right ideals of A are in one-to-one correspondence with the left ideals of Aop . ‡ 13. Let A be a ring. An anti-automorphism of A is an automorphism f : A → A of the additive group of A, such that f (ab) = f (b)f (a) for all a, b ∈ A and f (1) = 1. Show that A admits an anti-automorphism if and only if A and Aop are isomorphic. 14. We say that a ring is self-opposite if it admits an anti-automorphism. Show that commutative rings are self-opposite. 15. Let A be a self-opposite ring. Show that Mn (A) is self-opposite for all n ≥ 1. (Hint: See Problem 5 of Exercises 7.10.23.) 16. Let m, n ∈ Z. Calculate the generator of the product ideal (m)(n). 17. Let m, n ∈ Z. Calculate the generator of (m) ∩ (n). 18. Let n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes. Show that Zn is isomorphic as a ring to the direct product Zpr11 × · · · × Zprk . k

19. Use the preceding problem to give a quick proof that if n = pr11 . . . prkk , then Aut(Zn ) is isomorphic to Aut(Zpr11 ) × · · · × Aut(Zprk ). k

20. Suppose given an infinite family {ai | i ∈ I} of left ideals of A. Define the sum of this infinite family by ai = {ai1 + · · · + aik | k ≥ 1, i1 , . . . , ik ∈ I, and aij ∈ aij for 1 ≤ j ≤ k}. i∈I


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(a) Show that i∈I ai is the smallest left ideal that contains each ai . (b) If S is any subset of A, show that AS = x∈S Ax.

7.3

Polynomials

One of the most important constructions in ring theory is that of polynomial rings. We study the polynomials in one variable now and study the polynomials in several variables in a subsection (page 228). Definitions 7.3.1. Let A be a ring (not necessarily commutative) and let X be a variable. We define a ring A[X] as follows. The elements of A[X] are formal sums n i a X = an X n + · · · + a1 X + a0 , where n is a nonnegative integer and ai ∈ A for all i i=0 n i. We often use functional notation for these sums, writing, say, f (X) = i=0 ai X i for the element above. If an = 0, we say that f (X) has degree n and that an is its leading coefficient. Otherwise, m if m is the largest integer for which am = 0, we identify f (X) with the polynomial i=0 ai X i = am X m + · · · + a1 X + a0 , and hence it has degree m. (In the case of 0 = 0X 0 , there are no possible identifications with lower degree polynomials, and we say that 0 has degree −∞.) We say that a polynomial f (X) is monic if its leading coefficient is 1. As is implicit in the notations above, we write either a or aX 0 for the degree 0 polynomial whose 0-th coefficient is a. We write ι : A → A[X] for the map given by ι(a) = aX 0 for all a ∈ A. We also write X n for 1X n and write X for X 1 . Note that ntwo polynomials are equal if and only if their coefficients are equal. (Here, if f (X) = i=0 ai X i , then the coefficients of the X m for m > n are all implicitly n 0.) Addition of polynomials is obtained by adding the coefficients: If f (X) = i=0 ai X i n and g(X) = i=0 bi X i (where n is the larger of the degrees of f and g), then f (X) + g(X) =

n

(ai + bi )X i .

i=0

of polynomials is given as follows: if f (X) = mMultiplication i b X , then their product is given by i i=0 ⎛ ⎞ n+m i ⎝ aj bi−j ⎠ X i . f (X)g(X) = i=0

n i=0

ai X i and g(X) =

j=0

Note that if A is noncommutative, then the order of the products ai bi−j is important. Finally, we give a special name to some of the simplest polynomials: We shall refer to a polynomial of the form aX i for some a ∈ A and i ≥ 0 as a monomial. Note that every polynomial is a sum of monomials, so that the monomials generate A[X] as an abelian group. We will occasionally use other letters, such as T or Y for the variable of a polynomial ring. We shall refer to the variable as an indeterminate element over the ring A. Again, the reader should justify the definitions: Lemma 7.3.2. Let A be a ring. Then A[X] is a ring under addition and multiplication of polynomials, and ι : A → A[X] is a ring homomorphism. The element X ∈ A[X] commutes with every element of ι(A), and if A is commutative, so is A[X].


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Note that since every polynomial is a sum of monomials, the distributive law shows that the multiplication in A[X] is determined by the rule aX i · bX j = abX i+j . In particular, we see that if A is a subring of the real numbers R, then the multiplication formula is exactly the one we get by multiplying polynomial functions over R. So polynomial rings are familiar objects after all. Polynomial rings have a very important universal property. Proposition 7.3.3. Let g : A → B be a ring homomorphism and let b ∈ B be an element that commutes with every element of g(A). Then there is a unique ring homomorphism g b : A[X] → B such that g b (X) = b and the following diagram commutes. g /B AD DD z= z DD zz D zzgb ι DD z z ! A[X]

Moreover, g b satisfies the formula " n # n i gb ai X = g(ai )bi . i=0

i=0 0

In this formula, we adopt the convention that b = 1 for any b. Proof For the uniqueness statement, note that any ring homomorphism g b that carries X to b and makes the diagram commute must satisfy the stated formula: " n # n i = gb ai X g b (ι(ai ))(g b (X))i i=0

i=0

=

n

g(ai )bi .

i=0

Also, the stated formula does make the diagram commute and carry X to b. Thus, it suffices to show that g b , as defined by the formula, gives a ring homomorphism. That it is a homomorphism of additive groups follows from the distributive law in B. And g b (1) = ι(1) = 1, so it suffices to show that g b (f1 (X)f2 (X)) = g b (f1 (X))g b (f2 (X)) for all f1 (X), f2 (X) ∈ A[X]. Since the monomials generate A[X] additively, distributivity shows that it is sufficient to check this for the case that f1 (X) and f2 (X) are monomials. Thus, we must check that for a, a ∈ A and i, j ≥ 0, we have g(aa )bi+j = g(a)bi g(a )bj . Since b commutes with any element of g(A), so does bi by induction. Thus, g(a)bi g(a )bj = g(a)g(a )bi bj . That bi bj = bi+j follows from the power laws for monoids (Problem 5 of Exercises 2.2.18). Thus, the result follows from the fact that g is a ring homomorphism. If A is commutative, then so is A[X], and A[X] is an A-algebra via ι : A → A[X]. As such, it satisfies an important universal property. Corollary 7.3.4. Let A be a commutative ring and let B be an A-algebra, via ν : A → B. Then for each b ∈ B, there is a unique A-algebra homomorphism εb : A[X] → B with the property that εb (X) = b. It is given by the formula n n εb ( ai X i ) = ν(ai )bi , i=0

i=0


224

where we set b0 = 1 for any b. Proof Since ν : A → B is the structure map for B as an A-algebra, each element of ν(A) commutes with each b ∈ B. Thus, Proposition 7.3.3 provides a unique ring homomorphism ν b : A[X] → B with ν b ◦ ι = ν (i.e., ν b is an A-algebra map) and ν b (X) = b. So we set εb = ν b . The formula comes from Proposition 7.3.3. Definitions 7.3.5. Let B be an A-algebra and let b ∈ B. We shall refer to the unique A-algebra homomorphism, εb : A[X] → B, that carries X to b as the evaluation map obtained by evaluating X at b. In consonance with this language, for f (X) ∈ A[X], we sometimes write εb (f (X)) = f (b), as it is obtained by evaluating f (X) at X = b. We shall write A[b] for the image of εb . We call it the A-subalgebra of B generated by b, or the algebra obtained by adjoining b to A. Note that since εb : A[X] → A[b] is onto, A[b] must be commutative even if B is not. Examples 7.3.6. Recall that every ring is a Z-algebra in a unique way. Consider the complex numbers, C, as a Z-algebra, and form the Z-subalgebra, Z[i], obtained by adjoining i to Z. The ring Z[i] is known as the Gaussian integers. Note that i has order 4 in the unit group C× . Thus, the powers of i consist of i, −1, −i, and 1. Collecting terms, we see that every element of Z[i] has the form m + ni, with m, n ∈ Z, and hence Z[i] = {m + ni ∈ C | m, n ∈ Z}. As the addition and multiplication operations are inherited from those in C, we have (m + ni) + (k + li) = (m + k) + (n + l)i and (m + ni) · (k + li) = (mk − nl) + (ml + nk)i for m, n, k, l ∈ Z. Note also that since C is a Q-algebra, the same arguments may be used if we replace Z by Q: Adjoining i to Q, we obtain Q[i] = {a + bi ∈ C | a, b ∈ Q}. Recall from Lemma 7.1.35 that for n ≥ 1, the complex number ζn = ei2π/n has order n in C× . Also, ζ4 = i. Thus, the next example is a generalization of the previous one. Examples 7.3.7. The non-negative powers of ζn are 1, ζn , . . . , ζnn−1 . Thus, if A = Z or Q, the elements of A[ζn ] may all be written as sums a0 + a1 ζn + · · · + an−1 ζnn−1 with a0 , . . . , an−1 ∈ A. We shall refer to Z[ζn ] as the cyclotomic integers obtained by adjoining a primitive n-th root of unity and to Q[ζn ] as the cyclotomic extension of Q obtained by adjoining a primitive n-th root of unity. As we saw in the case of ζ4 = i, elements of A[ζn ], with A = Z or Q, do not have unique representations as sums of the form a0 + a1 ζn + · · · + an−1 ζnn−1 with a0 , . . . , an−1 ∈ A. We shall defer the determination of a unique representation for these elements until Section 8.8. We shall show in Proposition 8.2.5 that the rings Q[ζn ] are all subfields of C. Remarks 7.3.8. Let A be a commutative ring. Note that if A is a subring of the center of B and if we view B as an A-algebra whose structure map is the inclusion of A in B, then for b ∈ B, we have n n εb ( ai X i ) = ai bi , i=0 n

A[b] = {

i=0

and hence

i=0

ai bi | n ≥ 0 and ai ∈ A for 0 ≤ i ≤ n}.


225

We now give an important property of A[b]. Lemma 7.3.9. Let B be an A-algebra and let b ∈ B. Then A[b] is the smallest Asubalgebra of B that contains b. Proof An A-subalgebra of B is any subring of B containing ν(A), where ν : A → B is the structure map of B as an A-algebra. Thus, if C is an A-subalgebra of B that n contains b, it must contain every sum i=0 ν(ai )bi , and hence contains A[b]. Since A[b] is an A-subalgebra of B containing b, the result follows. We shall now compute the kernels of the evaluation maps εa : A[X] → A, for a ∈ A and give some applications. The next result is a version of the Euclidean algorithm. n i Proposition 7.3.10. Let A be any ring and let g(X) = i=0 ai X be a polynomial whose leading coefficient, an , is a unit in A. Then for any polynomial f (X) ∈ A[X], we may find polynomials q(X), r(X) ∈ A[X] such that f (X) = q(X)g(X) + r(X) and such that the degree of r(X) is strictly less than that of g(X). Proof We argue by induction on the degree of f (X). If f (X) has degree less than n, then we may simply take q(X) = 0 and take r(X) = f (X), and the statement is satisfied. m Thus, suppose that f (X) = i=0 bi X i with bm = 0 and that the result is true for all polynomials of degree less than m. In particular, we may assume that m ≥ n. But m−n then (bm a−1 )g(X) is a polynomial of degree m whose leading coefficient is the n X m−n )g(X) has degree less than m. By same as that of f (X), and hence f (X) − (bm a−1 n X m−n induction, we may write f (X) − (bm a−1 X )g(X) = q1 (X)g(X) + r(X), where r(X) n m−n ) and we has degree less than n. But then we may take q(X) = q1 (X) + (bm a−1 n X have f (X) = q(X)g(X) + r(X), as desired. As the reader may recall from calculus, the simplest application for division of polynomials is to the theory of roots. Definition 7.3.11. Let A be a commutative ring and let f (X) ∈ A[X]. Let B be an A-algebra. We say that b ∈ B is a root of f if f (X) lies in the kernel of the evaluation map εb : A[X] → B obtained by evaluating X at b. In particular, a ∈ A is a root of f if and only if f (a) = 0. Corollary 7.3.12. Let A be a commutative ring and let a ∈ A. Then the kernel of the evaluation map εa : A[X] → A is (X − a), the principal ideal generated by X − a. In particular, a ∈ A is a root of f (X) ∈ A[X] if and only if f (X) = (X − a)q(X) for some q(X) ∈ A[X]. Proof Since X − a has degree 1 and leading coefficient 1, we may write any polynomial f (X) ∈ A[X] as f (X) = q(X)(X − a) + r(X) where r(X) has degree ≤ 0. In other words, f (X) = q(X)(X − a) + r, where r ∈ A. But the evaluation map is a ring homomorphism, so that f (a) = q(a)(a − a) + r = r. Thus, f (X) ∈ ker εa if and only if r = 0. Thus, f (X) ∈ ker εa if and only if f (X) = q(X)(X − a) for some q(X) ∈ A[X]. But f (X) = q(X)(X − a) just says that f (X) is in the principal ideal generated by X − a. Corollary 7.3.13. Let A be an integral domain and let f (X) ∈ A[X] be a polynomial of degree n. Then f has at most n distinct roots in A.


226

Proof Suppose that a ∈ A is a root of f . Then f (X) = (X − a)g(X), where g(X) has degree n − 1. Thus, by induction, there are at most n − 1 roots of g. Thus, it suffices to show that if b = a is a root of f , then b is a root of g. But 0 = f (b) = (b − a)g(b). Since b − a = 0 and since A is an integral domain, b must be a root of g. We obtain a very useful consequence for field theory. Corollary 7.3.14. Let K be a field. Then any finite subgroup of K × is cyclic. Proof Recall from Problem 4 of Exercises 4.4.24 that a finite abelian group is cyclic if and only if it has at most d elements of exponent d for each d which divides its order. We claim that K × itself has at most d elements of exponent d for any integer d. To see this, note that 1 is the identity element of K × , so that if a ∈ K × has exponent d, we have ad = 1. But this says that a is a root of the polynomial X d − 1. Since X d − 1 has degree d, it can have at most d distinct roots, and hence K × can have at most d distinct elements of exponent d. As promised during our discussion of group theory, we obtain the following corollary. Corollary 7.3.15. Let K be a finite field. Then K × is a cyclic group. Corollary 7.3.15 is a useful consequence of the determination of the kernel of the evaluation map εa : A[X] → A. Later on, we shall derive useful consequences from the determinations of the kernels of the evaluation maps εb : A[X] → B for particular A-algebras B, and elements b ∈ B. Thus, Corollary 7.3.12 should be thought of as a first step in an important study. Exercises 7.3.16. 1. Give the proof of Lemma 7.3.2. 2. Suppose that A is noncommutative. What are the elements of the center of A[X]? 3. Let f (X), g(X) ∈ A[X]. Show that the degree of f (X) + g(X) is less than or equal to the larger of the degrees of f and g. 4. Let f (X), g(X) ∈ A[X]. Show that the degree of f (X) · g(X) is less than or equal to the sum of the degrees of f and g. 5. Show that A[X] is an integral domain if and only if A is. 6. Show that the units of A[X] are the image under ι of the units of A. 7. Let

M=

1 0

1 1

∈ M2 (R).

Consider the evaluation map εM : R[X] → M2 (R). Show that a polynomial f (X) ∈ R[X] lies in the kernel of εM if and only if f (1) = 0 and f (1) = 0, where f (X) is the derivative of f in the sense of calculus.


8. Let

M=

227

a 1 0 a

∈ M2 (R)

for some real number a. Consider the evaluation map εM : R[X] → M2 (R). Show that a polynomial f (X) ∈ R[X] lies in the kernel of εM if and only if f (a) = 0 and f (a) = 0, where f (X) is the derivative of f in the sense of calculus. 9. Let p be a prime. What are the elements of Z[1/p] ⊂ Q? 10. What are the elements of A[X 2 ] ⊂ A[X]? 11. What are the elements of A[X n ] ⊂ A[X]? 12. What are the elements of Z[2X] ⊂ Z[X]? 13. Show that C = R[i]. ‡ 14. Show that Q[i] is a subfield of C. √ 15. What are the elements of Z[ 2] ⊂ R? √ 16. What are the elements of Q[ 2] ⊂ R? 17. Let m be a positive, odd integer. Show that Z[ζm ] = Z[ζ2m ] as a subring of C. 18. Show that every element of Z[ζ8 ] may be written in the form a + bζ8 + cζ82 + dζ83 with a, b, c, d ∈ Z. Write out the formula for the product (a + bζ8 + cζ82 + dζ83 ) · (a + b ζ8 + c ζ82 + d ζ83 ). 19. Let A be a ring and let 1 = a ∈ A. Show that ak − 1 = (a − 1)(1 + a + · · · + ak−1 ) for k ≥ 1. ‡ 20. Cyclotomic units Let m and k be relatively prime to n. Show that the element (ζnm − 1)/(ζnk − 1) ∈ Q[ζn ] actually lies in Z[ζn ]. Deduce that (ζnm − 1)/(ζnk − 1) is a unit in Z[ζn ]. More generally, if (m, n) = (k, n) = n, show that (ζnm − 1)/(ζnk − 1) is a unit in Z[ζn ]. The elements (ζnm − 1)/(ζnk − 1) are often called the cyclotomic units. 21. Let n > 3 be odd. Show that the cyclotomic unit (ζn2 − 1)/(ζn − 1) has infinite order in Z[ζn ]× . 22. If (m, n) = 1 and m ≡ ±1 mod n, show that the cyclotomic unit (ζnm − 1)/(ζn − 1) has infinite order in Z[ζn ]× . (Hint: If 0 < φ < π/2, show that ei(θ−φ) + ei(θ+φ) = reiθ for some positive real number r. Show also that r > 1 if φ < π/3.) 23. Let n be an even number greater than 6. Show that Z[ζn ]× is infinite. 24. Show that in the category of rings, Z[X] is the free object on a one-element set.


228

25. Show that every element of Z6 is a root of X 3 − X ∈ Z6 [X]. Deduce that the hypothesis that A be an integral domain in Corollary 7.3.13 is necessary. 26. Show that X 2 + 1 has infinitely many roots in the quaternion ring, H. (The issue here is not zero-divisors, as H is a division ring. The point here is that evaluating polynomials at an element α ∈ H does not give a ring homomorphism from H[X] to H unless α lies in the center of H.) !∞ 27. Show that X p − X has infinitely many roots in i=1 Zp . 28. Let p1 , . . . , pk be distinct primes and let m be the least common multiple of p1 − 1, . . . , pk − 1. Show that every element of Zn is a root of X m+1 − X, where n = p1 . . . pk . 29. Let n > 1 be an integer and let (n, X) be the ideal of Z[X] generated by n and X. Show that Z[X]/(n, X) is isomorphic to Zn . 30. Let B be any ring and let A be the center of B. Let a ∈ A. Show that there is a unique homomorphism εa : B[X] → B such that εa (X) = a and εa ◦ ι = 1B , where ι : B → B[X] is the standard inclusion. Show that the kernel of εa is the two-sided ideal of B[X] generated by X − a. 31. Let A be a ring and X a variable. We define the ring of formal power series, A[[X]], with coefficients in A. ∞ The elements of A[[X]] are formal sums i=0 ai X i , with ai ∈ A for all i ≥ 0. We shall refer to them as formal power series with coefficients in A. Unlike the case of polynomial rings, we do not assume that only finitely many of the coefficients are nonzero; indeed, all of the coefficients may be nonzero. Here, the sum and product operations are given by ∞

"∞ i=0

i=0

ai X i +

∞

bi X i

i=0

=

∞

(ai + bi )X i ,

and

i=0

⎛ ⎞ # "∞ # ∞ i ⎝ = ai X i · bi X i aj bi−j ⎠ X i . i=0

i=0

j=0

We have an inclusion map η : A[X] → A[[X]], that identifies a polynomial, f (X), of degree n with the formal power series whose coefficients of index ≤ n agree with those of f , and whose higher coefficients are all 0. (a) Show that A[[X]] is a ring and that η is a ring homomorphism. (b) Show that ∞the polynomial 1 − X is a unit in A[[X]] (but not in A[X]) with inverse i=0 X i = 1 + X + X 2 + · · · + X n + · · · . ∞ (c) Let A be commutative. Show that a formal power series i=0 ai X i is a unit in A[[X]] if and only if a0 ∈ A× .

Polynomials in Several Variables We can also take polynomials in more than one variable. When considering polynomials in several variables, the most common case, which is the default if nothing more is said, is to assume that the variables commute with each other. In fact, we shall assume that


229

this is the case whenever we use the word polynomial, but some people would explicitly call these “polynomials in several commuting variables.” Definitions 7.3.17. Suppose given n variables X1 , . . . , Xn . A primitive7 monomial in these variables is a product X1i1 · · · Xnin , with ij ≥ 0 for 1 ≤ j ≤ n. We abbreviate X1i1 · · · Xnin = X I for I = (i1 , . . . , in ) and write I = {(i1 , . . . , in ) | i1 , . . . , in ∈ Z, i1 , . . . , in ≥ 0} for the set of all n-tuples of non-negative integers. (We call the elements of I multiindices.) For any ring A, we define a monomial in X1 , . . . , Xn with coefficient in A to be a product aX1i1 · · · Xnin where a ∈ A and X1i1 · · · Xnin is a primitive monomial in X1 , . . . , Xn . A polynomial in X1 , . . . , Xn with coefficients in A is a finite sum of monomials in X1 , . . . , Xn with coefficients in A. The elements of the polynomial ring A[X1 , . . . , Xn ] are the polynomials in X1 , . . . , Xn with coefficients in A. Since there’s no preferred order for the multi-indices of the monomials, we shall write the generic polynomial in the form f (X1 , . . . , Xn ) = aI X I , I∈I

where all but finitely many of the coefficients aI ∈ A are 0. So f is the sum of the finitely many monomials whose coefficients are nonzero. Addition of polynomials is obtained as in the one-variable case, by adding the co I efficients of the monomials: if f (X , . . . , X ) = a X and g(X1 , . . . , Xn ) = 1 n I I∈I I b X , then I I∈I f (X1 , . . . , Xn ) + g(X1 , . . . , Xn ) = (aI + bI )X I . I∈I

The multiplication in A[X1 , . . . , Xn ] is set up to have the following effect on monomials: If I = (i1 , . . . , in ) and J = (j1 , . . . , jn ), then aX I · bX J = abX I+J , where I + J = (i1 + j1 , . . . , in + jn ). For generic polynomials, this is expressed as follows: If f (X1 , . . . , Xn ) = I∈I aI X I and g(X1 , . . . , Xn ) = I∈I bI X I , then ⎛ ⎞ ⎝ aJ bI−J ⎠ X I . f (X1 , . . . , Xn ) · g(X1 , . . . , Xn ) = I∈I

J≤I

Here, for I = (i1 , . . . , in ) and J = (j1 , . . . , jn ) in I, J ≤ I means that jk ≤ ik for all indices k (so that ≤ is only a partial order on I), and we define I − J to be the expected multi-index (i1 − j1 , . . . , in − jn ). We write 0 for the multi-index whose coordinates are all 0, and identify a ∈ A with a · X 0 . We write ι : A → A[X1 , . . . , Xn ] for the map that takes each a ∈ A to a = aX 0 ∈ A[X1 , . . . , Xn ]. We identify each primitive monomial X I with 1X I and identify each variable Xi with the primitive monomial X (0,...,0,1,0,...,0) , where the 1 occurs in the i-th place. When n = 2, we shall customarily write A[X, Y ] in place of A[X1 , X2 ]. We require the usual justifications. 7 The use of the word primitive here has no connection whatever to the notion of a primitive element in a Hopf algebra.


230

Lemma 7.3.18. Suppose given a ring A and variables X1 , . . . , Xn . Then the polynomials A[X1 , . . . , Xn ] form a ring under the above addition and multiplication formulæ, and ι : A → A[X1 , . . . , Xn ] is a ring homomorphism. Each variable Xi commutes with every element of ι(A), and the variables Xi and Xj commute with each other for all i, j. If A is commutative, so is A[X1 , . . . , Xn ]. More generally, a polynomial f (X1 , . . . , Xn ) lies in the center of A[X1 , . . . , Xn ] if and only if its coefficients lie in the center of A. Polynomials in several variables satisfy universal properties similar to those of polynomials in one variable. We leave the proofs to Exercises 7.3.27. Proposition 7.3.19. Let g : A → B be a ring homomorphism and suppose given elements b1 , . . . , bn ∈ B such that the following two conditions hold. 1. Each bi commutes with every element of g(A). 2. For i, j ∈ {1, . . . , n}, bi commutes with bj . Then there is a unique ring homomorphism g b1 ,...,bn : A[X1 , . . . , Xn ] → B such that g b1 ,...,bn (Xi ) = bi for 1 ≤ i ≤ n and the following diagram commutes. g /B A LL LLL rr8 r r LLL r rrr ι LLL rrr gb1 ,...,bn % A[X1 , . . . , Xn ]

Moreover, g b1 ,...,bn satisfies the formula " # I = g b1 ,...,bn aI X g(aI )bI . I∈I I

Here, if I = (i1 , . . . , in ), we set b =

bi11

I∈I

. . . binn ,

where b0i is set equal to 1 for all i.

As in the case of a single variable, Proposition 7.3.19 admits a useful restatement in the context of algebras over a commutative ring. Corollary 7.3.20. Let A be a commutative ring and let B be an A-algebra, via ν : A → B. Suppose given b1 , . . . , bn ∈ B such that bi commutes with bj whenever i, j ∈ {1, . . . , n}. Then there is a unique A-algebra homomorphism εb1 ,...,bn : A[X1 , . . . , Xn ] → B with the property that εb (Xi ) = bi for 1 ≤ i ≤ n. It is given by the formula " # I = εb1 ,...,bn aI X ν(aI )bI . I∈I

I∈I

Here, if I = (i1 , . . . , in ), we set bI = bi11 . . . binn , where b0i is set equal to 1 for all i. We now codify some terms. Definitions 7.3.21. Let A be a commutative ring. Let B be an A-algebra and let b1 , . . . , bn ∈ B such that bi commutes with bj whenever i, j ∈ {1, . . . , n}. We shall refer to the A-algebra homomorphism εb1 ,...,bn as the evaluation map obtained by evaluating Xi at bi for 1 ≤ i ≤ n. We write A[b1 , . . . , bn ] for the image of εb1 ,...,bn . We call A[b1 , . . . , bn ] the A-subalgebra of B generated by b1 , . . . , bn . Suppose now that B is a commutative A-algebra. We say that B is finitely generated as an A-algebra (or has finite type as an A-algebra) if B = A[b1 , . . . , bn ] for some collection of elements b1 , . . . , bn ∈ B.


231

A special case of the evaluation maps is of fundamental importance in algebraic geometry. Proposition 7.3.22. Let A be a commutative ring and let a1 , . . . , an ∈ A. Let a = (X1 − a1 , . . . , Xn − an ), the ideal of A[X1 , . . . , Xn ] generated by X1 − a1 , . . . , Xn − an . Then a is the kernel of the evaluation map εa1 ,...,an : A[X1 , . . . , Xn ] → A. Proof Since each Xi −ai lies in the kernel of εa1 ,...,an , Corollary 7.2.19 shows that there is a unique ring homomorphism εa1 ,...,an : A[X1 , . . . , Xn ]/a → A such that the following diagram commutes. εa1 ,...,an

/ A[X1 , . . . , Xn ] q8 A RRR q q RRR qq RR qqq q π RRRR q ε a ,...,a n R( qq 1 A[X1 , . . . , Xn ]/a Here, π : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn ]/a is the canonical map. It suffices to show that εa1 ,...,an is an isomorphism. Let ι = π ◦ ι : A → A[X1 , . . . , Xn ]/a, where ι is the A-algebra structure map of A[X1 , . . . , Xn ]. Then εa1 ,...,an ◦ ι = εa1 ,...,an ◦ π ◦ ι = εa1 ,...,an ◦ ι = 1A , where the last equality comes from the fact that εa1 ,...,an is an A-algebra homomorphism. Thus, it suffices to show that ι is onto. Since every element of A[X1 , . . . , Xn ] is a sum of monomials and since π is onto, it suffices to show that π(aX1i1 . . . Xnin ) lies in the image of ι for all a ∈ A and all n-tuples (i1 , . . . , in ) of nonnegative integers. Since π(Xi − ai ) = 0, we have π(Xi ) = ι(ai ) for all i. Thus, π(aX1i1 . . . Xnin ) = ι(aai11 . . . ainn ), and the result follows. The next result could have been left as an exercise in universal properties, but it is sufficiently important that we give it here. Proposition 7.3.23. Let X1 , . . . , Xn be indeterminates over the ring A. Then there is a unique ring homomorphism α : (A[X1 , . . . , Xn−1 ])[Xn ] → A[X1 , . . . , Xn ] that carries Xi to Xi for 1 ≤ i ≤ n, and commutes with the natural inclusions of A in the two sides. Moreover, α is an isomorphism whose inverse is the unique ring homomorphism β : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn−1 ][Xn ] which carries Xi to Xi for 1 ≤ i ≤ n, and commutes with the natural inclusions of A in the two sides. Thus, A[X1 , . . . , Xn ] may be considered to be the polynomial ring in the variable Xn with coefficients in A[X1 , . . . , Xn−1 ]. In particular, every polynomial f (X1 , . . . , Xn ) in n variables over A may be written uniquely in the form f (X1 , . . . , Xn ) = fk (X1 , . . . , Xn−1 )Xnk + · · · + f0 (X1 , . . . , Xn−1 ) for some k ≥ 0, where fi (X1 , . . . , Xn−1 ) ∈ A[X1 , . . . , Xn−1 ] for all i.


232

Proof Here, the natural inclusion of A in A[X1 , . . . , Xn−1 ][Xn ] is the composite ι

ι

A− → A[X1 , . . . , Xn−1 ] − → A[X1 , . . . , Xn−1 ][Xn ], where ι is the standard inclusion of A in the polynomial ring in n − 1 variables and ι is the standard inclusion of A[X1 , . . . , Xn−1 ] in its polynomial ring in the variable Xn . Notice that the variables Xi lie in the centers of each of the rings under consideration. Thus, the existence and uniqueness of β is immediate from Proposition 7.3.19. For α, note that Proposition 7.3.19 provides a unique homomorphism α0 : A[X1 , . . . , Xn−1 ] → A[X1 , . . . , Xn ] that carries each Xi to Xi and commutes with the natural inclusions of A in both sides. But now the universal property of a polynomial ring in a single variable allows us to extend α0 to a map α carrying Xn to Xn . For the uniqueness of α, we claim that any two ring homomorphisms f, g : A[X1 , . . . , Xn−1 ][Xn ] → B that agree on A and agree on each of the variables Xi must be equal. To see this, note first that f and g must agree on A[X1 , . . . , Xn−1 ] by the uniqueness property of Proposition 7.3.19. But then, since f and g also agree on Xn , they must be equal by the uniqueness property of Proposition 7.3.3. But now notice that each of the composites α ◦ β and β ◦ α carries each Xi to Xi and commutes with the standard inclusions of A. Thus, α ◦ β is the identity by the uniqueness property of Proposition 7.3.19, while β ◦ α is the identity by the uniqueness property derived in the preceding paragraph. If A is commutative and f (X1 , . . . , Xn ) ∈ A[X1 , . . . , Xn ], then there is a function f : An → A which takes (a1 , . . . , an ) to f (a1 , . . . , an ), the effect of evaluating f at (a1 , . . . , an ). Indeed, much of our intuition about polynomials will have come from calculus, where polynomials are treated as functions. Of course, if A is finite, then there are only finitely many functions from An to A, but infinitely many polynomials in A[X1 , . . . , Xn ]. Thus, in this case, a polynomial is not determined by its effect as a function on An . Finiteness is not !∞the only difficulty in trying to treat polynomials as functions. For instance, if A = i=1 Z2 , then the polynomial X 2 is easily seen to induce the identity map of A, which is also induced by X. This suggests that if we wish to interpret polynomials as functions, then we might think about restricting attention to integral domains with infinitely many elements. First, we shall formalize the situation. Definition 7.3.24. Let X be any set and let A be a ring. The ring of functions from X to A, written Map(X, A), is the ring whose elements are the functions from X to A and whose operations are given by (f + g)(x) = f (x) + g(x)

and

(f · g)(x) = f (x)g(x)

for all f, g ∈ Map(X, A) and x ∈ X. The additive and multiplicative identities are the constant functions to 0 and 1, respectively.8 The next lemma follows from the fact that the evaluation maps are ring homomorphisms. 8 If X is a compact topological space, then the ring C(X) of continuous functions from X to C plays an important role in the theory of C ∗ algebras. Of course, C(X) is a subring of Map(X, C).


233

Lemma 7.3.25. Let A be a commutative ring and let φ : A[X1 , . . . , Xn ] → Map(An , A) be the map which takes each polynomial f to the function obtained by evaluating f on the elements of An . Then φ is a ring homomorphism. Thus, two polynomials, f and g, induce the same function from An to A if and only if f − g ∈ ker φ. Thus, the elements of A[X1 , . . . , Xn ] are determined by their effect as functions if and only if ker φ = 0. The next proposition gives a slightly stronger result when A is an infinite domain. Proposition 7.3.26. Let S be an infinite subset of the integral domain A. Suppose given a polynomial f (X1 , . . . , Xn ) in n variables over A with the property that f (a1 , . . . , an ) = 0 for all a1 , . . . , an ∈ S. Then f = 0. Thus, an element of A[X1 , . . . , Xn ] is determined by its effect on S × · · · × S ⊂ An for any infinite subset S of A. Proof We argue by induction on n, with the result being true for n = 1, as f would have infinitely many roots. Suppose, inductively, that the result is true for polynomials of n − 1 variables, and let a ∈ S. Then evaluating Xn at a gives a polynomial f (X1 , . . . , Xn−1 , a) in n − 1 variables over A, which vanishes on all (n − 1)-tuples, a1 , . . . , an−1 of elements of S. By the inductive assumption, f (X1 , . . . , Xn−1 , a) = 0. Write B = A[X1 , . . . , Xn−1 ]. By Proposition 7.3.23, we may identify A[X1 , . . . , Xn ] with B[Xn ]. Under this identification, f (X1 , . . . , Xn−1 , a) is the image of f under the evaluation map εa : B[Xn ] → B obtained by evaluating Xn at a. Thus, when f is thought of as an element of B[Xn ], each a ∈ S is a root of f . Since B is a domain and since f has infinitely many roots in B, f = 0. Exercises 7.3.27. 1. Give the proof of Lemma 7.3.18. 2. Give the proof of Proposition 7.3.19. 3. Give the proof of Corollary 7.3.20. ‡ 4. Let B be an A-algebra and let b1 , . . . , bn ∈ B such that bi commutes with bj whenever i, j ∈ {1, . . . , n}. Show that A[b1 , . . . , bn ] is the smallest A-subalgebra of B containing b1 , . . . , bn . 5. Let A be a commutative ring. What are the elements of A[X 2 , X 3 ] ⊂ A[X]? 6. Let A be a commutative ring. What are the elements of A[X 3 , X 4 ] ⊂ A[X]? 7. Let A be a commutative ring and let X and Y be indeterminates. Consider the A-algebra structure on A[X] × A[Y ] given by the structure map Δ

ι×ι

A −→ A × A −−→ A[X] × A[Y ] where Δ is the diagonal map Δ(a) = (a, a) and the maps ι are the standard structure maps for the polynomial rings A[X] and A[Y ]. What are the elements of the A-subalgebra of A[X] × A[Y ] generated by (X, 0) and (0, Y )?


234

8. Show that Z[1/2, 1/3] = Z[1/6] as subrings of Q. (Hint: By Problem 4, it suffices to show that 1/2, 1/3 ∈ Z[1/6] and that 1/6 ∈ Z[1/2, 1/3].) 9. Show that Q is not finitely generated as a Z-algebra. 10. Let m and n be relatively prime positive integers. Show that Z[ζm , ζn ] = Z[ζmn ] 2π as subrings of C. Here, ζk = ei· k is the standard primitive k-th root of unity. 11. Let A be a commutative ring. Show that A[X1 , . . . , Xn ] is the free commutative A-algebra on the set {X1 , . . . , Xn }.

7.4

Symmetry of Polynomials

We consider some issues relating to permutations of the variables in a polynomial ring. Throughout this section, A is a commutative ring. Suppose given n variables, X1 , . . . , Xn . Then the symmetric group on n letters, Sn , acts on the set of variables in the obvious way: σ · Xi = Xσ(i) . By Corollary 7.3.20, the induced map σ : {X1 , . . . , Xn } → {X1 , . . . , Xn } ⊂ A[X1 , . . . , Xn ] extends uniquely to an A-algebra homomorphism σ∗ : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn ] via σ∗ (f (X1 , . . . , Xn )) = f (Xσ(1) , . . . , Xσ(n) ). Since an A-algebra homomorphism out of A[X1 , . . . , Xn ] is determined by its effect on the variables, σ ·f = σ∗ (f ) is easily seen to give an action of Sn on A[X1 , . . . , Xn ] through A-algebra homomorphisms. We first consider the fixed points of this action. Definition 7.4.1. We write A[X1 , . . . , Xn ]Sn for the fixed points of the action of Sn on A[X1 , . . . , Xn ]. Thus, a polynomial f (X1 , . . . , Xn ) is in A[X1 , . . . , Xn ]Sn if and only if σ∗ (f ) = f for all σ ∈ Sn . We shall refer to A[X1 , . . . , Xn ]Sn as the symmetric polynomials in n variables over A. The reader may easily check that the symmetric polynomials form an A-subalgebra of A[X1 , . . . , Xn ]. We shall show that it is a polynomial ring on certain symmetric polynomials. We shall argue by an induction in which the next lemma will play a part. Lemma 7.4.2. Let π : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn−1 ] be the A-algebra map that takes Xi to Xi if i < n and takes Xn to 0. Then π carries the symmetric polynomials in n variables into the symmetric polynomials in n − 1 variables, inducing an A-algebra homomorphism π : A[X1 , . . . , Xn ]Sn → A[X1 , . . . , Xn−1 ]Sn−1 . Proof Recall from Proposition 7.3.23 that A[X1 , . . . , Xn ] may be identified with the polynomial ring on A[X1 , . . . , Xn−1 ] in the variable Xn . Thus, any nonzero element of A[X1 , . . . , Xn ] may be written uniquely in the form f (X1 , . . . , Xn ) =

m i=0

fi (X1 , . . . , Xn−1 )Xni


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for some m ≥ 0, with f0 , . . . , fm ∈ A[X1 , . . . , Xn−1 ], and fm = 0. Now Sn−1 acts as permutations on the first n − 1 variables of the polynomials in n variables, and if σ ∈ Sn−1 , then σ∗ (f ) =

m

σ∗ (fi )Xni .

i=0

But by the uniqueness of the representation of polynomials in n variables as polynomials in Xn with coefficients in A[X1 , . . . , Xn−1 ] we see that if f ∈ A[X1 , . . . , Xn ]Sn , then each fi must be fixed by each σ ∈ Sn−1 . Thus, if f is a symmetric polynomial in n variables, then each fi is a symmetric polynomial in n − 1 variables. But since π(Xn ) = 0, we see that π(f ) = f0 , and the result follows. We now define the elements we will show to be the polynomial generators of the symmetric polynomials. Definition 7.4.3. The elementary symmetric polynomials, or elementary symmetric functions, on n variables are defined by ⎧ 1 if i = 0 ⎪ ⎨ X . . . X if 1 ≤ i ≤ n j j 1 i si (X1 , . . . , Xn ) = ⎪ ⎩ j1 n. More generally, if B is a commutative ring and if b1 , . . . , bn ∈ B, then we write si (b1 , . . . , bn ) ∈ B for the element obtained by evaluating si (X1 , . . . , Xn ) at b1 , . . . , bn . Note that we may do this because the elementary symmetric functions have integer coefficients. Thus, for instance, s1 (X1 , . . . , Xn ) = X1 + · · · + Xn , or at the other extreme, sn (X1 , . . . , Xn ) = X1 · · · Xn . Note that the summands of si (X1 , . . . , Xn ) are the products of i distinct variables, and that each such product occurs exactly once. Thus, if σ ∈ Sn , then σ∗ permutes the summands of each si (X1 , . . . , Xn ), and hence the elementary symmetric functions are indeed symmetric polynomials. The elementary symmetric functions are ubiquitous in mathematics. One of the reasons for this is that they give the formula for computing the coefficients of a polynomial with a predetermined collection of roots. Lemma 7.4.4. Suppose given n elements, a1 , . . . , an ∈ A, not necessarily distinct. Then the following equality holds in A[X]. (X − a1 ) . . . (X − an ) = X n − s1 X n−1 + · · · + (−1)n sn , where si = si (a1 , . . . , an ) for i = 1, . . . , n. Proof In the expansion of (X − a1 ) . . . (X − an ) from the distributive law, the terms of degree k in X are those obtained by choosing k of the factors from which to choose an X, leaving n − k factors of the form (−ai ). Adding these up, we get ⎞ ⎛ ⎝ (−aj1 ) . . . (−ajn−k )⎠ · X k = (−1)n−k sn−k (a1 , . . . , an ) · X k j1 i, and write πi : A[X1 , . . . , Xn ]Sn → A[X1 , . . . , Xn−1 ]Sn−1 for the induced map. Thus, π = πn . Since h is in the kernel of π, symmetry shows that it must also be in the kernel of πi for 1 ≤ i ≤ n − 1. Since h(X1 , . . . , Xn ) = Xn · k(X1 , . . . , Xn ), we have 0 = πn−1 (h(X1 , . . . , Xn )) = πn−1 (Xn )πn−1 (k(X1 , . . . , Xn )). Since πn−1 (Xn ) = Xn−1 is not a zero divisor in A[X1 , . . . , Xn−1 ], k must lie in the kernel of πn−1 , and hence k(X1 , . . . , Xn ) = Xn−1 · p(X1 , . . . , Xn ) for a suitable polynomial p, by the argument above.


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An easy induction now shows that h(X1 , . . . , Xn ) = X1 · · · Xn · q(X1 , . . . , Xn ) for a suitable polynomial q. Since both h and X1 · · · Xn = sn (X1 , . . . , Xn ) are symmetric, and since sn (X1 , . . . , Xn ) is not a zero divisor, we see that q must be a symmetric polynomial. Note that by the known properties of π, h must have total degree less than or equal to that of f , while q has total degree less than that of h. Thus, q(X1 , . . . , Xn ) is in the image of αn by induction. But then f must also be in the image of αn . To see that αn is injective, we argue by contradiction. Let f (Y1 , . . . , Yn ) be an element of the kernel of αn with the lowest possible total degree. Write f (Y1 , . . . , Yn ) = m i f (Y i 1 , . . . , Yn−1 )Yn . Note that i=0 0 = (π ◦ αn )(f ) = f0 (s1 (X1 , . . . , Xn−1 ), . . . , sn−1 (X1 , . . . , Xn−1 )). Since αn−1 is injective, this forces f0 (Y1 , . . . , Yn−1 ) to be 0. But then f (Y1 , . . . , Yn ) = Yn · g(Y1 , . . . , Yn ),

and hence

αn (f ) = sn (X1 , . . . , Xn ) · αn (g).

Thus, g lies in the kernel of αn . Since its total degree is less than that of f , this contradicts the minimality of the degree of f . Another important topic regarding symmetries of polynomials concerns discriminants. Proposition 7.4.8. Consider the following element of A[X1 , . . . , Xn ]: . (Xj − Xi ). δ(X1 , . . . , Xn ) = i<j

Then for σ ∈ Sn , we have σ∗ (δ) = ε(σ) · δ, where ε(σ) is the sign of σ. Proof We may may identify the set of indices i < j in the above product with the set, S, of all two-element subsets of {1, . . . , n}. Of course, Sn acts on S, via σ({i, j}) = {σ(i), σ(j)}. Using this action to re-index the factors of δ, we see that . (Xmax(σ(i),σ(j)) − Xmin(σ(i),σ(j)) ). δ(X1 , . . . , Xn ) = i<j

For i < j, σ∗ (Xj − Xi ) = Xσ(j) − Xσ(i) = ±(Xmax(σ(i),σ(j)) − Xmin(σ(i),σ(j)) ), where the sign is positive if σ preserves the order of i and j and is negative otherwise. Thus, σ∗ (δ) = (−1)k δ, where k is the number of pairs i < j for which σ(j) < σ(i). Define ψ : Sn → ±1 by ψ(σ) = (−1)k where (−1)k is defined by the equation σ∗ (δ) = (−1)k δ. Then it follows immediately from the fact that the homomorphisms σ∗ define an action of Sn on A[X1 , . . . , Xn ] that ψ is a homomorphism. It suffices to show that ψ = ε. Since Sn is generated by transpositions, it suffices to show that if τ is a transposition, then ψ(τ ) = −1. Thus, let τ = (i j), with i < j. Then the pairs on which τ reverses order consist of the pairs i < k and k < j, where i < k < j, together with the pair i < j itself. The signs associated with i < k < j cancel each other out, leaving a single −1 from the pair i < j. Definition 7.4.9. The discriminant, Δ(X1 , . . . , Xn ), is defined by Δ = δ 2 , where δ = ! i<j (Xj − Xi ), as above.


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Since each σ∗ (δ) = ±δ, we obtain the following corollary. Corollary 7.4.10. The discriminant, Δ, is a symmetric polynomial in the variables X1 , . . . , Xn . Thus, there is a unique polynomial Fn (X1 , . . . , Xn ), such that Δ(X1 , . . . , Xn ) = Fn (s1 (X1 , . . . , Xn ), . . . , sn (X1 , . . . , Xn )).

Example 7.4.11. The discriminant on two variables is given by Δ(X, Y ) = (Y − X)2 = X 2 + Y 2 − 2s2 (X, Y ). Now note that (s1 (X, Y ))2 = (X + Y )2 = X 2 + Y 2 + 2s2 (X, Y ). Thus, Δ = s21 − 4s2 , and hence F2 (X, Y ) = X 2 − 4Y . We shall calculate F3 (X, Y, Z) in the next set of exercises. It takes considerably more work than F2 . Since the polynomials Fn (X1 , . . . , Xn ) have integer coefficients, one may save considerable time by using a computer algebra package to calculate them. Recall from Lemma 7.4.4 that if a polynomial f (X) factors as a product of monic polynomials of degree 1, then the coefficients of f are, up to sign, the elementary symmetric functions on the roots. Definition 7.4.12. Let f (X) = a0 X n + a1 X n−1 + · · · + an be a polynomial of degree n in A[X] and let Fn (X1 , . . . , Xn ) be the polynomial determined in Corollary 7.4.10. Then the discriminant, Δ(f ), of f is the following element of A: Δ(f ) = Fn (−a1 /a0 , a2 /a0 , . . . , (−1)n an /a0 ). Of course, we don’t know much at this point about the polynomials Fn , so the following corollary expresses the discriminant in a more familiar setting. It follows immediately from Lemma 7.4.4 and Corollary 7.4.10. Corollary 7.4.13. Let f (X) be a monic polynomial in A[X], and suppose that f factors as f (X) = (X − b1 ) . . . (X − bn ) in B[X], where B is a commutative ring that contains A as a subring. Then the discriminant, Δ(f ), may be calculated by Δ(f )

= =

Δ(b1 , . . . , bn ) . (bj − bi )2 . i<j

Exercises 7.4.14. 1. Write X12 + · · · + Xn2 as a polynomial in the elementary symmetric functions on X1 , . . . , Xn .


239

2. Show that πi : A[X1 , . . . , Xn ]Sn → A[X1 , . . . , Xn−1 ]Sn−1 is injective when restricted to the A-submodule consisting of the symmetric polynomials of total degree less than n. 3. Let f (X) = aX 2 + bX + c be a quadratic in A[X]. Show that Δ(f ) = (b2 − 4ac)/a2 . Deduce that a quadratic f (X) ∈ R[X] has a root in R if and only if Δ(f ) ≥ 0. 4. Calculate the polynomial F3 (X, Y, Z) that’s used to calculate the discriminant of a cubic. † 5. Show that if f (X) = X 3 + pX + q is a monic cubic with no X 2 term, then Δ(f ) = −4p3 − 27q 2 .

7.5

Group Rings and Monoid Rings

Polynomial rings are a special case of a more general construction called monoid rings. Definitions 7.5.1. Let M be a monoid, and write the product operation of M in multiplicative notation, even if M is abelian. We define A[M ], the monoid ring of M over A, as follows. The elements of A[M ] are formal sums m∈M am m, such that the coefficient am is zero for all but finitely many m in M . (We abbreviate this by saying that am = 0 almost polynomials, addition is given by adding everywhere.) As in the case of coefficients: ( m∈M am m) + ( m∈M bm m) = m∈M (am + bm )m. The multiplication is defined as follows. ⎛ ⎞ #" # " ⎝ am m bm m = ax by ⎠ m. m∈M

m∈M

m∈M

{(x,y) | xy=m}

The last sum ranges over all ordered pairs (x, y) of elements in M whose product is m. In the generic monoid, that collection of ordered pairs can be difficult to determine. But in the examples we are concerned with, there will be no difficulty. We identify each m ∈ M with the element of A[M ] whose m-th coefficient is 1 and whose other coefficients are all 0. Thus, we have a natural inclusion M ⊂ A[M ]. Similarly, we identify a ∈ A with the element of A[M ] whose m-th coefficient is 0 for all m = 1, and whose coefficient of 1 is equal to a. Here, we write 1 for the identity element of M . We write ι : A → A[M ] for the map that gives this identification. In the case where the monoid is a group G, we call A[G] the group ring (or group algebra, if A is commutative) of G over A. Example 7.5.2. Write Z2 = T = {1, T }. Then for any ring A, A[Z2 ] = {a + bT | a, b ∈ A}. Here, if a, b, c, d ∈ A, we have (a+bT )+(c+dT ) = (a+c)+(b+d)T and (a+bT )·(c+dT ) = (ac + bd) + (ad + bc)T . Lemma 7.5.3. Let A be a ring and let M be a monoid. Then A[M ] is a ring under the above operations, and ι : A → A[M ] is a ring homomorphism. The inclusion M ⊂ A[M ] is a monoid homomorphism from M to the multiplicative monoid of A[M ], and each m ∈ M commutes with every element of ι(A). If A is commutative, then A[M ] is an A-algebra, via ι : A → A[M ].


240

Proof Everything here is straightforward, with the possible exception of the associativity of multiplication in A[M ]. We shall treat that, and leave the rest to the reader. Thus, suppose given α, β, γ ∈ A[M ], with α= am m, β = bm m and γ = cm m. m∈M

m∈M

m∈M

Then αβ =

dm m, where dm =

ax by .

{(x,y) | xy=m}

m∈M

Thus, (αβ)γ

=

⎛

⎝

m∈M

=

⎛

⎛

⎝

{(w,z) | wz=m}

⎝

m∈M

dw cz ⎠ m

{(w,z) | wz=m}

⎝

m∈M

=

⎛

⎞

⎞

⎞

ax by ⎠ cz ⎠ m

{(x,y) | xy=w}

⎞

ax by cz ⎠ m.

{(x,y,z) | xyz=m}

Here, the last equality comes from distributing the terms cz through the sums {(x,y) | xy=w} ax by . For each ordered triple (x, y, z) with xyz = m, there is exactly one summand ax by cz thus obtained. As the reader may now check, the expansion of α(βγ) produces exactly the same result. Monoid rings have an important universal property. Proposition 7.5.4. Let f : A → B be a ring homomorphism. Let M be a monoid and let g : M → B be a homomorphism from M to the multiplicative monoid of B. Suppose that for each m ∈ M , g(m) commutes with every element of f (A). Then there is a unique ring homomorphism h : A[M ] → B such that the following diagram commutes: ⊃ ι / A[M ] o A? M ?? ?? ?? ?? h g f ?? B Explicitly, we have " h

m∈M

# am m

=

f (am )g(m).

m∈M

Proof The element m∈M am m is, in fact, a finite sum of terms am m. Moreover, if j : M ⊂ A[M ] is the inclusion, then am m = ι(am )j(m). Thus, if h : A[M ] → B is a ring


241

homomorphism making the diagram commute, then we must have h m∈M am m = m∈M f (am )g(m), and hence h is unique as claimed. Thus, it suffices to show that if f and g satisfy the stated hypotheses, then h, as defined by the formula above, is a ring homomorphism. By the distributive law in B and the fact that f is a homomorphism of abelian groups, we see that h is also a homomorphism of abelian groups. Also, h(1) = f (1)g(1) = 1, so it suffices to show that h(αβ) = h(α)h(β) for all α, β ∈ A[M ]. By the distributive laws in both A[M ] and B, it suffices to show that h(am · a m ) = h(am)h(a m ) for all a, a ∈ A and m, m ∈ M . But h(am · a m ) = h(aa mm ) = f (aa )g(mm ) = f (a)f (a )g(m)g(m ), while h(am)h(a m ) = f (a)g(m)f (a )g(m ), so the result follows from the fact that f (a ) commutes with g(m). In the case of algebras over a commutative ring, the statement becomes simpler. Corollary 7.5.5. Let A be a commutative ring and let B be an A-algebra. Let M be a monoid and let g : M → B be a monoid homomorphism from M to the multiplicative monoid of B. Then there is a unique A-algebra homomorphism h : A[M ] → B such that h

the composite M ⊂ A[M ] − → B is equal to g. Explicitly, " # h am m = ν(am )g(m), m∈M

m∈M

where ν : A → B is the structure map of B as an A-algebra. Proof If B is an A-algebra with structure map ν, then ν(A) is contained in the center of B, and hence every element of ν(A) commutes with every element of g(M ). Thus, Proposition 7.5.4 provides a unique ring homomorphism h, given by the displayed formula, such that the diagram in the statement of Proposition 7.5.4 commutes, with f replaced by ν. But the left-hand triangle of the diagram commutes if and only if h is an A-algebra homomorphism, while the right-hand triangle commutes if and only if the h composite M ⊂ A[M ] − → B is equal to g. Note that if G is a group, then a monoid homomorphism g : G → B from G to the multiplicative monoid of B always takes value in B × , the group of invertible elements in the multiplicative monoid of B. Thus, g restricts to give a group homomorphism g : G → B × . We obtain the following corollary. Corollary 7.5.6. Let A be a commutative ring and let B be an A-algebra. Let G be a group and let f : G → B × be a homomorphism. Then there is a unique A-algebra h homomorphism h : A[G] → B such that the composite G ⊂ A[G] − → B is equal to f . Explicitly, ⎛ ⎞ h⎝ ag g ⎠ = ν(ag )f (g), g∈G

g∈G

where ν : A → B is the structure map of B as an A-algebra. There is an important homomorphism from a group ring A[G] to A. Definitions 7.5.7. Let A be a ring and let G be a group. Then the standard augmentation of the group ring A[G] is the unique ring homomorphism (which exists by


242

Proposition 7.5.4) ε : A[G] → A with the property that ε(g) = 1 for all g ∈ G and such that ε◦ι = 1A , where ι is the standard inclusion of A in A[G]. Explicitly, Proposition 7.5.4 gives ⎛ ⎞ ε⎝ ag g ⎠ = ag . g∈G

g∈G

We define the augmentation ideal, I(G) (or IA (G), if the ring A is allowed to vary), to be the kernel of ε. Thus, we can obtain information about A[G] from information about A and information about I(G). We shall study augmentation ideals in the next set of exercises. In the case that A is commutative, augmentation ideals admit a nice generalization to the study of certain A-algebras. Definition 7.5.8. Let A be a commutative ring. An augmented A-algebra consists of an A-algebra, B, together with an A-algebra homomorphism ε : B → A. Here, since A is given the A-algebra structure map 1A , this simply means that ε is a ring homomorphism such that ε ◦ ν = 1A , where ν is that A-algebra structure map of B. The ideal of the augmentation ε is its kernel. Examples 7.5.9. 1. Let A be a commutative ring and let a ∈ A. Then the evaluation map εa : A[X] → A is an augmentation. By Corollary 7.3.12, the ideal of this augmentation is the principal ideal (X − a). 2. Similarly, if A is commutative and a1 , . . . , an ∈ A, then the evaluation map εa1 ,...,an : A[X1 , . . . , Xn ] → A is an augmentation. Here, Proposition 7.3.22 tells us the augmentation ideal is (X1 − a1 , . . . , Xn − an ). 3. Let M be any monoid and let A be a commutative ring. Then the trivial monoid homomorphism from M tothe multiplicative monoid of A induces an augmentation map ε : A[M ] → A via ε( m∈M am m) = m∈M am . Exercises 7.5.10. 1. Complete the proof of Lemma 7.5.3. 2. Show that in the group ring of a group G over A, we may write ax by = ax bx−1 g . {(x,y) | xy=g}

x∈G

3. Let Z2 = T = {1, T }. For any ring A, show that (1 − T )(1 + T ) = 0 in A[Z2 ], and hence A[Z2 ] has zero-divisors. 4. Show that A[Zn ] has zero-divisors for any ring A and any integer n > 1. Deduce that if A is any ring and if G is a group with torsion elements of order > 1, then A[G] has zero-divisors. 5. Let A be a ring and let G be a finite group. Let Σ ∈ A[G] be the sum of all the elements of G: Σ = g∈G g. Show that for any x ∈ A[G], xΣ = (ι ◦ ε(x))Σ, where ε : A[G] → A is the standard augmentation and ι : A → A[G] is the standard inclusion.


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‡ 6. Let ζn = ei·2π/n ∈ C be the standard primitive n-th root of unity. Let A be any subring of C and let Zn = T = {1, T, . . . , T n−1 }. Show that there is a unique A-algebra homomorphism f : A[Zn ] → A[ζn ] such that f (T ) = ζn . Show that f is onto. ‡ 7. Let Zn = T = {1, T, . . . , T n−1 }. Let f : Z[Zn ] → Z[ζn ] be the homomorphism of Problem 6. (a) Show that Σ = 1 + T + · · · + T n−1 lies in the kernel of f , and hence there’s an induced, surjective ring homomorphism f : Z[Zn ]/(Σ) → Z[ζn ], where (Σ) is the principal ideal generated by Σ. (b) Suppose given integers m, k which are relatively prime to n. Let l > 0 be an integer such that m ≡ kl mod n, and let umk = 1 + T k + · · · + (T k )l−1 , where the right-hand side is interpreted as lying in Z[Zn ]/(Σ). Show that if we chose a different integer l with m ≡ kl mod n, then the above formula gives rise to precisely the same element of Z[Zn ]/(Σ). Show that umk is a unit in Z[Zn ]/(Σ), and that f (umk ) is the cyclotomic unit (ζnm − 1)/(ζnk − 1) from Problem 20 of Exercises 7.3.16. 8. Show that there is a commutative diagram of ring homomorphisms Z[Zn ]

π / Z[Zn ]/(Σ)

ε

ε

Z

/ Zn

π

where ε is the standard augmentation, the maps π are the canonical maps to the respective quotient rings, and Σ = 1 + T + · · · + T n−1 as in Problem 7. (a) Show that the above diagram induces an isomorphism ∼ =

h : Z[Zn ] −→ Z ×Zn (Z[Zn ]/(Σ)) , via h(α) = (ε(α), π(α)) for α ∈ Z[Zn ]. (See Problem 17 of Exercises 7.1.27.) For this reason, we call the diagram a pullback square. Deduce that there is a pullback square: Z[Zn ]×

π / × (Z[Zn ]/(Σ))

ε Z×

ε π

/ Z× n


244

(b) Let m1 , . . . , ms , k1 , . . . , ks be relatively prime to n and let r1 , . . . , rs > 0. Let u = urm11 k1 . . . urmss ks , the unit of Z[Zn ]/(Σ) constructed in Problem 7. Show that u lifts to a unit of Z[Zn ] if and only if mr11 . . . mrss ≡ ±k1r1 . . . ksrs mod n. 9. Let X = {X k | k ≥ 0}. Show that the operation X k · X l = X k+l turns X into a monoid isomorphic to the monoid, N, of non-negative integers under addition. Show that the monoid ring A[X] is isomorphic to the polynomial ring A[X] for any ring A. 10. Let X be as in the previous exercise and let Xn = X × · · · × X be the direct n times

product of X with itself n times. For any ring A, show that the polynomial ring A[X1 , . . . , Xn ] is isomorphic to the monoid ring A[Xn ]. ‡ 11. Let A be a ring and let f : M → M be a monoid homomorphism. Show that there is a ring homomorphism f : A[M ] → A[M ] defined by f ( ∗ ∗ m∈M am m) = ( a )m . −1 m m ∈M m∈f (m ) ‡ 12. Show that if f : A → B is a ring homomorphism and M is a monoid, then there is a ring homomorphism f∗ : A[M ] → B[M ] defined by f∗ ( am m) = f (am )m. m∈M

m∈M

13. Let G be a group and let A be a commutative ring. Let α = g∈G ag g be an element of A[G]. Show that α is in the center of A[G] if and only if axgx−1 = ag for all x, g ∈ G. The reader who has studied module theory should now show that the center of A[G] is free as an A-module, and give a basis for it. 14. Show that if A is a ring and if M1 and M2 are monoids, then there is an isomorphism between (A[M1 ])[M2 ] and A[M1 × M2 ]. (Hint: One could either argue directly, or via universal mapping properties. The latter method is quicker, and less cumbersome notationally.) 15. Let A and B be rings and let M be a monoid. Show that (A × B)[M ] is isomorphic to A[M ] × B[M ]. † 16. Let A be a ring and let a be the principal two-sided ideal in the polynomial ring A[X] generated by X n − 1. Show that the quotient ring A[X]/a is isomorphic to the group ring A[Zn ] of Zn over A. (Hint: There is a direct proof, that uses module theory, and there is an easier one using universal mapping properties.) 17. Let a be a two-sided ideal of A and let M be a monoid. Let π : A → A/a be the canonical map and let π∗ : A[M ] → (A/a)[M ] be the ring homomorphism given by π∗ ( m∈M am m) = m∈M π(am )m (see Problem 12). Show that π∗ is a surjection whose kernel is the two-sided ideal of A[M ] generated by a. (Here, we identify a with its image under ι : A → A[M ].) 18. Let A be any ring and let Z2 = T = {1, T }. Determine precisely which elements a+bT of A[Z2 ], a, b ∈ A, lie in the augmentation ideal IA (Z2 ). Deduce that IA (Z2 ) is the principal two-sided ideal generated by T − 1.


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19. Let A be a commutative ring and let G be a group. Let S ⊂ G be a generating set for G (i.e., G = S as in Definitions 2.2.11). Show that the augmentation ideal IA (G) is the two-sided ideal of A[G] generated by {x − 1 | x ∈ S}. (Hint: Consider the proof of Proposition 7.3.22.) 20. Let f : G → K be a split surjection of groups. Find a set of generators (as an ideal) for the kernel of the ring homomorphism f∗ : A[G] → A[K] given by f∗ ( g∈G ag g) = k∈K ( g∈f −1 k ag )k, where A is any commutative ring. (This map was shown to be a ring homomorphism in Problem 11.) 21. Show that a group ring A[G] over a commutative ring A is self-opposite. 22. Show that a group ring A[G] over a self-opposite ring A is self-opposite. 23. Show that if A is commutative, then the passage from monoids M to the monoid rings A[M ] gives a left adjoint to the forgetful functor from A-algebras to their multiplicative monoids. 24. Let A be commutative. Construct the free A-algebra on an arbitrary set X. 25. Let A be commutative. Construct the free commutative A-algebra on an arbitrary set X. (So far, we’ve only handled the case where X is finite.) 26. Show that if A is commutative, then the passage from groups G to the group rings A[G] provides a left adjoint to the functor that takes an A-algebra to its group of units.

7.6

Ideals in Commutative Rings

We learned a lot about groups by studying their factor groups. Similarly, quotient rings will be useful for studying rings. The drawback here, of course, is that while the kernel of a surjective group homomorphism is again a group, the kernel of a surjective ring homomorphism is only an ideal and not a ring. Thus, there is no good analogue for a composition series for a ring. However, we can mimic the first step by which the composition series for a finite group was constructed. In that step, we constructed a surjective homomorphism from our finite group to a simple group. The analogue for rings of a simple group is a ring with no two-sided ideals other than 0 and the ring itself, and hence has no useful quotient rings. Definition 7.6.1. A nonzero ring is Jacobson simple if it has no two-sided ideals other than 0 and the ring itself. Problem 11 of Exercises 7.2.24 shows that if D is a division ring, then Mn (D) is a Jacobson simple ring for all n ≥ 1. (We shall give a slicker argument for this in Chapter 12.) There are other examples of noncommutative Jacobson simple rings more complicated than these. But in the commutative case, Jacobson simple rings are as nice as we could ask. The next corollary is immediate from Lemma 7.2.7. Corollary 7.6.2. A commutative ring is Jacobson simple if and only if it is a field. We wish to study those quotient rings A/a that are not the 0 ring, and hence, we are concerned with those ideals a that are not equal to A. We shall give them a name.


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Definition 7.6.3. An ideal a of A is proper if a ⊂ A is a proper inclusion (i.e., if a = A). We now introduce two important kinds of proper ideals. Definitions 7.6.4. Let A be a commutative ring. A proper ideal, p, of A is prime if ab ∈ p implies that either a or b must be in p. A proper ideal, m, of A is maximal if the only proper ideal containing m is m itself. Examples 7.6.5. Let A be a commutative ring. Then it is immediate from the definitions that the ideal 0 of A is prime if and only if A is an integral domain. Note that the ideal 0 is maximal if and only if A has no proper ideals other than 0. But we’ve just seen in Corollary 7.6.2 that this occurs if and only if A is a field. These two examples give a characterization of prime and maximal ideals in terms of the quotient rings they determine. Lemma 7.6.6. Let A be a commutative ring. Then a proper ideal p of A is prime if and only if the quotient ring A/p is an integral domain. Also, a proper ideal m of A is maximal if and only if the quotient ring A/m is a field. Thus, every maximal ideal is prime. Proof Let p be a prime ideal of A. We must show that A/p has no zero-divisors. Thus, let a and b be elements of A/p such that a · b = 0. But this says that ab = 0, and hence ab ∈ p. But then one of a or b is in p, and hence that element goes to 0 in A/p. Conversely, if A/p is an integral domain and if ab ∈ p, then ab = 0, so that one of a and b must be 0 in A/p. But then one of a and b must be in p. For the case of maximal ideals, the correspondence between the ideals in a quotient ring A/a and the ideals in A containing a (Problem 5 of Exercises 7.2.24) shows that a proper ideal m of A is maximal if and only if the 0 ideal is maximal in A/m. As shown in the preceding example, this occurs if and only if A/m is a field. Examples 7.6.7. 1. For any prime p, the ideal (p) ⊂ Z is maximal, as Zp is a field. In fact, Lemma 7.1.12 shows that if n > 1, then (n) is a prime ideal of Z if and only if n is a prime number. Thus, every nonzero prime ideal of Z is maximal. 2. Let K be a field and let a ∈ K. Then the evaluation map εa : K[X] → K is surjective, so its kernel is a maximal ideal. By Corollary 7.3.12, ker εa is the principal ideal generated by X − a. p → Zp be the ring homomorphism from the p-adic integers onto Zp 3. Let π1 : Z that’s induced by the projection map. (See Lemma 7.1.38.) Then ker π1 , which by Problem 4 of Exercises 7.1.41 is the principal ideal generated by p, is maximal. The next lemma is an easy, but important observation. Lemma 7.6.8. Let f : A → B be a ring homomorphism between commutative rings and let p be a prime ideal of B. Then f −1 (p) is a prime ideal of A. Note that the analogue of the preceding lemma for maximal ideals is false: Let i : Z → Q be the inclusion. Then 0 is a maximal ideal of Q, while f −1 (0) = 0, which is prime in Z, but not maximal. Prime ideals satisfy an important primality principle with respect to the ideal product of Definitions 7.2.21. This would have made a good exercise, but its uses are ubiquitous enough that we give it here.


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Lemma 7.6.9. Let p be a prime ideal in the commutative ring A and let a and b be ideals with ab ⊂ p. Then at least one of a and b must be contained in p. Proof If ab ⊂ p and a ⊂ p, let a ∈ a with a ∈ p. Then ab ∈ p for all b ∈ b. Since a ∈ p, we must have b ∈ p for all b ∈ b. By Lemma 7.2.22, ab ⊂ a ∩ b for any ideals a, b of a commutative ring. Corollary 7.6.10. Let p be a prime ideal in the commutative ring A and let a and b be ideals such that a ∩ b ⊂ p. Then at least one of a and b must be contained in p. Prime ideals may be used to define a notion of dimension for a commutative ring. Definition 7.6.11. Let A be a commutative ring. We say that the Krull dimension of A is ≤ n if whenever we have a sequence p0 ⊂ p1 ⊂ · · · ⊂ pk of proper inclusions (i.e., inclusions that are not onto) of prime ideals of A, we have k ≤ n. Thus, the Krull dimension of A is finite if there is a maximum length for such sequences of proper inclusions of prime ideals in A, and if p 0 ⊂ p1 ⊂ · · · ⊂ pn is the longest such sequence that can occur in A, then the Krull dimension is n. Examples 7.6.12. 1. Since a field has only one prime ideal, 0, every field has Krull dimension 0. 2. The first example in Examples 7.6.7 shows that the longest chains of proper inclusions of prime ideals in Z are those of the form 0 ⊂ (p), for p prime. Thus, Z has Krull dimension 1. Let us return to the analogy with group theory. As noted above, we’ve made use of the fact that any nontrivial finite group has a nontrivial simple group as a factor group. By analogy, we would like to show that any commutative ring has a Jacobson simple ring as a quotient ring. Since the only commutative rings that are Jacobson simple are fields, this is equivalent to showing that every commutative ring has a maximal ideal. To show this, we shall make use of Zorn’s Lemma, which is equivalent to the Axiom of Choice. The reader can find the proof of this equivalence in any good book on set theory.9 First we shall need some setup. Recall that a partially ordered set is a set X together with a relation, ≤, which is reflexive (x ≤ x for all x ∈ X), transitive (if x ≤ y and y ≤ z, then x ≤ z), and antisymmetric (if x ≤ y and y ≤ x, then x = y). In a partially ordered set, it is not required that an arbitrary pair of elements in X be comparable by the relation. In other words, there may exist pairs x, y such that neither x ≤ y nor y ≤ x is true. On the other hand, a partially ordered set in which any pair of elements is comparable is called a totally ordered set. Let S be a subset of a partially ordered set. We say that x ∈ X is an upper bound for S if s ≤ x for all s ∈ S. Note that the upper bound x need not lie in S. Finally, a maximal element of X is an element y such that if x ∈ X with y ≤ x, then x and y must be equal. 9 There’s a good exposition of Zorn’s Lemma and the Well Ordering Principle in Topology, by James Dugundji, Allyn and Bacon, 1966.


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Lemma 7.6.13. (Zorn’s Lemma) Let X be a nonempty partially ordered set such that every totally ordered subset of X has an upper bound in X. Then X has at least one maximal element. Corollary 7.6.14. Let A be a commutative ring. Then every proper ideal of A is contained in a maximal ideal. Proof Let a be a proper ideal of A, and let X be the set of all proper ideals of A that contain a. Put a partial ordering on X by b ≤ b if b ⊂ b . Then a maximal element of X with respect to this ordering is precisely a maximal ideal containing a. Since a ∈ X, X is nonempty, so it suffices to show that any totally ordered subset of X has an upper bound in X. Let S be a totally ordered subset of X. We claim that the union b∈S b of the ideals in S is such an upper bound. Of course, unions of ideals are not generally ideals, but if we’re dealing with a totally ordered set of ideals, this problem disappears: For x, y ∈ b∈S b, we can find b, b ∈ S such that x ∈ b and y ∈ b . But since any two elements of S are comparable, either b ≤ b or b ≤ b. But in either case, x+ y is in the larger of the two, as is ax for any a ∈ A. But then x + y and ax are in b∈S b, and hence the union is an ideal. Clearly, it is an upper bound for the ideals in S, so it suffices to show that it is proper. But if 1 ∈ b∈S b, then 1 must be an element of some b ∈ S. But this is impossible, since the ideals in X are all proper. If we make extra assumptions on our ring, we can obtain an even stronger sort of maximality principle for ideals. Definitions 7.6.15. Let A be a commutative ring. We say that A is Noetherian, or has the ascending chain condition (a.c.c.) if any ascending sequence of inclusions of ideals of A is eventually constant. In other words, given a sequence a 1 ⊂ a2 ⊂ · · · ⊂ ak ⊂ · · · there is some n ≥ 1 such that an = ak for all k ≥ n. Alternatively, we say that A is Artinian, or has the descending chain condition (d.c.c.) if any decreasing sequence of ideals is eventually constant. This means that given a sequence a1 ⊃ a2 ⊃ · · · ⊃ ak ⊃ · · · there is some n ≥ 1 such that an = ak for all k ≥ n. It turns out that the Artinian condition (in a commutative ring) implies that the ring is Noetherian. The Artinian condition is a very strong one, and fails to hold in many rings of interest. For instance, as we shall see in the exercises below, the only Artinian integral domains are fields. The descending chain condition for left or right ideals turns out to be extremely useful for certain topics in noncommutative ring theory. For instance, descending chain conditions are important in the study of group rings K[G], where K is a field and G is a finite group. Here is the strengthened maximality principle that holds in Noetherian rings. Proposition 7.6.16. Let A be a Noetherian ring and let X be the partially ordered set of ideals in A (ordered by inclusion). Let S be any nonempty subset of X. Then S, with the partial ordering inherited from X, has maximal elements.


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Proof We argue by contradiction. Suppose that S has no maximal elements, and let a1 be an element of S. Since a1 is not a maximal element of S, we can find a proper inclusion a1 ⊂ a2 , with a2 in S. Since no element of S is maximal, induction now shows that we may choose a sequence a 1 ⊂ a2 ⊂ · · · ⊂ ak ⊂ · · · of elements of S, such that each inclusion ak−1 ⊂ ak is proper.10 But this contradicts the hypothesis that A is Noetherian, and hence S must have maximal elements after all. We shall study Noetherian rings in much greater detail in the context of module theory. Exercises 7.6.17. 1. Let f : A → B be a surjective ring homomorphism. Show that there is a one-to-one correspondence between the prime ideals of B and those primes of A that contain ker f . Show also that there is a one-to-one correspondence between the maximal ideals of B and those maximal ideals of A containing ker f . 2. Let n > 1. Show that every prime ideal of Zn is maximal. Given the prime decomposition of n, list all the maximal ideals, m, of Zn and calculate the quotient fields Zn /m. 3. Show that an integral domain has Krull dimension 0 if and only if it is a field. 4. Let p be a prime ideal in the commutative ring A and let A[X] · p be the ideal generated by p in A[X]. Show that A[X] · p is a prime ideal of A[X]. 5. Let A be an integral domain which is not a field. Show that the principal ideal (X − a) of A[X] (here, a ∈ A) is a prime ideal that is not maximal. Deduce that A[X] has Krull dimension ≥ 2. 6. Let p be a prime and let n be any integer. Show that the ideal (p, X − n) of Z[X], generated by p and X − n, is maximal. 7. Let f : Q[X] → Q[i] ⊂ C be the unique Q-algebra map that carries X to i. Recall from Problem 14 of Exercises 7.3.16 that Q[i] is a field. Show that ker f = (X 2 +1), and hence (X 2 + 1) is a maximal ideal in Q[X]. 8. Let Z4 = T = {1, T, T 2 , T 3 }. Recall from Problem 6 of Exercises 7.5.10 that there is a unique, surjective, Q-algebra homomorphism f : Q[Z4 ] → Q[i] with f (T ) = i. Show that ker f = (T 2 + 1), and hence (T 2 + 1) is a maximal ideal in Q[Z4 ]. 9. Let X1 , . . . , Xn be indeterminates over a field K. Let a1 ∈ K and for 2 ≤ i ≤ n, let fi (X) ∈ K[X1 , . . . , Xi−1 ] ⊂ K[X1 , . . . , Xn ]. Show that (X1 − a1 , X2 − f2 (X), . . . , Xn − fn (X)) is a maximal ideal of K[X1 , . . . , Xn ]. 10. Let X1 , . . . , Xn be indeterminates over a field K. Find a chain p0 ⊂ · · · ⊂ pn of proper inclusions of prime ideals of K[X1 , . . . , Xn ]. Deduce that K[X1 , . . . , Xn ] has Krull dimension ≥ n. 10 Note that the infinite induction here makes use of the Axiom of Choice. Thus, while our conclusion is stronger than that of the Zorn’s Lemma argument for the existence of maximal ideals, the Noetherian property has not permitted us to dispense with the use of the choice axiom.


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11. Let A and B be commutative rings. What are the prime ideals of A × B? Suppose that A and B have finite Krull dimension. Calculate the Krull dimension of A × B in terms of the dimensions of A and B. 12. Show that Z is Noetherian but not Artinian. 13. Show that Zn is both Noetherian and Artinian. 14. Let A be an Artinian integral domain and let a be a nonzero element of A. Show that the descending chain condition on (a) ⊃ (a2 ) ⊃ · · · ⊃ (an ) ⊃ · · · implies that a is a unit. Deduce that A must be a field. Deduce that a finite domain is a field. 15. Let A be an Artinian commutative ring. Show that every prime ideal of A is maximal.

7.7

Modules

Modules play a role in ring theory analogous to the role of G-sets in understanding groups. Among other things, module theory will enable us to obtain a better grasp of the theory of ideals, and hence of the structure of the rings themselves. We give some general concepts and constructions here, and study free modules and exact sequences in subsections. We then close the section with a subsection on the notion of rings without identity elements. Definitions 7.7.1. Let A be a ring. A left A-module is an abelian group M together with an operation that assigns to each a ∈ A and m ∈ M an element am ∈ M subject to the following rules: a(m + m )

= am + am (a + a )m = am + a m 1 · m = m, a(a m) = (aa )m,

for all a, a ∈ A and all m, m ∈ M . If M is an abelian group, then an operation A × M → M making M an A-module is called an A-module structure on M . A submodule of M is a subgroup M ⊂ M of the additive group of M , such that am ∈ M for all a ∈ A and m ∈ M (i.e., M is closed under multiplication by elements of A). An A-module homomorphism between the left A-modules M and N is a homomorphism f : M → N of abelian groups such that f (am) = af (m) for all a ∈ A and all m ∈ M. Right modules and their homomorphisms are defined by reversing the order of the a’s and m’s in the above definitions.


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Note that if K is a field, then a K-module is nothing other than a vector space over K, and a K-module homomorphism between vector spaces is nothing other than a K-linear map. Another case where we understand what it means to be a module is that of Z-modules. The proof of the next lemma is similar to that of Lemma 7.1.6, which we left to the reader, so we shall leave this to the reader as well. Lemma 7.7.2. Every abelian group G is a Z-module. Moreover, the Z-module structure on G is unique: for n ∈ Z and g ∈ G, ng is the n-th power of g in the group structure of G. (Thus, if n > 0, ng = g + · · · + g, the sum of n copies of g.) Finally, every group homomorphism between abelian groups is a Z-module homomorphism. When A is commutative, any left A-module may be made into a right A-module by setting m · a = am. This process is clearly reversible, so if A is commutative, then there is a one-to-one correspondence between the left and right A-modules. Not only does the above correspondence fail for noncommutative rings, but the theories of left modules and of right modules can be quite different for some rings. Example 7.7.3. Let A be a ring. Then A is a left A-module via the structure map a · a = aa . Indeed, the equalities that demonstrate that this is an A-module structure are all part of the definition of a ring. The submodules of A under this structure are precisely the left ideals of A. Note that A is also a right A-module, via a · a = a a. The submodules under this right module structure are the right ideals of A. In particular, module theory will help us study ideals, and hence will provide important information about the ring A. Here is another source of A-modules. Example 7.7.4. Let f : A → B be a ring homomorphism. Then B is a left A-module via a · b = f (a)b. Also, B is a right A-module via b · a = bf (a). For simplicity, we shall restrict attention to left modules for the rest of this section. Analogues of all the results will be true for right modules. Thus, for the duration, “Amodule” will mean left A-module. We shall, however, use the word “left” to emphasize certain one-sided properties. Given the modules from the examples above, we can construct yet more via factor groups. Lemma 7.7.5. Let N be a submodule of the A-module M . Then the factor group M/N is an A-module via a·m = am. Moreover, if f : M → M is an A-module homomorphism, then there is a factorization / M ME EE x< x EE xx E xx π EE " xx f M/N f

(i.e., an A-module homomorphism f making the diagram commute) if and only if N ⊂ ker f . The factorization f , if it exists, is unique. Finally, the kernel of an A-module homomorphism is always a submodule, and hence ∼ = we obtain an isomorphism of A-modules f : M/(ker f ) −→ im f from any A-module homomorphism f : M → M .


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We now analyze the A-module homomorphisms out of A. Lemma 7.7.6. Let M be a left A-module. Then for each m ∈ M there is a unique A-module homomorphism fm : A → M with fm (1) = m. Proof We must have fm (a) = fm (a · 1) = afm (1) = am, so uniqueness follows, and it suffices that this formula defines an A-module homomorphism. But that follows immediately from the definition of module. In category theoretic terms, this says that A is the free A-module on one generator. There’s another way to express the result which introduces a useful piece of notation. Definition 7.7.7. Let M and N be A-modules. We write HomA (M, N ) for the set of A-module homomorphisms from M to N .11 Lemma 7.7.8. Let M and N be A-modules. Then HomA (M, N ) is an abelian group under the operation that sets (f + g)(m) = f (m) + g(m) for f, g ∈ HomA (M, N ). Under certain circumstances (e.g., if A is commutative), HomA (M, N ) is naturally an A-module. We shall discuss the details in Section 9.7. We now give an easy consequence of Lemma 7.7.6. Corollary 7.7.9. Let M be an A-module. Then there is an isomorphism of abelian ∼ = groups, ε : HomA (A, M ) −→ M , given by ε(f ) = f (1). ε is an example of what’s called an evaluation map, as it is obtained by evaluating the homomorphisms at 1. As shown in Proposition 9.7.4, ε is an A-module isomorphism under the appropriate conventions on the group of homomorphisms. We return to the study of the homomorphisms from A to M . Definitions 7.7.10. Let m ∈ M and let fm : A → M be the unique A-module homomorphism that carries 1 to m. Then we write Am for the image of fm and write Ann(m) (or AnnA (m), if there’s more than one ring over which M is a module) for the kernel. Thus, Am Ann(m)

= {am | a ∈ A} and = {a ∈ A | am = 0}.

We shall refer to Ann(m) as the annihilator of m, and say that a kills (or annihilates) m if am = 0. In general, we say that N is a cyclic module if N = An for some n ∈ N . Corollary 7.7.11. Let M be a left A-module and let m ∈ M . Then Ann(m) is a left ideal of A, and the cyclic module Am is isomorphic to A/Ann(m). There is also the notion of the annihilator of a module. Definition 7.7.12. Let M be an A-module. Then the annihilator of M , written as Ann(M ) or AnnA (M ), is given by Ann(M )

= {a ∈ A | am = 0 for all m ∈ M } = Ann(m). m∈M

11 We

use the same notation for the set of A-module homomorphisms between a pair of right A-modules. This can, at times, require some specificity of language.


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Unlike the case of the annihilator of an element, we have the following. Lemma 7.7.13. The annihilator of an A-module is a two-sided ideal of A. Warning: If A is not commutative, then Ann(Am) may not be equal to Ann(m). In addition, Ann(m) may not be a two-sided ideal, while Ann(Am) always is. Definitions 7.7.14. Let f : A → B be a ring homomorphism. Let M be a B-module. The A-module structure on M induced by f is given by the structure map a · m = f (a)m for a ∈ A and m ∈ M . Let N be an A-module. We say that a B-module structure on N is compatible with the original A-module structure if the A-module structure induced by f from the Bmodule structure is the same as the one we started with (i.e., f (a) · m = am for all a ∈ A and m ∈ M ). We wish to study these relationships in the case where f is the canonical map π : A → A/a, where a is a two-sided ideal of A. Here, if M is an A/a module, then we shall implicitly consider M to be an A-module under the structure induced by π. Proposition 7.7.15. Let a be a two-sided ideal in the ring A. Then an A-module M admits an A/a-module structure compatible with the given A-module structure if and only if a ⊂ Ann(M ). Such an A/a-module structure is unique if it exists. Finally, if M and N are A/a-modules, then every A-module homomorphism from M to N is an A/a-module homomorphism as well. Proof If M is an A/a-module, then, in the induced A-module structure, if x ∈ a and m ∈ M , then xm = π(x)m = 0, because π(x) = 0. Thus, a ⊂ Ann(M ). Conversely, if a ⊂ Ann(M ), given a ∈ A, x ∈ a, and m ∈ M , (a + x)m = am. Thus, setting a · m = am gives a well defined correspondence from A/a × M to M . And this correspondence is easily seen to satisfy the conditions required for an A/a-module structure on M . Two A/a-module structures that induce the same A-module structure must be equal, because π : A → A/a is surjective. Finally, given an A-module homomorphism f : M → N between A/a-modules, we have f (am) = f (am) = af (m) = af (m), since the A-module structures on both M and N are induced by the given A/a-module structures. For A = Z, this gives a characterization of Zn -modules. Corollary 7.7.16. An abelian group M admits a Zn -module structure if and only if M has exponent n as a group. The Zn -module structure on M , if it exists, is unique. Finally, if M and N are Zn -modules, then every group homomorphism f : M → N is a Zn -module homomorphism. Proof We use the fact that Zn = Z/(n). We know that every abelian group M has a unique Z-module structure, in which k · m is the k-th power of m in the abelian group structure on M for all k ∈ Z and m ∈ M . Thus, k annihilates M if and only if M has exponent k. But since (n) is the smallest ideal of Z containing n, we see that n annihilates M if and only if (n) ⊂ Ann(M ). The result now follows directly from Proposition 7.7.15. Now we consider operations on submodules. The constructions that work are similar to those that work for ideals.


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Definitions 7.7.17. Let N1 and N2 be submodules of M and let a be a left ideal of A. We define N1 + N2 and aN1 by N1 + N2 aN1

= {n + m | n ∈ N1 , m ∈ N2 } = {a1 n1 + · · · + ak nk | k ≥ 1, ai ∈ a, and ni ∈ N1 for i = 1, . . . , k}

Lemma 7.7.18. Let N1 and N2 be submodules of the A-module M and let a be a left ideal of A. Then N1 + N2 , aN1 , and N1 ∩ N2 are all submodules of M . The modules aM have the following virtue. Lemma 7.7.19. Let M be an A-module and let a be a two-sided ideal of M . Then M/aM is an A/a-module. Moreover, if N is any A/a-module and if f : M → N is an A-module homomorphism, then there is a unique factorization /N M G x; GG x x GG xx G xx π GG xx f # M/aM f

where f : M/aM → N is an A/a-module homomorphism. Proof Clearly, a ⊂ Ann(M/aM ), and hence M/aM is an A/a-module by Proposition 7.7.15. If f : M → N is an A-module homomorphism, with N an A/a-module, then a ⊂ Ann(N ), and hence aM must lie in ker f . Thus, there is a unique A-module homomorphism f : M/aM → N making the diagram commute, and f is an A/a-module homomorphism by Proposition 7.7.15. In categorical terms, this says that the passage from M to M/aM is a left adjoint to the forgetful functor from A/a-modules to A-modules. We can also consider external operations on modules. Recall that the direct sum M1 ⊕ · · · ⊕ Mk of a finite collection of abelian groups is an alternative notation for the direct product. Definition 7.7.20. Let M1 , . . . , Mk be A-modules. By the direct sum, M1 ⊕ · · · ⊕ Mk , of M1 , . . . , Mk , we mean their direct sum as abelian groups, endowed with the A-module structure obtained by setting a(m1 , . . . , mk ) = (am1 , . . . , amk ), for each a ∈ A and each k-tuple (m1 , . . . , mk ) ∈ M1 ⊕ · · · ⊕ Mk . Note that the canonical inclusions ιi : Mi → M1 ⊕ · · · ⊕ Mk are A-module maps, as are the projection maps πi : M1 ⊕ · · · ⊕ Mk → Mi . The proof of the following lemma is identical to that of its analogue for abelian groups (see Proposition 2.10.6). Lemma 7.7.21. Let fi : Mi → N be A-module homomorphisms for 1 ≤ i ≤ k. Then there is a unique A-module homomorphism f : M1 ⊕ · · · ⊕ Mk → N such that f ◦ ιi = fi for 1 ≤ i ≤ n. Explicitly, f (m1 , . . . , mk ) = f1 (m1 ) + · · · + fk (mk ).


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The next lemma will be useful, for instance, in our understanding of the properties of free modules. Lemma 7.7.22. Let a be a two-sided ideal in A and let M1 , . . . , Mk be A-modules. Let M = M1 ⊕ · · · ⊕ Mk . Then there is a natural isomorphism of A/a-modules between M/aM and (M1 /aM1 ) ⊕ · · · ⊕ (Mk /aMk ). Proof Recall from Proposition 7.7.15 that any A-module map between A/a-modules is an A/a-module map. Thus, it suffices to show that aM = (aM1 ) ⊕ · · · ⊕ (aMk ). Note that the inclusion aM ⊂ (aM1 ) ⊕ · · · ⊕ (aMk ) follows from the fact that if a ∈ a, then a(m1 , . . . , mk ) = (am1 , . . . , amm ) lies in (aM1 ) ⊕ · · · ⊕ (aMk ) for each k-tuple (m1 , . . . , mk ). Since (aM1 ) ⊕ · · · ⊕ (aMk ) is closed under addition, the stated inclusion follows. For the opposite inclusion, note that (m1 , . . . , mk ) = ι1 (m1 ) + · · · + ιk (mk ). Since each ιi is an A-module homomorphism, we have ιi (mi ) ∈ aM whenever mi ∈ aMi . Since aM is closed under addition, we see that (aM1 ) ⊕ · · · ⊕ (aMk ) ⊂ aM , as claimed. We shall also make use of* infinite direct sums. Recall that if Mi is !an abelian group for i ∈ I, then the direct sum i∈I Mi is the subgroup of the product i∈I Mi consisting of those I-tuples (mi | i ∈ I) such that *all but finitely many of the mi are 0. The homomorphisms ιi : Mi → i∈I Mi are defined by setting ιi (m) to be the Ituple whose i-th coordinate is m and whose other coordinates are all 0. Making use of m as an alternate notation for (m | i ∈ I), Lemma 6.5.9 gives i i i∈I mi = ιi (mi ), i∈I

where the right-hand side is the sum in for which mi = 0.

*

i∈I i∈I

Mi of the finite collection of terms ιi (mi )

Definition 7.7.23. Let Mi be an A-module for each i ∈ I. We define the direct sum * abelian group i∈I Mi to be the A-module whose underlying is the direct sum of the Mi and whose A-module structure is given by a i∈I mi = i∈I ami . * The maps ιi : Mi → i∈I Mi are easily seen to be A-module homomorphisms. The proof of the next result is identical to that of its analogue for Z-modules (Proposition 6.5.10). Proposition 7.7.24. Suppose given A-modules Mi for i ∈ I. Then for any collection of A-module homomorphisms {fi : Mi → N | i ∈ I}, there is a unique A-module homo* morphism f : i∈I Mi → N such that f ◦ ιi = fi for each i. Explicitly, # " mi = fi (mi ). f i∈I

i∈I

We also have an infinite sum operation for submodules of a given module. Definition 7.7.25. Suppose given a set {Ni | i ∈ I} of submodules of an A-module M . Then the sum of the Ni is the submodule Ni = {ni1 + · · · + nik | k ≥ 1, ij ∈ I and nij ∈ Nij for j = 1, . . . , k}. i∈I

Thus,

i∈I

Ni is the set of all finite sums of elements from the various submodules Ni .


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Sums of submodules satisfy the following properties. The proof is left as an exercise. Proposition 7.7.26. Let M be an A-module. If N1 and N2 are submodules of M , then N1 + N2 is the smallest submodule of M containing both N1 and N2 . Similarly, if Ni is a submodule of M for i ∈ I, then i∈I Ni is the smallest submodule of M that contains each of the Ni . If fi : Mi → M * is an A-module homomorphism for each i ∈ I (with I either finite or infinite) and if f : i∈I Mi → M is the unique homomorphism such that f ◦ ιi = fi for all i, then im fi . im f = i∈I

Exercises 7.7.27. 1. Give the proof of Lemma 7.7.13. 2. Give the proof of Lemma 7.7.18. 3. Give the proof of Corollary 7.7.26. 4. Let M be an abelian group, considered as a Z-module. Show that the annihilator Ann(M ) is nonzero if and only if M has some positive exponent. If Ann(M ) = (n), with n > 0, show that n is the smallest positive number which is an exponent for M. 5. Let N1 and N2 be submodules of M . Show that Ann(N1 + N2 ) = Ann(N1 ) ∩ Ann(N2 ). 6. Let N1 and N2 be submodules of M . Show that Ann(N1 ) + Ann(N2 ) ⊂ Ann(N1 ∩ N2 ). Can you give an example where this inclusion is proper? 7. Consider Z8 as a Z-module. What is Ann((2)Z8 )? 8. Let f : A → B be a ring homomorphism. In the left A-module structure on B that’s induced by f , what is Ann(B)? 9. Let A be a commutative ring. Let m be an A-module and let m ∈ M . Show that Ann(m) = Ann(Am). 10. Let A be any ring and let x ∈ Mn (A) be the matrix whose 11-entry is 1 and whose other entries are all 0. ⎛ ⎞ 1 0 ... 0 ⎜ 0 0 ... 0 ⎟ ⎜ ⎟ a=⎜ ⎟ .. ⎝ ⎠ . 0

0

...

(a) What are the elements of AnnMn (A) (x)?

0


257

(b) What are the elements of the principal left ideal Mn (A)x? (c) What are the elements of AnnMn (A) (Mn (A)x)? 11. Let A be a ring and let G be a finite group. Let Σ ∈ A[G] be the sum of all the elements of G: Σ = g∈G g. Show that AnnA[G] (Σ) = I(G), the augmentation ideal of G. 12. Let A be a ring. Let G be a finite group and let x ∈ G. Show that an element g∈G ag g ∈ A[G] lies in Ann(x − 1) if and only if for each left coset gx of x, the coefficients of the elements in gx are all equal (i.e., agxi = agx j for all i, j ∈ Z). Deduce that if G = x, then Ann(x − 1) = A[G]Σ, where Σ = g∈G g. 13. Let (m) be an ideal in Z. Show that for n ≥ 1, Zn /(m)Zn is isomorphic to Z(m,n) , where (m, n) is the greatest common divisor of m and n. (Recall that Z1 is the trivial group.) 14. Let M be an abelian group and let r ≥ 0. Show that (pr )M/(pr+1 )M is a vector space over Zp . 15. Let M be a finite abelian group of exponent pr . Show that (pr−1 )M is a vector space over Zp . 16. Let M be a finite abelian group whose order is prime to n. Show that M/(n)M = 0. 17. Let M = Zpr1 × · · · × Zprk , where 1 ≤ r1 ≤ · · · ≤ rk . (a) Show that M/(p)M is the product of k copies of Zp . (b) Calculate, in similar terms, the Zp -vector spaces (pr )M/(pr+1 )M for r ≥ 1. (c) Calculate Ann(M ). 18. Suppose given a family of submodules {Ni | i ∈ I} of M . Show that the intersection i∈I Ni is a submodule of M . 19. Let N be a submodule of M . Show that N = x∈N Ax. 20. What is Ann(M1 ⊕ · · · ⊕ Mk )? † 21. Let A be a ring. Show that a left A-module structure on an abelian group M determines and is determined by a ring homomorphism from A to EndZ (M ). 22. Use the preceding problem to give a quick proof of Proposition 7.7.15. 23. Let A be a ring. Show that a right A-module structure on an abelian group M determines and is determined by a ring homomorphism from the opposite ring Aop to EndZ (M ). † 24. Let A be a ring and let M be either a right or a left A-module. Let EndA (M ) ⊂ EndZ (M ) be the collection of A-module homomorphisms from M to itself. Show that EndA (M ) is a subring of EndZ (M ). 25. Let A be a commutative ring and let M be an A-module. Show that EndA (M ) is an A-algebra via the map ι : A → EndA (M ) defined by ι(a)(m) = am for all a ∈ A and m ∈ M .


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26. Let A be a commutative ring. Suppose given an A-algebra B and an A-module M . Show that there is a one-to-one correspondence between the A-algebra homomorphisms from B to EndA (M ) and the B-module structures on M that are compatible with the original A-module structure. 27. For any ring A and any A-module M , we write AutA (M ) for the group of units in EndA (M ). Suppose that A is commutative and that G is a group. Show that there is a one-to-one correspondence between the group homomorphisms from G to AutA (M ) and the A[G]-module structures on M that are compatible (with respect to the standard inclusion ι : A → A[G]) with the original A-module structure on M.

Free Modules and Generators Finite generation is as useful a concept in module theory as it is in group theory. Definitions 7.7.28. Let m1 , . . . , mk be elements of the A-module M . By the submodule generated by m1 , . . . , mk , we mean the sum Am1 + · · · + Amk . Clearly, this is the smallest submodule of M containing the elements m1 , . . . , mk . The elements of Am1 + · · · + Amk all have the form a1 m1 + · · · + ak mk , for some a1 , . . . , ak ∈ A. Such a sum is called a linear combination of m1 , . . . , mk . Similarly, the submodule of M generated by an infinite family {mi | i ∈ I} is the infinite sum, i∈I Ami , of the submodules Ami . Here, a linear combination of the mi is a finite sum ai1 mi1 + · · · + aik mik , with i1 , . . . , ik ∈ I. If M is generated by a finite set m1 , . . . , mk , we say that M is a finitely generated A-module. Similarly, we say that a left ideal is finitely generated if it is finitely generated as an A-module. Clearly, A is generated by 1, while the cyclic module Ax is generated by x. A very important collection of finitely generated modules is given by the free modules An : Definitions 7.7.29. We write An for the direct sum of n copies of A, and write ei = (0, . . . , 0, 1, 0, . . . , 0). for the element of An whose i-th coordinate is 1 and whose other coordinates are all 0, with 1 ≤ i ≤ n. Note that (a1 , . . . , an ) = a1 e1 + · · · + an en , so that An is generated by e1 , . . . , en . We call e1 , . . . , en the canonical basis of An . (We adopt the convention that A0 = 0, the 0-module, whose canonical basis is taken to be the null set.) We call An the standard free A-module of rank n. The next lemma shows that An is the free A-module (in the sense of category theory) on the set {e1 , . . . , en }. Lemma 7.7.30. Let M be an A-module and let m1 , . . . , mn ∈ M . Then there is a unique A-module homomorphism f : An → M with f (ei ) = mi for i = 1, . . . , n. Explicitly, f (a1 , . . . , an ) = a1 m1 + · · · + an mn . Proof Let fi : A → M be given by fi (a) = ami . Thus, fi is the unique A-module homomorphism with fi (1) = mi . By Lemma 7.7.21 there is a unique A-module homomorphism from An to N with the property that f ◦ ιi = fi for 1 ≤ i ≤ n, where ιi : A → An


259

is the inclusion of the i-th summand. Since ei = ιi (1), this gives f (ei ) = fi (1) = mi . The explicit formula of Lemma 7.7.21 gives f (a1 , . . . , an ) = f1 (a1 ) + · · · + fn (an ) = a1 m1 + · · · + an mn , as claimed. Note that the uniqueness statement, while embedded in the above, also follows from the fact that the e1 , . . . , en generate An : Since (a1 , . . . , an ) = a1 e1 + · · · + an en , we must have f (a1 , . . . , an ) = a1 f (e1 ) + · · · + an f (en ), and hence any A-module homomorphism on An is determined by its values on e1 , . . . , en . We can generalize this to the study of arbitrary direct sums of copies of A. * Definition 7.7.31. Let * I be any set, and let ei = ιi (1) ∈ i∈I A for i ∈ I. Thus, in the I-tuples notation for i∈I A, ei is the element whose i-th coordinate is 1 and whose other * coordinates are all 0. The elements ei , i ∈ I are called the canonical basis elements of i∈I A. * Note that an element of i∈I A may be written uniquely as a sum i∈I ai ei , where all but finitely many of the ai are equal to 0. The proof of the next lemma is identical to that of Lemma 7.7.30, using Proposition 7.7.24 in place of Lemma 7.7.21. Lemma 7.7.32. Let I be any set. Let M be an A-module and let mi ∈ M for all i ∈ I. * Then there is a unique A-module homomorphism f : i∈I A → M such * that f (ei ) = mi for all i ∈ I, where the elements ei are the canonical basis elements of i∈I A. Explicitly, # " f ai ei = ai mi . i∈I

i∈I

For some purposes it is useful to get away from the use of canonical bases. Definitions 7.7.33. An A-module M is a free module if it is isomorphic to a direct sum of copies of A. We say that M is free of finite rank if it is isomorphic to An for some integer n ≥ 0. We say that a finite sequence x1 , . . . , xn of elements of M is a basis for M if the unique A-module homomorphism f : An → M with f (ei ) = xi for i = 1, . . . , n is an isomorphism.12 If I is an infinite set, we say * that a set {mi | i ∈ I} is a basis for M if the unique A-module homomorphism f : i∈I A → M with f (ei ) = mi for i ∈ I is an isomorphism. Note that in the case of infinitely generated free modules, a basis comes equipped with an indexing set I. In the finitely generated case, we insist that a basis, if non-null, be indexed by one of the standard ordered sets {1, . . . , n}. Examples 7.7.34.

12 By abuse of language, we shall speak of a sequence x , . . . , x of elements of M as an ordered subset n 1 of M . In the case of a basis, the elements x1 , . . . , xn must all be distinct, so that the sequence in question does specify an ordering on an n-element subset of M .


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1. Let G be a finite group, and consider A[G] as a left A-module via the canonical inclusion ι : A → A[G]. Then the elements of G, in any order, form a basis for A[G]. The point is that every element of A[G] may be written uniquely as g∈G ag g, so the A-module homomorphism from A|G| to A[G] induced by an ordering of the elements of G is a bijection. 2. For the same reason, if G is an infinite group, then {g | g ∈ G} forms a basis for the infinitely generated free module A[G]. 3. Consider Mn (A) as a left A-module via the canonical inclusion ι : A → Mn (A). Let eij be the matrix whose ij-th coordinate is 1 and whose other coordinates are all 0, where 1 ≤ i, j ≤ n. Then the matrices eij , in any order, form a basis for Mn (A). Here, we make use of the obvious fact that (aij ) = aij eij 1≤i,j≤n

for any matrix (aij ) ∈ Mn (A). The very same elements form bases for A[G] and Mn (A) when we consider them as right modules via the inclusions ι. Another very important example of free modules comes from quotients of polynomial algebras. Proposition 7.7.35. Let A be a commutative ring and let f (X) ∈ A[X] be a polynomial of degree n > 0 whose leading coefficient is a unit. Then A[X]/(f (X)) is a free Amodule with basis given by the images of 1, X, . . . , X n−1 under the canonical map from A[X]. Here, (f (X)) is the principal ideal of A[X] generated by f (X), and the A-module structure on the quotient ring A[X]/(f (X)) comes from the A-algebra structure induced by the standard algebra structure on A[X]. Proof Write h : An → A[X]/(f (X)) for the A-module homomorphism that carries ei to X i−1 for i = 1, . . . , n. We show that h is an isomorphism. For g(X) ∈ A[X], the Euclidean algorithm for polynomials (Proposition 7.3.10) allows us to write g(X) = q(X)f (X) + r(X), where deg r(X) < deg f (X). But then the images of g(X) and r(X) under the canonical map π : A[X] → A[X]/(f (X)) are equal, and, if r(X) = an−1 X n−1 + · · · + a0 , then π(r(X)) = h(a0 , . . . , an−1 ). Thus, h is onto. Now suppose that (a0 , . . . , an−1 ) ∈ ker h and let g(X) = an−1 X n−1 + · · · + a0 . Then π(g(X)) = 0, and hence g(X) ∈ (f (X)). Thus, g(X) = q(X)f (X) for some q(X) ∈ A[X]. Since the leading coefficient of f (X) is a unit, the degree of q(X)f (X) must be ≥ deg f unless q(X) = 0. Thus, q(X) = g(X) = 0, and hence ai = 0 for 0 ≤ i ≤ n − 1. Thus, h is injective. The next lemma is almost immediate from the definition, but is useful nonetheless, especially in understanding matrix groups. Lemma 7.7.36. Let f : M → N be an A-module homomorphism between the free modules M and N , and let x1 , . . . , xn be any basis of M . Then f is an isomorphism if and only if f (x1 ), . . . , f (xn ) is a basis for N . Proof Let g : An → M be the A-module homomorphism with g(ei ) = xi for i = 1, . . . , n. By the definition of basis, g is an isomorphism. The result now follows from the fact that f ◦ g is the unique A-module homomorphism from An to N which takes ei to f (xi ) for i = 1, . . . , n.


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Surprisingly, rank is not well defined for free modules over arbitrary rings: There are rings for which An ∼ = Am for n = m. But such rings turn out to be exotic. We shall see shortly that rank is well defined over commutative rings. The following corollary, which is immediate from Lemma 7.7.22 and the definition of An , will help us do so. Corollary 7.7.37. Let a be a two-sided ideal in the ring A. Then An /aAn is isomorphic to (A/a)n as an A/a-module. While free modules of finite rank may not always have a uniquely defined rank, we can always tell the difference between the cases of finite and infinite rank. * Lemma 7.7.38. Let I be a set. Suppose that the free module M = i∈I A is finitely generated. Then I must be a finite set. Thus, a finitely generated free module has finite rank. Proof Let x1 , . . . , xn be a generating set for M , and write xj = i∈I aij ei for j = 1, . . . , n. Then for each j, all but finitely many of the aij are 0. Thus, if S = {i ∈ I | aij = 0 for some j}, then S must be finite. But then, the submodule of M generated by x1 , . . . , xn must lie in the submodule generated by {ei | i ∈ S}. But this says that M itself must be generated by {ei | i ∈ S}. But clearly, if i is not in S, then ei is not in the submodule of M generated by {ei | i ∈ S}. Thus, S = I, and hence I is finite. It is useful to understand what it means to be a basis. Definitions 7.7.39. Let m1 , . . . , mn be elements of the A-module M . We say that m1 , . . . , mn are linearly dependent if there are elements a1 , . . . , an of A, not all 0, such that a1 m1 + · · · + an mn = 0. Similarly, we say that an infinite set {mi | i ∈ I} is linearly dependent if there is a relation of the form ai1 mi1 + · · · + aik mik = 0, where k > 0, i1 , . . . , ik are distinct elements of I and not all of the aij are 0. We say that a collection of elements is linearly independent if it is not linearly dependent. Proposition 7.7.40. Let m1 , . . . , mn be elements of the A-module M and let f : An → M be the unique A-module homomorphism such that f (ei ) = mi for 1 ≤ i ≤ n. Then the following conditions hold. 1. The elements m1 , . . . , mn are linearly independent if and only if f is an injection. 2. The image of f is the submodule generated by m1 , . . . , mn . Thus, m1 , . . . , mn generate M if and only if f is a surjection. Thus, m1 , . . . , mk is a basis for M if and only if it is linearly independent and generates M. * Similarly, if {mi | i ∈ I} is a family of elements of M , and if f : i∈I A → M is the unique homomorphism such that f (ei ) = mi for all i, then f is injective if and only if {mi | i ∈ I} are linearly independent, and is surjective if and only if {mi | i ∈ I} generate M.


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Proof First, consider the case of a finite subset m1 , . . . , mk . Here, an element of the kernel of f is precisely an n-tuple (a1 , . . . , an ) such that a1 m1 + · · · + an mn = 0. So linear independence is equivalent to f being injective. Also, f (a1 , . . . , an ) = a1 m1 + · · · + an mn . Since this gives the generic element of the submodule generated by m1 , . . . , mn , the second condition also holds. In the infinitely generated case, a nonzero element of i∈I A may be written uniquely in the form ai1 ei1 + · · · + aik eik , where k > 0, i1 , . . . , ik are distinct elements of I, and ai = 0 for all i. And f (ai1 ei1 + · · · + aik eik ) = ai1 mi1 + · · · + aik mik . Thus, there is a nonzero element of the kernel if and only if {mi | i ∈I} is linearly dependent. Proposition 7.7.26 shows that the image of f is i∈I Ami , so the result follows. The following corollary is immediate. Corollary 7.7.41. An A-module M is finitely generated if and only if it is a quotient module of An for some integer n. Exercises 7.7.42. 1. Both Z2 and Z3 are Z6 -modules, as they both have exponent 6. Show that Z2 ⊕ Z3 is a free Z6 -module, but that neither Z2 nor Z3 is free over Z6 . † 2. Show that every A-module is a quotient module of a free module. 3. Show that every quotient module of a finitely generated module is finitely generated.

Exact Sequences Definition 7.7.43. Suppose given a “sequence,” either finite or infinite, of A-modules and module homomorphisms: f

g

· · · → M − →M − → M → · · · We say that the sequence is exact at M if the kernel of g is equal to the image of f . We call the sequence itself exact if it is exact at every module of the sequence that sits between two other modules. One sort of exact sequence has special importance. Definition 7.7.44. A short exact sequence of A-modules is an exact sequence of the form f

g

0 → M − →M − → M → 0. Note that exactness at M shows that g has to be onto, while exactness at M shows that f is an injection. Thus, M/g(M ) is isomorphic to M , and hence the short exact sequence presents M as an extension of M by M . Example 7.7.45. For any pair M, N of A-modules, the sequence ι

π

0→M − →M ⊕N − →N →0 is exact, where ι is the standard inclusion map and π is the standard projection map. We shall make extensive use of exact sequences, both short and otherwise. For instance, we can use exactness to measure the deviation of a map from being an isomorphism.


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Definition 7.7.46. Let f : M → N be an A-module homomorphism. Then the cokernel of f is the module C = N/(im f ). In particular, f is onto if and only if the cokernel of f is 0. Lemma 7.7.47. Let f : M → N be an A-module homomorphism. Then there is an exact sequence f

i

π

0→K− →M − →N − → C → 0, where i is the inclusion of the kernel of f and π is the canonical map onto the cokernel of f . Conversely, given a six term exact sequence f

0 → K → M − → N → C → 0, f is injective if and only if K = 0 and is surjective if and only if C = 0. The next result can be very useful for recognizing isomorphisms. The technique used in the proof is called a diagram chase. Lemma 7.7.48. (Five Lemma) Suppose given a commutative diagram with exact rows M1

α1 / M2

f1 N1

β1

α2 / M3

α3 / M4

f2

f3

f4

/ N2

/ N3

/ N4

β2

β3

α4 / M5 f5 β4 / N5

Then the following properties hold. 1. If f2 and f4 are injective and if f1 is surjective, then f3 is injective. 2. If f2 and f4 are surjective and if f5 is injective, then f3 is surjective. In particular, if f2 and f4 are isomorphisms, f1 is surjective, and f5 is injective, then f3 is an isomorphism. Proof Suppose first that f2 and f4 are injective and f1 is surjective. Let m ∈ ker f3 . Then (f4 ◦ α3 )(m) = 0 by the commutativity of the diagram. In other words, α3 (m) is in the kernel of f4 . Since f4 is injective, m must be in the kernel of α3 . By exactness, m = α2 (m ) for some m ∈ M2 . Since the diagram commutes and m ∈ ker f3 , this shows that f2 (m ) ∈ ker β2 . By the exactness of the lower sequence, f2 (m ) = β1 (n) for some n ∈ N1 . Since f1 is surjective, n = f1 (m ) for some m ∈ M1 . But then f2 (α1 (m )) = β1 (n). But n was chosen so that β1 (n) = f2 (m ), so α1 (m ) has the same image as m under f2 . Since f2 is injective, m = α1 (m ). But m was chosen so that α2 (m ) = m. Since α2 ◦ α1 = 0, m must be 0, and hence f3 is injective as claimed. Now suppose that f2 and f4 are surjective and f5 is injective, and let n ∈ N3 . Since f4 is surjective, β3 (n) = f4 (m) for some m ∈ M4 . Since β4 ◦ β3 = 0, commutativity of the diagram shows that (f5 ◦ α4 )(m) = 0. But f5 is injective, and hence α4 (m) = 0. By the exactness of the upper sequence, m = α3 (m ) for some m ∈ M3 . Recall that m was chosen so that f4 (m) = β3 (n). By commutativity of the diagram, β3 carries n and f3 (m ) to the same element, so that n − f3 (m ) = β2 (n ) for some n ∈ N2 by exactness. But f2 is surjective, so that n = f2 (m ) for some m ∈ M2 . So f3 (m + α2 (m )) = n − f3 (m ) + f3 (m ) = n, and hence f3 is surjective.


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A useful illustration of the utility of the Five Lemma comes from an analysis of when a short exact sequence may be used to express the middle term as the direct sum of its surrounding terms. Recall that a section for a homomorphism g : M → N is a homomorphism s : N → M such that g ◦ s = 1N . Recall also that a retraction for a homomorphism f : N → M is a homomorphism r : M → N such that r ◦ f = 1N . We assume that all homomorphisms under discussion are A-module homomorphisms. Proposition 7.7.49. Suppose given a short exact sequence f

g

→M − → M → 0. 0 → M − Then the following additional conditions are equivalent. 1. f admits a retraction. 2. g admits a section. 3. There is an isomorphism h : M ⊕ M → M such that the following diagram commutes. 0

/ M

ι

/ M ⊕ M

π / M

/ 0

/ M

/ 0

h

0

/ M

f

/ M

g

Proof We first show that the existence of a retraction for f implies the existence of a section for g. Thus, let r be a retraction for f . We claim then that the restriction, g , of g to the kernel of r is an isomorphism of ker r onto M . Note that this is sufficient to produce a section, s, for g by setting s to be the composite M

(g )−1 / ker r ⊂ M.

To prove the claim, let m ∈ ker r. If g(m) = 0, then m ∈ ker g = im f , and hence m = f (m ) for some m ∈ M . But since r ◦ f = 1M , r(m) = m . But m = 0, since m ∈ ker r, and hence m = 0 as well. Thus, the restriction of g to ker r is injective, and it suffices to show that g : ker r → M is onto. Let m ∈ M . Then m = g(m) for some m ∈ M . Consider the element x = m − f (r(m)). Then r(x) = r(m) − r(m), since r ◦ f = 1M . Thus, x ∈ ker r. But clearly, ∼ = g(x) = m , so g : ker r −→ M as claimed, and hence g admits a section. Now suppose given a section s for g. We define h : M ⊕ M → M by h(m , m ) = f (m ) + s(m ). This is easily seen to make the diagram that’s displayed in the third condition commute. But then h is an isomorphism by the Five Lemma, and hence the third condition holds. It suffices to show that the third condition implies the existence of a retraction for f . But this is immediate: just take r = π1 ◦ h−1 , where π1 is the projection of M ⊕ M onto its first factor. Note that the specific choice of r, s, or h above determines the other two maps precisely.


265

Definition 7.7.50. If a short exact sequence g

f

0 → M − →M − → M → 0 satisfies the three properties listed in Proposition 7.7.49, then we say the sequence splits. A specification of a specific retraction for f , a specific section of g, or a specific isomorphism h as in the third condition is called a splitting for the sequence. Note that the situation with regard to splitting extensions of A-modules is cleaner than that for nonabelian groups. Here, there is no analogue of a semidirect product, and every split extension is a direct sum. If the third term of a short exact sequence is free, then it always splits. Proposition 7.7.51. Suppose given a short exact sequence g

f

0 → M − →M − → M → 0 with M free. Then the sequence splits. Proof Let {mi | i ∈ I} be a basis for M , and, for each i ∈ I, choose mi ∈ M such that π(mi ) = mi . We claim that there is an A-module map s : M → M such that s(mi ) = mi for all i ∈ I. Once we show this we’ll be done, as then g ◦ s is the identity on {mi | i ∈ I}. But any two A-module homomorphisms that agree on a generating set for a module must be equal, so g ◦ s = 1M , and hence s is a section for g. We now show that such an s exists. Recall from Lemma 7.7.32 that if N is any A-module and if* {ni | i ∈ I} are elements of N , then there’s a unique A-module homomorphism from i∈I A to N which carries the canonical basis vector ei to ni for each i ∈ I. * Let f : i∈I A → M be the unique A-module homomorphism that carries * ei to mi for each i ∈ I. Then f is an isomorphism by the definition of a basis. Let s : i∈I A → M be the unique A-module homomorphism that carries ei to mi for each i ∈ I. Then the desired homomorphism s is the composite s ◦ f −1 . Exercises 7.7.52. 1. Let A be any ring and let 0 → M → M → M → 0 be an exact sequence of A-modules. Show that if M and M are finitely generated, then so is M . † 2. Snake Lemma Suppose given a commutative diagram with exact rows: M

i

/ M

f 0

/ N

p

/ N

/ 0

f

f j

/ M

q

/ N

Show that there is an exact sequence ker f → ker f → ker f − → coker f → coker f → coker f , δ


266

where “coker” stands for cokernel, and δ is defined as follows. For m ∈ ker f , let m = p(m). Then f (m) = j(n ) for some n ∈ N , and we let δ(m ) be the element of coker f represented by n . Show also that if i is injective, so is ker f → ker f , and if q is surjective, so is coker f → coker f . (Hint: The key here is showing that δ, as described, is a well defined homomorphism.) ‡ 3. Let A be an integral domain and let (x) be the principal ideal generated by x ∈ A. Show that there is a short exact sequence fx

π

→ A/(x) → 0 0 → A −→ A − of A-modules, where fx (a) = ax. 4. Let Zn = T and let A be any commutative ring. Show that there is an exact sequence of A[Zn ]-modules · · · → Ck → · · · → C0 → A → 0, where Ck = A[Zn ] for all k ≥ 0, and the maps are given as follows. The map C0 → A is the standard augmentation. Each map C2k+1 → C2k with k ≥ 0 is multiplication by T − 1, and each map C2k → C2k−1 with k ≥ 1 is multiplication by Σ = 1 + T + · · · + T n−1 . (In the parlance of homological algebra, the sequence · · · → Ck → · · · → C0 is called a resolution of A over A[Zn ].

Rings Without Identity Not all algebraists require rings to have multiplicative identity elements. Here, we discuss the relationship between such a theory and that of rings as we have defined them. Definition 7.7.53. A ring without identity is an abelian group R together with an associative binary operation, to be thought of as multiplication, that satisfies the distributive laws: a(b + c) = ab + ac

and

(a + b)c = ac + bc.

When we speak of a ring, we will continue to assume that it has a multiplicative identity. The new notion must be named in full: rings without identity. Example 7.7.54. Let a be a left or right ideal of a ring. Then, under the multiplication inherited from A, a is a ring without identity. We shall see presently that every ring without identity may be embedded as an ideal in a ring. But it may be possible to embed it as an ideal in many different rings. We define homomorphisms, modules, and ideals for rings without identity in the obvious ways. Since rings with identity may also be considered in this broader context, we shall, at the risk of redundancy, refer to the modules as “nonunital modules.” Definitions 7.7.55. A homomorphism f : R → S of rings without identity is a homomorphism of additive groups with the additional property that f (rs) = f (r)f (s) for all r, s ∈ R.


267

A nonunital module M over a ring R without identity is an abelian group together with a multiplication R × M → M such that the following conditions hold: r(m + m )

= rm + rm

(r + r )m = rm + r m r(r m) = (rr )m, for all r, r ∈ R and all m, m ∈ M . A homomorphism f : M → N of nonunital modules over a ring R without identity is a group homomorphism such that f (rm) = rf (m) for all r ∈ R and m ∈ M . The basics of module theory, such as direct sums, exact sequences, quotient modules, etc. go over to this context without change. Free modules, on the other hand, are problematic, as R itself is not necessarily free in the categorical sense. There is no good notion of basis here. Left, right, and two-sided nonunital ideals may now be defined in the obvious way, and the quotient by a two-sided nonunital ideal satisfies the analogue of Proposition 7.2.9. There is a natural way to adjoin an identity to a ring without identity. ) of R is Definition 7.7.56. Let R be a ring without identity. The associated ring, R, ) given by R = Z ⊕ R, as an abelian group, with the multiplication (m, r) · (n, s) = (mn, ms + nr + rs), where ms and nr denote the m-th power of s and the n-th power of r, respectively, in the additive group of R. ) for the map ι(r) = (0, r) for r ∈ R. Write ι : R → R The reader may check the details of the following proposition. ) is Proposition 7.7.57. Let R be a ring without identity. Then the associated ring R ) a ring with identity element (1, 0), and ι : R → R is a homomorphism of rings without ) identity, whose image is a two-sided ideal of R. Moreover, if A is a ring, then for each homomorphism f : R → A of rings without ) → A such that f) ◦ ι = f . identity, there is a unique ring homomorphism f) : R ) is a left adjoint to In categorical language, this says that the passage from R to R the forgetful functor that forgets that a ring has an identity element. Given the relationship between module structures and homomorphisms into endomorphism rings, the following proposition should not be a surprise. Proposition 7.7.58. Let M be a nonunital module over the ring R without identity. ) Then there is a unique R-module structure on M whose restriction to the action by elements in the image of ι gives the original nonunital R-module structure on M : (n, r) · m = nm + rm, where nm is the n-th power of m in the additive group of M . ) ) The nonunital R-submodules of M and the R-submodules of this induced R-module structure coincide. This correspondence is natural in M in that it gives a functor from the category of ) nonunital R-modules to the category of R-modules.


268

Exercises 7.7.59. 1. Suppose the ring R without identity has exponent n as an additive group. Show that there is a ring structure on Zn ⊕R that extends the multiplication on R = 0⊕R. Show that the analogue of Proposition 7.7.57 holds for homomorphisms into rings of characteristic n. 2. Let A be a commutative ring and let R be a ring without identity which is also an A-module, such that r · as = ar · s = a(r · s) for r, s ∈ R, and a ∈ A. Construct an A-algebra associated to R and state and prove an analogue of Proposition 7.7.57 for appropriate homomorphisms into A-algebras.

7.8

Chain Conditions

We return to the discussion of chain conditions that was begun in Section 7.6. The module theory we’ve now developed will be very useful in this regard. First, let us generalize the chain conditions to the study of modules and of left or right ideals. Definitions 7.8.1. A ring is left Noetherian if it has the ascending chain condition for left ideals. In other words, any ascending sequence a1 ⊂ a2 ⊂ · · · ⊂ ak ⊂ · · · of inclusions of left ideals is eventually constant in the sense that there’s an n such that an = ak for all k ≥ n. Similarly, left Artinian rings are those with the descending chain condition for left ideals. Here, any descending sequence a1 ⊃ a2 ⊃ · · · ⊃ ak ⊃ · · · of inclusions of left ideals is eventually constant in the sense that there’s an n such that an = ak for all k ≥ n. Right Noetherian and right Artinian rings are defined analogously via chain conditions on right ideals. An A-module M is a Noetherian A-module if it has the ascending chain condition for its submodules. It is an Artinian module if it has the descending chain condition for its submodules. Clearly, A is left Noetherian (resp. left Artinian) if and only if it is Noetherian (resp. Artinian) as a left A-module. We shall not make use of chain conditions on two-sided ideals. When we speak of a Noetherian or Artinian ring, without specifying left or right, we shall mean a commutative ring with the appropriate chain condition on its ideals. Examples 7.8.2. 1. Recall that a division ring has no left (or right) ideals other than 0 and the ring itself. Thus, division rings are left and right Noetherian and Artinian. 2. Recall that the left ideals of a product A × B all have the form a × b, with a a left ideal of A and b a left ideal of B. Thus, if A and B are both left Noetherian (resp. left Artinian), so is A × B.


269

3. Regardless *∞ of any chain conditions that hold for a nonzero ring A, the infinite direct sum i=1 A is neither Noetherian nor Artinian as an A-module. 4. The integers, Z, are not Artinian, but Zn is an Artinian Z-module for all n > 0. !∞ 5. The infinite product i=1 Zp satisfies neither chain condition on its ideals, but the module structure on Zp obtained from the projection onto the first factor satisfies both chain conditions. 6. Let

A=

a b 0 c

a ∈ Q, b, c ∈ R .

Let a ⊂ A be the set of all elements of the form ( 00 0b ) with b ∈ R. Then a is a two-sided ideal of A. Note that a b 0 d 0 ad = . 0 c 0 0 0 0 Thus, the left submodules of a may be identified with the Q-submodules of R. Since R is infinitely generated as a vector space over Q, it satisfies neither chain condition by Example 3. (See Corollary 7.9.14 for a demonstration that R is free as a Q-module.) Thus, A itself is neither left Artinian nor left Noetherian. But 0 d a b 0 dc = , 0 0 0 c 0 0 so the right submodules of a may be identified with the R-submodules of R, which consist only of 0 and R itself. Thus, a is both right Noetherian and right Artinian as an A-module. There is a surjective ring homomorphism f : A → Q × R, given by a b f = (a, c). 0 c The kernel of f is a. Since Q × R is commutative, the left and right A-module structures on it coincide. Note that the only proper A-submodules of Q × R are Q × 0, 0 × R, and 0. Thus, Q × R satisfies both chain conditions as an A-module. Moreover, we have a short exact sequence 0→a→A→Q×R→0 of two-sided A-modules. As right modules, both a and Q × R are both Noetherian and Artinian. As we shall see in Lemma 7.8.7, this implies that A itself is both right Noetherian and right Artinian. Thus, A satisfies both chain conditions on one side, and satisfies neither chain condition on the other. The Noetherian property can be characterized in terms of finite generation. Proposition 7.8.3. An A-module M is a Noetherian module if and only if each of its submodules is finitely generated.


270

Proof Suppose that every submodule is finitely generated and that we are given an ascending chain N1 ⊂ N2 ⊂ · · · ⊂ Nk ⊂ · · · of submodules of M . Let N = k≥0 Nk . Then it is easy to see that N is a submodule of M . Suppose that N is generated by n1 , . . . , nk . Then we may choose m to be large enough that n1 , . . . , nk ∈ Nm . But then the submodule of M generated by n1 , . . . , nk must be contained in Nm . Since n1 , . . . , nk generate N , we must have Nm = Nm+l = N for all l ≥ 0. Conversely, suppose that M is a Noetherian module and let N be one of its submodules. Suppose by contradiction that N is not finitely generated. Then by induction we may choose a sequence of elements ni ∈ N such that n1 = 0 and for each i > 1, ni is not in the submodule generated by n1 , . . . , ni−1 . Write Nk for the submodule generated by n1 , . . . , nk . Then each inclusion Nk−1 ⊂ Nk is proper, and hence the sequence N1 ⊂ N2 ⊂ · · · ⊂ Nk ⊂ · · · contradicts the hypothesis that M is a Noetherian module. Since a ring is left Noetherian if and only if it is Noetherian as a left module over itself, we obtain the following corollary. Corollary 7.8.4. A ring A is left Noetherian if and only if every left ideal is finitely generated. Note that every ideal in a principal ideal ring is a cyclic module, and hence is finitely generated. Corollary 7.8.5. Principal ideal rings are Noetherian. The next very useful lemma shows that chain conditions are inherited by submodules and quotient modules. Lemma 7.8.6. Let A be a ring and let M be a Noetherian (resp. Artinian) A-module. Then any submodule or quotient module of M is Noetherian (resp. Artinian) as well. Proof An ascending (resp. descending) chain in a submodule of M is a chain in M . But then an application of the appropriate chain condition in M shows that it must hold in N as well. For quotient modules M/N , the result follows from the usual one-to-one correspondence between submodules of the quotient and the submodules of M containing N . We wish to show that the free modules An share whatever chain conditions A has. The next lemma will help us to do so. Lemma 7.8.7. Suppose given a short exact sequence f

g

0 → M − →M − → M → 0 of A-modules. Then M is Noetherian (resp. Artinian) if and only if both M and M are.


271

Proof If M satisfies the stated chain condition, then so do M and M by Lemma 7.8.6. For the converse, we restrict attention to the Noetherian case. The Artinian case is similar. Thus, suppose that M and M are Noetherian modules, and suppose given an ascending chain N1 ⊂ N2 ⊂ · · · ⊂ N k ⊂ · · · of submodules of M . Then we obtain ascending chains g(N1 ) ⊂ g(N2 ) ⊂ · · · ⊂ g(Nk ) ⊂ · · · in M and f −1 (N1 ) ⊂ f −1 (N2 ) ⊂ · · · ⊂ f −1 (Nk ) ⊂ · · · in M . Thus, since M and M are Noetherian, for k large enough, we have g(Nk ) = g(Nk+l ) and f −1 (Nk ) = f −1 (Nk+l ) for all l ≥ 0. But we claim then that Nk = Nk+l . To see this, consider the following diagram, in which the vertical maps are the inclusion maps from the sequences above. Since f (f −1 (Ni )) = Ni ∩ ker g, the rows are exact. f g / f −1 (Nk ) / Nk / g(Nk ) / 0 0

0

/ f −1 (Nk+1 )

f

/ Nk+1

g

/ g(Nk+1 )

/ 0

Since the inclusions f −1 Nk ⊂ f −1 Nk+l and g(Nk ) ⊂ f (Nk+l ) are isomorphisms, the inclusion of Nk in Nk+l is also by the Five Lemma. Since sequences of the form 0→M →M ⊕N →N →0 are exact, a quick induction on n gives the next corollary. Corollary 7.8.8. If A is left Noetherian (resp. left Artinian), then An is Noetherian (resp. Artinian) as a left A-module for all n ≥ 0. Now, a finitely generated module is a quotient module of An for some n. Since passage to quotient modules preserves chain conditions, the next proposition is immediate. Proposition 7.8.9. Let A be a left Noetherian (resp. left Artinian) ring. Then any finitely generated left A-module is a Noetherian (resp. Artinian) module over A. This now allows us to characterize the Noetherian property. Corollary 7.8.10. A ring A is left Noetherian if and only if it has the property that every submodule of every finitely generated left A-module is finitely generated. Proof If every left submodule of A itself is finitely generated, then A is left Noetherian by Corollary 7.8.4. Conversely, if A is left Noetherian and M is a finitely generated A-module, then M is Noetherian, and hence each of its submodules is finitely generated by Proposition 7.8.3.


272

Proposition 7.8.9 also has immediate applications to establishing chain conditions in some of the examples we’ve seen of rings. Corollary 7.8.11. Let f : A → B be a ring homomorphism and suppose that B is finitely generated as a left A-module. Then if A is left Noetherian or left Artinian, so is B. Proof Every B-module is an A-module via f . So a chain of ideals in B is a chain of A-submodules of the finitely generated A-module B. We obtain two immediate corollaries. Corollary 7.8.12. Let G be a finite group. Then if A is left or right Noetherian or Artinian, then so is the group ring A[G]. In particular the group ring D[G] over a division ring D satisfies all four chain conditions. Corollary 7.8.13. If A is left or right Noetherian or Artinian, then so are the matrix rings Mn (A) for n ≥ 1. In particular the matrix rings Mn (D) over a division ring D satisfy all four chain conditions. Of course, A[X] is infinitely generated as an A-module, so we shall have to work harder to show that it inherits the Noetherian property from A. As a result, however, we shall obtain a considerable strengthening of the Noetherian case of Corollary 7.8.11 when A is commutative. Theorem 7.8.14. (Hilbert Basis Theorem) Let A be a left Noetherian ring. Then the polynomial ring A[X] is also left Noetherian. Proof We show that every left ideal of A[X] is finitely generated. Let a be a left ideal of A[X] and define b ⊂ A to be the set of all leading coefficients of elements of a. Note that if a and b are the leading coefficients of f and g, respectively, and if f and g have degrees m and n, respectively, with m ≥ n, then the leading coefficient of f + X m−n g is a + b. Thus, b is a left ideal of A. Since A is left Noetherian, there is a finite generating set x1 , . . . , xk for b. But then we can find polynomials f1 , . . . , fk whose leading coefficients are x1 , . . . , xk , respectively. After stabilizing by multiplying by powers of X, if necessary, we may assume that the polynomials f1 , . . . , fk all have the same degree, say m. Let M be the A-submodule of A[X] generated by 1, X, X 2 , . . . , X m−1 . Thus, the elements of M are the polynomials of degree less than m. Now a ∩ M is a submodule of a finitely generated A-module. Since A is Noetherian, a ∩ M is a finitely generated A-module. We claim that every element of a may be written as the sum of an element of A[X]f1 + · · · + A[X]fk (i.e., the ideal generated by f1 , . . . , fk ) with an element of a ∩ M . This will be sufficient, as then a is generated as an ideal by f1 , . . . , fk together with any set of A-module generators for a ∩ M . But the claim is an easy induction on degree. Let f ∈ a have degree n. If n < m, then f ∈ a ∩ M , and there’s nothing to show. Otherwise, let a be the leading coefficient of f and write a = a1 x1 + · · · + ak xk with ai ∈ A for all i. Then a is the leading coefficient of a1 X n−m f1 + · · · + ak X n−m fk , which has degree n. But then f − (a1 X n−m f1 + · · · + ak X n−m fk ) has degree less than n, and hence the claim follows by induction. Recall that an A-algebra of finite type is one which is finitely generated as an Aalgebra.


273

Corollary 7.8.15. Let A be a commutative Noetherian ring and let B be a commutative A-algebra of finite type. Then B is Noetherian. Proof The hypothesis on B is that there is a surjective ring homomorphism f : A[X1 , . . . , Xn ] → B for some n ≥ 0. By Corollary 7.8.11, it suffices to show that A[X1 , . . . , Xn ] is Noetherian. But Proposition 7.3.23 shows that A[X1 , . . . , Xn ] is a polynomial ring in one variable on A[X1 , . . . , Xn−1 ], so the result follows from the Hilbert Basis Theorem and induction. Exercises 7.8.16. 1. Let A be a nonzero ring. Show that A[X] is not left Artinian. 2. In multiplicative notation, we can write the abelian group Z as {X n | n ∈ Z}, with 1 = X 0 as the identity element. For a ring A, the group ring A[Z] is sometimes called the ring of Laurent polynomials in A, and is sometimes denoted by A[X, X −1 ]. (This is absolutely not to be confused with a polynomial ring in two variables.) Note that A[X, X −1 ] contains the polynomial ring A[X] as a subring. Show that for any f ∈ A[X, X −1 ] there is a unit u such that uf ∈ A[X]. Use this, together with either the statement or the proof of the Hilbert Basis Theorem, to show that if A is left Noetherian, so is A[X, X −1 ]. 3. Since an abelian group is precisely a Z-module, any finitely generated abelian group is a quotient group of Zn for some n. Show that if G is a finitely generated abelian group and if A is a left Noetherian ring, then A[G] is also left Noetherian. 4. Suppose there is a finitely generated group, say with n generators, such that Z[G] fails to be left Noetherian. Show, then, that if F is a free group on n or more generators, then Z[F ] must also fail to be left Noetherian.

7.9

Vector Spaces

We give some of the basic properties of vector spaces, which we shall use in studying modules over more general rings. In particular, we shall show that if A is a commutative ring and if we have an Amodule isomorphism Am ∼ = An , then m = n. Thus, using vector spaces, we shall show that the rank of a finitely generated free module over a commutative ring is well defined, independently of the choice of an isomorphism to one of the canonical free modules An (i.e., independently of the choice of a basis). For the sake of generality, we will work over a division ring, rather than a field. (In particular, we shall not treat determinants here.) Thus, “vector space” here will mean a left module over a division ring D, and a D-linear map is a D-module homomorphism. The results, of course, will all extend to right modules as well. Definition 7.9.1. We say that a vector space V over D is finite dimensional if it is finitely generated as a D-module. Recall that a finite sequence, v1 , . . . , vk , of elements of a D-module V is a basis for V if the unique D-module homomorphism f : Dk → V for which f (ei ) = vi for i = 1, . . . , k is an isomorphism, where ei is the i-th canonical basis vector of Dk . Recall also (from Proposition 7.7.40) that v1 , . . . , vk is a basis if and only if v1 , . . . , vk are linearly independent and generate V .


274

Proposition 7.9.2. Let V be a finite dimensional vector space over D, with generators v1 , . . . , vn . Then there is a subset vi1 , . . . , vik of v1 , . . . , vn which is a basis for V . In addition, if v1 , . . . , vm are linearly independent for some m ≤ n, then we may choose the subset vi1 , . . . , vik so that it contains v1 , . . . , vm . In particular, every finite dimensional D-vector space is free as a D-module. Proof If V = 0, then the null set is a basis. Otherwise, we argue by induction on n − m. If v1 , . . . , vn is linearly independent, there is nothing to prove. Otherwise, we can find x1 , . . . , xn ∈ D, not all 0, such that x1 v1 + · · · + xn vn = 0. Now at least one of xm+1 , . . . , xn must be nonzero, as otherwise the elements v1 , . . . , vm would be linearly dependent. For simplicity, suppose that xn = 0. Then vn = (−x−1 n x1 )v1 + x )v . Thus, v is in the submodule generated by v , . . . , v · · · + (−x−1 n−1 n−1 n 1 n−1 . Since n v1 , . . . , vn generate V , V must in fact be generated by v1 , . . . , vn−1 . But the result now follows by induction. Corollary 7.9.3. Let V be a finite dimensional vector space and let v1 , . . . , vm be linearly independent in V . Then there is a basis of V containing v1 , . . . , vm . Proof Let v1 , . . . , vn be obtained from v1 , . . . , vm by adjoining any finite generating set for V , and apply Proposition 7.9.2. Proposition 7.9.4. Let v1 , . . . , vn be a basis for the vector space V and let w1 , . . . , wk be linearly independent in V . Then k ≤ n. Proof We shall show by induction on i ≥ 0 that we may relabel the v1 , . . . , vn in such a way that w1 , . . . , wi , v1 , . . . , vn−i generate V . If k were greater than n, this would say that w1 , . . . , wn is a generating set for V , and hence that wn+1 is a linear combination of w1 , . . . , wn , say wn+1 = x1 w1 +· · ·+xn wn , with x1 , . . . , xn ∈ D. But then x1 w1 + · · · + xn wn + (−1)wn+1 = 0, contradicting the linear independence of w1 , . . . , wk . Thus, for i ≥ 1, we suppose inductively that for some relabelling of the v’s, the set w1 , . . . , wi−1 , v1 , . . . , vn−i+1 generates V . Thus, we may write wi as a linear combination of these elements, say wi = x1 w1 + · · · + xi−1 wi−1 + xi v1 + · · · + xn vn−i+1 . Now it cannot be the case that xj = 0 for all j ≥ i, as then the elements w1 , . . . , wi would be linearly dependent. After relabelling, if necessary, we may assume that xn = 0, in which case vn−i+1 may be written as a linear combination of w1 , . . . , wi , v1 , . . . , vn−i , which then must generate V . Corollary 7.9.5. Let V be a finite dimensional vector space. Then any two bases of V have the same number of elements. Proof Since any basis of V is linearly independent, the number of elements in each basis must be less than or equal to the number in the other, by Proposition 7.9.4. Thus, we can make the following definition. Definition 7.9.6. Let V be a finite dimensional vector space. Then the dimension of V , written dim V , is the number of elements in any basis of V . By the definition of a basis, V has a basis with n elements (and hence has dimension n) if and only if V is isomorphic to Dn . Corollary 7.9.7. Two finite dimensional vector spaces over D have the same dimension if and only if they are isomorphic. In particular, Dn ∼ = Dm if and only if n = m.


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We can now deduce an important fact about free modules over commutative rings. Theorem 7.9.8. Let A be a commutative ring. Then An is isomorphic to Am if and only if m = n. Thus, finitely generated free modules over a commutative ring have a well defined rank, independent of the choice of an isomorphism to some An . Proof Suppose that An is isomorphic to Am and let m be a maximal ideal of A. Then An /mAn is isomorphic as a vector space over the field A/m to Am /mAm . But for k arbitrary, Corollary 7.7.37 provides an isomorphism of A/m-vector spaces between Ak /mAk and (A/m)k . Thus, An /mAn and Am /mAm have dimensions n and m, respectively, as A/m-vector spaces, and hence n = m. The following lemma is useful in detecting the difference between finite and infinite dimensional vector spaces. Lemma 7.9.9. Let V be an infinite dimensional vector space over D. Then there are linearly independent subsets of V of arbitrary length. By passage to the vector subspaces that they generate, we see that V has subspaces of every finite dimension. Proof We argue by contradiction. Suppose that the maximal length of a linearly independent subset of V is k, and let v1 , . . . , vk be linearly independent. Let W be the subspace of V generated by v1 , . . . , vk . Then W is finite dimensional, so there must be an element vk+1 of V that is not in W . We claim that v1 , . . . , vk+1 is linearly independent, contradicting the maximality of k. To see this, suppose that x1 v1 + · · · + xk+1 vk+1 = 0, with x1 , . . . , xk+1 ∈ D. If −1 xk+1 = 0, then vk+1 = (−x−1 k+1 x1 )v1 + · · · + (−xk+1 xk )vk . But this says that vk+1 ∈ W , contradicting the choice of vk+1 , so we must have xk+1 = 0. But this gives x1 v1 + · · · + xk vk = 0. Since v1 , . . . , vk are linearly independent, we must have x1 = · · · = xk = 0, and hence v1 , . . . , vk+1 are indeed linearly independent. This now gives a quick proof of the following fact, which we already knew from the fact that D is left Noetherian. Corollary 7.9.10. Let V be a finite dimensional vector space. Then any subspace of V is finite dimensional as well. Proof If V ⊂ V , then any linearly independent subset of V is a linearly independent subset of V . By Proposition 7.9.4, the maximal length of a linearly independent subset of V is dim V . So V must be finite dimensional by Lemma 7.9.9. We can now analyze the behavior of extensions with respect to dimension. Proposition 7.9.11. Suppose given a short exact sequence 0→V − →V − → V → 0 i

π

of vector spaces over D. Then V is finite dimensional if and only if both V and V are finite dimensional, in which case dim V = dim V + dim V . Proof Suppose V is finite dimensional. Since π : V → V is surjective, any set of generators of V is carried onto a set of generators of V by π. Since i(V ) is a subspace of V that is isomorphic to V , finite generation of V was just shown in Corollary 7.9.10. Now suppose that V and V are finite dimensional. We identify V with the image of i. Let v1 , . . . , vk be a basis for V and let z1 , . . . , zl be a basis for V . We claim that


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if we’re given any choice of w1 , . . . , wl ∈ V such that π(wi ) = zi for 1 ≤ i ≤ k, then v1 , . . . , vk , w1 , . . . , wl is a basis for V . First, we show that v1 , . . . , vk , w1 , . . . , wl is linearly independent. Suppose given x1 , . . . , xk+l ∈ D with x1 v1 + · · · + xk vk + xk+1 w1 + · · · + xk+l wl = 0. Then applying π, we see that xk+1 z1 + · · · + xk+l zl = 0. Since z1 , . . . , zl are linearly independent, we must have xk+1 = · · · = xk+l = 0. But then the original equation gives x1 v1 +· · ·+xk vk = 0. Since v1 , . . . , vk are linearly independent, we must have x1 = · · · = xk = 0 as well. The proof that v1 , . . . , vk , w1 , . . . , wl generate V works in an exact sequence of finitely generated modules over any ring. Let v ∈ V . Then π(v) = xk+1 z1 + · · · + xk+l zl for some l l xk+1 , . . . , xk+l ∈ D. But then π(v − i=1 xk+i wi ) = 0, and hence (v − i=1 xk+i wi ) ∈ ker π. The result now follows, since ker π = V is generated by v1 , . . . , vk . We obtain a characterization of isomorphisms of finite dimensional vector spaces. Corollary 7.9.12. Let V and W be finite dimensional vector spaces over D with the same dimension. Let f : V → W be a D-linear map. Then the following conditions are equivalent. 1. f is an isomorphism. 2. f is injective. 3. f is surjective. Proof Clearly, an isomorphism is both injective and surjective. We shall show that if f is either injective or surjective, then it is an isomorphism. If f is injective, then we get an exact sequence f

π

0→V − →W − → C → 0, where C is the cokernel of f . But then dim C = dim W − dim V = 0. But then C = 0, as otherwise, C would have at least one linearly independent element in it. If f is surjective, then a similar argument shows that the kernel of f is 0. For infinite dimensional vector spaces, we can give an analogue of Proposition 7.9.2. Proposition 7.9.13. Let V be an infinite dimensional vector space over D. Let {vi | i ∈ I} be linearly independent in V , and suppose given a family {vi | i ∈ J} such that V is generated by {vi | i ∈ I ∪ J}. Then there is a subset, K, of I ∪ J containing I such that {vi | i ∈ K} is a basis for V . Proof Let S = {K ⊂ I ∪ J | I ⊂ K and {vi | i ∈ K} is linearly independent } . Then S is nonempty, and has a partial ordering given by inclusion of subsets. We claim that S satisfies the hypothesis of Zorn’s Lemma. = Thus, let T = {Kα | α ∈ A} be a totally ordered subset of S and let K α∈A Kα . We claim that K is linearly independent, and hence K is an upper bound for T in S. To see this, suppose that xi1 vi1 + · · · + xik vik = 0, where vi1 , . . . , vik are distinct Then for each j, vi ∈ Kα for some αj ∈ A. But since T is totally elements of K. j j ordered, there is an r ∈ {1, . . . , k} such that Kαj ⊂ Kαr for j = 1, . . . , k. But then


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vi1 , . . . , vik all lie in Kαr . Since Kαr is linearly independent, this forces xij = 0 for is linearly independent as well. j = 1, . . . , k. Thus, K Thus, by Zorn’s Lemma, there is a maximal element, K, in S. We claim that {vi | i ∈ K} is a basis for V . To see this, it suffices to show that {vi | i ∈ K} generates V , which in turn will follow if we show that each vi for i ∈ J is contained in the vector subspace generated by {vi | i ∈ K}. Thus, suppose that r is contained in the complement of K in I ∪ J, and let K = K ∪{r}. Then {vi | i ∈ K } must be linearly dependent by the maximality of K. Suppose given a relation of the form xi1 vi1 +· · ·+xik vik = 0, where vi1 , . . . , vik are distinct elements of K , and xij is nonzero for j = 1, . . . , k. Since {vi | i ∈ K} is linearly independent, then at least one of the ij must equal r, −1 say r = ik . But then vr = −x−1 r xi1 vi1 − · · · − xr xik−1 vik−1 lies in the vector subspace of V generated by {vi | i ∈ K}. Taking the set {vj | j ∈ J} to be the set of nonzero elements of V , if necessary, we see that every vector space over D has a basis. Corollary 7.9.14. Every module over a division ring is free. The following corollary is now immediate from Proposition 7.7.51. Corollary 7.9.15. Every short exact sequence 0 → V → V → V → 0 of modules over a division ring splits.

7.10

Matrices and Transformations

Recall that an m × n matrix is one with m rows and n columns. We review the basic definitions and properties of nonsquare matrices. Definitions 7.10.1. Addition of nonsquare matrices is defined coordinate-wise: If M1 = (aij ) and M2 = (bij ) are both m × n matrices with coefficients in A, then M1 + M2 is the m × n matrix whose ij-th coefficient is aij + bij . Multiplication of non-square matrices is defined analogously to that of square matrices, except that we may form the product M1 M2 whenever the number of columns of M1 is equal to the number of rows of M2 . Thus, if M1 = (aij ) and M2 = (bij ) are m × n and n × l matrices, respectively, then we define M1 M2 to be the m × l matrix whose ij-th n entry is k=1 aik bkj for 1 ≤ i ≤ m and 1 ≤ j ≤ l. Nonsquare matrices behave analogously to square ones. The proof of the next lemma is left to the reader. Lemma 7.10.2. Let A be a ring and let M1 , M2 , and M3 be m × n, n × k, and k × l matrices over A, respectively. Then (M1 M2 )M3 = M1 (M2 M3 ). In addition, the distributive laws hold: with the Mi as above, if M1 and M2 are m × n and n × k matrices, respectively, then (M1 + M1 )M2 = M1 M2 + M1 M2 and M1 (M2 + M2 ) = M1 M2 + M1 M2 . Finally, for a ∈ A, let aIn ∈ Mn (A) be the matrix whose diagonal entries are all a and whose off-diagonal entries are all 0. Then if M1 = (bij ) and M2 = (cij ) for M1 and M2 above, then M1 · aIn is the matrix whose ij-th entry is bij a, and aIn · M2 is the matrix whose ij-th entry is acij .


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The following corollary is now immediate. Corollary 7.10.3. Right multiplication by the matrices aIn gives a right A-module structure to the set of m × n matrices over A, and left multiplication by aIm gives it a left A-module structure. In both cases the underlying abelian group structure of the module is given by matrix addition, but the two module structures have no direct relationship to each other if A is not commutative. (What we have instead is an example of an A-A-bimodule, as we shall encounter in our discussions of tensor products in Section 9.4. We shall not make use of bimodules here.) Let us first establish notation: Definitions 7.10.4. For m, n ≥ 1, we write Mm,n (A) for the set of m × n matrices over A. We write eij ∈ Mm,n (A) for the matrix whose ij-th coordinate is 1 and whose other coordinates are all 0, for 1 ≤ i ≤ m and 1 ≤ j ≤ n. The next lemma generalizes a result we’ve seen for square matrices. Lemma 7.10.5. Let A be a ring. Then as either a right or a left A-module, Mm,n (A) is a free module with basis given by any ordering of {eij | 1 ≤ i ≤ m, 1 ≤ j ≤ n}. In the study of classical linear algebra over a field, it is shown that if K is a field, then the K-linear maps from K n to K m are in one-to-one correspondence with Mm,n (K) We shall generalize this here to show that if A is a commutative ring, then the A-module homomorphisms from An to Am are in one-to-one correspondence with Mm,n (A). Thus, matrices correspond to transformations of free modules. The situation for noncommutative rings is more complicated, but only by a very little. The (usually) standard situation, where an m × n matrix acts from the left on an n × 1 column vector, gives the right A-module homomorphisms from An to Am . Left module homomorphisms, on the other hand, are given by the right action of the n × m matrices on the 1 × n row vectors. We now give formal treatments of these correspondences, beginning with the case of right modules. Thus, for r ≥ 1, identify Ar with the right A-module of r × 1 column matrices (i.e., column vectors) for all r ≥ 1. A typical element is ⎞ ⎛ a1 ⎟ ⎜ x = ⎝ ... ⎠ . ar As shown above, the standard right module structure on Ar is given by setting xa equal to the matrix product x · aI1 for a ∈ A. We write ei for the basis element ei1 of Lemma 7.10.5, so that ⎛ ⎞ 0 ⎜ .. ⎟ ⎜ . ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎟ ei = ⎜ ⎜ 1 ⎟. ⎜ 0 ⎟ ⎜ ⎟ ⎜ . ⎟ ⎝ .. ⎠ 0


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where nthe 1 is in the i-th place, with 1 ≤ i ≤ n. Thus, with x as above, we have x = i=1 ei ai . Regarding An and Am are right A-modules, recall that HomA (An , Am ) denotes the set of right A-module homomorphisms from An to Am . Recall also that HomA (An , Am ) is an abelian group under the operation that sets (f + g)(x) = f (x) + g(x) for f, g ∈ HomA (An , Am ) and x ∈ An . Proposition 7.10.6. Let m, n ≥ 1. Then there is an isomorphism of abelian groups ∼ = Φm,n : Mm,n (A) −→ HomA (An , Am ) defined as follows. For an m × n matrix M , we define Φm,n (M ) : An → Am by setting (Φm,n (M ))(x) to be equal to the matrix product M x for all x ∈ An . Proof Φm,n (M ) is a group homomorphism from An to Am by the distributive law of matrix multiplication, and is an A-module homomorphism by associativity: M · (xa) = M (x · aI1 ) = (M x)aI1 = (M x)a. Thus, Φm,n : Mm,n (A) → HomA (An , Am ) is a well defined function, and by the distributive law, it is a group homomorphism. It is easy to see, by direct calculation of matrix products, that M ei is the i-th column of M . Thus, if Φm,n (M ) and Φm,n (M ) have the same effect on the canonical basis vectors, the matrices M and M must be equal. Thus, Φm,n is injective. Let f ∈ HomA (An , Am ), and let M be the matrix whose i-th column is f (ei ) for i = 1, . . . , n. Then Φm,n (M ) and f have the same effect on the canonical basis elements. Since the canonical basis generates An as an A-module, Φm,n (M ) and f must have the same effect on every element of An , and hence Φm,n (M ) = f . Thus, Φm,n is surjective. As we shall see in Section 9.7, there is a natural left A-module structure on HomA (An , Am ) coming from the left action of A on Am . Similarly, there is a natural right A-module structure on HomA (An , Am ) coming from the left action of A on An . As the interested reader may check upon reading Section 9.7, Φm,n gives an A-module isomorphism from the standard left module structure on matrices to the standard left module structure on homomorphisms, as well as from the standard right module structure on matrices to the standard right module structure on homomorphisms. Recall from Problem 24 of Exercises 7.7.27 that HomA (An , An ) forms a ring, otherwise denoted EndA (An ), called the endomorphism ring of An over A. The addition is the same one we’ve just considered: (f + g)(x) = f (x) + g(x). Multiplication is given by composition of functions: (f · g)(x) = f (g(x)). Proposition 7.10.7. For n ≥ 1, Φn,n induces an isomorphism of rings from the matrix ring Mn (A) to the endomorphism ring EndA (An ). (Here, An is considered as a free right A-module.) More generally, if M is an m × n matrix and M is an n × k matrix, then the A-module homomorphism induced by the matrix product M M is the composite of the homomorphism induced by M and the homomorphism induced by M : Φm,k (M M ) = (Φm,n (M )) ◦ (Φn,k (M )) . Proof We’ve seen that Φn,n is a group homomorphism. Moreover, the multiplicative identity element In of Mn (A) is known to induce the identity transformation of An . Thus, Φn,n is a ring homomorphism if and only if it respects multiplication, and hence the first statement follows from the second. But the second statement is immediate from the associativity of matrix multiplication: (Φm,n (M ) ◦ Φn,k (M )) (x) = M (M x) = (M M )x = Φm,k (M M )(x).


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Recall that a matrix is invertible if it is a unit in Mn (A), and that the group of units there is Gln (A), the n-th general linear group of A. Note that M ∈ Gln (A) if and only if there is a matrix M ∈ Mn (A) with M M = M M = In . There is a module theoretic analogue of the general linear group. Definition 7.10.8. If N is an A-module, we shall write AutA (N ), the group of A-module automorphisms of N , for the unit group of EndA (N ). Since composition is the ring product in EndA (An ) and the identity function is the multiplicative identity element, an A-module automorphism is precisely an isomorphism f : N → N of A-modules. Since Φn,n is an isomorphism of rings, we obtain the following corollary. Corollary 7.10.9. An n × n matrix M over A is invertible if and only if the right Amodule homomorphism that it induces (i.e., Φn,n (M ) : An → An ) is an isomorphism. In particular, Φn,n restricts to an isomorphism ∼ =

Φn,n : Gln (A) −→ AutA (An ).

Recall from Lemma 7.7.36 that if f : N → N is an A-module homomorphism between free modules and if x1 , . . . , xn is a basis for N , then f is an isomorphism if and only if f (x1 ), . . . , f (xn ) is a basis for N . Since M ei is the i-th column of M , we obtain the following corollary. Corollary 7.10.10. An n × n matrix M over A is invertible if and only if its columns form a basis of An . It is useful to be able to do manipulations on the matrix level that match standard constructions in module theory. One such construction is the block sum (sometimes called Whitney sum13 ) of matrices: given an m × n matrix M and an m × n matrix M , the block sum M ⊕ M of M and M is the matrix M 0 , 0 M where the 0’s are 0-matrices of the appropriate size. In coordinates, the ij-th coordinate of M ⊕ M is given as follows. If 1 ≤ i ≤ m and 1 ≤ j ≤ n, then the ij-th coordinate of M ⊕ M is equal to the ij-th coordinate of M . If 1 ≤ i ≤ m and j > n, or if i > m and j ≤ n, then the ij-th coordinate of M ⊕ M is 0. If m + 1 ≤ i ≤ m + m and n + 1 ≤ j ≤ n + n , then the ij-th coordinate of M ⊕ M is equal to the (i − m)(j − n)-th coordinate of M . The use of the symbol “⊕” for block sum is meant to be suggestive. In moduletheoretic terms, we use it as follows: If f1 : N1 → N1 and g : N2 → N2 are A-module homomorphisms, we write f ⊕ g : N1 ⊕ N2 → N1 ⊕ N2 for the cartesian product of f and g. Thus, (f ⊕ g)(n1 , n2 ) = (f (n1 ), g(n2 )). 13 After

the topologist Hassler Whitney.


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Proposition 7.10.11. Let M be an m × n matrix and let M be an m × n matrix over A. Then if we identify An ⊕ An with An+n in the standard way (i.e., identifying n n n n ( i=1 ei ai , i=1 ei ai ) with i=1 ei ai + i=1 ei+n ai ) and identify Am ⊕Am with Am+m similarly, then the A-module homomorphism represented by the block sum M ⊕ M is equal to the direct sum of the A-module homomorphisms represented by M and by M . Symbolically, we have Φm+m ,n+n (M ⊕ M ) = Φm,n (M ) ⊕ Φm n (M ).

Proof Both sides agree on the canonical basis elements of An+n . So far we have only treated the module homomorphisms from An to Am , rather than treating the module homomorphisms between arbitrary finitely generated free modules. The reason is that there is no way to get from module homomorphisms to matrices that doesn’t depend on making explicit choices of bases for the two modules: a different choice of basis results in a different matrix to represent it. Thus, we should think of An or Am in this discussion as a finitely generated free module with a specific preferred basis. Of course, the point is that if N is a finitely generated free module, then an ordered subset B = n1 , . . . , nk of N is a basis if and only if the unique homomorphism fB : Ak → N with fB (ei ) = ni for 1 ≤ i ≤ k is an isomorphism. Thus, the bases of N with k elements are in one-to-one correspondence with the isomorphisms from Ak to N . Definition 7.10.12. Suppose given finitely generated free right A-modules N and N , and let B = n1 , . . . , nk and B = n1 , . . . , nm be bases for N and N , respectively. Write fB : Ak → N and fB : Am → N for the isomorphisms induced by B and B . Then if g : N → N is an A-module homomorphism, we define the matrix of g with respect to the bases B to be the m × k matrix defined as follows. For 1 ≤ j ≤ k, write m B and g(nj ) = i=1 ni aij . Then the ij-th entry of the matrix of g is defined to be aij . Note that the j-th column of the matrix of g with respect to the bases B and B is precisely fB−1 g(nj ). Thus, the following lemma is immediate. Lemma 7.10.13. The matrix of g : N → N with respect to the bases B and B is equal −1 to the matrix of (fB−1 ◦ g ◦ fB ) : Ak → Am , i.e., it is the matrix Φ−1 m,k (fB ◦ g ◦ fB ). Of particular interest is the use of matrices to study the A-module homomorphisms from a free right A-module N to itself (i.e., to study elements of EndA (N )). Since the source and target of these homomorphisms are the same, we need only one basis to define a matrix. Definition 7.10.14. Let B = x1 , . . . , xn be a basis for the free right A-module N and let fB : An → N be the A-module homomorphism that takes ei to xi for each i. Let g : N → N be an A-module homomorphism. Then the matrix of g with respect to the −1 −1 basis B is defined to be Φ−1 n,n (fB ◦ g ◦ fB ), the matrix whose i-th column is fB (g(xi )). The next result makes it clearer what is going on. Lemma 7.10.15. Let B = x1 , . . . , xn be a basis for the free right A-module N and let fB : An → N be the A-module homomorphism that takes ei to xi for each i. Then there is a ring isomorphism fB∗ : EndA (N ) → EndA (An ) defined by fB∗ (g) = fB−1 ◦ g ◦ fB .


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Moreover, the matrix of g with respect to the basis B is the image of g under the composite ring isomorphism f∗

Φ−1 n,n

B EndA (An ) −−−→ Mn (A). EndA (N ) −→

In particular, the passage from a transformation to its matrix with respect to B gives an isomorphism of rings from EndA (N ) to Mn (A). Proof The function fB∗ is easily seen to be a ring homomorphism. It is an isomorphism because the correspondence that takes g ∈ EndA (An ) to fB ◦ g ◦ fB−1 gives an inverse for it. The rest follows from the definition of the matrix of g with respect to B. It is valuable to discern which properties of a transformation may be detected by its matrix regardless of the basis used. Thus, it is important to understand what happens to the matrix of g when we change the basis used to define it. Of course, as we’ve stated above, the rank of a free module is not well defined over every ring A (though it is for the ones that we care about, as we’ll see below). We shall not consider the base-change problem between bases with a different number of elements, but shall restrict attention to understanding what happens when we change between bases with the same number of elements. Definition 7.10.16. Suppose given bases B = x1 , . . . , xn and B = x1 , . . . , xn of the free right A-module N and let fB : An → N and fB : An → N be the A-module homomorphisms that take ei to xi and xi , respectively, for 1 ≤ i ≤ n. Then the base change matrix MBB is defined to be the matrix of the identity map 1N : N → N , where B is used as the basis for the domain of 1N , and B is used as the basis for the target −1 copy of N . Notationally, we have MBB = Φ−1 n,n (fB fB ). Lemma 7.10.17. Suppose given bases B, B , and B for the free right A-module N , all of which have n elements. Then the base change matrices satisfy MB B · MBB = MBB . Additionally, MBB is the identity matrix, and hence MBB is invertible, with inverse given by MB B . Proof We first show that the displayed formula holds. By definition, it suffices to show that Φn,n (MB B · MBB ) = fB−1 fB . But Φn,n is a ring homomorphism, so that Φn,n (MB B · MBB ) = Φn,n (MB B ) ◦ Φn,n (MBB ) = fB−1 ◦ fB ◦ fB−1 ◦ fB , and hence the formula holds. Now MBB = Φ−1 n,n (1) = In , and hence MB B MBB = In by setting B = B in the displayed equation. Exchanging the roles of B and B now gives the desired result. We may now express the difference between the matrices of a given homomorphism with respect to different bases. Corollary 7.10.18. Suppose given bases B = x1 , . . . , xn and B = x1 , . . . , xn of the free right A-module N and let MBB be the base change matrix defined above. Then if g ∈ EndA (N ) and if M is the matrix of g with respect to the basis B, then the matrix of −1 g with respect to B is the matrix product MBB M MBB .


283

Proof Since Φn,n is a ring homomorphism, we have −1 −1 −1 Φn,n (MBB M MBB ◦ g ◦ fB ) ◦ (fB−1 ◦ fB ). ) = (fB ◦ fB ) ◦ (fB

The result now follows. Definition 7.10.19. We say that two n × n matrices, M, M ∈ Mn (A) are similar if there is an invertible n × n matrix M such that M = M M (M )−1 . Corollary 7.10.20. Let N be a finitely generated free right A-module and let g ∈ EndA (N ). Suppose we’re given two bases of N with the same number of elements. Then the matrices of g with respect to these two bases are similar. Of course, we’d like to treat left modules as well. When A is commutative, the usual thing is to identify left modules with right modules, and to study homomorphisms of free left modules by the methods above. For noncommutative rings, this doesn’t work, and hence further argumentation is needed. In this case, we identify the free left module An with the 1 × n row matrices (i.e., row vectors). This has the advantage that the notation agrees with the notation we’re used to for free modules: A typical element is x = (a1 , . . . , an ). Again we write ei = (0, . . . , 0, 1, 0, . . . , 0), where the 1 is in the i-th place. Here, the key is that if M is an n × m matrix, then the matrix product ei M is precisely the i-th row of M . Given this, and the fact that the left A-module structure on An is obtained by setting ax equal to the matrix product (aI1 )x, the proof of the next proposition is entirely analogous to that of Proposition 7.10.6, and is left to the reader. Proposition 7.10.21. Write HomA (An , Am ) for the set of left A-module homomorphisms from An to Am . Then there is a group isomorphism Ψn,m : Mn,m (A) → HomA (An , Am ) defined by setting (Ψn,m (M ))(x) equal to the matrix product xM . In this case, there is a reversal between composition of functions and matrix multiplication: if M is n × m and M is k × n, then (Ψn,m (M ) ◦ Ψk,n (M )) (x) = (xM )M = x(M M ) = Ψk,m (M M )(x). We obtain the following proposition. Proposition 7.10.22. For n ≥ 1, Ψn,n induces an isomorphism from the matrix ring Mn (A) to the opposite ring of the endomorphism ring of the free left A-module An . As in the case of right modules, the block sum of matrices represents the direct sum of the associated A-module homomorphisms. The treatment of matrices associated to different bases is analogous to that given for right modules. We leave the details to the reader. Exercises 7.10.23. 1. Let N be a free right A-module with basis B = x1 , . . . , xn . Let g ∈ EndA (N ) and let M be the matrix of g with respect to the basis B. Let M be an n × n matrix that is similar to M . Show that M is the matrix of g with respect to some other basis of N .


2. Suppose given matrices M1 M= M3

M2 M4

284

and

M =

M1 M3

M2 M4

,

where M1 is m × n, M2 is m × s, M3 is r × n, and M4 is r × s, while M1 is n × k, M2 is n × t, M3 is s × k, and M4 is s × t. Show that M1 M1 + M2 M3 M1 M2 + M2 M4 . MM = M3 M1 + M4 M3 M3 M2 + M4 M4 3. Let M be an (n + k) × (n + k) matrix over A. Show that M has the form M1 M 2 M= 0 M3 with M1 n × n, M2 n × k, and M3 k × k if and only if (Φn,n (M ))(An ) ⊂ An (i.e., the standard right A-module homomorphism induced by M carries An into itself). Here An is the submodule of An+k generated by e1 , . . . , en . Show also that the matrices of the above form (where n is fixed) give a subring of Mn+k (A). 4. Let B = Mn (A). Show that the matrix ring Mk (B) is isomorphic as a ring to Mnk (A). ‡ 5. If A is commutative, then, since left and right A-modules may be identified with each other, Ψ−1 n,n ◦ Φn,n provides an isomorphism from Mn (A) to its opposite ring for each n ≥ 1. Explicitly write down the effect of Ψ−1 n,n ◦ Φn,n on a matrix (aij ).

7.11

Rings of Fractions

Think for a moment about how our number system evolved. First, there were the nonnegative integers. Then negative numbers were adjoined, producing the integers. This process of adjoining inverses has since been generalized to a construction that produces an abelian group from any abelian monoid. It is called the Grothendieck construction, and will be studied below. The next stage in the development of our number system was the creation of the rational numbers, by adjoining multiplicative inverses to the nonzero integers. We shall generalize this last process in this section. One of the consequences will be that we can then embed any integral domain in a field. Definition 7.11.1. Let A be a commutative ring. A multiplicative subset of A is a subset S ⊂ A such that 0 ∈ S and such that S is closed under multiplication: For s, t ∈ S, the product st is also in S.14 Note: A ring called “A” in this section will be implicitly assumed to be commutative. Other rings, however, will not. Let S be a multiplicative subset of the commutative ring A. We shall construct a new ring S −1 A, in which the elements of S are invertible. 14 Some people permit 0 to be an element of a multiplicative subset. Indeed, the entire theory of rings of fractions generalizes to that context, but, if 0 ∈ S, then the ring of fractions S −1 A is the 0 ring. So all we’re doing is excluding this rather boring case.


285

The elements of S −1 A will be equivalence classes of ordered pairs (a, s) with a ∈ A and s ∈ S. We say that (a, s) is equivalent to (b, t) (written (a, s) ∼ (b, t)) if there is an s ∈ S such that s (ta − sb) = 0. Notice that if A is an integral domain, then this is just the familiar condition we use to test when two fractions a/s and b/t are equal in Q. Lemma 7.11.2. Let S be a multiplicative subset of the commutative ring A. Then the above relation ∼ is an equivalence relation. Proof The relation is obviously symmetric and reflexive, so it suffices to show transitivity. Suppose that (a, s) ∼ (b, t) and that (b, t) ∼ (c, u). Then there are elements s , s ∈ S with s (ta − sb) = 0 and s (ub − tc) = 0. We claim now that s s t(ua − sc) = 0, and hence (a, s) ∼ (c, u). Now s s tua = s s usb, because s (ta − sb) = 0. Similarly, s s usb = s s stc, because s (ub − tc) = 0. Putting this together, we see that s s t(ua − sc) = 0 as desired. Definition 7.11.3. Let S be a multiplicative subset of the commutative ring A. We define the ring of fractions S −1 A as the set of equivalence classes of pairs (a, s), and, to emphasize the analogy with fractions, we write a/s, or as , for the element of S −1 A represented by (a, s). The addition and multiplication operations in S −1 A are given by ta + sb a b + = s t st

and

a b ab · = . s t st

We define the canonical map η : A → S −1 A by setting η(a) = as/s for any element s ∈ S. The proof of the following proposition is tedious but straightforward. We leave it to the reader. Proposition 7.11.4. Let S be a multiplicative subset of A. Then S −1 A is a commutative ring under the above operations. The additive identity is 0/s for any s ∈ S, and multiplicative identity is s/s for any s ∈ S. The canonical map η : A → S −1 A is a ring homomorphism whose kernel is the set of all a ∈ A that are annihilated by some s ∈ S. For s ∈ S, the element η(s) is a unit in S −1 A, with inverse 1/s. The generic element of S −1 A may then be characterized by a = η(a) · (η(s))−1 . s The reader may very well ask: Why don’t we insist that 1 ∈ S? The answer, it turns out, is that there isn’t a good reason other than the fact that it isn’t necessary. We’ll see in Exercises 7.11.27 that if 1 ∈ S, and if we take S = S ∪ {1}, then the natural inclusion −1 of S −1 A in S A is an isomorphism. The principal examples of rings of fractions are important enough to warrant names. Definitions 7.11.5. 1. Let A be an integral domain and let S be the set of all nonzero elements in A. The resulting ring S −1 A is called the field of fractions, A(0) , of A. 2. Let A be any commutative ring and let p be a prime ideal of A. For S = {a ∈ A | a ∈ p} and call it the localization of A at p.

we write

S −1 A = Ap ,


286

3. Let A be any commutative ring and let a ∈ A be any element that is not nilpotent. Let S = {ak | k ≥ 0}. We write S −1 A = A [1/a] , and call it the ring obtained by adjoining an inverse to a. It is easy to see that Q is the field of fractions of Z, so that we have, as promised, generalized the construction of the rationals from the integers. Note that the field of fractions of an integral domain A really is a field, as η(a)(η(s))−1 is a unit in A(0) whenever a = 0. We obtain the following corollary. Corollary 7.11.6. The field of fractions, A(0) , of an integral domain A is a field, and η : A → A(0) an embedding. Thus, a ring is an integral domain if and only if it is a subring of a field. The following examples are important in field theory and number theory, respectively. Examples 7.11.7. 1. Let K be a field and let X1 , . . . , Xn be variables. We write K(X1 , . . . , Xn ) for the field of fractions of the polynomial ring K[X1 , . . . , Xn ]. We call it the field of rational functions over K in the variables X1 , . . . , Xn . p form an 2. Recall from Problem 1 of Exercises 7.1.41 that the p-adic integers Z integral domain. We write Qp for its field of fractions, known as the field of p-adic rationals. Note that a field of fractions is a special case of localization, as the ideal (0) is prime in an integral domain. If p ∈ Z is prime, then the ring Z [1/p] is sometimes called the “localization of Z away from p.” As we shall see below, this is a definite abuse of the word localization. Lemma 7.11.8. Let B be a ring and let a and b be elements of B that commute with each other, where b ∈ B × . Then a commutes with b−1 . Proof Clearly, b annihilates ab−1 − b−1 a. Since b is a unit, the result follows. We return to the general case of a multiplicative subset of a commutative ring A. The ring of fractions S −1 A satisfies a useful universal property. Proposition 7.11.9. Let f : A → B be a ring homomorphism such that f (s) is a unit for all s ∈ S. Then there is a unique ring homomorphism f : S −1 A → B such that the following diagram commutes. f / B A? ? ?? ?? η ?? f S −1 A Moreover, if f gives an A-algebra structure on B, then f , if it exists, induces an S −1 A-algebra structure on B. We obtain a one-to-one correspondence between the S −1 Aalgebra structures on B and the A-algebra structures f : A → B with the property that f (S) ⊂ B × . And if g : B → C is an A-algebra homomorphism between S −1 A-algebras, then it is also an S −1 A-algebra homomorphism.


287

Proof Suppose there is a ring homomorphism f that makes the diagram commute. Since η(s) is a unit in S −1 A for each s ∈ S, f (s) = f (η(s)) must be a unit in B. Moreover, since a/s = η(a)(η(s))−1 , f must satisfy the formula f (a/s) = f (a)f (s)−1 . Thus, f is uniquely defined by the commutativity of the diagram. Conversely, if f (s) is a unit for each s ∈ S, we define f by the same formula: f (a/s) = f (a)f (s)−1 . This is well defined, since if s (ta − sb) = 0, we have f (ta − sb) = 0 by the invertibility of f (s ). Making use of Lemma 7.11.8, if B is noncommutative, it is easy to see that f is a ring homomorphism. The statements about existence and uniqueness of algebra structures are straightforward. To see that every A-algebra homomorphism f : B → C between S −1 A-algebras is an S −1 A-algebra homomorphism, write ν and μ for the S −1 A-algebra structures on B and C, respectively. We want to show that f ◦ ν = μ. Our assumption is that f is an A-algebra homomorphism, which means that f ◦ ν agrees with μ if we precompose with η : A → S −1 A. But two ring homomorphisms out of S −1 A that agree when precomposed with η must be equal. Now Q is the field of fractions of Z. Recall that for any ring B, there is a unique ring homomorphism f : Z → B, where f (n) = n · 1, the n-th “power” of 1 in the additive group structure on B. The homomorphism f gives B a unique Z-algebra structure, and B has characteristic 0 if and only if f is injective. We obtain the following corollary. Corollary 7.11.10. A ring B is a Q-algebra if and only if n · 1 is a unit in B for each 0 = n ∈ Z. The Q-algebra structure on B, if it exists, is unique, and any ring homomorphism between Q-algebras is a Q-algebra homomorphism. In particular, there is a unique embedding (as a ring) of Q in any field of characteristic 0. We also have the following. Corollary 7.11.11. Let f : A → K be an embedding of an integral domain A into a field K. Then there is a unique extension of f to an embedding f : A(0) → K, where A(0) is the field of fractions of A. The image of f is the smallest subfield of K containing the image of f . We now have a framework to classify the subrings of Q. Proposition 7.11.12. Let A be a subring of Q and let S = A× ∩ Z. Then A is isomorphic as a ring to S −1 Z. Proof The elements of S become units in A. Thus, Proposition 7.11.9 provides a unique ring homomorphism f : S −1 Z → A. We claim that f is an isomorphism. f First, note that the composite S −1 Z − → A ⊂ Q takes m/s to m/s. Since the equivalence relations defining S −1 Z and Q are the same (because Z is an integral domain), the above composite is injective, and hence so is f . Now suppose given an element of A. Since A ⊂ Q, we may write it as m/n for m, n ∈ Z with n = 0. By reducing the fraction, we may assume that m and n are relatively prime. It suffices to show that n is a unit in A, i.e., that the element 1/n of Q lies in A. Since m and n are relatively prime, we can find a, b ∈ Z with am + bn = 1. Dividing both sides by n, we see that a(m/n) + b = 1/n. Since a, m/n and b are in A, so is 1/n.


288

The universal property for homomorphisms out of rings of fractions gives the simplest proof of the following characterization of modules over these rings. The situation is similar to that given by Proposition 7.7.15. Because S −1 A is an A-algebra, every S −1 A-module has an underlying A-module structure. So an immediate question is: Which A-modules admit a compatible S −1 Amodule structure? And if a compatible S −1 A-module structure exists, is it unique? (Compatibility here means that the A-module structure induced by the S −1 A-module structure is the same as the one we started with.) A good answer will give us a complete understanding of S −1 A-modules. Proposition 7.11.13. Let M be an A-module. Then M admits a compatible S −1 A module structure if and only if multiplication by s induces an isomorphism (of abelian groups) from M to M for each s ∈ S. Moreover, there is at most one S −1 A-module structure compatible with the original A-module structure on M . Finally, if the A-modules M and N admit compatible S −1 A-module structures, then every A-module homomorphism from M to N is an S −1 A-module homomorphism as well. Proof We use the correspondence given by Problem 21 of Exercises 7.7.27 between A-module structures on M and ring homomorphisms from A to EndZ (M ). Note that if f : A → EndZ (M ) and f : S −1 A → EndZ (M ) are ring homomorphisms, then the S −1 A-module structure on M induced by f is compatible with the A-module structure coming from f if and only if the diagram f / EndZ (M ) A JJ JJ t: JJ ttt J t J t η J$ tt f −1 S A commutes. For a given homomorphism f : A → EndZ (M ), the homomorphisms f : S −1 A → EndZ (M ) that make the diagram commute are characterized by Proposition 7.11.9: For a given f there is at most one f with this property, and such an f exists if and only if f (s) is a unit in EndZ (M ) (i.e., a group isomorphism of M ) for each s ∈ S. It remains to show that if f : M → N is an A-module homomorphism between the S −1 A-modules M and N , then f is in fact an S −1 A-module homomorphism. In particular, it suffices to show that f ((1/s)m) = (1/s)f (m) for s ∈ S and m ∈ M . Write s : M → M and s : N → N for the homomorphisms induced by multiplication by s. Then the statement that f ((1/s)m) = (1/s)f (m) for all m ∈ M amounts to the −1 statement that f ◦ −1 s = s ◦ f . Now, since f is an A-module homomorphism, we have f ◦ s = s ◦ f . So the result now follows as in the proof of Lemma 7.11.8. We’d like now to study ideals in rings of fractions. Indeed, the properties of the ideals in a localization form one of the primary motivations for studying rings of fractions. Given the relationship between ideals and modules, it seems reasonable to discuss modules of fractions first. Let M be an A-module and let S be a multiplicative subset of A. We shall define an S −1 A-module S −1 M as follows. The elements of S −1 M are equivalence classes of ordered pairs (m, s) with m ∈ M and s ∈ S, where (m, s) is equivalent to (n, t) (written (m, s) ∼ (n, t)) if there is an s ∈ S with s (tm − sn) = 0. As expected, we write m/s for the element of S −1 M represented by the ordered pair (m, s). The proof of the following lemma is straightforward and is left to the reader.


289

Lemma 7.11.14. The relation ∼ above is an equivalence relation. The resulting set S −1 M of equivalence classes is an abelian group and an S −1 A-module via m n tm + sn + = s t st

a m am · = , s t st

and

respectively. There is a canonical A-module homomorphism η : M → S −1 M given by η(m) = ms/s for any s ∈ S. If N is an S −1 A-module and f : M → N is an A-module homomorphism, then there is a unique S −1 A-module homomorphism f : S −1 M → N that makes the following diagram commute. f / N M? ? ?? ?? η ??? f S −1 M We obtain an isomorphism η∗

HomS −1 A (S −1 M , N ) −→ HomA (M, N ) ∼ =

via η ∗ (f ) = f ◦ η. Localization forms an important special case of modules of fractions: Definition 7.11.15. Let p be a prime ideal of A and let S be the complement of p in A. Then we write Mp for S −1 M . We call it the localization of M at p. Let S be an arbitrary multiplicative subset of A and let f : M → N be an A-module homomorphism. Let S −1 f : S −1 M → S −1 N be defined by (S −1 f )(m/s) = f (m)/s. Then S −1 f is clearly the unique S −1 A-module homomorphism that makes the following diagram commute. f

M η

/ N η

S

S −1 M

−1

f/

S −1 N

Clearly, passage from M to S −1 M and from f to S −1 f provides a functor from Amodules to S −1 A-modules (i.e., S −1 f ◦S −1 g = S −1 (f ◦g), and S −1 (1M ) = 1S −1 M , where 1N denotes the identity map of the module N ). Proposition 7.11.16. Suppose given an exact sequence f

g

· · · → M − →M − → M → · · · of A-modules. Then the induced sequence S −1 f

S −1 g

· · · → S −1 M −−−→ S −1 M −−−→ S −1 M → · · · of S −1 A-modules is also exact.


290

Proof A sequence is exact if it is exact at every module in it. We show exactness at M . First, note that S −1 g ◦ S −1 f = S −1 (g ◦ f ) = S −1 0, since g ◦ f is the 0-homomorphism 0 (since im f ⊂ ker g). But S −1 0 = 0, so im(S −1 f ) ⊂ ker(S −1 g). Now let m/s ∈ ker S −1 g. Then a glance at the equivalence relation defining S −1 M shows that tg(m) = 0 for some t ∈ S. But then g(tm) = tg(m) = 0, so tm ∈ ker g = im f . −1 f is onto. Let tm = f (m ). Then m/s = (S −1 f )( m st ). Thus, S This behavior may be abstracted as follows. Definition 7.11.17. A functor which takes exact sequences to exact sequences is called an exact functor. In particular, the passage from M to S −1 M is an exact functor. This has a number of uses, since certain kinds of relationships can be described by exactness properties. For ⊂ instance, N is a submodule of M if and only if the sequence 0 → N −→ M is exact. We immediately obtain the following corollary. Corollary 7.11.18. Let S be a multiplicative subset of A and let M be an A-module. Then if N is a submodule of M , the natural map from S −1 N to S −1 M is an inclusion of S −1 A-modules. Thus, if a is an ideal of A, then S −1 a is an ideal of S −1 A. We shall show that every ideal of S −1 A arises this way: Proposition 7.11.19. Let S be a multiplicative subset of A and let b be an ideal of S −1 A. Let η : A → S −1 A be the canonical homomorphism and let a = η −1 b. Then b = S −1 a. Proof Let x ∈ a. Then xs/s ∈ b for any s ∈ S, and hence so is x/t = (xs)/(ts) for any t ∈ S, since b is an ideal of S −1 A. Thus, S −1 a ⊂ b. Suppose given an arbitrary element x/s ∈ b with x ∈ A and s ∈ S. It suffices to show that x ∈ a. But this follows, since η(s) · (x/s) = η(x) (i.e., η(x) ∈ b). Corollary 7.11.20. Let S be a multiplicative subset of A. Then the prime ideals of S −1 A are those of the form S −1 p, where p is a prime ideal of A such that p ∩ S = ∅. Proof Since the preimage under a ring homomorphism of a prime ideal is prime, every prime ideal of S −1 A has the form S −1 p for some prime ideal p of A. Also, since S −1 p must be a proper ideal of S −1 A in order to be prime, S −1 p must not contain a unit of S −1 A, and hence p ∩ S must be empty. Thus, it suffices to show that if p is any prime ideal of A that doesn’t meet S, then S −1 p is prime in S −1 A. Suppose given such a p, and suppose that (a/s) · (b/t) ∈ S −1 p. Say (ab)/(st) = x/s , with x ∈ p and s ∈ S. This says there is an s ∈ S with s (s ab − stx) = 0. But then s s ab ∈ p. Since s and s are not in p, one of a and b must be. Thus, one of a/s and b/t is in S −1 p. Thus, it suffices to show that S −1 p is proper. Suppose that x/s = 1 in S −1 A, with x ∈ p and s ∈ S. Since 1 = s /s for any s ∈ S, we can find s , s ∈ S with s (s x − s s) = 0. But this places s s s in p, which is impossible, and hence S −1 p cannot be all of S −1 A. Of course, if S −1 p and S −1 p are prime ideals of S −1 A, then S −1 p ⊂ S −1 p if and only if p ⊂ p, by Proposition 7.11.19. We now are in a position to explain the term “localization.”


291

Definition 7.11.21. A local ring is a commutative ring that has only one maximal ideal. Fields, of course, are local rings, as are localizations, as we shall see shortly. Another p. example of a local ring that we’ve seen is Z Example 7.11.22. Recall from Proposition 7.1.40 that the ring homomorphism p → Zp π1 : Z p is a unit if and only if π1 (a) is nonzero in Zp . Thus, any has the property that a ∈ Z p p which doesn’t lie in ker π1 is a unit. Therefore, any proper ideal of Z element of Z must be contained in ker π1 , and hence ker π1 is the unique maximal ideal of Zp . Note that Problem 4 of Exercises 7.1.41 shows that ker π1 = (p), the principal ideal p is a local ring with maximal ideal (p). generated by p. Thus, Z Proposition 7.11.23. Let p be a prime ideal in the commutative ring A. Then the localization Ap is a local ring whose maximal ideal is pp . Proof Let S be the complement of p in A. Then p ∩ S = ∅, so that pp = S −1 p is a prime ideal of Ap . Since maximal ideals are prime, it suffices to show that any other prime ideal of Ap must be contained in pp . But the prime ideals of Ap all have the form pp , where p ∩ S = ∅. By the definition of S, this forces p ⊂ p, and the result follows. This, of course, begs the question of the relationship of the field Ap /pp to A itself. Proposition 7.11.24. Let p be a prime ideal of A. Then Ap /pp is isomorphic to the field of fractions of A/p. In particular, if m is a maximal ideal of A, then Am /mm is just A/m. Proof We have a commutative diagram A π A/p

η

η

/ Ap π

/ Ap /pp

where π and π are the canonical maps. It is easy to check that p = η −1 (pp ), and hence η is an embedding. By Corollary 7.11.11, η extends uniquely to an embedding, η : K → Ap /pp of the field of fractions, K, of A/p. But η is surjective, as π (a/s) = η(π(a)/π(s)). If p is maximal, then A/p is a field, and hence A/p = K by Corollary 7.11.11. Localization turns out to be a very important technique in understanding A-modules. The point is that modules over a local ring are easier to classify than those over a non-local ring. This will become especially clear in our study of projective modules in Section 9.8. So it is valuable to be able to deduce information about a module M from information about the localizations Mp . Proposition 7.11.25. Let M be a module over the commutative ring A. Then the following conditions are equivalent.


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1. M = 0. 2. Mp = 0 for each prime ideal p of A. 3. Mm = 0 for each maximal ideal m of A. Proof Clearly, the first condition implies the second, which implies the third. We shall show that the third implies the first. Let M be an A-module such that Mm = 0 for each maximal ideal m of A. Let m ∈ M . It suffices to show that the annihilator of m is A. Thus, let m be a maximal ideal of A. Then m is in the kernel of the canonical map from M to Mm . Thus, m/1 = 0/1 in Mm , so there exists s in A which is not in m such that sm = 0. In other words, Ann(m) is not contained in m. But since Mm = 0 for every maximal ideal m of A, we see that Ann(m) is not contained in any of the maximal ideals of A. Since Ann(m) is an ideal, it must be all of A. Corollary 7.11.26. Let f : M → N be an A-module homomorphism. Then the following conditions are equivalent. 1. f is injective. 2. fp is injective for each prime ideal p of A. 3. fm is injective for each maximal ideal m of A. The same holds if we replace injective by surjective. Proof Consider the exact sequence i

f

π

→M − →C→0 →N − 0→K− of Lemma 7.7.47. Since localization is an exact functor, we see that Kp is the kernel of fp and Cp is the cokernel of fp for each prime ideal p of A. Thus, the result follows immediately from Proposition 7.11.25. Exercises 7.11.27. 1. Show that Zn is a local ring if and only if n is a prime power. 2. Show that the localization, Z(p) , of Z at (p) embeds uniquely as a subring of the p. p-adic integers Z 3. Let a be any non-nilpotent element of the commutative ring A. Show that A[1/a] is isomorphic as an A-algebra to A[X]/(aX − 1), where (aX − 1) is the principal ideal generated by aX − 1 in the polynomial ring A[X]. 4. Let A be an integral domain and let K be its field of fractions. Show that if 0 = a ∈ A, then the ring of fractions A[1/a] may be identified with the image of the evaluation map ε1/a : A[X] → K obtained by evaluating X at 1/a. Thus, the notation A[1/a] is consistent with that for adjoining an element to A. 5. Let A be a P.I.D. and let S be a multiplicative subset of A. Show that the ring of fractions S −1 A is also a P.I.D.


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‡ 6. Let S be a multiplicative subset of A with 1 ∈ S. Suppose that S is generated as a monoid by a finite set s1 , . . . , sk , i.e., an element is in S if and only if it has the form sr11 . . . srkk where ri ≥ 0 for 1 ≤ i ≤ k. Show that S −1 A is isomorphic as an A-algebra to A[X1 , . . . , Xk ]/(s1 X1 − 1, . . . , sk Xk − 1). Here, X1 , . . . , Xk are indeterminates over A, and (s1 X1 − 1, . . . , sk Xk − 1) is the ideal generated by s1 X1 − 1, . . . , sk Xk − 1. It is common to write A[1/s1 , . . . , 1/sk ] for S −1 A in this context, a usage which may be of assistance in designing the proof. 7. Let A be a commutative ring and let S ⊂ A[X] be given by S = {X k | k ≥ 0}. Show that S −1 A[X] is isomorphic to the group ring A[Z] of Z over A. 8. We say that a multiplicative subset S ⊂ A is saturated if for any s ∈ S and any a ∈ A dividing s, we must have a ∈ S. Show that if S is any multiplicative subset, then T = {a ∈ A | a divides s for some s ∈ S} is a multiplicative subset. Show also that T is saturated. We call T the saturation of S. † 9. Let S be a multiplicative subset of A and let T be its saturation. Show that for −1 any multiplicative subset S with S ⊂ S ⊂ T , the natural maps S −1 A → S A → −1 T A are isomorphisms of rings. Show also that if M is an A-module, then the −1 natural maps S −1 M → S M → T −1 M are isomorphisms of S −1 A-modules. 10. To complete the classification of the subrings of Q, it suffices to find all of the saturated multiplicative subsets of Z. Show that there is a one-to-one correspondence between the saturated multiplicative subsets of Z and the subsets of the set of all primes in Z. Here, if T is a set of primes in Z, then the associated multiplicative subset is given by S = {±pr11 . . . prkk | k ≥ 0, p1 , . . . , pk ∈ T and ri ≥ 0 for i = 1, . . . , k}. 11. Let A be an integral domain and let B be any commutative ring. Show that the ideal p = (0 × B) of A × B is prime and that the localization (A × B)p is isomorphic to the field of fractions of A. 12. Let p be a prime number and let M be a finite abelian group. Show that the localization M(p) is isomorphic to the subgroup Mp of M consisting of those elements whose order is a power of p. (Hint: Consider the restriction of the natural map η : M → M(p) to Mp .) 13. Let S be a multiplicative subset of Z and let M be a finite abelian group. Show that S −1 M is isomorphic to the subgroup of M consisting of those elements whose order is relatively prime to every element in S. 14. Let S be a multiplicative subset of A. Show that if S ⊂ A× , then the natural map η : A → S −1 A is an isomorphism. 15. Let S be a multiplicative subset of A and let M be an A-module that admits a compatible S −1 A-module structure. Show that the natural map η : M → S −1 M is an isomorphism. 16. Let F be a functor from A-modules to B-modules. Show that F is an exact functor if and only if for any short exact sequence f

g

→M − → M → 0 0 → M −


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of A-modules, the induced sequence F (f )

F (g)

0 → F (M ) −−−→ F (M ) −−−→ F (M ) → 0 of B-modules is also exact.

Chapter 8

P.I.D.s and Field Extensions The theories of fields and of P.I.D.s are inextricably intertwined. If K is a field, then the polynomial ring K[X] is a P.I.D. We shall show in Section 8.1 that the elements of a P.I.D. have prime decompositions similar to the ones in the integers. This is used to study the extension fields L of K, via the evaluation maps εα : K[X] → L obtained by evaluating X at an element α of L. Also, we give a classification of the finitely generated modules over a P.I.D. in Section 8.9. We shall use this to classify the n × n matrices over a field K up to similarity in Section 10.6. The point is that an n × n matrix induces a K[X]-module structure on K n in which X acts on K n by multiplication by the matrix in question. The theory of prime factorization in P.I.D.s also gives some elementary examples of the behavior of primes under finite extensions of the rational numbers, Q. The point is that if K is a finite extension field of Q, then there is a subring O(K), called the ring of integers of K, that plays a role in K similar to the role of the integers in Q. An important topic in number theory is the way in which the prime numbers in Z factor in O(K). In the general case, O(K) is a Dedekind domain, but not a P.I.D. In that case, we study the factorizations of the principal ideal (p) of O(K) as a product of ideals. But in some nice cases, which we shall illustrate in the exercises to Section 8.1, the ring of integers is a P.I.D., and we may study this factorization in terms of factorizations of prime elements. The theory of prime factorization is developed in Section 8.1 in a generalization of P.I.D.s called unique factorization domains, or U.F.D.s. In Section 8.2, we study the algebraic extensions of a field K. These are the extension fields with the property that the evaluation maps εα : K[X] → L have nontrivial kernels for each α ∈ L. They will be the primary objects of study in Chapter 11. We begin to analyze them here. Section 8.3 studies extension fields which are not algebraic. There is a notion of transcendence degree, which measures how far a given extension is from being algebraic. There is a theory of transcendence bases, which behave toward the transcendence degree in a manner analogous to the relationship between a basis for a vector space and its dimension. In Section 8.4, we construct an algebraic closure for a field K. Uniqueness statements regarding algebraic closures are deferred to Section 11.1. We shall see that every algebraic extension of a field K may be embedded in its algebraic closure. So algebraic closures will be an important tool in the material on Galois theory. Algebraic closures are also important in matrix theory, as the Jordan canonical form

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may be defined for n × n matrices over an algebraically closed field. Thus, we may study matrices over K by passage to the matrix ring of its algebraic closure. The various applications of prime factorization in the P.I.D. K[X] depend strongly on our ability to actually compute these factorizations. In particular, we need to be able to test polynomials for primality. In Section 8.5, we give criteria for primality in Q[X], or, more generally, in K[X], where K is the field of fractions of a unique factorization domain. In the process, we show that if A is a U.F.D., then so is A[X]. In particular, the polynomial ring K[X1 , . . . , Xn ] is a U.F.D. for any field K. In Section 9.3, we make use of unique factorization in K[X1 , . . . , Xn ], the study of transcendence bases, and the theory of Noetherian rings to prove Hilbert’s Nullstellensatz, which states that if the extension field L of K is finitely generated as a K-algebra, then L is a finite extension of K. In consequence, we determine all of the maximal ideals of K[X1 , . . . , Xn ] when K is an algebraically closed field. This is the starting point for the study of algebraic varieties. We proceed to define algebraic varieties and to describe the Zariski topology on affine n-space, K n , as well as on the prime spectrum of a commutative ring. This material forms an introduction to the study of algebraic geometry. The next three sections study field extensions, culminating in the determination of the degree of the cyclotomic extension Q[ζn ] over Q, as well as the determination of the minimal polynomial of ζn over Q. This establishes the most basic facts about a family of examples that has vital connections to a number of branches of mathematics. For instance, we shall see that the cyclotomic rationals play a vital role in computations of Galois theory. And their rings of integers, Z[ζn ], are of crucial importance in issues related to Fermat’s Last Theorem, as well as to calculations of the K-theory of integral group rings. The latter plays an important role in topology. Section 8.6 introduces the Frobenius homomorphism and discusses perfect fields. Section 8.7 introduces the notion of repeated roots, which forms the basis for the study of separable extensions. We shall return to these ideas in Chapter 11.

8.1

Euclidean Rings, P.I.D.s, and U.F.D.s

Recall that the Euclidean algorithm was the main tool we used to analyze the structure of the integers. There is an analogue of it which holds in a variety of other rings, including the polynomial ring K[X] over a field K. The consequences, for rings that satisfy this property, are very similar to the consequences that hold in Z. Definition 8.1.1. A Euclidean domain is an integral domain A that admits a function ϕ : A → Z with the following properties. 1. If b divides1 a and a = 0, then ϕ(b) ≤ ϕ(a). 2. For any a, b ∈ A with b = 0, there are elements q, r ∈ A such that a = qb + r and ϕ(r) < ϕ(b). We call the function ϕ : A → Z a Euclidean structure map for A. 1 I.e.,

b.

a = bc for some c ∈ A. Once again, this just says that a ∈ (b), the principal ideal generated by


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Examples 8.1.2. 1. Z is Euclidean, with the structure map given by the absolute value function | | : Z → Z. 2. Let K be a field, and define ϕ : K[X] → Z by setting ϕ(f (X)) equal to the degree of f (X) if f (X) = 0, and setting ϕ(0) = −1. Then Proposition 7.3.10 shows that ϕ puts a Euclidean structure on K[X]. 3. In Problem 5 of Exercises 7.1.41, it is shown that every nonzero element of the p may be written as a product pk u with k ≥ 0 and u ∈ Z × . Define p-adic integers Z p p → Z by setting ϕ(pk u) = k and ϕ(0) = −1. Since p is not a unit in the ϕ:Z p , ϕ is a well defined function. Note that if a and b are nonzero elements domain Z of Zp , then ϕ(b) ≤ ϕ(a) if and only if b divides a. It follows that ϕ is a Euclidean p. structure map for Z p is also a local ring, meaning Recall from the discussion in Example 7.11.22 that Z p is an that it has only one maximal ideal, in this case the principal ideal (p). Thus, Z example of the following notion. Definition 8.1.3. A discrete valuation ring, or D.V.R., is a Euclidean domain that is also a local ring. The Euclidean structure on Z was crucial in our analysis of the subgroups of Z. It will have similar value for the study of the ideals in other Euclidean domains. There is an additional property one could ask for from the function ϕ: Definition 8.1.4. We say that a Euclidean structure map ϕ : A → Z is positive definite if ϕ(a) > 0 for all 0 = a ∈ A and ϕ(0) = 0. Lemma 8.1.5. Let A be a Euclidean domain. Then there is a Euclidean structure map for A which is positive definite. Proof We claim that if ϕ : A → Z is any Euclidean structure map for A, then ϕ(0) < ϕ(a) for all a = 0. Given this, it will suffice to replace ϕ by a function ϕ given by ϕ (a) = ϕ(a) − ϕ(0). To prove the claim, let a = 0 and write 0 = qa + r with ϕ(r) < ϕ(a). But then r = (−q)a is divisible by a, which would force ϕ(a) ≤ ϕ(r) unless r = 0. But then r must equal 0, displaying ϕ(0) < ϕ(a) as desired. Recall that a principal ideal domain, or P.I.D., is an integral domain in which every ideal is principal. Proposition 8.1.6. A Euclidean domain is a principal ideal domain. Indeed, let ϕ : A → Z be a positive definite Euclidean structure map for A. Let a be a nonzero ideal of A and let a ∈ a be such that ϕ(a) is the smallest positive value taken on by the restriction of ϕ to a. Then a = (a). Proof Let 0 = b ∈ a and write b = qa + r with ϕ(r) < ϕ(a). But r = b − qa is in a, so by the minimality of ϕ(a), we must have ϕ(r) = 0. But then r = 0 and b ∈ (a).


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Lemma 8.1.7. Let A be an integral domain and let a, b ∈ A. Then (a) = (b) if and only if a = ub for some unit u of A. Indeed, if a = 0 and (a) = (b), and if a = bc, then c must be a unit. Thus, if b = 0 and c is not a unit, then the inclusion (bc) ⊂ (b) is proper. Proof The case a = 0 is immediate, so we assume a = 0. Here, if (a) = (b), we can find c, d such that a = bc and b = ad. Then a = adc. By the cancellation property in a domain, dc = 1. Since the units in K[X] (K a field) are the nonzero elements of K, we obtain an immediate corollary. Corollary 8.1.8. Let K be a field and let a be a nonzero ideal in K[X]. Let n be the smallest non-negative integer such that a contains an element of degree n. Then there is a unique monic polynomial, f (X), in a of degree n, and a is the principal ideal generated by f (X). Moreover, f (X) is the unique monic polynomial generating a. Definition 8.1.9. Let B be an algebra over the field K and let b ∈ B. Let εb : K[X] → B be the K-algebra map obtained by evaluating X at b. If εb is injective, we say that the minimal polynomial of b over K is 0. Otherwise, the minimal polynomial of b over K is the monic polynomial of smallest degree in ker εb . We denote it by minb (X), if the context is obvious, or more formally by minb/K (X). Corollary 8.1.10. Let B be an algebra over the field K and let b ∈ B. Then K[b] is isomorphic as a K-algebra to K[X]/(minb (X)). In particular, K[b] is finite dimensional as a K-vector space if and only if minb (X) = 0, and in this case, the dimension of K[b] over K is equal to the degree of minb (X). Proof If minb (X) = 0, then the determination of the dimension of K[b] follows from Proposition 7.7.35. Otherwise, K[b] is isomorphic to K[X], which is easily seen to be an infinite dimensional vector space over K, with basis {X i | i ≥ 0}. It turns out that P.I.D.s have almost all the nice properties exhibited by the integers, including those shown in Sections 2.3 and 4.4. We shall treat the material from the former in this section and the latter in Section 8.9. Definitions 8.1.11. Let A be a commutative ring. Then p ∈ A is irreducible if p is not a unit, but whenever p = ab, then either a or b must be a unit. We say that p ∈ A is a prime element if p = 0 and the principal ideal (p) is a prime ideal. Note that if A is a domain, then (0) is a prime ideal, but 0 is neither a prime element nor an irreducible element. There is a basic ideal theoretic consequence of irreducibility. Lemma 8.1.12. Let p be an irreducible element in the commutative ring A and suppose that a divides p. Then either a is a unit, or p divides a. In ideal theoretic terms, if (p) ⊂ (a), then either (a) = (p) or (a) = A. Thus, the ideal generated by an irreducible element is a maximal element in the partially ordered set of proper principal ideals, ordered by inclusion.


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Proof If a divides p, then p = ab for some b ∈ A. Since p is irreducible, either a or b is a unit. If b is a unit, then a = pb−1 , and hence p divides a. In terms of ideals, (p) ⊂ (a) if and only if p ∈ (a), which is equivalent to a dividing p. If a is a unit, (a) = A. Otherwise, p divides a, and hence (a) ⊂ (p). In a domain, Lemma 8.1.12 has a converse. Lemma 8.1.13. Let A be an integral domain. Then a nonzero nonunit p ∈ A is irreducible if and only if the principal ideal (p) satisfies the property that (p) ⊂ (a) implies that either (a) = (p) or (a) = A. Proof Suppose that (p) satisfies the stated condition and suppose that p = ab. Then (p) ⊂ (a). If (a) = A, then a is a unit. If (p) = (a), then Lemma 8.1.7 shows b to be a unit. Thus, p is irreducible. If every ideal is principal, then Lemma 8.1.13 may be restated as follows. Corollary 8.1.14. Let A be a P.I.D. Then a nonzero element p ∈ A is irreducible if and only if the principal ideal (p) is a maximal ideal of A. Thus, since maximal ideals are prime, every irreducible element in a P.I.D. is prime. But this does not hold in a general domain. Surprisingly, the reverse implication does hold. Lemma 8.1.15. Let p be a prime element in the integral domain A. Then p is irreducible. Proof Suppose that p = ab. Then ab ∈ (p), a prime ideal, so either a or b must lie in (p). Say a ∈ (p). But p is also in (a), since p = ab, so (a) = (p), and hence b is a unit by Lemma 8.1.7. Since maximal ideals are prime, the next corollary is immediate from Lemma 8.1.15 and Corollary 8.1.14. Corollary 8.1.16. Let A be a P.I.D. Then an element p ∈ A is irreducible if and only if it is a prime element. In particular, if p ∈ A is irreducible and if p divides ab, then it must divide either a or b. Moreover, since every nonzero prime ideal of A is generated by a prime element, every nonzero prime ideal of A is maximal, and has the form (p), where p is an irreducible element of A. Thus, the longest possible chain of proper inclusions of prime ideals in a P.I.D. has the form 0 ⊂ (p) where p is a prime element of A. This may only occur if A is not a field. Corollary 8.1.17. A P.I.D. that is not a field has Krull dimension 1. The statement in the first paragraph of Corollary 8.1.16 has a more classical proof, which is modelled on the one that we used for the integers. We shall give it to illustrate that it really uses exactly the same ideas as the proof we just gave.


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Definition 8.1.18. Let A be a P.I.D. Then the greatest common divisor of a and b is the ideal (c) = (a) + (b). We write (a, b) = (c). If (a) + (b) = A, then we say that a and b are relatively prime, and write (a, b) = 1. Note that the notation of (a, b) for the ideal (a) + (b) is consistent with the notation that (a, b) is the ideal generated by a and b. We will occasionally refer to c as the greatest common divisor of a and b when (a, b) = (c). But it is really more correct for g.c.d. to refer to the ideal (c), rather than the generator c. Lemma 8.1.19. Let A be a P.I.D. and let (a, b) = (c). Then x divides both a and b if and only if x divides c. Proof Since a and b are both in (c), c divides a and b. Thus, it suffices to show that an element dividing both a and b must divide c. But since (c) = (a) + (b), c = ra + sb for some r, s ∈ A. Thus, any element that divides a and b will divide c. The entire argument above reduces to the statement that (a) + (b) is the smallest ideal containing both a and b, in the context that every ideal is principal. Second Proof of Corollary 8.1.16 Let p be an irreducible element of A. We shall show that (p) is a prime ideal. Suppose that p divides ab but does not divide a. Since p is irreducible, Lemma 8.1.12 shows that any divisor of p is either a unit or is divisible by p. Thus, a and p cannot have any common divisors other than units. So a and p are relatively prime, and hence there are elements r, s ∈ A with ra + sp = 1. But then rab + spb = b. Since p divides ab, it divides the left-hand side, so it must divide b. Note the role that the maximality of (p) plays in the above argument. Now we would like to show that every nonzero element of a P.I.D. may be factored in some sense uniquely as a product of primes. This may be read as a statement that every P.I.D. is an example of a more general type of ring called a unique factorization domain, or U.F.D. Definition 8.1.20. A unique factorization domain, or U.F.D., is an integral domain in which every nonzero nonunit may be written as a product of prime elements. The next lemma has already been seen to hold in P.I.D.s. Lemma 8.1.21. Irreducible elements in a U.F.D. are prime. Thus, in a U.F.D., the prime elements and the irreducible elements coincide. Proof Let p be an irreducible element in the U.F.D. A. Then p is a product of prime elements, and hence is divisible by a prime element, q. Say p = qr. Since q is prime, it is not a unit. But p is irreducible, so r must be a unit. Thus, (p) = (q), a prime ideal, and hence p is prime. We now wish to show that every P.I.D. is a U.F.D. Since we’ve shown that every irreducible element in a P.I.D. is prime, it suffices to show that every nonzero nonunit is a product of irreducible elements. If our P.I.D. is actually a Euclidean domain, this may be shown by a quick induction on ϕ(a), much as the proof for the integers worked. For the general P.I.D., we need to work a little bit harder. The main tool comes from the Noetherian property.


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Proposition 8.1.22. Every P.I.D. is a U.F.D. Proof Let A be a P.I.D. Since every ideal is principal, the ideals are certainly finitely generated, and hence A is Noetherian by Corollary 7.8.4. Let 0 = a ∈ A be a nonunit, and suppose by contradiction that a is not a product of primes. Since a is not a unit, (a) is proper, and hence (a) is contained in a maximal ideal (p1 ). Thus, a = p1 a1 for some a1 ∈ A. Now p1 is irreducible by Corollary 8.1.14, and hence prime, by Corollary 8.1.16. Since p1 is prime and a = p1 a1 is not a product of primes, a1 cannot be a product of primes. Also, the inclusion (a) ⊂ (a1 ) must be proper by Lemma 8.1.7, as p1 and a1 are nonzero nonunits. We may continue by induction, obtaining elements ak ∈ A for k ≥ 1, such that ak−1 = pk ak with pk prime, and hence ak cannot be a product of primes. Thus, we obtain an infinite sequence of proper inclusions (a) ⊂ (a1 ) ⊂ · · · ⊂ (ak ) ⊂ · · · contradicting the fact that A is Noetherian. We shall show in Section 8.5 that if A is a U.F.D., so is A[X]. Taking A to be a P.I.D. that is not a field, this will give a family of examples of U.F.D.s that are not P.I.D.s. We shall now discuss the uniqueness properties of the factorizations of the elements in a U.F.D. as products of primes. Definition 8.1.23. Let p and q be prime elements in the U.F.D. A. We say that p and q are equivalent if (p) = (q). In other words, p and q are equivalent if p = uq for some unit u. In the integers, this doesn’t say much, since the only units are ±1. Thus, it is easy to choose a representative for each equivalence class of primes in Z: Just pick the positive one. That’s what we did in Section 2.3. In the generic P.I.D. there can be infinitely many units, so choosing a representative for each equivalence class of primes could get cumbersome. For this reason, we shall give a uniqueness statement first that uses equivalence classes, and then follow it up with another in which actual representatives are chosen. We say that a collection p1 , . . . , pk of primes is pairwise inequivalent if pi and pj are inequivalent for i = j. Proposition 8.1.24. Let A be a U.F.D., and suppose given an equality upr11 . . . prkk = vq1s1 . . . qlsl where u and v are units, p1 , . . . , pk are pairwise inequivalent primes, q1 , . . . , ql are pairwise inequivalent primes, and the exponents ri and si are all positive. Then k = l, and after reordering (and relabelling) the q1 , . . . , ql if necessary, pi is equivalent to qi and ri = si for 1 ≤ i ≤ k. Proof We argue by induction on r = r1 + · · · + rk . Here, we take the empty product to be 1, so if r = 0, then the left side is just u. But then vq1s1 . . . qlsl is a unit, and hence q1s1 . . . qlsl must be the empty product, too. If r ≥ 1, then p1 divides the left-hand side and hence divides the right as well. In particular, it must divide qi for some i, since the ideal (p1 ) is prime. But since p1 is not a unit and qi is irreducible, p1 and qi must be equivalent, say qi = wp1 , with w a unit.


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Reorder the q’s so that i = 1. Then p1 (up1r1 −1 . . . prkk − vwq1s1 −1 . . . qlsl ) = 0. Since A is an integral domain, we have upr11 −1 . . . prkk = vwq1s1 −1 . . . qlsl , and the result follows by induction. By a choice of representatives for the primes in A, we mean a collection {pi | i ∈ I} of primes in A such that for each prime p of A there is a unique i ∈ I such that p is equivalent to pi . Thus, for example, the positive primes form a choice of representatives for the primes in Z. Corollary 8.1.25. Let A be a U.F.D. and let {pi | i ∈ I} be a choice of representatives for the primes in A. Then any nonzero nonunit in A may be written uniquely in the form upri11 . . . prikk , where k > 0, u is a unit, i1 , . . . , ik are distinct elements of I, and ri > 0 for 1 ≤ i ≤ k. Proof Every element may be written in the above form because equivalent primes differ by multiplication by a unit. Thus, the u term can absorb the differences between a given product of primes and a product of pi ’s. The uniqueness statement now follows from that in Proposition 8.1.24, since pi is not equivalent to pj unless i = j. Thus, if two elements of the form upri11 . . . prikk are equal, the primes in them, together with their exponents, must be identical. Thus, we’re reduced to the case where upri11 . . . prikk = vpri11 . . . prikk , with u and v units. But since we’re in an integral domain, we can cancel off the primes in the above equation, leaving u = v. Corollary 8.1.26. Let A be a U.F.D. and let a ∈ A. Suppose that a = upr11 . . . prkk , where u is a unit, p1 , . . . , pk are pairwise inequivalent primes, and ri ≥ 0 for i = 1, . . . , k. Then an element b divides a if and only if b = vps11 . . . pksk , where v is a unit and 0 ≤ si ≤ ri for 1 ≤ i ≤ k. Proof If b divides a, then any prime that divides b will also divide a. If a = bc, then c also divides a, and the same reasoning applies. Thus, b = vps11 . . . pskk , and c = wpt11 . . . ptkk , where v and w are units and the exponents are all non-negative. But then a = bc = vwps11 +t1 . . . pskk +tk . By the uniqueness of factorization, si + ti = ri for i = 1, . . . , k, and hence si ≤ ri as claimed. Conversely, if si ≤ ri for i = 1, . . . , k and if v is a unit, then vps11 . . . pskk is easily seen to divide a. We’d like to be able to talk about greatest common divisors in a U.F.D., but the definition that we used in P.I.D.s fails for U.F.D.s that are not P.I.D.s. The next corollary will help us. Corollary 8.1.27. Let A be a U.F.D. and let 0 = a, b ∈ A. Let p1 , . . . , pk be a set of representatives for the set of primes dividing either a or b. Thus, p1 , . . . , pk are pairwise inequivalent, and we may write a = upr11 . . . prkk and b = vps11 . . . pskk , where u and v are units, and the exponents are non-negative. Let ti = min(ri , si ) for 1 ≤ i ≤ k and let c = pt11 . . . ptkk . (Note that c = 1 if no prime element divides both a and b.) Then c divides both a and b. In addition, if d is any element that divides both a and b, then d divides c.


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In the language of ideals, this says that (a) + (b) ⊂ (c), and if d is any other element such that (a) + (b) ⊂ (d), then (c) ⊂ (d). In particular, (c) is the smallest principal ideal that contains a and b. Proof This is almost immediate from Corollary 8.1.26: If d divides both a and b, then d = wpl11 . . . plkk , where, if 1 ≤ i ≤ k, then 0 ≤ li ≤ ri , because d divides a, and 0 ≤ li ≤ si , because d divides b. But then li ≤ ti for all i, so d divides c. Clearly, c divides a and b, so the result follows. Definitions 8.1.28. Let A be a U.F.D. and let a, b ∈ A. Then the greatest common divisor of a and b is the smallest principal ideal that contains (a) + (b). (Such an ideal exists by Corollary 8.1.27.) If the greatest common divisor of a and b is c, we write gcd(a, b) = (c), or sometimes, by abuse of notation, gcd(a, b) = c. We say that a and b are relatively prime if their greatest common divisor is (1). Notice that if A is a P.I.D., then these definitions agree with the old ones. Of course, Corollary 8.1.27 tells us exactly how to calculate greatest common divisors in a U.F.D. We state the result formally here: Corollary 8.1.29. Let A be a U.F.D. and let 0 = a, b ∈ A. Let p1 , . . . , pk be a set of representatives for the set of primes dividing either a or b. Thus, p1 , . . . , pk are pairwise inequivalent, and we may write a = upr11 . . . prkk and b = vps11 . . . pskk , where u and v are units, and the exponents are non-negative. Let ti = min(ri , si ) for 1 ≤ i ≤ k. Then the greatest common divisor of a and b is (c), where c = pt11 . . . ptkk . In particular, a and b are relatively prime if and only if no prime element of A divides both a and b. More generally, we can talk about the greatest common divisor of a set of elements in a U.F.D. Definition 8.1.30. Let a1 , . . . , an be elements of the U.F.D. A. Then the greatest common divisor, gcd(a1 , . . . , an ) of these elements is the smallest principal ideal containing (a1 ) + · · · + (an ). By abuse of notation, we also write gcd(a1 , . . . , an ) for any generator of this ideal. We say that a1 , . . . , an are relatively prime if gcd(a1 , . . . , an ) = 1. Notice the difference between saying that a1 , . . . , an are relatively prime and saying that they are pairwise relatively prime. The reader should supply the proof of the next corollary. Corollary 8.1.31. Let a1 , . . . , an be nonzero elements of the U.F.D. A. Let p1 , . . . , pk be a set of representatives for the set of all primes that divide the elements a1 , . . . , an , and write ai = ui pr1i1 . . . prkik for 1 ≤ i ≤ n, with ui a unit in A. Let tj = min(r1j , . . . , rnj ) for 1 ≤ j ≤ k. Then

gcd(a1 , . . . , an ) = pt11 . . . ptkk .


304

Another construction which makes sense in a U.F.D. is the least common multiple. Once again we shall need a computation to set it up. Proposition 8.1.32. Let A be a U.F.D. and let a1 , . . . , an ∈ A. Define an element a ∈ A as follows. If any of the ai = 0, then set a = 0. Otherwise, let p1 , . . . , pk be a set of representatives for the primes that divide at least one of the ai , and let ai = ui pr1i1 . . . prkik for 1 ≤ i ≤ n, where ui is a unit and the exponents rij are all non-negative. Then define tj = max(r1j , . . . , rnj ), and set a = pt11 . . . pktk . Then the intersection of the principal ideals generated by the ai is the principal ideal generated by a: n

(ai ) = (a).

i=1

n Proof Here, i=1 (ai ) is the collection of elements divisible by each of the ai , so the result follows immediately from Corollary 8.1.26. Definition 8.1.33. Let A be a U.F.D. and let a 1 , . . . , an ∈ A. By the least common n multiple of a1 , . . . , an we mean the intersection, i=1 n(ai ), of the principal ideals they generate, or, by abuse of notation, any generator of i=1 (ai ). We also obtain a sharpening of the Chinese Remainder Theorem. Corollary 8.1.34. (Chinese Remainder Theorem, second form) Let A be a P.I.D. and let a1 , . . . , ak ∈ A, such that ai and aj are relatively prime for i = j. Let a = a1 . . . ak . Then there is an A-algebra isomorphism ∼ =

f : A/(a) −→ A/(a1 ) × · · · × A/(ak ) given by f (x) = (x, . . . , x) for all x ∈ A. Proof Since ai and aj are relatively prime for i = j, (ai )+(aj ) = A for i = j. Thus, the first form of the Chinese Remainder Theorem (Proposition 7.2.23) applies, showing that the A-algebra homomorphism f : A → A/(a1 ) × · · · × A/(ak ) given by f (a) = (a, . . . , a) k k is a surjection whose kernel is i=1 (ai ). But i=1 (ai ) = (a) by Proposition 8.1.32. If a U.F.D. A is not a P.I.D., then its prime elements generate prime ideals which will not, in general, be maximal ideals. In fact, they turn out to be minimal nonzero prime ideals of A. Definition 8.1.35. Let A be a domain. A minimal nonzero prime ideal of A is a nonzero prime ideal with the property that the only prime ideal which is properly contained in it is 0. Proposition 8.1.36. Let A be a U.F.D. Then a prime ideal p of A is a minimal nonzero prime ideal if and only if p = (p) for a prime element p of A.


305

Proof To show that every minimal nonzero prime ideal of A is principal, it suffices to show that every nonzero prime ideal q contains a prime element. But if 0 = a ∈ q, let a = p1 . . . pk , a product of (not necessarily distinct) prime elements of A. Because a ∈ q and q is prime, at least one of the pi must lie in q. Conversely, let p be a prime element and suppose that q is a prime ideal that’s properly contained in (p). Let K be the fraction field of A and let a = (1/p)q = {(1/p)a | a ∈ q} ⊂ K. Since every element of q is divisible by p, a ⊂ A, and is easily seen to be an ideal of A. But now q = pa = (p)a. Since q is prime, either (p) ⊂ q or a ⊂ q by Lemma 7.6.9. Since q is properly contained in (p), we must have a ⊂ q. But the fact that q = (p)a shows that q ⊂ a, so q = a. In other words, q = (p)q = pq. But by induction, we see that q = pn q for all n, and hence each a ∈ q is divisible by pn for all n. By the uniqueness of factorization in A, no element but 0 can be divisible by all powers of a prime element, so q must be the 0 ideal. Thus, the only prime ideal that can be properly contained in a principal prime ideal is 0. Exercises 8.1.37. 1. Let A be a U.F.D. Suppose that a divides bc in A, where a and b are relatively prime. Show that a divides c. 2. Let p1 , . . . , pk be pairwise inequivalent primes in the U.F.D. A and let a = upr11 . . . prkk for some unit u. Show that a divides b if and only if pri i divides b for 1 ≤ i ≤ k. 3. Let K be a field. Show that every degree 1 polynomial of K is irreducible in K[X]. If f (X) has degree 1, what is K[X]/(f (X))? 4. Let K be a field and let f (X) ∈ K[X] be a polynomial whose degree is either 2 or 3. Show that f (X) is irreducible if and only if it has no roots. 5. Let K be the field of fractions of the U.F.D. A. Let a/b ∈ K be a root of f (X) = am X m + · · · + a0 ∈ A[X], where am = 0 and where a, b ∈ A are relatively prime. Show that b must divide am and a must divide a0 . 6. The simplest procedure for finding primes in Z is to proceed inductively, testing each number by dividing it by all primes smaller than its square root. If it has no prime divisor less than or equal to its square root, it is prime. One could adopt a similar strategy in F [X] for a finite field F . Here, a polynomial f (X) is irreducible if it has no irreducible factors of degree ≤ 12 deg f . Carry out this procedure for F = Z2 to find all irreducible polynomials of degree ≤ 4. 7. Let A be a P.I.D. and let B be a subring of the field of fractions of A with A ⊂ B. Show that B = S −1 A for a suitable multiplicative subset S of A. 8. Let A be a U.F.D. and let S be a saturated multiplicative subset of A. Let T be the set of prime elements of A that lie in S. Show that S = {up1r1 . . . prkk | k ≥ 0, u ∈ A× , p1 , . . . , pk ∈ T and ri ≥ 0 for i = 1, . . . , k}. 9. Let A be a U.F.D. with only one prime element, p, up to equivalence of primes. Show that there is a Euclidean structure map ϕ : A → Z obtained by setting ϕ(0) = −1 and setting ϕ(a) = k if pk is the highest power of p that divides the nonzero element a. Deduce that a ring is a discrete valuation ring if and only if it is a U.F.D. with at most one prime element up to equivalence of primes.


306

10. Let A be a U.F.D. and let S be a multiplicative subset of A. Show that S −1 A is a U.F.D. 11. Let A be a U.F.D. and let p be a prime element of A. Show that the localization A(p) of A at the prime ideal (p) is a discrete valuation ring. 12. Let A be a U.F.D. that has infinitely many equivalence classes of prime elements. Show that the field of fractions of A is infinitely generated as an A-algebra. 13. Show that if A is not a field, then A[X] is not a P.I.D. 14. Recall the ring of Gaussian integers, Z[i], introduced in Example 7.3.6: Z[i] is the subring of C consisting of the elements a + bi with a, b ∈ Z. (a) Show that Z[i] is Euclidean, via the function ϕ(a + bi) = a2 + b2 . (b) Show that complex conjugation restricts to an isomorphism of rings from Z[i] to itself. Show also that ϕ(z) = zz for z ∈ Z[i], where z is the complex conjugate of z. Deduce that ϕ(zw) = ϕ(z)ϕ(w) for z, w ∈ Z[i]. (c) Show that the only units in Z[i] are ±i and ±1. (d) Show that anelement z ∈ Z[i] is prime if and only if it has no divisors w with 1 < ϕ(w) ≤ ϕ(z). (e) Show that 1 + i is a prime and that 1 + i = −i · (1 + i). Show also that ϕ(1 + i) = 2, so that 2 = (1 + i) · (1 + i) = −i · (1 + i)2 . (In the language of algebraic number theory, this implies that 2 ramifies in Z[i].) Deduce that 2 is not prime in Z[i]. (f) Show that if z = a+bi ∈ Z[i] with a, b ∈ Z is a prime element such that neither a nor b is zero and (z) = (1 + i), then z is a prime that is not equivalent to z. Deduce that every nonzero element n ∈ Z has a prime decomposition in Z[i] of the form n = u · (1 + i)2r · pr11 . . . prkk · z1s1 z s11 . . . zlsl z sl l , where u is a unit, the pi are primes in Z that remain prime in Z[i], and the zi are primes in Z[i] of the form a + bi, where a and b are positive integers with b < a. (g) Let z = a + bi where a and b are nonzero integers. Show that z is prime in Z[i] if and only if ϕ(z) is prime in Z. (h) Let p be a prime in Z that is not prime in Z[i]. Show that p = ϕ(z) for some z ∈ Z[i]. Deduce that there are integers a and b with p = a2 + b2 . (i) Show that a prime in Z that is congruent to 3 mod 4 remains prime in Z[i]. (j) Suppose that a prime p in Z remains prime in Z[i]. Deduce that Z[i]/(p) is a field of order p2 , which is additively a vector space over Zp of dimension 2, with a basis given by the images of 1 and i under the canonical map from Z[i]. Deduce that the only elements of order 4 in the unit group (Z[i]/(p))× must lie outside the image of the unique ring homomorphism from Zp to Z[i]/(p). Deduce that Zp × has no elements of order 4, and hence that p must be congruent to 3 mod 4.


307

(k) Let p be a prime in Z that is congruent to 1 mod 4. Deduce from the results above that p = zz for some prime z of Z[i]. Show that the field Z[i]/(z) is of characteristic p, and is generated additively by the images of 1 and i under the canonical map from Z[i]. Deduce from the fact that Zp has units of order 4 that the image of i under the canonical map must lie in the image of the unique ring homomorphism from Zp to Z[i]/(z). Deduce that Z[i]/(z) is isomorphic to Zp . 15. Characterize the collection of integers that may be written as the sum of two squares. 16. We study the subring Z[ζ3 ] of C. (a) Show that the kernel of the evaluation map εζ3 : Z[X] → Z[ζ3 ] is the principal ideal of Z[X] generated by X 2 + X + 1. Deduce from Proposition 7.7.35 that the elements of Z[ζ3 ] may be written uniquely in the form m + nζ3 with m, n ∈ Z. Write down the multiplication formula for the product of two such elements. (b) Using the fact that ζ 3 = ζ3−1 = ζ32 , where ζ 3 is the complex conjugate of ζ3 , show that setting ϕ(α) = αα defines a Euclidean structure map ϕ : Z[ζ3 ] → Z. (c) Show that Z[ζ3 ]× = {±1, ±ζ3 , ±ζ32 }, where ζ32 = −1−ζ3 and −ζ32 = ζ6 = 1+ζ3 . (d) Show that 3 ramifies in Z[ζ3 ] in the sense that 3 = up2 , where u ∈ Z[ζ3 ]× and p is a prime of Z[ζ3 ]. (e) Show that an element of the form m + nζ3 , where m and n are both nonzero elements of Z, is prime in Z[ζ3 ] if and only if its image under ϕ is a prime in Z. (f) Show that a prime in Z remains prime in Z[ζ3 ] if and only if it is not in the image of ϕ : Z[ζ3 ] → Z. (g) Show that a prime p = 3 in Z remains prime in Z[ζ3 ] if and only if p is not congruent to 1 mod 3.


308

√ 17. Show that Z[i 2] is a Euclidean√domain via the square of the complex norm, as above. What are the units of Z[i√ 2]? Find congruences that will guarantee that a prime in Z remains prime in Z[i 2]. √ 18. Show that Z[ 2] admits a Euclidean structure map.

8.2

Algebraic Extensions

Definitions 8.2.1. Let K be a field. By an extension of K, or, more formally, an extension field of K, we mean a field L that contains K as a subfield. We say that L is a finite extension of K if it is finite dimensional as a vector space over K. If L is a finite extension of K, then the degree of L over K, written [L : K], is the dimension of L as a K-vector space. If L is an infinite extension of K, we write [L : K] = ∞. Throughout this section, K is a field. Lemma 8.2.2. Let L be an extension of K and let L be an extension of L. Then L is finite over K if and only if both L is finite over L and L is finite over K, in which case [L : K] = [L : L] · [L : K]. Proof If L is finite over K, then the other two extensions are clearly finite as well. Conversely, if y1 , . . . , yn is a basis for L as a vector space over L, and x1 , . . . , xm is a basis for L as a vector space over K, then any ordering of {xi yj | 1 ≤ i ≤ m, 1 ≤ j ≤ n} will give a basis for L over K. We now consider the elements of an extension field L of K. For each α ∈ L, there is a unique K-algebra homomorphism εα : K[X] → L that takes X to α. Here, εα evaluates polynomials at α: " n # n i εα ai X = ai αi . i=0

i=0

Thus, f (X) ∈ ker εα if and only if α is a root of f . In particular, ker εα = 0 if and only if α is a root of some nonzero polynomial over K. Recall from Corollary 8.1.10 that the kernel of εα is the principal ideal generated by minα (X), the minimal polynomial of α. Here, minα (X) = 0 if εα is injective, and is the monic polynomial of lowest degree in ker εα otherwise. In particular, minα (X) divides any polynomial that has α as a root. The image of εα , of course, is K[α], the K-subalgebra of L generated by α. Definition 8.2.3. Let L be an extension field of K and let α ∈ L. We say that α is algebraic over K if α is a root of a nonzero polynomial over K, and hence minα (X) = 0. If α is not a root for any nonzero polynomial over K, we say that α is transcendental over K. Examples 8.2.4. 2π 1. Let ζn = ei· n ∈ C be the standard primitive n-th root of unity in C. Since ζn has order n in C× , it is a root of the polynomial X n − 1 ∈ Z[X], and hence is algebraic over Q.


309

√ n

a is a root of the integer polynomial

2. For positive integers n and a, the real number X n − a, and hence is algebraic over Q.

3. The real numbers π and e are known to be transcendental over Q. The standard proofs of this use analytic number theory. Let α be an element of the extension field L of K. Then K[X]/(minα (X)) ∼ = K[α], which, as a subring of L, is an integral domain. Thus, (minα (X)) is a prime ideal in K[X]. But by Corollary 8.1.16, the nonzero prime ideals of the P.I.D. K[X] are all maximal, generated by irreducible elements of K[X]. Thus, if α is algebraic over K, then K[α] ∼ = K[X]/(minα (X)) is a field. The following proposition is now immediate from Corollary 8.1.10. Proposition 8.2.5. Let L be an extension field of K and let α ∈ L be algebraic over K. Then K[α] is a finite extension field whose degree over K is equal to the degree of the minimal polynomial of α over K: [K[α] : K] = deg minα (X).

Recall that K[α] is the smallest K-subalgebra of L that contains α. Definition 8.2.6. Let L be an extension field of K and let α ∈ L. We write K(α) for the smallest subfield of L that contains both K and α. More generally, if {αi | i ∈ I} is any family of elements in L, we write K(αi | i ∈ I) for the smallest subfield of L containing both K and {αi | i ∈ I}. For a finite set α1 , . . . , αn of elements of L, the notation for this subfield, of course, becomes K(α1 , . . . , αn ). In particular, Proposition 8.2.5, when α is algebraic over K, or Corollary 7.11.11, when α is transcendental, now gives the following corollary. Corollary 8.2.7. Let L be an extension field of K and let α ∈ L. If α is algebraic over K, then K(α) = K[α]. If α is transcendental over K, then K(α) is isomorphic to the field of fractions of K[α], and consists of the elements of the form f (α)/g(α), where f (X), g(X) ∈ K[X], with g(α) = 0. Since α is transcendental over K, this just says that g(X) = 0, and hence K(α) is isomorphic to K(X), the field of rational functions over K. The extensions K(α) with α algebraic over K are of sufficient interest to warrant a name. Definitions 8.2.8. An extension field L of K is simple if L = K(α) for some α ∈ L that is algebraic over K. An element α that is algebraic over K and satisfies L = K(α) is called a primitive element for L over K. An alternative name for a simple extension is a primitive extension. We shall see in Chapter 11, using the Fundamental Theorem of Galois Theory, that if K has characteristic 0, then every finite extension of K is simple. For fields of nonzero characteristic, we shall see that every finite separable extension is simple. The degree of an extension is a useful number to calculate. Thus, to study a simple extension K(α) of K, it is useful to be able to determine the minimal polynomial of α


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over K. In practice, what we’re often given is some polynomial f (X) ∈ K[X] of which α is a root (e.g., ζn is a root of X n − 1, but we don’t yet know the minimal polynomial of ζn over Q, nor, indeed, the degree [Q(ζn ) : Q]). All we know is that minα (X) is one of the prime divisors of f (X). Thus, it is quite valuable to be able to compute prime decompositions in K[X], and, in particular, to test the elements of K[X] to see if they are irreducible. Tests of irreducibility will be a topic in Section 8.5. Even in the case of a simple extension, it is sometimes difficult to find a primitive element. Thus, it is valuable to be able to handle extensions that are generated as algebras by more than one element. Recall that if B is a commutative K-algebra and if b1 , . . . , bn ∈ B, then we write K[b1 , . . . , bn ] for the image of the K-algebra homomorphism εb1 ,...,bn : K[X1 , . . . , Xn ] → B obtained by evaluating Xi at bi for i = 1, . . . , n. It is easy to see that K[b1 , . . . , bn ] is the smallest K-subalgebra of B containing b1 , . . . , bn . The next lemma gives a partial generalization of Proposition 8.2.5. Lemma 8.2.9. Let L be an extension field of K and let α1 , . . . , αn ∈ L be algebraic over K. Then K[α1 , . . . , αn ] is a field. Indeed, K[α1 , . . . , αn ] is a finite extension of K. Proof We argue by induction on n, with the case n = 1 being given by Proposition 8.2.5. Suppose then by induction that K[α1 , . . . , αn−1 ] is a finite extension field of K. Since K[α1 , . . . , αn ] is the smallest K-subalgebra of L that contains the elements α1 , . . . , αn , it must be the case that K[α1 , . . . , αn ] = K[α1 , . . . , αn−1 ][αn ]. Now αn is algebraic over K, and hence is a root of a nonzero polynomial f (X) ∈ K[X]. But f may be regarded as a polynomial over K[α1 , . . . , αn−1 ], so αn is algebraic over K[α1 , . . . , αn−1 ] as well. But Proposition 8.2.5 now shows that K[α1 , . . . , αn ] is a finite extension of K[α1 , . . . , αn−1 ], and hence also of K by Lemma 8.2.2. Corollary 8.2.10. Let L be an extension of K and let α1 , . . . , αn ∈ L be algebraic over K. Then K(α1 , . . . , αn ) = K[α1 , . . . , αn ] is a finite extension of K. We would like to study the situation of adjoining infinitely many algebraic elements to a field K. Definition 8.2.11. Let B be a commutative A-algebra and let {bi | i ∈ I} be a family of elements of B. We write A[bi | i ∈ I] for the set of all elements in B lying in one of the subalgebras A[bi1 , . . . , bik ] for a finite subset {i1 , . . . , ik } ⊂ I. Clearly, A[bi | i ∈ I] is the smallest A-subalgebra of B that contains {bi | i ∈ I}. Proposition 8.2.12. Let L be an extension field of K and let {αi | i ∈ I} be elements of L that are algebraic over K. Then K[αi | i ∈ I] is a field, and hence K(αi | i ∈ I) = K[αi | i ∈ I] is the smallest K-subalgebra of L containing {αi | i ∈ I}.


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Proof Every nonzero element β ∈ K[αi | i ∈ I] lies in an extension K[αi1 , . . . , αik ], which is a field, since the α’s are algebraic over K. So β is a unit in K[αi | i ∈ I]. We shall be especially interested in extension fields that are algebraic over K: Definitions 8.2.13. An extension L of K is algebraic if every element of L is algebraic over K. An extension that is not algebraic is said to be transcendental. The next proposition characterizes the algebraic elements in an extension field L. Proposition 8.2.14. Let L be an extension field of K and let α ∈ L. Then the following conditions are equivalent. 1. α is algebraic over K. 2. K[α] is a finite extension of K. 3. There is a field L ⊂ L, containing both K and α, such that L is a finite extension of K. Proof The first condition implies the second by Proposition 8.2.5, while the second implies the first by Corollary 8.1.10. The second implies the third by taking L = K[α], so it suffices to show that the third implies the second. But if L is a finite extension of K containing α, then K[α] ⊂ L , and hence K[α] is also finite over K. Corollary 8.2.15. An extension L of K is algebraic if and only if every element of L is contained in a finite extension of K. In particular, every finite extension of K is algebraic. Corollary 8.2.16. Let L be an extension field of K. Then L is a finite extension of K if and only if L = K(α1 , . . . , αn ), where α1 , . . . , αn are algebraic over K. Proof If L = K(α1 , . . . , αn ), where α1 , . . . , αn are algebraic over K, then L is finite over K by Corollary 8.2.10. Conversely, if L is finite over K, then every element of L is algebraic over K by Proposition 8.2.14. If α1 , . . . , αn is a basis for L over K, then clearly, L = K[α1 , . . . , αn ], so the result follows from Corollary 8.2.10. Thus, if α1 , . . . , αn are algebraic over K, then K(α1 , . . . , αn ) is an algebraic extension of K. Indeed, we may extend this to infinitely generated extensions. Corollary 8.2.17. An extension L of K is algebraic if and only if L = K(αi | i ∈ I) for some collection of elements {αi | i ∈ I} that are algebraic over K. Proof Let L = K(αi | i ∈ I), where each αi is algebraic over K. Then L = K[αi | i ∈ I] by Proposition 8.2.12. In particular, every element of L lies in a finitely generated K-subalgebra of the form K[αi1 , . . . , αik ]. Since the αi are algebraic over K, these subalgebras are finite extension fields of K, and hence L is algebraic over K by Proposition 8.2.14. Conversely, if L is algebraic over K and if {αi | i ∈ I} is any basis for L as a vector space over K, then L = K(αi | i ∈ I).


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We can also use Proposition 8.2.14 to obtain an important closure property for algebraic extensions. Corollary 8.2.18. Let L be an algebraic extension of K and let L1 be any extension of L. Suppose that α ∈ L1 is algebraic over L. Then α is algebraic over K as well. In particular, if L is an algebraic extension of K, then any algebraic extension of L is an algebraic extension of K. Proof Let α ∈ L1 be algebraic over L. Then α is a root of a nonzero polynomial f (X) over L. Say f (X) = αn X n + · · · + α0 , with αi ∈ L for 0 ≤ i ≤ n. Since L is algebraic over K, so is αi for 0 ≤ i ≤ n, and hence K(α0 , . . . , αn ) is a finite extension of K by Corollary 8.2.10. But α is algebraic over K(α0 , . . . , αn ). Thus, K(α0 , . . . , αn )(α) is a finite extension of K(α0 , . . . , αn ), and hence of K as well. So α is contained in a finite extension of K, and hence is algebraic over K by Proposition 8.2.14. We shall be primarily interested in finite extensions, but some results regarding algebraic closures will require information about infinite algebraic extensions. The next result is trivial for finite extensions and false for transcendental extensions (see Corollary 8.6.7). Thus, it shows the strength of the condition that an extension is algebraic. Proposition 8.2.19. Let L be an algebraic extension of K and let ν : L → L be a homomorphism of fields that restricts to the identity map on K. Then ν is an automorphism of L. Proof Every homomorphism of fields is injective. So it suffices to show ν is onto. If L is a finite extension, this is immediate from a dimension argument. Otherwise, let α ∈ L. Let f (X) ∈ K[X] be the minimal polynomial of α over K, and let α1 , . . . , αn be the set of all roots of f in L, with α = α1 . Then K(α1 , . . . , αn ) is a finite extension of K, and it suffices to show that ν carries K(α1 , . . . , αn ) into itself. Since K(α1 , . . . , αn ) is the smallest K-subalgebra of L containing α1 , . . . , αn , it suffices to show that ν carries each αi into {α1 , . . . , αn }. Write f (X) = am X m + · · · + a0 , with a0 , . . . , am ∈ K. Then f (αi ) = 0, so 0 = ν(f (αi )) = am (ν(αi ))m + · · · + a0 = f (ν(αi )), since ν acts as the identity on K. In particular, ν(αi ) is a root of f in L. And the set of all such roots is precisely {α1 , . . . , αn }. We shall next show that in any extension L of K, there is a largest subfield K1 of L such that K1 is an algebraic extension of K. Definition 8.2.20. Let L be an extension of K. Then the algebraic closure of K in L is the set of all elements of L that are algebraic over K. Proposition 8.2.21. Let L be an extension of K. Then the algebraic closure of K in L is a subfield of L containing K. In particular, it is an algebraic extension of K. Proof If α is algebraic over K, then K(α) is a finite extension of K, and hence every element of K(α) must be algebraic over K by Proposition 8.2.14. In particular, −α and α−1 are algebraic over K. If α and β are both algebraic over K, then β is a root of a nonzero polynomial over K, and hence also over K(α). Thus, β is contained in a finite extension, L , of K(α),


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which is then finite over K by Lemma 8.2.2. But α + β and αβ lie in L , and hence α + β and αβ are algebraic over K. Thus, the elements of L that are algebraic over K form a subfield of L. This subfield includes K, as each a ∈ K is a root of X − a. Definition 8.2.22. Let L be an extension of K. We say that K is algebraically closed in L if the only elements of L that are algebraic over K are those of K itself. Proposition 8.2.23. Let L be an extension of K and let K1 be the algebraic closure of K in L. Then K1 is algebraically closed in L. Proof Let α ∈ L be algebraic over K1 . Then α is algebraic over K by Corollary 8.2.18, and hence an element of K1 , as K1 is the algebraic closure of K in L. Exercises 8.2.24. 1. Let α ∈ C with α ∈ R. Show that both α + α and α · α are real. Deduce that minα/R (X) = (X − α)(X − α) = X 2 − (α + α)X + αα. 2. Find the minimal polynomial of i over Q. 3. Find the minimal polynomial of ζ3 over Q. 4. Show that minζ8 /Q (X) = X 4 + 1. (Hint: Show that Q(ζ8 ) is a degree 2 extension of Q(i).) 5. Let n be any √ integer that is not a perfect cube (i.e., not the cube of an integer). Show that Q( 3 n) is a degree 3 extension of Q. 6. Let L be an extension of K of degree p, with p prime. Show that every element α of L that is not in K is a primitive element for L over K (i.e., that L = K(α)). 7. Let K be a field of characteristic = 2 and let L be a quadratic extension of K (i.e., [L : K] = 2). Show that there is a primitive element, α, for L over K such that the minimal polynomial of α over K has the form X 2 − d with d ∈ K. Thus, we may √ write L = K( d). (Hint: Suppose that L has a primitive element whose minimal polynomial is aX 2 + bX + c. Show that d = b2 − 4ac has a square root, α, in L and that L = K(α).) 8. Let K be the field of fractions of a domain A of √ characteristic = 2, and let L be a quadratic extension of K. Show that L = K( d) for some d ∈ A. (Hint: If L = K( a/b), with a, b ∈ A, take d = ab.) If A is a U.F.D., show that we may additionally assume that the element d ∈ A is square-free, in the sense that it is not divisible by the square of any prime element of A. 9. Let K be any field and let K(X) be the field of fractions of K[X]. Show that K is algebraically closed in K(X). 10. Deduce from the preceding problem that if K is finite, then the only finite subfields of K(X) are those contained in K itself.


8.3

314

Transcendence Degree

Some extensions are more transcendental than others. We shall discuss this measurement here. There’s a notion of transcendence basis analogous to that of a basis for a vector space. We obtain a notion of finite transcendence degree, measured by the number of elements in a transcendence basis. Definitions 8.3.1. Let A be a subring of the commutative ring B. We say that the elements b1 , . . . , bn of B are algebraically independent over A if the evaluation map εb1 ,...,bn : A[X1 , . . . , Xn ] → B that evaluates each Xi at bi is injective. Let L be an extension field of K We say that α1 , . . . , αn is a transcendence basis for L if they are algebraically independent and L is algebraic over K(α1 , . . . , αn ). (Infinite transcendence bases may be defined analogously.) We say that an extension L of K has finite transcendence degree over K if there exists a finite transcendence basis for L over K. Examples 8.3.2. 1. A single element α ∈ L is algebraically independent if and only if it is transcendental over K. 2. The null set forms a transcendence basis for an algebraic extension of K. In particular, algebraic extensions have finite transcendence degree. 3. Let α = f (X)/g(X) be any element of K(X) that’s not in K. Then α is a transcendence basis for K(X) over K. The point is that X is algebraic over K(α), as it is a root of the polynomial αg(T ) − f (T ) in K(α)[T ]. Thus, α is transcendental over K, as otherwise, X would be algebraic over K. 4. Of course, X1 , . . . , Xn is a transcendence basis for the ring of rational functions K(X1 , . . . , Xn ). Recall the isomorphism A[X1 , . . . , Xn ] ∼ = A[X1 , . . . , Xn−1 ][Xn ]. Under this isomorphism, we may write any polynomial f (X1 , . . . , Xn ) uniquely as a polynomial in Xn with coefficients in A[X1 , . . . , Xn−1 ]: f (X1 , . . . , Xn ) = fk (X1 , . . . , Xn−1 )Xnk + · · · + f0 (X1 , . . . , Xn−1 ). Definitions 8.3.3. Suppose given a polynomial f (X1 , . . . , Xn ) = fk (X1 , . . . , Xn−1 )Xnk + · · · + f0 (X1 , . . . , Xn−1 ) in A[X1 , . . . , Xn ]. We say that f has degree k in Xn if fk (X1 , . . . , Xn−1 ) = 0 in the equation above. Let A be a subring of the commutative ring B and b1 , . . . , bn ∈ B. Suppose given a polynomial f (X1 , . . . , Xn ) = fk (X1 , . . . , Xn−1 )Xnk + · · · + f0 (X1 , . . . , Xn−1 ) in A[X1 , . . . , Xn ]. We say that f has degree m in bn when evaluated at b1 , . . . , bn if the polynomial fk (b1 , . . . , bn−1 )Xnk + · · · + f0 (b1 , . . . , bn−1 ) has degree m as a polynomial in Xn over A[b1 , . . . , bn−1 ]. Here, by abuse of notation, we shall say that f (b1 , . . . , bn ) has degree m in bn .


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Lemma 8.3.4. Let L be an extension of K and let α1 , . . . , αn ∈ L be algebraically independent over K. Then an element β ∈ L is algebraic over K(α1 , . . . , αn ) if and only if α1 , . . . , αn , β are algebraically dependent over K. Proof Suppose that β is algebraic over K(α1 , . . . , αn ), so it satisfies an equation of the form fm (α1 , . . . , αn ) m f0 (α1 , . . . , αn ) β + ··· + = 0, gm (α1 , . . . , αn ) g0 (α1 , . . . , αn ) with positive degree in β. Now, multiplying both sides by the product g0 (α1 , . . . , αn ) . . . gm (α1 , . . . , αn ), we obtain an algebraic dependence relation between α1 , . . . , αn , β. Conversely, if α1 , . . . , αn , β are algebraically dependent, let f (X1 , . . . , Xn+1 ) be a nonzero polynomial in A[X1 , . . . , Xn+1 ] that vanishes when evaluated at α1 , . . . , αn , β. Write k f (X1 , . . . , Xn+1 ) = fk (X1 , . . . , Xn )Xn+1 + · · · + f0 (X1 , . . . , Xn ).

Since α1 , . . . , αn are algebraically independent, fi (α1 , . . . , αn ) is nonzero whenever fi is a nonzero polynomial. Thus, f has positive degree in β, and hence β is algebraic over K(α1 , . . . , αn ). A new characterization of transcendence bases is now immediate. Corollary 8.3.5. Let L be an extension of K and let α1 , . . . , αn ∈ L be algebraically independent over K. Then α1 , . . . , αn is a transcendence basis for L over K if and only if there is no element β in L such that α1 , . . . , αn , β are algebraically independent over K. The theory of transcendence bases now proceeds much like the theory of bases in ordinary linear algebra. Proposition 8.3.6. Let L be an extension of K and let α1 , . . . , αn in L such that L is algebraic over K(α1 , . . . , αn ). Suppose in addition that α1 , . . . , αm are algebraically independent for some m < n. Then we can find i1 , . . . , ik ∈ {m + 1, . . . , n} such that α1 , . . . , αm , αi1 , . . . , αik is a transcendence basis for L over K. Proof By induction on n − m, we may choose i1 , . . . , ik so that α1 , . . . , αm , αi1 , . . . , αik is algebraically independent, but is not properly contained in any other algebraically independent subset of α1 , . . . , αn . For simplicity, we reorder the α’s if necessary so that α1 , . . . , αm+k is our maximal algebraically independent subset containing α1 , . . . , αm . Now, for i > m+k, α1 , . . . , αm+k , αi is algebraically dependent, so Lemma 8.3.4 shows αi to be algebraic over K(α1 , . . . , αm+k ). Corollary 8.2.10 now shows that K(α1 , . . . , αn ) is algebraic over K(α1 , . . . , αm+k ), and hence L is also. Recall that the total degree of a monomial aX1i1 . . . Xnin is i1 + · · · + in and that the total degree of a polynomial f (X1 , . . . , Xn ) is the largest of the total degrees of the nonzero monomials in it. Definition 8.3.7. Let A be a subring of a commutative ring B and let b1 , . . . , bn ∈ B. We say that a nonzero polynomial f (X1 , . . . , Xn ) ∈ A[X1 , . . . , Xn ] gives a minimal algebraic dependence relation between b1 , . . . , bn if f (b1 , . . . , bn ) = 0 and if no nonzero polynomial whose total degree is less than that of f vanishes when evaluated on b1 , . . . , bn .


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Lemma 8.3.8. Let A be a subring of a commutative ring B and let b1 , . . . , bn ∈ B. Suppose that f (X1 , . . . , Xn ) ∈ A[X1 , . . . , Xn ] gives a minimal algebraic dependence relation between b1 , . . . , bn . Suppose also that f has degree k > 0 in Xn . Then f (b1 , . . . , bn ) has degree k in bn . Proof Suppose that f (X1 , . . . , Xn ) = fk (X1 , . . . , Xn−1 )Xnk + · · · + f0 (X1 , . . . , Xn−1 ). Then fk (X1 , . . . , Xn−1 ) is a nonzero polynomial whose total degree is k less than that of f . If fk (b1 , . . . , bn−1 ) = 0, then fk may be regarded as an algebraic dependence relation for b1 , . . . , bn , contradicting the minimality of f . Thus, fk (b1 , . . . , bn−1 ) = 0. Proposition 8.3.9. Let α1 , . . . , αn be a transcendence basis for L over K and let β1 , . . . , βk ∈ L be algebraically independent over K. Then k ≤ n, and if k = n, then β1 , . . . , βn is a transcendence basis for L over K. Proof Suppose that k ≥ n. We shall show by induction on i that we may reorder the α’s in such a way that L is algebraic over K(β1 , . . . , βi , α1 , . . . , αn−i ) for each i ≤ n. When i = n, this says that L is algebraic over K(β1 , . . . , βn ), so that β1 , . . . , βn is a transcendence basis for L over K. But if k > n, Lemma 8.3.5 then shows that β1 , . . . , βk cannot be algebraically independent over K, so the result will follow from our induction on i. Suppose, inductively, that L is algebraic over K(β1 , . . . , βi−1 , α1 , . . . , αn−i+1 ). Then Proposition 8.3.6 shows that we may reorder the α’s if necessary so that β1 , . . . , βi−1 , α1 , . . . , αr is a transcendence basis for L over K for some r ≤ n − i + 1. Thus, Lemma 8.3.4 shows that β1 , . . . , βi , α1 , . . . , αr are algebraically dependent. Let f (X1 , . . . , Xr+i ) be a minimal algebraic dependence relation between β1 , . . . , βi , α1 , . . . , αr . By Lemma 8.3.8, f must have positive degree in at least one of the α’s, as otherwise, f only involves the variables X1 , . . . , Xi , and hence gives an algebraic dependence relation for the β’s. Renumbering the α’s, if necessary, we may assume that f has positive degree in αr . But then αr is algebraic over K(β1 , . . . , βi , α1 , . . . , αr−1 ). Since L is algebraic over K(β1 , . . . , βi , α1 , . . . , αr ), we have established the inductive step. Corollary 8.3.10. Let L be an extension of finite transcendence degree over K. Then any two transcendence bases of L over K have the same number of elements. Thus, we may define transcendence degree. Definition 8.3.11. Let L be an extension of finite transcendence degree over K. Then the transcendence degree of L over K is the number of elements in any transcendence basis of L over K. Exercises 8.3.12. 1. Suppose that L has transcendence degree n over K and that L is algebraic over K(α1 , . . . , αn ). Show that α1 , . . . , αn is a transcendence basis for L over K. 2. Suppose given extensions K ⊂ L ⊂ L1 such that L has finite transcendence degree over K and L1 has finite transcendence degree over L. Show that L1 has finite transcendence degree over K, and that the transcendence degree of L1 over K is the sum of the transcendence degree of L over K and the transcendence degree of L1 over L.


8.4

317

Algebraic Closures

Definition 8.4.1. A field K is algebraically closed if every polynomial in K[X] of degree ≥ 1 has a root in K. We shall show in Section 11.6, in what’s known as the Fundamental Theorem of Algebra, that the complex numbers, C, are an algebraically closed field. Every argument for this relies at some point on topology. Perhaps the simplest proof makes use of fundamental groups, but requires no ring or field theory beyond the definitions of complex multiplication and of polynomials. However, the topology would require too much development to present here. So we shall give a proof that is more algebraic, and relies on the Fundamental Theorem of Galois Theory, in Chapter 11. Lemma 8.4.2. Let K be an algebraically closed field. Then every polynomial f (X) of degree ≥ 1 in K[X] breaks up as a product f (X) = a(X − a1 ) . . . (X − an ) in K[X], where a, a1 , . . . , an ∈ K. Proof Since every polynomial of degree ≥ 1 has a root in K, it has a degree 1 factor. So no polynomial of degree > 1 is irreducible. So the irreducibles are precisely the degree 1 polynomials, and the result follows from unique factorization in K[X]. A simpler argument, by induction on degree, would have sufficed, of course. We shall make use of the next lemma in Section 9.3. Lemma 8.4.3. Every algebraically closed field is infinite. Proof First note that Z2 is not algebraically closed, as X 2 + X + 1 has no roots in Z2 . Let K be a finite field with more than two elements and let a be a non-identity element of K × . Let n be the order of K × . Then every element of K × has exponent n, and hence X n − a has no roots in K. Often, problems over R are most easily solved by passing to C and first solving the problem there. In general, it can be very useful to study a given field by making use of an algebraically closed field in which it embeds. Definition 8.4.4. Let K be a field. An algebraic closure of K is an algebraic extension, L, of K, such that L is algebraically closed. Notice that this is a quite different notion from the algebraic closure of K in some extension, L, of K (Definition 8.2.20), though that concept can be useful in constructing some algebraic closures. Proposition 8.4.5. Suppose L is an extension of K that is algebraically closed. Let K1 ⊂ L be the algebraic closure of K in L (i.e., the set of elements of L that are algebraic over K.) Then K1 is an algebraic closure of K. Proof We’ve seen in Proposition 8.2.21 that K1 is an algebraic extension of K. So it suffices to show that K1 is algebraically closed. Thus, suppose that f (X) ∈ K1 [X] has positive degree. Since L is algebraically closed, f has a root, α, in L. It suffices to show that α is in K1 . But α is algebraic over K1 , and by Proposition 8.2.23, the only elements of L that are algebraic over K1 are those of K1 .


318

Thus, once we’ve established the Fundamental Theorem of Algebra, we’ll know that the algebraic closure of Q in C is an algebraic closure of Q. In particular, it will be our preferred model for the algebraic closure of Q, as we’ve become accustomed to visualizing the elements of C. It also allows the use of complex analysis to study the algebraic extensions of Q, and forms the basis for the field of analytic number theory. Here, we shall give an abstract proof that algebraic closures exist. We shall treat the uniqueness of algebraic closures in Chapter 11. Lemma 8.4.6. Let f1 (X), . . . , fk (X) be polynomials over K. Then there exists an extension of K in which each of the polynomials fi has a root. Proof By induction, it suffices to treat the case of a single polynomial f (X). And by factoring f , if necessary, we may assume that f (X) is irreducible in K[X]. Thus, L = K[X]/(f (X)) is a field, containing K as a subfield. Moreover, if α ∈ L is the image of X under the canonical map from K[X], then f (α) = 0, and hence α is a root of f in L. The construction we shall use relies on polynomial rings in infinitely many variables. Here, if A is a ring and S is a set of variables, the elements of the polynomial ring A[S] are simply polynomials in finitely many of the variables of S with coefficients in A. We add and multiply these as we would in a polynomial ring in finitely many variables, and may do so because for any finite collection of elements of A[S], there is a finite subset {X1 , . . . , Xn } such that each of the polynomials in this collection lies in A[X1 , . . . , Xn ] ⊂ A[S]. (In categorical language, A[S] is the direct limit of the polynomial rings on the finite subsets of S.) Polynomials in infinitely many variables behave analogously to polynomials in finitely many variables. In particular, there is an inclusion map i : S → A[S] which, as the reader may verify, is universal in the following sense. Proposition 8.4.7. Let A be a commutative ring and let B be a commutative A-algebra. Let S be a set of variables and let f : S → B be any function. Then there is a unique A-algebra homomorphism εf : A[S] → B such that εf ◦ i = f . The image of εf is A[f (Y ) | Y ∈ S], the smallest A-subalgebra of B containing the image of f . We shall refer to εf as the A-algebra homomorphism obtained by evaluating each Y ∈ S at f (Y ). The next lemma is the key step in constructing algebraic closures. Lemma 8.4.8. Let K be a field. Then there is an algebraic extension K1 of K such that every polynomial in K[X] has a root in K1 . Proof It suffices to construct K1 so that each monic irreducible polynomial of degree > 1 in K[X] has a root in K1 . Thus, for each monic irreducible polynomial f (X) of degree > 1 in K[X], we define a variable Xf , and we let S be the set of all such Xf . Then substituting Xf for the variable X, we obtain the polynomial f (Xf ) ∈ K[S]. Let a be the ideal of K[S] generated by the set {f (Xf ) | Xf ∈ S}. We claim that a is a proper ideal of K[S]. To see this, we argue by contradiction. Suppose that 1 ∈ a. Then there are monic irreducible polynomials f1 (X), . . . , fk (X) of degree > 1 in K[X] and polynomials g1 , . . . , gk ∈ K[S] such that 1 = g1 f1 (Xf1 ) + · · · + gk fk (Xfk ).


319

By Lemma 8.4.6, there is an extension L of K in which each of f1 (X), . . . , fk (X) has a root. Say αi ∈ L is a root of fi (X) for 1 ≤ i ≤ k. Let h : S → L be any function such that h(Xfi ) = αi for 1 ≤ i ≤ k and let εh : K[S] → L be the K-algebra map obtained by evaluating each Xf in S at h(Xf ). Applying εh to the displayed equation above, we get 1

= εh (g1 )f1 (α1 ) + · · · + εh (gk )fk (αk ) =

0,

as αi is a root of fi for 1 ≤ i ≤ k. Thus, a is a proper ideal of K[S], and hence there is a maximal ideal m of K[S] that contains it. So let K1 = K[S]/m. Now write βf for the image in K1 of Xf under the canonical map π : K[S] → K1 for each Xf ∈ S. Then f (βf ) = π(f (Xf )) = 0, and hence each monic irreducible polynomial f (X) ∈ K[X] of degree > 1 has a root in K1 , as claimed. Also, each βf is algebraic over K. Since K1 = K[βf | Xf ∈ S], it is algebraic over K by Proposition 8.2.12 and Corollary 8.2.17. Now, we can construct algebraic closures. Proposition 8.4.9. Let K be a field. Then K has an algebraic closure. Proof By induction, using Lemma 8.4.8, construct a sequence K = K 0 ⊂ K1 ⊂ · · · ⊂ K n ⊂ · · · of field extensions such that for i ≥ 0, Ki+1 is an algebraic extension of Ki with the property that each polynomial in Ki [X] has a root in Ki+1 . Note that an inductive application of Corollary 8.2.18 now shows that Ki is an algebraic extension of K for all i. Using set theory, we may construct a set in which the Ki all embed, compatibly with their embeddings in each other. For instance, the direct limit − lim → Ki constructed in Section 6.8 has that property. We set L equal to the union of the Ki in such a set. Thus, each element of L lies in Ki for some i, and for any finite set of elements in L, we can find a Ki that contains them all. Thus, the addition and multiplication operations on the Ki induce addition and multiplication operations in L, and the fact that each Ki is a field implies that L is a field under these operations. Since every element of L lies in some Ki , each element of L is algebraic over K. Also, if f (X) ∈ L[X], then there is some Ki that contains all of the coefficients of f . In particular, f (X) ∈ Ki [X] ⊂ L[X], and hence has a root in Ki+1 , and hence in L. Thus, L is algebraically closed, and hence an algebraic closure for K. Exercises 8.4.10. 1. Show that the algebraic closure of Q in C (and hence any algebraic closure of Q, once we have the uniqueness statement) is countable. 2. Show that if K is an algebraically closed field and if L is any extension of K, then K is algebraically closed in L. 3. Let K be an algebraically closed field and let D be a division ring that is also a Kalgebra. Suppose that D is finite dimensional as a vector space over K, under the vector space structure induced by its K-algebra structure. Show that the inclusion K ⊂ D given by the K-algebra structure map is a bijection, so that K = D.


8.5

320

Criteria for Irreducibility

We wish to study the irreducibility of polynomials over Q. For this purpose, it is useful to note that if f (X) = αn X n + · · · + α0 with α0 , . . . , αn ∈ Q, and if m is divisible by the least common multiple of the denominators (when placed in lowest terms) of α0 , . . . , αn , then mf (X) has coefficients in Z. We can use this technique to reduce questions about Q[X] to questions about Z[X]. Along the way, we shall show that Z[X] is a U.F.D. Let p be a prime number in Z. Then the divisors of p in Z[X] all lie in Z, and hence p is irreducible in Z[X]. Thus, the following result is necessary for Z[X] to be a U.F.D. Lemma 8.5.1. Let p be a prime number and let f (X), g(X) ∈ Z[X] be such that p divides f (X) · g(X). Then p must divide at least one of f and g. In other words, the principal ideal generated by p in Z[X] is a prime ideal. n m i i Proof Write f (X) = i=0 ai X and g(X) = i=0 bi X . Suppose that p does not divide f . Then there is a smallest index i such that ai is not divisible by p. Similarly, if g is not divisible by p, there is a smallest index j such that bj is not divisible by p. i+j Now the i + j-th coefficient of f (X)g(X) is k=0 ak bi+j−k . If k < i, ak is divisible by p. If k > i, bi+j−k is divisible by p. Thus, the i + j-th coefficient of f (X)g(X) is divisible by p if and only if ai bj is divisible by p. Since neither ai nor bj is divisible by p, this cannot be. Thus, either f or g must have been divisible by p. The greatest common divisor of a collection of integers is the largest integer that divides all of them. It can be computed in terms of primes in the usual way (Corollary 8.1.31). Definitions 8.5.2. The content, c(f ), of a polynomial f (X) ∈ Z[X] is the greatest common divisor of the coefficients of f . We say that f (X) is primitive if it has content 1. The next result is a corollary of Lemma 8.5.1. Corollary 8.5.3. Let f (X), g(X) ∈ Z[X]. Then c(f (X)g(X)) = c(f ) · c(g). Proof Dividing, we can write f (X) = c(f )f1 (X) and g(X) = c(g)g1 (X), where f1 (X) and g1 (X) are primitive. Multiplying these, we see that it suffices to show that the product of two primitive polynomials is primitive. But if the product of two polynomials is not primitive, then it must be divisible by some prime number p, in which case p must divide one of the two polynomials. This now sets up the very useful Gauss Lemma. Lemma 8.5.4. (Gauss Lemma) Let f (X) ∈ Z[X] be a polynomial of degree > 1. Then f (X) is reducible in Q[X] if and only if f factors in Z[X] as the product of two polynomials of degree ≥ 1. Proof Suppose that f (X) = g(X)h(X) in Q(X), with both g and h of positive degree. By first clearing the denominators and then factoring out contents, we can write m k and h(X) = h1 (X), g1 (X) n l where g1 and h1 are primitive polynomials in Z[X] and k, l, m, n ∈ Z. Thus, nlf (X) = mkg1 (X)h1 (X). Equating the contents of both sides, we see nlc(f ) = ±mk, and hence nl divides mk. In particular, mk = nlr for some r ∈ Z, and hence f (X) = rg1 (X)h1 (X) does indeed factor in Z[X] as claimed. g(X) =


321

A variant of the above argument is used in the proof of the next lemma. Lemma 8.5.5. Let f (X) ∈ Z[X] be a primitive polynomial that is irreducible as an element of Q[X]. Then f (X) is a prime element in Z[X]. Proof Suppose given g(X), h(X) ∈ Z[X] such that f divides g(X)h(X) in Z[X]. Then f also divides g(X)h(X) in Q[X]. Since f (X) is a prime element in the P.I.D. Q[X], it must divide one of g and h there. Let’s say that g(X) = f (X)g1 (X) in Q[X]. Write g1 (X) = m n g2 (X), where g2 (X) is a primitive polynomial in Z[X] and m, n ∈ Z with (m, n) = 1. Then ng(X) = mf (X)g2 (X). Comparing contents, we see that nc(g) = ±m. Since (m, n) = 1, this gives n = ±1, and hence g(X) = ±mf (X)g2 (X) is divisible by f (X) in Z[X]. The next lemma adds a little clarity. Lemma 8.5.6. Let f (X) ∈ Q[X]. Then up to sign, there is a unique primitive polynomial in Z[X] that generates the same principal ideal of Q[X] as f does. In particular, if f is irreducible, then up to sign there is a unique primitive polynomial in Z[X] equivalent to f as a prime in Q[X]. Proof It suffices to show that if g(X) and h(X) are primitives in Z[X] that differ by a unit in Q[X], then g = ±h. But if g and h differ by a unit in Q[X], then there are integers m, n such that mg(X) = nh(X). Comparing contents, we see that m = ±n, and hence g = ±h. To get a unique choice of primitive polynomial in Z[X] equivalent to a given prime in Q[X], we may choose the one whose leading coefficient is positive. We now obtain the prime decompositions in Z[X]. Proposition 8.5.7. Let f (X) be a polynomial of degree > 0 over Z and suppose that the prime decomposition of f (X) in Q[X] is f (X) =

m r r (p1 (X)) 1 . . . (pk (X)) k , n

where p1 (X), . . . , pk (X) are pairwise inequivalent prime elements in Q[X]. Then if qi (X) is the unique primitive polynomial in Z[X] with positive leading coefficient that is equivalent to pi in Q[X], we have r

rk

f (X) = ±c(f ) (q1 (X)) 1 . . . (qk (X)) in Z[X].

Proof By uniqueness of prime decomposition in Q[X], we have f (X) =

k r r (q1 (X)) 1 . . . (qk (X)) k l

for some k, l ∈ Z, and hence r

rk

lf (X) = k (q1 (X)) 1 . . . (qk (X))

.

Comparing contents, we see that lc(f ) = ±k, so the result follows by dividing l into k.


322

This leads directly to the verification that Z[X] is a U.F.D. Corollary 8.5.8. The polynomial ring Z[X] is a U.F.D. whose prime elements are, up to sign, the primes in Z, together with the primitive polynomials that are irreducible in Q[X]. Proof We’ve already seen that the above elements are prime elements of Z[X]. By factoring c(f ) in Z, Proposition 8.5.7 shows that every nonzero nonunit in Z[X] may be written as a product of these primes. Thus, Z[X] is a U.F.D. The uniqueness statement for prime decompositions in a U.F.D. now shows that every prime element in Z[X] must be equivalent to one of these primes. As the units in Z[X] are ±1, the primes here are determined up to sign. Actually, the proofs given apply to a much more general setting. The reader may verify the following proposition. Proposition 8.5.9. Let A be a U.F.D. with fraction field K. Then A[X] is a U.F.D. whose prime elements are the primes in A, together with the primitive polynomials of A[X] that are irreducible in K[X]. The following criterion for irreducibility shows the value of the Gauss Lemma. n Proposition 8.5.10. (Eisenstein’s Criterion) Let f (X) = i=0 ai X i ∈ Z[X] and let p be a prime. Suppose that the leading coefficient an is relatively prime to p and that p divides ai for 0 ≤ i < n. Suppose also that p2 does not divide the constant term a0 . Then f (X) is irreducible in Q[X]. Proof If f (X) is reducible in Q[X], then the Gauss Lemma provides a factorization f (X) = g(X)h(X) in Z[X], where g and h both have positive degree. Write g(X) = k n−k i i i=0 bi X and h(X) = i=0 ci X . Since p divides the constant term of f but p2 does not, p divides the constant term of exactly one of the polynomials g and h. Let’s say that it divides the constant term of g. We claim now by induction that p must divide every coefficient of g. Thus, suppose i that p divides b0 , . . . , bi−1 for i ≤ k. Then we have ai = j=0 bj ci−j . Now p divides ai , since i ≤ k and k < n. Also, for j < i, p divides bj , and hence divides the summand bj ci−j . Thus, p must divide the remaining summand bi c0 . Since p does not divide c0 , it must divide bi . In particular, we obtain that p divides the leading coefficient bk . But an = bk cn−k , contradicting the fact that p does not divide an . Thus, f must not be reducible. We shall see that there are lots of irreducible polynomials that do not satisfy Eisenstein’s criterion, but many do. Corollary 8.5.11. Let m be a positive integer that’s divisible by a prime p but is not divisible by p2 . Then X n − m is irreducible in Q[X] for all n ≥ 1. In particular, √ [Q( n m) : Q] = n. Proof The polynomial X n − m clearly √ satisfies Eisenstein’s criterion for the prime p, and hence is irreducible over Q. Since n m is a root of X n −m, its minimal polynomial is a prime divisor of X n − m. But X n − m is prime, so it must be the minimal polynomial √ n of m.


323

Corollary 8.5.12. For each n ≥ 2, there is a simple extension of Q of degree n. In some cases, we may apply Eisenstein’s criterion via a change of variables. Corollary 8.5.13. For r ≥ 1, X 2 particular,

r−1

+ 1 is the minimal polynomial of ζ2r over Q. In

[Q(ζ2r ) : Q] = 2r−1 . Proof Write f (X) = X 2 + 1. Since ζ2r has order 2r in C× , and since −1 is the × unique element of C of order 2, ζ2r is a root of f . Thus, it suffices to show that f (X) is irreducible over Q. Let f (X + 1) be the polynomial obtained by evaluating f at X + 1. Note that for any polynomial g(X), g(X + 1) has the same degree as g(X). Thus, if f (X) is reducible, f (X + 1) will be reducible also. So it suffices to show that f (X + 1) is irreducible. r−1 r−1 Now f (X + 1) = (X + 1)2 + 1, and (X + 1)2 can be expanded by the Binomial Theorem (Theorem 4.5.16). We have r−1

2r−1

(X + 1)

=

r−1 2

i=0

2r−1 X i. i

r−1

In particular, (X + 1)2 is monic, and hence so is f (X + 1), while the constant term of 2r−1 (X + 1) is 1. Thus, the constant term of f (X +1) is 2. Thus, the result will follow from Eisenstein’s r−1 criterion for p = 2, provided that the coefficients 2 i are even for 0 < i < 2r−1 . But this follows immediately from Lemma 4.5.17. Exercises 8.5.14. 1. Show that X n − 1 = (X − 1)(X n−1 + X n−2 + · · · + 1) for all n > 1. Deduce that the minimal polynomial of ζn divides (X n−1 + X n−2 + · · · + 1). † 2. Let p be an odd prime and let f (X) = X p−1 + X p−2 + · · · + 1. The preceding problem gives (X + 1)p − 1 = X · f (X + 1). Deduce from Eisenstein’s criterion that f (X + 1) is irreducible, and hence f (X) is the minimal polynomial of ζp . Deduce that [Q(ζp ) : Q] = p − 1. 3. Suppose that f (X) ∈ Z[X] has a leading coefficient prime to p, and that f (X) is irreducible in Zp [X]. Show that f (X) must be irreducible in Q[X]. 4. Show that X 4 + 1 is not irreducible in Z2 [X]. Deduce that the converse of the preceding problem is false. 5. Let f (X) ∈ Z[X] be primitive, and write Z[X] · f (X) and Q[X] · f (X) for the principal ideals of Z[X] and Q[X], respectively, that are generated by f . Show that (Q[X] · f (X)) ∩ Z[X] = Z[X] · f (X). 6. Show that Eisenstein’s criterion holds in the field of fractions of any U.F.D.


8.6

324

The Frobenius

The material here is very basic, but it is important enough to deserve emphasis. Let p be a prime and let A be any commutative ring of characteristic p. By Proposition 7.2.14, this means that the unique ring homomorphism Z → A factors through an embedding of Zp in A. But since Zp is a field, any ring homomorphism out of Zp is injective, and hence the rings of characteristic p are precisely the Zp -algebras. Notice that the Zp -algebra structure on a ring A of characteristic p is unique, and that A has a unique subring isomorphic to Zp . We identify this subring with Zp . Definition 8.6.1. Let p be a prime and let A be a commutative ring of characteristic p. Then the Frobenius map ϕ : A → A is defined by ϕ(a) = ap for all a ∈ A. Proposition 8.6.2. Let A be a commutative ring of characteristic p. Then the Frobenius map ϕ : A → A is a ring homomorphism. Proof This is an immediate application of the Binomial Theorem. We have (a + b)p =

p p i=0

i

ai bp−i .

But for 0 < i < p, Lemma 4.5.17 shows pi to be divisible by p. But every element in a ring of characteristic p has additive exponent p (Corollary 7.7.16), and hence the summands with 0 < i < p all vanish. We are left with (a + b)p = ap + bp , so the Frobenius is a homomorphism of additive groups. But the rest is immediate. Corollary 8.6.3. Let K be a field of characteristic p. Then the Frobenius map ϕ : K → K is an embedding. In particular, if K is finite, then ϕ is an automorphism of the field K. For K either finite or infinite, the elements fixed by ϕ (i.e., the elements x ∈ K with ϕ(x) = x are precisely the elements of Zp . Proof Of course, every homomorphism of fields is an embedding. As far as the fixed points are concerned, note that x is a fixed point of ϕ if and only if x is a root of X p − X. As a polynomial of degree p over the field K, X p − X has at most p roots in K. So it suffices to show that every element of Zp is fixed by ϕ. p−1 Now Z× = 1 for all 0 = x ∈ Zp . So xp = x p is a (cyclic) group of order p − 1, so x for all x ∈ Zp . Definition 8.6.4. Let K be a field of characteristic p. We say that K is perfect if the Frobenius map ϕ : K → K is onto, and hence is an automorphism of the field K. We also say that every field of characteristic 0 is perfect. The next corollary is now immediate from Corollary 8.6.3. Corollary 8.6.5. All finite fields are perfect. Since Zp [X] has characteristic p, it has a Frobenius homomorphism, ϕ. Lemma 8.6.6. The Frobenius homomorphism ϕ : Zp [X] → Zp [X] is given by ϕ(f (X)) = f (X p ), the effect of evaluating f at X p .


325

Proof Since ϕ is a homomorphism, we have # " n n i = ai X ϕ(ai )ϕ(X)i ϕ i=0

i=0

=

n

ai (X p )i

i=0

since ϕ is the identity on Zp . But this last expression is f (X p ). We now show that the field of rational functions, Zp (X), of Zp is imperfect. Corollary 8.6.7. The Frobenius map on Zp (X) gives an isomorphism from Zp (X) onto Zp (X p ), the field of fractions of the subring Zp [X p ]. In particular, Zp (X) is not perfect. Proof The elements of Zp (X) have the form f (X)/g(X), where f (X), g(X) ∈ Zp [X] with g(X) = 0. Clearly, ϕ (f (X)/g(X)) = ϕ (f (X)) /ϕ (g(X)) = f (X p )/g(X p ), so the image of ϕ is Zp (X p ), as claimed. To see that this is not all of Zp (X), note that if f (X p )/g(X p ) = X/1 in Zp (X), then f (X p ) = Xg(X p ), which is impossible due to the congruences mod p of the exponents that appear. Exercises 8.6.8. 1. Show that every algebraically closed field is perfect. 2. Let k be a perfect field and let f (X) ∈ k[X]. Show that there is a polynomial g(X) ∈ k[X] such that f (X p ) = (g(X))p . 3. Let K be any field of characteristic p = 0. Show that K(X) is not perfect and calculate the image of ϕ. 4. Let A be a ring of characteristic 2. Show that the Frobenius function ϕ(a) = a2 gives a ring homomorphism from A to A if and only if A is commutative.

8.7

Repeated Roots

The material here is essential in understanding Galois theory for fields of nonzero characteristic. We shall make use of it in studying separable extensions. For now, we look at the elementary properties involved in the issues of repeated roots. Definitions 8.7.1. Let L be an extension of K and let α ∈ L be a root of the polynomial f (X) ∈ K[X]. We say that α is a repeated root, or multiple root, of f if (X − α)2 divides f in L[X]. We say that α is a root of multiplicity k for f (X) if f (X) = (X − α)k g(X) in L[X], where g(α) = 0.


326

On the surface, this would seem to depend on the field L, but in fact, it only depends on f and α. Lemma 8.7.2. Let L be an extension field of K and let f (X) ∈ K[X]. Suppose that α ∈ L is a root of multiplicity k ≥ 1 of f (X). Then there is a factorization f (X) = (X − α)k g(X) in K(α)[X], with g(α) = 0. In particular, α is a root of multiplicity k of f when considered as an element of any extension field of K(α). Proof Since f (α) = 0, the Euclidean algorithm in K(α)[X] gives a factorization f (X) = (X − α)f1 (X) in K(α)[X]. And unique factorization in L[X] says that α is a root of multiplicity k − 1 of f1 (X). The result follows by induction. Example 8.7.3. Over Q(i), X 4 + 2X 2 + 1 factors as (X − i)2 (X + i)2 . So i and −i are both multiple roots of the rational polynomial X 4 + 2X 2 + 1. It is a little more difficult to produce an example of an irreducible polynomial with repeated roots. Example 8.7.4. Let L = Zp (T ) be the field of fractions of the polynomial ring Zp [T ]. Let K = Zp (T p ) be the field of fractions of the subring Zp [T p ] of Zp [T ]. Then T ∈ L is a root of X p −T p ∈ K[X]. Moreover, by consideration of the Frobenius homomorphism in L[X], we obtain a factorization X p − T p = (X − T )p in L[X]. So T is a repeated root of X p − T p . Since X − T is prime in L[X], an examination of constant terms shows that no proper factor of (X − T )p can have coefficients in K. So X p − T p must be irreducible in K[X]. We shall see later that this shows L to be an inseparable extension of K. There is a formal derivative for polynomials, used to detect multiple roots. n Definition 8.7.5. Let A be a commutative ring and let f (X) = i=0 ai X i be a polynomial in A. The formal derivative of f is defined as follows: d (f (X)) = f (X) = iai X i−1 . dX i=1 n

Lemma 8.7.6. Let A be a commutative ring. Then the formal derivative A[X] is an A-module homomorphism satisfying the product rule:

d dX

: A[X] →

d (f (X)g(X)) = f (X) · g(X) + f (X) · g (X). dX

A morphism with these properties is often called an A-linear derivation. The formal derivative detects multiple roots in the following manner. Lemma 8.7.7. Let L be an extension field of K and let f (X) ∈ K[X]. Then an element α ∈ L is a repeated root of f if and only if α is a root of both f (X) and f (X).


327

Proof If α is a repeated root of f , then we have f (X) = (X − α)2 g(X) in L[X]. The product rule then gives f (X) = (X − α) [2g(X) + (X − α)g (X)] and hence α is a root of f (X) also. Conversely, if α is a root of both f (X) and f (X), then f (X) = (X − α)g(X) in L[X], and it suffices to show that α is a root of g(X). Here, the product rule gives f (X) = g(X) + (X − α)g (X), and hence f (α) = g(α). Since α is a root of f , the result follows. This allows us to characterize the irreducible polynomials with multiple roots. Proposition 8.7.8. Let L be an extension field of K and let f (X) be an irreducible element of K[X] with a root in L. Then f has a repeated root in L if and only if the formal derivative f (X) is the zero polynomial, in which case every root of f (X) is a repeated root. If K has characteristic 0, this cannot happen at all. If K has characteristic p, then the derivative f (X) vanishes if and only if f (X) = g(X p ) for some irreducible polynomial k g(X) in K[X]. Inductively, if f (X) has a repeated root, we may write f (X) = h(X p ) for some k > 0, where h(X) is irreducible in K[X] and h (X) = 0. Proof If α ∈ L is a repeated root of f , then α must also be a root of f (X). But f is irreducible, and hence must be, up to multiplication by a scalar, the minimal polynomial of α. Since α is a root of f , f must divide f . But the degree of f is less than that of f , so this can only happen if f is the zero polynomial. Note that if f = 0, then every root of f is also a root of f and hence is a multiple root of f . n Since f is irreducible, it must have degree at least one. Say f (X) = i=0 ai X i with n > 0 and an = 0. So if K has characteristic 0, the term nan X n−1 in the expansion of f (X) is nonzero, and hence f (X) is nonzero. Thus, α could not have been a repeated root of f . If K has characteristic p, then f (X) vanishes if and only if the only nonzero coeffik cients of f (X) are those in degrees divisible by p. This then gives f (X) = i=0 bi X pi = k g(X p ), where bi = api , k = np and g(X) = i=0 bi X i . Since any factorization of g would induce a factorization of f (X) = g(X p ), we see that g is irreducible. For the final statement, note that if g does not have repeated roots, we may take h(X) = g(X) and k = 1. Otherwise, the result follows by induction on degree, as the degree of g is less than that of f .

8.8

Cyclotomic Polynomials

We characterize the minimal polynomial of ζn over Q, for each n. First, consider the subgroup of Q(ζn )× generated by ζn : ζn = {1, ζn , . . . , ζnn−1 }, a cyclic group of order n. Thus, the order of ζnk is n/(n, k), and hence ζnk generates ζn if and only if (n, k) = 1.


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Definitions 8.8.1. The primitive n-th roots of unity in Q(ζn ) (or in C) are the generators of ζn . We define the Euler φ function by setting φ(n) equal to the number of primitive n-th roots of unity for n ≥ 1. Thus, φ(n) is the number of generators of a cyclic group of order n. As noted above, m generates Zn if and only if (m, n) = 1. We see (e.g., from Lemma 4.5.3) that φ(n) is the order of the group of units Z× n of the ring Zn . As such, a calculation of φ(n) is given in Section 4.5, where it is shown that φ(n) is the order of the automorphism group of the cyclic group Zn , which is then calculated in Corollary 4.5.9: Corollary 8.8.2. Let n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes and the exponents are all positive. Then φ(n) = p1r1 −1 (p1 − 1) . . . prkk −1 (pk − 1).

Notice that since ζn ⊂ Q(ζn )× has order n, every element in ζn is a root of X n −1. In particular, this means that there are n distinct roots of X n − 1 in Q(ζn ). Factoring out the associated degree one polynomials and equating leading coefficients, we obtain a factorization of X n − 1. Lemma 8.8.3. In Q(ζn )[X], we have Xn − 1 =

n−1 .

(X − ζnk ).

k=0

Corollary 8.8.4. In Z[X], there is a decomposition X n − 1 = p1 (X) . . . pk (X) where p1 (X), . . . , pk (X) are distinct monic polynomials in Z[X] that are irreducible in Q[X]. Proof By Proposition 8.5.7, we may write X n − 1 = ±p1 (X) . . . pk (X) in Z[X], where p1 (X), . . . , pk (X) are (not necessarily distinct) primitive polynomials in Z[X] that are irreducible in Q[X]. We may assume that the pi (X) all have positive leading coefficients. But in this case, since X n − 1 is monic, each pi (X) must be monic as well, and the sign in the displayed equation must be positive. So it suffices to show that the pi are all distinct. Suppose that pi (X) = pj (X) for some i = j. As a factor of X n − 1 of positive degree, pi (X) must have at least one root, say α, in Q(ζn ) by Lemma 8.8.3. But then α is also a root of pj , and hence is a repeated root of X n − 1. Since X n − 1 has n distinct roots in Q(ζn ), this is impossible. Since each pi (X) is monic and is irreducible in Q[X], it must be the minimal polynomial over Q of any of its roots in Q(ζn ). Since each ζnk is a root of X n − 1, its minimal polynomial must be one of the pi .


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Definition 8.8.5. We write Φn (X) for the minimal polynomial over Q of ζn . We call Φn (X) the n-th cyclotomic polynomial over Q. By the above, we see that Φn (X) is a monic polynomial in Z[X]. Examples 8.8.6. 1. Clearly, Φ1 (X) = X − 1, and Φ2 (X) = X + 1. 2. We’ve seen in Corollary 8.5.13 that Φ2r (X) = X 2

r−1

+ 1.

3. In Problem 2 of Exercises 8.5.14, it is shown that if p is an odd prime, then Φp (X) = 1 + X + · · · + X p−1 .

We now obtain our main result. Proposition 8.8.7. The roots of Φn (X) in Q(ζn ) are precisely the primitive n-th roots of unity. In particular, [Q(ζn ) : Q] = φ(n), and . (X − ζnk ) Φn (X) = 0 0. Show that Φpr (1) = p. Deduce that . r−1 p= (1 − ζpkr ) = u(1 − ζpr )(p−1)p 0 0, M and N have the same number of summands isomorphic to A/(pr ). Proof The existence of such a decomposition may be obtained by applying the Chinese Remainder Theorem to break up the cyclic modules of the decomposition from Theorem 8.9.15 into direct sums of primary torsion modules. Thus, it suffices to show uniqueness. Note that Lemma 8.9.19 shows that if f : M → N is an isomorphism of A-modules, then it restricts to an isomorphism f : Mp → Np for each prime p of A. Thus, we may assume that M = A/(pr1 ) ⊕ · · · ⊕ A/(prk ), where r1 ≥ · · · ≥ rk ≥ 1, and N = A/(ps1 ) ⊕ · · · ⊕ A/(psl ), where s1 ≥ · · · ≥ sl ≥ 1. We are given an isomorphism from M to N , and must deduce that k = l and ri = si for 1 ≤ i ≤ k. But now (pr1 ) ⊂ · · · ⊂ (prk ) and (ps1 ) ⊂ · · · ⊂ (psl ), so we may apply the uniqueness statement from the first form of the Fundamental Theorem. Exercises 8.9.21. 1. Show that Q/Z is a torsion module over Z, but that no element of Z annihilates all of Q/Z. (Of course, Q/Z is not finitely generated as a Z-module.) 2. Show that a finitely generated torsion module M over the P.I.D. A is cyclic if and only if there are no primes p of A for which M contains a submodule isomorphic to A/(p) ⊕ A/(p).

Chapter 9

Radicals, Tensor Products, and Exactness In this chapter, we develop some important tools for studying rings and modules. We shall use a number of these tools in Chapter 12, where we shall study some particular types of rings in detail. More generally, these tools are essential for anyone who works in fields involving rings and modules. In Section 9.1, we develop the Jacobson and nil radicals and the radical of an ideal in a commutative ring. We also give Nakayama’s Lemma, an essential ingredient in certain exactness arguments and in the study of local rings. Section 9.2 develops the theory of primary decomposition, a weak analogue of the unique factorization of ideals occurring in a P.I.D. In Section 9.3, Hilbert’s Nullstellensatz is proven, using many of the tools so far developed, and is used to define affine algebraic varieties. Section 9.4 gives the basic theory of tensor products. Extension of rings is emphasized as an application. Then, in Section 9.5, we use extension of rings and the theory of radicals to extend some of the basic results regarding homomorphisms of finite dimensional vector spaces to the study of free modules over more general rings. In particular, if A is a ring that admits a ring homomorphism to a division ring, we show that Am ∼ = An m n if and only if m = n. We also show that if f : A → A , then the following conditions hold. 1. If f is surjective, then m ≥ n. 2. If A is commutative, f is surjective, and m = n, then f is an isomorphism. 3. If A is commutative and f is injective, then m ≤ n. We also discuss flat modules and characterize the flat modules over a P.I.D. Section 9.6 gives the tensor product of algebras and verifies its universal property. Then, in Section 9.7, we develop the module structures and exactness properties of the Hom functors and describe their relationship to tensor products. In Sections 9.8 and 9.9, we develop the theory of projective modules. These are the modules P for which the functor HomA (P, −) preserves exact sequences, and are a generalization of free modules. They play a crucial role in our analysis of semisimple rings and Dedekind domains in Chapter 12, and hence are important in the study of

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342

group representations and in number theory. They are also of fundamental importance in homological algebra. Section 9.9 introduces algebraic K-theory with the functor K0 , which classifies the finitely generated projective A-modules modulo a sort of stabilization process. It plays an important role in numerous mathematical problems. Section 9.10 introduces the tensor algebra, TA (M ), on an A-module M . This is the free A-algebra on M . We use it to construct the symmetric algebra and exterior algebra on M , and to construct the Clifford algebra on a finite set of generators. The exterior algebras are related to determinant theory, while the Clifford algebras have applications in both topology and analysis. In the process of analyzing these algebras, we develop the theory of skew commutative graded algebras, which has applications in homological algebra and topology. We shall assume a familiarity here with the notions of categories and functors, as developed in the first two sections of Chapter 6. We shall also make use of some other ideas from that chapter, but will not assume a familiarity with them.

9.1

Radicals

We return to the study of maximality considerations for ideals, but this time in the context of noncommutative rings. Definitions 9.1.1. A maximal left ideal is a maximal element of the partially ordered set of proper left ideals. Maximal right ideals are defined analogously, and a maximal two-sided ideal is a maximal element in the partially ordered set of proper two-sided ideals. The next lemma is proven by the argument that was given for Corollary 7.6.14. Lemma 9.1.2. Let A be a nonzero ring. Then any proper left ideal of A is contained in a maximal left ideal, any proper right ideal is contained in a maximal right ideal, and any proper two-sided ideal is contained in a maximal two-sided ideal. Maximal left ideals have important consequences to module theory. They relate to the analogue for modules of simple groups. Definition 9.1.3. A nonzero left A-module is called simple, or irreducible, if it has no submodules other than 0 and itself. The maximal left ideals are the key to understanding simple modules. Lemma 9.1.4. Let M be a simple left A-module and let 0 = m ∈ M . Then Ann(m) is a maximal left ideal m of A, and there is a module isomorphism between A/m and M . Conversely, if m is a maximal left ideal of A, then A/m is a simple left A-module. Proof Given 0 = m ∈ M , the submodule Am generated by m is nonzero, and hence equal to M , since M is simple. But if m is the annihilator of m, we have an isomorphism fm : A/m → Am = M , inducing a one-to-one correspondence between the submodules of M and the left ideals of A containing m. Thus, the only left ideals containing m are m and A, so that m is a maximal left ideal. The converse comes precisely from the fact that there are only two left ideals containing a maximal left ideal m: m itself and A. We shall also make use of the following fact about maximal left ideals.


343

Lemma 9.1.5. Let A be a ring and let x ∈ A. Then x lies outside all the maximal left ideals if and only if x has a left inverse. Proof Consider the principal left ideal Ax generated by x. Then x lies in no maximal left ideal if and only if Ax lies in no maximal left ideal. But this is the case if and only if 1 ∈ Ax, which happens if and only if 1 = yx for some y ∈ A. Definition 9.1.6. Let A be a ring. The Jacobson radical, R(A), is the intersection of the maximal left ideals of A. Many of the rings we’re familiar with have R(A) = 0. Thus, it is valuable to keep in mind that if A is a local ring, with maximal ideal m, then R(A) = m. The next proposition gives a useful characterization of elements in R(A). Proposition 9.1.7. Let A be a ring and let x ∈ A. Then the Jacobson radical of A is a two-sided ideal, and the following statements are equivalent. 1. x is in the Jacobson radical, R(A). 2. For every simple left A-module M , x ∈ Ann(M ). 3. For any a, b ∈ A, 1 − axb is a unit in A. Proof Since the annihilator of a module is a two-sided ideal, the fact that R(A) is two-sided will follow from the equivalence of Conditions 1 and 2. We shall establish this first. We first show that the first condition implies the second. Let x ∈ R(A) and let M be a simple left A-module. It suffices to show that for 0 = m ∈ M , xm = 0. But the annihilator of m is a maximal left ideal of A because M is simple. Since R(A) is the intersection of the maximal left ideals, x must annihilate m as required. But now we can see that the second condition implies the first: if x annihilates A/m, with m a maximal left ideal, then it must annihilate the element 1 ∈ A/m. Since the annihilator of 1 is m, x ∈ m. We now show that the first two conditions imply the third. Let x ∈ R(A) and let a, b ∈ A. Since R(A) is a two-sided ideal, axb ∈ R(A) as well. But then 1 − axb cannot lie in any maximal left ideal, as that would force 1 to be in the ideal as well. By Lemma 9.1.5, this says that 1 − axb has a left inverse, y. But then y − yaxb = 1, and hence −yaxb = 1 − y. But −yaxb is in R(A), and hence lies in every maximal left ideal of A. But since 1−y is in every maximal left ideal, y can be in no maximal left ideal, and hence y has a left inverse z. But a standard argument now shows that z = 1 − axb, and hence y is a two-sided inverse for 1 − axb. Finally, we show that the third condition implies the first. Suppose that 1 − axb is a unit for each a, b ∈ A and let m be a maximal left ideal of A. Suppose that x is not in m. Then we must have Ax + m = A. But then 1 = ax + m for some a ∈ A and m ∈ m, so that 1 − ax ∈ m. But this contradicts the fact that 1 − ax is a unit, so x must have been in m. The following use of radicals is valuable in the study of local rings. Proposition 9.1.8. (Nakayama’s Lemma) Let A be a ring and let a be a left ideal contained in R(A). Suppose that aM = M for some finitely generated left A-module M . Then M = 0.


344

Proof Since a is contained in R(A), we may as well use R(A) in place of a. We argue by contradiction. Assume that M = 0, and let m1 , . . . , mk be a set of A-module generators for M , which is minimal in the sense that no proper subset will generate M . Since R(A)M = M , we have m1 ∈ R(A)M , and hence m1 = a1 n1 + · · · + ar nr , with ai ∈ R(A) and ni ∈ M for all i. But since since each ni is a linear combination of the generators m1 , . . . , mk , and since R(A) is a right ideal, we may collect terms, writing m1 = b1 m1 + · · · + bk mk , with bi ∈ R(A) for all i. Rearranging terms, we get (1 − b1 )m1 = b2 m2 + · · · + bk mk . But since b1 ∈ R(A), 1 − b1 is a unit, so that m1 is in the submodule of M generated by m2 , . . . , mk . But then m2 , . . . , mk generate M , contradicting the minimality of the generating set m1 , . . . , mk . In the commutative case, there is another radical which is of interest. Lemma 9.1.9. Let A be a commutative ring. Then the set of all nilpotent elements in A forms an ideal, N(A), called the nil radical of A. Proof Let x ∈ N(A) and let a ∈ A. Then xn = 0 for some n ≥ 1, and hence (ax)n = an xn = 0, also. It suffices to show that N(A) is closed under addition. Thus, let x, y ∈ N(A), with xn = y m = 0 for n, m ≥ 0. We claim that (x + y)n+m−1 = 0. To see this, recall that the Binomial Theorem (Theorem 4.5.16) holds in any commun+m−1 n+m−1 i n+m−i−1 xy tative ring. Thus, (x + y)n+m−1 = i=0 . But if xi = 0, then i n+m−i−1 i < n, and hence n + m − i − 1 ≥ m, so that y = 0. The nil radical also has a characterization related to that of the Jacobson radical. Proposition 9.1.10. Let A be a commutative ring. Then the nil radical of A is the intersection of all the prime ideals of A. Proof Let x ∈ N(A) and let p be a prime ideal of A. We claim that x ∈ p. To see this, let n > 0 be the smallest exponent such that xn = 0. Then xn ∈ p. If n = 1, there’s nothing to show. Otherwise, either x ∈ p or xn−1 ∈ p, so in either case, xn−1 ∈ p. But then a downward induction shows that x ∈ p. It suffices to show that if x ∈ A is not nilpotent, then there is a prime ideal that doesn’t contain x. Let X = {xk | k ≥ 1} and let S be the set of all ideals a such that a ∩ X = ∅. Since x is not nilpotent, 0 ∈ S, and hence S is nonempty. By an argument similar to the one that shows the existence of maximal ideals, Zorn’s Lemma implies that there are maximal elements in S. We claim that if p is a maximal element of S, then p is prime. It suffices to show that if a and b are in the complement of p, then so is ab. Since a is not in p, p + (a) strictly contains p, and hence xm = p + ya for some p ∈ p, y ∈ A, and m > 0. Similarly, there are p ∈ p, y ∈ A, and n > 0, with xn = p + y b. But multiplying these two elements together, we get xn+m = p + yy ab for some p ∈ p. But since p does not meet X, ab cannot lie in p. We can generalize the nil radical to define a radical for any ideal of a commutative ring. Definition 9.1.11. Let a be an ideal in the commutative ring A. The radical rad(a) is given by rad(a) = {x ∈ A | xn ∈ a for some n > 0}.


345

Notice that if π : A → A/a is the canonical map, then rad(a) = π −1 (N(A/a)). Thus, rad(a) is an ideal of A. The following proposition is now immediate from the one-to-one correspondence between the prime ideals of A/a and the primes of A that contain a. Proposition 9.1.12. Let a be a proper ideal in the commutative ring A. Then rad(a) is the intersection of the prime ideals of A containing a. It is useful to be able to characterize the ideals a in a commutative ring with rad(a) = a. As the reader may easily check, rad(rad(a)) = rad(a) for any ideal a. We obtain the following lemma. Lemma 9.1.13. An ideal a of a commutative ring A is its own radical if and only if a = rad(b) for some ideal b of A. Of course, if p is prime, then the intersection of the prime ideals containing it is precisely p. Proposition 9.1.12 gives the following corollary. Corollary 9.1.14. A prime ideal is its own radical. The next lemma is easy and its proof is left as an exercise. It use is sufficiently common to warrant a formal statement. n Lemma 9.1.15. Let a1 , . . . , an be ideals in a commutative ring A. Then rad( i=1 ai ) = n i=1 rad(ai ). Corollary 9.1.14 now gives a corollary. Corollary 9.1.16. A finite intersection of prime ideals is its own radical. We shall show in Corllary 9.2.11 that a proper ideal in a Noetherian commutative ring is its own radical if and only if it is a finite intersection of prime ideals. Exercises 9.1.17. 1. What is R(A × B)? What is N(A × B)? 2. Let A be a P.I.D. Show that R(A) = 0 if and only if A has infinitely many prime ideals. 3. Let A be the ring of Example 6 of Examples 7.8.2. What is R(A)? 4. Let A be a ring. Show that R(A) is equal to the intersection of the maximal right ideals in A. 5. Let A be a ring. Show that R(A) is contained in the intersection of the maximal two-sided ideals in A. 6. Give the proof of Lemma 9.1.15. 7. Let n > 1 in Z. What is rad((n))? 8. Let A be a commutative ring with only one prime ideal. Show that any injective A-module homomorphism from A to itself is an isomorphism.


9.2

346

Primary Decomposition

Primary decomposition of ideals is a weak analogue of the unique factorization of ideals occurring in a P.I.D. The theory was first developed for polynomial rings by the chess master Emanuel Lasker. Emmy Noether then showed that primary decompositions exist for all proper ideals of a Noetherian commutative ring. We shall give her argument here. The theory works as follows: There is a generalization of prime ideals called primary ideals. In a Noetherian commutative ring, every proper ideal is a finite intersection of primary ideals. But the decomposition is not unique, even if we insist that none of the primary ideals contain the intersection of the others. However, under that assumption, the set of ideals which are radicals of the primary ideals in the decomposition is unique. As an application of this theory, we shall show that an ideal a in a Noetherian commutative ring satisfies a = rad(a) if and only if a is a finite intersection of prime ideals. We shall make use of this in Section 9.3 to give decompositions of affine varieties over an algebraically closed field. Definition 9.2.1. Let A be a commutative ring. A proper ideal q of A is primary if xy ∈ q implies that either x ∈ q or y ∈ rad(q) (i.e., y n ∈ q for some n > 0). If a is an ideal of A, then a primary decomposition of a is a presentation a=

m

qi

i=1

of a as a finite intersection of primary ideals qi for i = 1, . . . , m. Clearly, any prime ideal is primary. (Recall from Corollary 9.1.14 that a prime ideal is its own radical.) We shall give further examples below. Throughout this section, A denotes a commutative ring. Since rad(q) is the inverse image under the canonical map of the nil radical of A/q, the next lemma is immediate. Lemma 9.2.2. A proper ideal q of A is primary if and only if every zero-divisor in A/q is nilpotent. The reader should verify the following lemma. Lemma 9.2.3. Let q be a primary ideal of A. Then rad(q) is prime. Definition 9.2.4. If q is primary with radical p, we say that q is p-primary, or that p is the prime associated to q. The converse to Lemma 9.2.3 is false: there are nonprimary ideals whose radical is prime. Having a maximal ideal as radical is stronger. Proposition 9.2.5. Let q be an ideal of A whose radical is a maximal ideal. Then q is primary. Proof Let m = rad(q). Since the radical is the intersection of the prime ideals containing q (Proposition 9.1.12), any prime ideal containing q must contain m. Thus, m is the only prime ideal of A containing q, and hence m/q is the only prime ideal in A/q. Thus, any element of A/q outside m/q is a unit. So every zero-divisor of A/q lies in m/q. But since m = rad(q), m/q is the nil radical of A/q, so all its elements are nilpotent. Thus, q is primary by Lemma 9.2.2.


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Examples 9.2.6. 1. Let m be a maximal ideal of A and let mk = m . . . m be the ideal product of m with itself k times. (See Definitions 7.2.21.) Let q be any proper ideal containing mk . Then m ⊂ rad(q). Since q is proper, so is its radical. So rad(q) = m, and q is primary by Proposition 9.2.5. 2. If K is a field and r1 , . . . , rn > 0, let q = (X1r1 , . . . , Xnrn ) ⊂ K[X1 , . . . , Xn ], the ideal generated by X1r1 , . . . , Xnrn . Then rad(q) contains the maximal ideal m = (X1 , . . . , Xn ). Since q is proper, it is m-primary. 3. Let A be a P.I.D. and let a = pr11 . . . prkk , where p1 , . . . , pk are pairwise inequivalent primes of A. Then an element lies in rad((a)) if and only if it is divisible by p1 , . . . , pk . Thus, rad((a)) = (p1 . . . pk ). Thus, a nonzero ideal (a) is primary if and only if (a) = (pr ) = (p)r for some prime element p of A and r > 0. The ideal 0 is also primary, being prime. We shall now show that every proper ideal in a Noetherian commutative ring has a primary decomposition. We shall make use of the following operation on ideals. Definition 9.2.7. Let a and b be ideals of A, Then the ideal quotient (a : b) is the ideal (a : b) = {x ∈ A | xb ⊂ a}. Thus, x ∈ (a : b) if and only if xb ∈ a for all b ∈ b. For a principal ideal (x) ⊂ A, we write (a : (x)) = (a : x). Note that y ∈ (a : x) if and only if xy ∈ a. We shall actually show that every proper ideal in a Noetherian commutative ring is a finite intersection of irreducible ideals. Definition 9.2.8. A proper ideal a of A is irreducible if a = b ∩ c implies that either a = b or a = c. Lemma 9.2.9. If A is Noetherian, then every irreducible ideal of A is primary. Proof Let q be an irreducible ideal of A and let xy ∈ q with y ∈ q. Then we have an ascending chain of ideals (q : x) ⊂ (q : x2 ) ⊂ · · · ⊂ (q : xk ) ⊂ · · · . Since A is Noetherian, the sequence is eventually constant, and hence (q : xn ) = (q : xn+1 ) for some n > 0. We claim that (q + (y)) ∩ (q + (xn )) = q. To see this, note that xy ∈ q implies that q + (y) ⊂ (q : x). Thus, if z + axn ∈ q + (y) with z ∈ q, then zx + axn+1 = w ∈ q. So a ∈ (q : xn+1 ) = (q : xn ). But then z + axn ∈ q, as desired. Thus, (q+(y))∩(q+(xn )) = q. Since q is irreducible and y ∈ q, this forces q+(xn ) = q, and hence x ∈ rad(q). Thus, q is primary. We can now show the existence of primary decompositions in a Noetherian commutative ring. The technique we shall use is called Noetherian induction. We shall make further use of it in Chapter 12. Theorem 9.2.10. Every proper ideal in a Noetherian commutative ring A is a finite intersection of irreducible ideals. In particular, Lemma 9.2.9 now shows that any proper ideal of A has a primary decomposition.


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Proof Recall from Proposition 7.6.16 that if S is any nonempty set of ideals in a Noetherian ring, ordered by inclusion, then S has maximal elements. Here, a is maximal in S if a ∈ S and any ideal in S containing a must equal a. Thus, we may argue by contradiction: Assuming the theorem false, let S be the set of proper ideals which are not finite intersections of irreducible ideals and let a be a maximal element in S. Since a is in S, it cannot be irreducible, so we may write a = b ∩ c, where b and c both properly contain a (and hence b and c are both proper ideals). Since a is a maximal element in S, b and c cannot lie in S, and hence each of them is a finite intersection of irreducible ideals. But a = b ∩ c, so this contradicts the assumption that a is not a finite intersection of irreducible ideals. Corollary 9.2.11. A proper ideal a in a Noetherian commutative ring A is its own radical if and only if a is a finite intersection of prime ideals. Proof Corollary 9.1.16 shows that a finite intersection of prime ideals is its own radical, so it suffices to show the converse. As noted in Lemma 9.1.13, this amounts to showing that the radical of any proper ideal of A is a finite intersection of prime ideals. n Let a be a proper ideal of A. Theorem 9.2.10 n gives a decomposition a = i=1 qi where qi is primary for all i. But then rad(a) = i=1 rad(qi ) (Lemma 9.1.15). The result follows, since the radical of a primary ideal is prime (Lemma 9.2.3). We shall now discuss the uniqueness properties of primary decompositions. As we shall assume the existence of such a decomposition, A need not any longer be Noetherian. Theorem 9.2.10 itself provides an example of nonuniqueness, in that not every primary ideal is irreducible. Example 9.2.12. Let K be a field and let m = (X, Y ), the ideal of K[X, Y ] generated by X and Y . Then m2 = (X 2 , XY, Y 2 ) is primary by Proposition 9.2.5. As the reader may check, m2 = (m2 + (X)) ∩ (m2 + (Y )), the intersection of two ideals properly containing it. Also, both m2 + (X) and m2 + (Y ) have radical m, and hence are primary. We can eliminate this particular sort of nonuniqueness to our decompositions if we insist that each of the primary ideals being intersected have a distinct radical. The following lemma allows us to do this. Lemma 9.2.13. Let q1 , . . . , qm be p-primary ideals m in A for some prime ideal p (i.e., each qi is primary with rad(qi ) = p). Then q = i=1 qi is p-primary as well. m Proof By Lemma 9.1.15, rad(q) = i=1 rad(qi ) = p, so it suffices to show q is primary. Suppose xy ∈ q and y ∈ rad(q). It suffices to show that x ∈ q. Since q is the intersection of the qi , xy ∈ qi for all i. But y ∈ rad(qi ) = rad(q). Since each qi is primary, x must lie in qi for all i, and hence in the intersection, q. m Definition 9.2.14. A primary decomposition a = i=1 qi of an ideal a is reduced if 1. The prime ideals rad(q1 ), . . . , rad(qm ) are all distinct. 2. No qi contains the intersection of the other q’s.


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Using Lemma 9.2.13 to obtain the first condition and then induction to obtain the second, Theorem 9.2.10 gives the following corollary. Corollary 9.2.15. Every proper ideal in a Noetherian commutative ring has a reduced primary decomposition. Even reduced primary decompositions are not unique. Example 9.2.16. Let a = (X 2 , XY ) ⊂ K[X, Y ] with K a field. Let m = (X, Y ). Then (X 2 , Y ) and m2 = (X 2 , XY, Y 2 ) both have radical m, and hence are primary by Proposition 9.2.5. Thus, the following equalities, which the reader should verify, give two distinct primary decompositions for a. a a

and = (X) ∩ m2 2 = (X) ∩ (X , Y ).

Note that the radicals of the primary ideals in these two decompositions are given by (X) and (X, Y ). Note that (X) ⊂ (X, Y ), even though there is no analogous inclusion between the corresponding primary ideals in these decompositions. We shall now develop some positive results regarding uniqueness of primary decompositions. First note that prime ideals are better behaved than primary ideals with respect to irreducibility. Lemma 9.2.17. Prime ideals are irreducible. Proof Let p = a∩b with p prime. Since p contains a∩b, either a or b must be contained in p by Corollary 7.6.10. But p is contained in both a and b, and hence must equal one of them. Lemma 9.2.18. Let p be a prime ideal of A and let the ideal quotient (q : x) is given by ⎧ ⎨ A q (q : x) = ⎩ a p-primary ideal

q be p-primary. Then for x ∈ A, if x ∈ q if x ∈ p otherwise.

Proof The case x ∈ q is true for any ideal, while that of x ∈ p is immediate from the definition of a primary ideal: If xy ∈ q and x ∈ p = rad(q), then y must lie in q. Thus, it suffices to show that if x ∈ q then (q : x) is p-primary. Note, then that if y n x ∈ q, then y n ∈ rad(q) = p, since x ∈ q and q is primary. But then y ∈ p, since p is prime. Thus, rad((q : x)) ⊂ p. But q ⊂ (q : x), and hence p = rad(q) ⊂ rad((q : x)), so p = rad((q : x)), and it suffices to show that (q : x) is primary. Thus, suppose that ab ∈ (q : x) with a ∈ rad((q : x)) = p. We wish to show b ∈ (q : x). We have abx ∈ q, so bx ∈ (q : a) = q by the second case above. But then b ∈ (q : x). We can use the ideal quotients to show that the primes associated to the primary ideals in a primary decomposition are unique. m Proposition 9.2.19. Let a = i=1 qi be a reduced primary decomposition of the ideal a and let pi = rad(qi ) for i = 1, . . . , m. Then {p1 , . . . , pm } is the set of prime ideals which occur among the ideals {rad((a : x)) | x ∈ A}. k Thus, if a = i=1 qi is another reduced primary decomposition of a, then k = m, and we may reindex the qi so that rad(qi ) = pi for i = 1, . . . , m.


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m m Proof As the reader may easily check, ( i=1 qi : x) = i=1 (qi : x). Since radicals also commute with intersections (Lemma 9.1.15), Lemma 9.2.18 shows that m if x ∈ a = i=1 qi {i | x ∈qi } pi rad((a : x)) = A if x ∈ a. If rad((a : x)) is prime, then x ∈ a, and rad((a : x)) is the finite intersection x ∈qi pi . But prime ideals are irreducible (Lemma 9.2.17), so rad((a : x)) must equal one of the pi . Conversely, since the decomposition is reduced, for each i we can find an element xi which lies in j =i qj but not in qi . But then the displayed equation shows that rad((a : xi )) = pi . Definitions 9.2.20. Let a be an ideal admitting a primary decomposition. Then the primes associated to a (or, more simply, the primes of a) are the radicals of the primary ideals in any reduced primary decomposition of a. Let p1 , . . . , pm be the primes of a. If pi contains pj for some j = i, we say that pi is an embedded m prime of a. Otherwise, we say that pi is an isolated prime of a. If a = i=1 qi is a reduced primary decomposition of a, with pi the associated prime of qi , we call the qi the primary components of the decomposition, and say that qi is isolated (resp. embedded) if pi is isolated (resp. embedded). Thus, Example 9.2.16 shows that (X) is an isolated prime of (X 2 , XY ) ⊂ K[X, Y ] and (X, Y ) is embedded. The word “embedded” expresses the fact that if K is an algebraically closed field and if p ⊂ p is a proper inclusion of prime ideals of K[X1 , . . . , Xn ], then the affine variety V(p ) induced by p embeds as a subvariety of V(p). We shall investigate this situation in Section 9.3. We can obtain a uniqueness result for the isolated primary components of a decomposition. m Proposition 9.2.21. Let a = i=1 qi be a reduced primary decomposition for a with associated primes p1 , . . . , pm . Suppose that pi is an isolated prime of a. Then qi = {x ∈ A | (a : x) ⊂ pi }. Since the right hand side is independent of the choice of reduced primary decomposition, so is qi . Proof Write S = {x ∈ A | (a : x) ⊂ pi }. We first show that S ⊂ qi . Since a ⊂ qi , (a : x) ⊂ (qi : x) for all x. If x ∈ qi , then (qi : x) ⊂ pi by Lemma 9.2.18. But then (a : x) ⊂ pi , and hence x ∈ S. So S mustbe contained in qi . Since pi is isolated, it cannot contain j =i pj : If it did, then it would have to contain some pj for j = i by Corollary 7.6.10. Let y be an element of j =i pj not in pi . Since pj = rad(qj ) for all j, we can choose an k > 0 such that y k ∈ j =i qj . Since pi is a prime not containing y, y k ∈ p. Let x ∈ qi . Then xy k lies in all of the q’s, and hence in a. Thus, y k ∈ (a : x). Since k y ∈ pi , (a : x) ⊂ pi , and hence x ∈ S. Thus, qi ⊂ S, and hence qi = S. The situation for embedded primary components is much more complicated. Indeed, if A is Noetherian, then one may make infinitely many different choices for each embedded component.1 1 See

p. 231 of Zariski and Samuel, Commutative Algebra, Vol. I , Van Nostrand, 1958.


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But reduced primary decompositions are unique if all primes are isolated. The most obvious case is for a p-primary ideal, q. Here, there is only one associated prime, p, and hence q = q is the only reduced primary decomposition of q. A more interesting case is that of a finite intersection of prime ideals. (Recall from Corollary 9.2.11 that if A isNoetherian, then an ideal has this form if and only if it is its m own radical.) Here, if a = i=1 pi , then we can obtain a reduced primary decomposition for a by inductively eliminating primes containing the intersection of all the other primes. The result is a primary decomposition in which every primary ideal is prime and every prime is isolated. Corollary 9.2.22. Let a be a finite intersection of prime ideals. Then a may be written uniquely as a finite intersection of prime ideals none of which contain the intersection of the others. Here, the uniqueness is up to permutation of the primes.

9.3

The Nullstellensatz and the Prime Spectrum

Recall that an A-algebra of finite type is one which is finitely generated as an A-algebra. Lemma 9.3.1. Suppose given inclusions A ⊂ B ⊂ C of commutative rings, where A Noetherian and C is an A-algebra of finite type. Suppose that C is finitely generated as a B-module. Then B is finitely generated as an A-algebra. Proof Suppose that C = A[c1 , . . . , cn ] and that d1 , . . . , dm generate C as a B-module. Then we can find elements bij and bijk in B such that cj dj dk

= =

m i=1 m

bij di bijk di

for j = 1, . . . , n

and

for 1 ≤ j, k ≤ m.

i=1

Write A = A[bij , birs | 1 ≤ i, r, s ≤ m, 1 ≤ j ≤ n] ⊂ B. Since A is finitely generated as an A-algebra, it is Noetherian by Corollary 7.8.15. Using the displayed equations and induction, it is easy to see that any monomial in the cj with coefficients in A lies in the A-submodule of C generated by 1, d1 , . . . , dm . Thus, C is finitely generated as an A-module. But then C is a Noetherian A-module (Proposition 7.8.9), so its submodule B is also finitely generated as an A-module (Proposition 7.8.3). Since A is finitely generated as an A-algebra, B must be, also. We obtain a recognition principle for finite extensions. Theorem 9.3.2. (Hilbert’s Nullstellensatz) Suppose the extension field L of K is finitely generated as a K-algebra. Then L is a finite extension of K. Proof Let L = K[α1 , . . . , αn ]. If each αi is algebraic over K, then we are done by Lemma 8.2.9. Otherwise, we wish to derive a contradiction. By Proposition 8.3.6, we may reorder the α’s so that α1 , . . . , αr is a transcendence basis for L over K for some r ≤ n. Thus, L is algebraic over K1 = K(α1 , . . . , αr ). Since L is obtained by adjoining finitely many algebraic elements to K1 , Lemma 8.2.9 shows that L is a finite extension of K1 .


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By Lemma 9.3.1, K1 is finitely generated as a K-algebra. We write K1 = K[β1 , . . . , βs ]. Since α1 , . . . , αr are algebraically independent, K[α1 , . . . , αr ] is a unique factorization domain, by Proposition 8.5.9. Since K1 is its field of fractions, we may write βi = fi (α1 , . . . , αr )/gi (α1 , . . . , αr ), where fi and gi are relatively prime for i = 1, . . . , s. But then any polynomial in β1 , . . . , βs has a denominator whose prime factors all divide the product of the gi . Thus, it suffices to show that K[α1 , . . . , αr ] has infinitely many primes, as, if h has a prime factor that does not divide any of the gi , then the element 1/h of K(α1 , . . . , αr ) cannot lie in K[β1 , . . . , βs ]. But if p1 , . . . , pk are primes in K[α1 , . . . , αr ], then p1 . . . pk + 1 is relatively prime to each pi , and hence has a prime factor distinct from the pi . The Nullstellensatz has a number of important consequences for commutative algebras of finite type over a field. Corollary 9.3.3. Let K be a field and let f : A → B be a K-algebra homomorphism between commutative K-algebras of finite type. Then for each maximal ideal m of B, the inverse image f −1 (m) is maximal in A. Proof Let m be a maximal ideal of B and let p = f −1 (m). Then there is an inclusion of K-algebras A/p ⊂ B/m. But the Nullstellensatz shows B/m to be a finite extension of K. Since A/p is a subalgebra of B/m, every element α ∈ A/p is algebraic over K (Proposition 8.2.14). Thus, (Proposition 8.2.5) K[α] ⊂ A/p is a field, and hence α is invertible in A/p. Since α was arbitrary, A/p is a field, and hence p = f −1 (m) is maximal in A. This allows us to deduce an important fact about radicals in commutative algebras of finite type over a field. Proposition 9.3.4. Let K be a field and let a be a proper ideal in K[X1 , . . . , Xn ]. Then the radical, rad(a), of a is the intersection of the maximal ideals of K[X1 , . . . , Xn ] containing a. In consequence, if A is a commutative algebra of finite type over a field K, then the Jacobson radical and nil radical of A coincide: N(A) = R(A). Proof The second assertion is immediate from the first: if A is a commutative algebra of finite type over K and if f : K[X1 , . . . , Xn ] → A is a surjective K-algebra homomorphism, then f −1 (N(A)) = rad(ker f ), while f −1 (R(A)) is the intersection of the maximal ideals of K[X1 , . . . , Xn ] containing ker f . Thus, let a be a proper ideal of K[X1 , . . . , Xn ]. By Proposition 9.1.12, rad(a) is the intersection of the prime ideals of K[X1 , . . . , Xn ] containing a. It suffices to show that if f (X1 , . . . , Xn ) is not in rad(a), then there is a maximal ideal containing a which doesn’t contain f . Since f is not in rad(a), there is a prime ideal p containing a but not f . Thus, f represents a nonzero element f of the domain A = K[X1 , . . . , Xn ]/p. Let B = A[1/f ], the subring of the field of fractions of A obtained by adjoining 1/f to A. Write g : K[X1 , . . . , Xn ] → B for the composite π

→ K[X1 , . . . , Xn ]/p = A ⊂ B, K[X1 , . . . , Xn ] − where π is the canonical map. Then g is a K-algebra homomorphism between commutative K-algebras of finite type. Thus, if m is a maximal ideal of B, then g −1 (m) is a


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maximal ideal of K[X1 , . . . , Xn ] by Corollary 9.3.3. Since g(f ) is a unit in B, f ∈ g −1 (m). Since a ⊂ ker g ⊂ g −1 (m), the result follows. In a domain A, there are no nonzero nilpotent elements, and hence N(A) = 0. Corollary 9.3.5. Let A be an integral domain finitely generated as an algebra over a field K. Then R(A) = 0. We can use the Nullstellensatz to give a complete determination of the maximal ideals of a polynomial ring over an algebraically closed field. This consitutes the starting point for algebraic geometry. Corollary 9.3.6. (Nullstellensatz, weak version) Let K be an algebraically closed field and let m be maximal ideal of K[X1 , . . . , Xn ]. Then m is the kernel of the evaluation map εa1 ,...,an : K[X1 , . . . , Xn ] → K for some a1 , . . . , an ∈ K. Thus, by Proposition 7.3.22, m = (X1 − a1 , . . . , Xn − an ), the ideal generated by X1 − a1 , . . . , Xn − an . Proof Let L = K[X1 , . . . , Xn ]/m. Since the polynomial ring maps onto L, L is a K-algebra of finite type, and hence is a finite extension of K, by the Nullstellensatz. Since K is algebraically closed, it has no irreducible polynomials of degree > 1. Thus, any element algebraic over K has a minimal polynomial of degree 1, and hence lies in K. Thus, the inclusion of K in L is an isomorphism. ∼ = We obtain a K-algebra isomorphism f : K[X1 , . . . , Xn ]/m −→ K. Note that m is the kernel of the composite π

f

→ K[X1 , . . . , Xn ]/m − K[X1 , . . . , Xn ] − → K, where π is the canonical map. But if f ◦ π(Xi ) = ai for i = 1, . . . , n, then f ◦ π must be the evaluation map εa1 ,...,an : K[X1 , . . . , Xn ] → K by the universal property of a polynomial algebra (Corollary 7.3.20). Definitions 9.3.7. Let A be a commutative ring. We write Spec(A) for the set of prime ideals of A and write Max(A) for the set of maximal ideals of A. We call them the prime spectrum and the maximal ideal spectrum of A, respectively. If K is a field and (a1 , . . . , an ) ∈ K n , we write ma1 ,...,an ∈ Max(A) for the kernel of the evaluation map εa1 ,...,an : K[X1 , . . . , Xn ] → K: ma1 ,...,an = {f ∈ K[X1 , . . . , Xn ] | f (a1 , . . . , an ) = 0}. By Proposition 7.3.22, ma1 ,...,an = (X1 − a1 , . . . , Xn − an ), the ideal generated by X1 − a1 , . . . , Xn − an . We write ν : K n → Max(K[X1 , . . . , Xn ]) for the map given by ν(a1 , . . . , an ) = ma1 ,...,an . Proposition 9.3.8. The maps ν : K n → Max(K[X1 , . . . , Xn ]) are injective for any field K and are surjective if K is algebraically closed. Proof If ma1 ,...,an = mb1 ,...,bn , then Xi − ai and Xi − bi lie in this ideal for each i. But (Xi − ai ) − (Xi − bi ) is a unit unless ai = bi , so ν is injective. If K is algebraically closed, then ν is surjective by the weak Nullstellensatz.


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If K is algebraically closed, we can think of ν as a way of providing a geometry for the maximal ideals of K[X1 , . . . , Xn ]. In particular, when K = C,2 this provides a standard sort of geometry to associate to algebraically significant subsets of the set of maximal ideals. Some subsets of K n which have algebraic significance are the affine varieties. Definitions 9.3.9. Let K be an algebraically closed field. We shall refer to K n as affine n-space over K. If S ⊂ K[X1 , . . . , Xn ] is any subset, then the affine variety, V(S), determined by S is the subset of affine n-space consisting of the common zeroes of the elements of S: V(S) = {(a1 , . . . , an ) ∈ K n | f (a1 , . . . , an ) = 0

for all f ∈ S}.

As we shall verify in Lemma 9.3.12, the affine variety V(S) is mapped by ν : K n ∼ = Max(K[X1 , . . . , Xn ]) to the set of maximal ideals of K[X1 , . . . , Xn ] containing S. There is a geometry to affine varieties intrinsic to the algebra. It does not coincide with ordinary complex geometry when K = C. It is called the Zariski topology, and may be defined not only on affine varieties, but also on Spec(A) for any commutative ring A. Definitions 9.3.10. Let A be a commutative ring and let S ⊂ A. The Zariski open set U (S) of Spec(A) is given by U (S) = {p ∈ Spec(A) | S ⊂ p}. The Zariski topology on Spec(A) is the one in which the open sets are precisely the Zariski open sets. The Zariski topology on Max(A) is the subspace topology of that on Spec(A). Let K be an algebraically closed field. Then Proposition 9.3.8 shows that ν : K n → Max(K[X1 , . . . , Xn ]) is a bijection. We define the Zariski topology on K n by declaring ν to be a homeomorphism. That this gives a topology, of course, requires justification. Proposition 9.3.11. Let A be a commutative ring. Then the Zariski open sets form a topology on Spec(A). Moreover, they have the following properties. 1. U ({1}) = Spec(A). 2. U ({0}) = ∅. 3. If S ⊂ T ⊂ A, then U (S) ⊂ U (T ). 4. If a is the ideal generated by S, then U (S) = U (a). 5. If Si ⊂ A for i ∈ I, then i∈I U (Si ) = U ( i∈I Si ). 6. U ({s}) ∩ U ({t}) = U ({st}) for all s, t ∈ A. 7. For S, T ⊂ A, U (S) ∩ U (T ) = U ({st | s ∈ S and t ∈ T }). 2 We

shall show in Theorem 11.6.7 that C is algebraically closed.


355

Proof The first condition holds because the elements of Spec(A) are proper ideals. The next four conditions are clear, and the sixth is a restatement of the definition of a prime ideal: st ∈ p if and only if either s ∈ p or t ∈ p. The last condition follows from the others. If K is algebraically closed, then the Zariski topology on affine n-space over K is determined by the affine varieties. Lemma 9.3.12. Let K be an algebraically closed field. Then the affine varieties are precisely the closed subspaces in the Zariski topology on the affine n-space K n . Specifically, if S ⊂ K[X1 , . . . , Xn ], then the homeomorphism ν : K n → Max(K[X1 , . . . , Xn ]) carries the affine variety V(S) onto the complement of the Zariski open set U (S). Proof An n-tuple (a1 , . . . , an ) lies in V(S) if and only if f (a1 , . . . , an ) = 0 for all f ∈ S. But f (a1 , . . . , an ) = 0 if and only if f lies in the maximal ideal ν(a1 , . . . , an ) = ker εa1 ,...,an , so ν(V(S)) is the set of maximal ideals of K[X1 , . . . , Xn ] containing S. But this is precisely the complement of U (S). By analogy, for an arbitrary commutative ring A and a subset S ⊂ A, we shall write V (S) for the complement of U (S) in either Spec(A) or Max(A). In this more general context, the spaces are called schemes, rather than varieties. From a topologist’s point of view, the topologies on Spec(A) and Max(A) are bizarre. Note that a point p in Spec(A) is closed if and only if there is a set S such that p contains S, but no other prime ideal contains S. Clearly, such an S exists if and only if p is maximal. Thus, Max(A) is a T1 space, but Spec(A) is not. So why bother with Spec(A)? For one thing, prime ideals have some interest in their own right. For another, as we shall see presently, Spec is a functor of commutative rings, but Max is not. However, Corollary 9.3.3 shows that Max is a functor on algebras of finite type over a field K: Definition 9.3.13. Let f : A → B be a homomorphism between commutative rings. Then the induced map f ∗ : Spec(B) → Spec(A) is given by f ∗ (p) = f −1 (p). If K is a field and f : A → B is a K-algebra homomorphism between commutative K-algebras of finite type, then the induced map f ∗ : Max(B) → Max(A) is given by f ∗ (m) = f −1 (m). The next lemma is left as an exercise. Lemma 9.3.14. Let f : A → B be a homomorphism of commutative rings. Then f ∗ : Spec(B) → Spec(A) is continuous. Moreover, if g : B → C is another homomorphism of commutative rings, then f ∗ ◦ g ∗ = (g ◦ f )∗ . Thus, Spec is a contravariant functor from the category of commutative rings to the category of topological spaces. The same conclusions hold for the maps on Max induced by K-algebra homomorphisms between commutative algebras of finite type over a field K. Let a be an ideal of the commutative ring A and let π : A → A/a be the canonical map. Then π ∗ : Spec(A/a) → Spec(A) is a closed map, inducing a homeomorphism of Spec(A/a) onto the closed subset V (a) of Spec(A). Moreover, since π −1 (p) is maximal in A if and only if p is maximal in A/a, π ∗ restricts to a homeomorphism of Max(A/a) onto the closed subset V (a) of Max(A). The next lemma is obvious, but important.


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Lemma 9.3.15. Let a be the ideal generated by S in A, then V (S) = V (a) in Spec(A). Thus, the closed subspaces of Spec(A) are the homeomorphic images of the prime spectra of the quotient rings of A. Of course, the ideals in a Noetherian ring are all finitely generated. Corollary 9.3.16. Let A be a Noetherian ring. Then every closed subset of Spec(A) or Max(A) has the form V ({a1 , . . . , ak }) for some finite subset {a1 , . . . , ak } of A. Similarly, if K is an algebraically closed field, then every affine variety has the form V({f1 , . . . , fk }) for some finite subset {f1 , . . . , fk } of K[X1 , . . . , Xn ]. Examples 9.3.17. 1. Let A be a P.I.D. and let 0 = a ∈ A. Then V ({a}) is the set of equivalence classes of prime elements that divide a, and hence is finite. In fact, we can see that the proper closed subsets of Spec(A) are precisely the finite subsets of Max(A). p ), then 2. If A is a P.I.D. with only one nonzero prime ideal (e.g., A = Z(p) or A = Z Spec(A) is a space consisting of two points, one of which is open and the other not. This space is sometimes called Sierpinski space. 3. Let K be algebraically closed. Then in affine n-space K n , we have V({Xn }) = {(a1 , . . . , an ) ∈ K n | an = 0}, the image of the standard inclusion of K n−1 in K n . In particular, not all Zariski closed sets are finite. Remarks 9.3.18. There is an important interplay between the study of affine varieties over the complex numbers and differential topology. Given a finite collection f1 , . . . , fk of C ∞ functions from Rm to R, let f : Rm → Rk be given by f (x) = (f1 (x), . . . , fk (x)) for x ∈ Rn . Then the set of common zeros of f1 , . . . , fk is precisely f −1 (0). Using the Implicit Function Theorem of advanced calculus, one can show that f −1 (0) is a smooth (i.e., C ∞ ) submanifold of Rm , provided that the Jacobian matrix of f has rank k at every point of f −1 (0). This construction is generic for smooth manifolds: Every smooth manifold is diffeomorphic (i.e., equivalent as a smooth manifold) to a manifold obtained by the above procedure. Now, a complex affine variety is the set of common zeros of a finite collection of complex polynomials f1 , . . . , fk ∈ C[X1 , . . . , Xn ], and hence is the zero set of the function f : Cn → Ck whose i-th component function is fi for i = 1, . . . , n. Regarding f as a function from R2n to R2k , we can compute the Jacobian matrix of f . If it has rank 2k at each point of V(f1 , . . . , fk ), then V(f1 , . . . , fk ) is a smooth manifold. (Here, we use the subspace topology from R2n , of course, and not the Zariski topology.) In this manner, complex varieties have been an important source of examples of manifolds. Such examples have been useful in a number of contexts, especially in real dimension 4, where some of the standard techniques of smooth handle theory fail. Complex varieties that are not manifolds are also interesting topologically as a sort of generalized manifolds. From the other side, topological techniques provide invariants for complex varieties. These invariants are generally weaker than their structure as algebraic varieties, but can be useful nevertheless. Returning to the study of algebraic varieties, let K be an algebraically closed field and let a be an ideal in K[X1 , . . . , Xn ]. By Lemma 9.3.14, we may identify the affine


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variety V(a) with Max(K[X1 , . . . , Xn ]/a), the maximal ideal spectrum of a commutative K-algebra of finite type. Since Max is a functor on such algebras, we see that the structure of the affine variety V(a) depends only on the K-algebra K[X1 , . . . , Xn ]/a. Indeed, if A is a commutative K-algebra of finite type, we can think of Max(A) as an abstract affine variety: Any choice of K-algebra generators, a1 , . . . , an , of A induces a surjective K-algebra homomorphism εa1 ,...,an : K[X1 , . . . , Xn ] → A, and hence induces a homeomorphism ε∗a1 ,...,an : Max(A) → V (ker εa1 ,...,an ) ∼ = V(ker εa1 ,...,an ). Abstract varieties may be compared by means of algebraic maps. Definitions 9.3.19. Let K be an algebraically closed field and let A and B be commutative K-algebras of finite type. An algebraic map Max(B) → Max(A) is a function of the form f ∗ for some K-algebra homomorphism f : A → B. Here, the map depends only on its effect as a function Max(B) → Max(A), and not on the homomorphism inducing it. We need to take a slightly different approach in order to define mappings between affine varieties. The point is that while there’s a standard identification of V(a) ⊂ K n with Max(K[X1 , . . . , Xn ]/a), the variety V(a) does not determine the ideal a: There exist proper inclusions a ⊂ b of ideals such that V(b) = V(a). To resolve this, we shall develop the notions of polynomial mappings and coordinate rings. Definition 9.3.20. Let K be an algebraically closed field and let a be an ideal of K[X1 , . . . , Xn ]. Write X ⊂ K n for the affine variety determined by a: X = V(a) = {(a1 , . . . , an ) | f (a1 , . . . , an ) = 0

for all f ∈ a}.

We define the ideal of X, written IX , to be the set of all polynomials which vanish on every element of X: IX = {f ∈ K[X1 , . . . , Xn ] | f (a1 , . . . , an ) = 0

for all (a1 , . . . , an ) ∈ X}.

Then the coordinate ring, O(X), of X is defined by O(X) = K[X1 , . . . , Xn ]/IX . The coordinate ring catalogues the polynomial functions from X to K. Recall from Lemma 7.3.25 that the passage from polynomials f (X1 , . . . , Xn ) ∈ K[X1 , . . . , Xn ] to the induced functions f : K n → K gives a ring homomorphism φ : K[X1 , . . . , Xn ] → Map(K n , K), where Map(K n , K) is the ring of all functions from K n to K. Indeed, restricting the effect of these polynomials to the affine variety X ⊂ K n gives a ring homomorphism φX : K[X1 , . . . , Xn ] → Map(X, K). The ideal, IX , of X is easily seen to be the kernel of φX . Thus, we may identify O(X) with the image of φX , the ring of functions from X to K induced by polynomials. Thus, two polynomials, f and g, are equal in O(X) if and only if they agree as functions when restricted to X. By definition of X = V(a), we see that a ⊂ IX . Proposition 9.3.21. Let X = V(a) be an affine variety over the algebraically closed field K and let IX be the ideal of X. Then IX is the intersection of all the maximal ideals of K[X1 , . . . , Xn ] containing a. Thus, by Proposition 9.3.4, IX = rad(a).


358

Moreover, V(IX ) = V(a) = X, and the canonical map π : K[X1 , . . . , Xn ]/a → K[X1 , . . . , Xn ]/IX = O(X) induces a homeomorphism from Max(O(X)) to Max(K[X1 , . . . , Xn ]/a). Proof Recall that the element (a1 , . . . , an ) ∈ K n corresponds to the maximal ideal ma1 ,...,an ⊂ K[X1 , . . . , Xn ], which is the kernel of the evaluation map εa1 ,...,an : K[X1 , . . . , Xn ] → K. An element (a1 , . . . , an ) lies in X if and only if every element of a vanishes on (a1 , . . . , an ), meaning that a ⊂ ma1 ,...,an . In particular, since K is algebraically closed, the points of X correspond to the maximal ideals of K[X1 , . . . , Xn ] containing a. But an element of K[X1 , . . . , Xn ] lies in IX if and only if it vanishes on each point of X, and hence is contained in every maximal ideal of K[X1 , . . . , Xn ] containing a. Since a ⊂ IX , V(IX ) ⊂ V(a). But since every element of IX kills every element of X = V(a), the two varieties coincide. The result now follows from Lemma 9.3.14. In particular, an ideal b has the form IX for an affine variety X if and only if b = rad(a) for some ideal a. Since rad(rad(a)) = rad(a), this is equivalent to saying that b is its own radical (Lemma 9.1.13). Theorem 9.3.22. Let K be an algebraically closed field. There is a one-to-one correspondence between the affine varieties in K n and the ideals in K[X1 , . . . , Xn ] which are their own radicals. The correspondence takes a variety X to IX and takes an ideal a that is its own radical to V(a). Since each direction of this correspondence is order reversing, we see that X ⊂ Y if and only if IY ⊂ IX , and if either inclusion is proper, so is the other. Similarly, if each of a and b is its own radical, then a ⊂ b if and only if V(b) ⊂ V(a), and if either inclusion is proper, so is the other. Proof If a is its own radical, then IV(a) = rad(a) = a. Conversely, if X is a variety in K n , then X = V(IX ) by Proposition 9.3.21. It is easy to see (Corollary 9.1.16) that a finite intersection of prime ideals is its own radical. Corollary 9.3.23. Let K be an algebraically closed field and let a be a finite intersection of prime ideals of K[X1 , . . . , Xn ]. Then O(V(a)) = K[X1 , . . . , Xn ]/a. In particular, affine n-space K n is V(0). Since 0 is prime, we obtain a calculation of O(K n ). Corollary 9.3.24. O(K n ) = K[X1 , . . . , Xn ] for any algebraically closed field K. Thus, polynomials in n variables over K are determined by their effect as functions on K n . We could have given simpler argument using Proposition 7.3.26 and Lemma 8.4.3. Since K[X1 , . . . , Xn ] is Noetherian, the theory of primary decomposition shows (Corollary 9.2.11) that an ideal a of K[X1 , . . . , Xn ] is its own radical if and only if it is a finite intersection of prime ideals. Using this, we can show that every affine variety over an algebraically closed field has a unique decomposition as a union of irreducible varieties. Definition 9.3.25. An affine variety X over an algebraically closed field is irreducible if whenever we have affine varieties Y and Z with X = Y ∪ Z, either X = Y or X = Z. Recall from Corollary 7.6.10 that a prime ideal contains an intersection a ∩ b of ideals if and only if it contains at least one of a and b. We obtain the following lemma.


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Lemma 9.3.26. Let a and b be ideals in the commutative ring A. Then V (a ∩ b) = V (a) ∪ V (b) in either Spec(A) or Max(A). If a and b are ideals in K[X1 , . . . , Xn ] for K algebraically closed, then the affine varieties V(a ∩ b) and V(a) ∪ V(b) are equal. Proposition 9.3.27. An affine variety X over an algebraically closed field K is irreducible if and only if IX is prime. Proof Suppose that IX is prime and that X = Y ∪ Z. Clearly, IY∪Z = IY ∩ IZ . But prime ideals are irreducible ideals (Lemma 9.2.17), so either IX = IY or IX = IZ . By Theorem 9.3.22, either X = Y or X = Z, so X is irreducible. Conversely, supposethat X is irreducible. By Corollary 9.2.11 and Theorem 9.3.22, m X = V(a),where a = i=1 pi , where pi is prime for i = 1, . . . , m. Now Lemma 9.3.26 m gives X = i=1 V(pi ). Since X is irreducible, X = V(pi ) for some i, and hence IX = pi by Theorem 9.3.22. Corollary 9.2.22 shows that mevery self-radical ideal of K[X1 , . . . , Xn ] may bewritten uniquely as an intersection i=1 pi of prime ideals such that no pi contains j =i pj . Theorem 9.3.22 and Lemma 9.3.26 give the following proposition. Proposition 9.3.28. Every affine variety X over the algebraically closed field has a unique decomposition X=

m

Xi

i=1

where Xi is irreducible for i = 1, . . . , m, and no Xi is contained in uniqueness is up to a reordering of the Xi .

j =i

Xj . Here, the

We now define the mappings appropriate for comparing two affine varieties. Definition 9.3.29. Let K be an algebraically closed field and let X ⊂ K n be an affine variety. A polynomial function f : X → K m is a function of the form f (x) = (f1 (x), . . . , fm (x)) for all x ∈ X, where f1 , . . . , fm ∈ O(X). If the image of f is contained in an affine variety Y ⊂ K m , we may consider f to be a polynomial function from X to Y. The next lemma is quite elementary, but occurs frequently in the subsequent material. Lemma 9.3.30. Let A be a commutative ring and let g(X1 , . . . , Xm ) ∈ A[X1 , . . . , Xm ]. Suppose given polynomials f1 , . . . , fm ∈ A[X1 , . . . , Xn ]. Then evaluating each Xi at fi gives a polynomial g(f1 , . . . , fm ) ∈ A[X1 , . . . , Xn ]. The evaluation of this polynomial at an element x ∈ An is given by g(f1 , . . . , fm )(x) = g(f1 (x), . . . , fm (x)), where the right hand side is the evaluation of g at the element (f1 (x), . . . , fm (x)) ∈ Am . In other words, if εf1 ,...,fm : A[X1 , . . . , Xm ] → A[X1 , . . . , Xn ] is the A-algebra homomorphism obtained by evaluating Xi at fi for i = 1, . . . , m, then for g ∈ A[X1 , . . . , Xm ], the function from An to A induced by εf1 ,...,fm (g) is given by (εf1 ,...,fm (g))(x) = (g ◦ f )(x), where f : An → Am is the function given by f (x) = (f1 (x), . . . , fm (x)) for all x ∈ An .


360

Proof If f : B → C is a homomorphism of commutative A-algebras and b1 , . . . , bm ∈ B, then the composite εb

,...,bm

f

1 A[X1 , . . . , Xm ] −−− −−−→ B − →C

is just the evaluation map εf (b1 ),...,f (bm ) . Taking f to be the map from A[X1 , . . . , Xn ] obtained from evaluation at x ∈ An , the result follows. A first application is to composites of polynomial functions. Corollary 9.3.31. Let K be an algebraically closed field and let X ⊂ K n and Y ⊂ K m be affine varieties. Then for any polynomial mapping f : X → Y, there is a polynomial mapping f : K n → K m making the following diagram commute. X ∩ Kn

f

/ Y ∩ f / m K

If Z ⊂ K k is another affine variety and if g : Y → Z is a polynomial mapping, then the composite g ◦ f is a polynomial mapping as well. Finally, let ξi be the image of Xi under the canonical map π : K[X1 , . . . , Xn ] → O(X) for i = 1, . . . , n. Then ξi : X → K is the restriction to X of the projection of K n onto its i-th factor. Thus, the polynomial mapping ι : X → K n given by ι(x) = (ξ1 (x), . . . , ξn (x)) is just the inclusion map of X in K n . The image of ι is, of course X, so the identity map of X is a polynomial mapping. Proof Since f is a polynomial mapping, there are elements f1 , . . . , fm ∈ O(X) such that f (x) = (f1 (x), . . . , fm (x)) for all x ∈ X. But O(X) = K[X1 , . . . , Xn ]/IX , so there are polynomials fi ∈ K[X1 , . . . , Xn ] representing fi for i = 1, . . . , n. Setting f(x) = (f1 (x), . . . , fm (x)) for all x ∈ K n , we obtain a polynomial mapping f : K n → K m making the diagram commute. Similarly, there are polynomials gi ∈ K[X1 , . . . , Xm ] for i = 1, . . . , k such that the polynomial mapping g : K m → K k given by g(y) = ( g1 (y), . . . , gk (y)) for y ∈ K m restricts on Y to g : Y → Z. The component functions of g ◦ f are the maps gi ◦ f, which satisfy ( gi ◦ f)(x) = gi (f1 (x), . . . , fm (x)) for all x ∈ K n . By Lemma 9.3.30, this is the map induced by the polynomial gi (f1 , . . . , fm ) ∈ K[X1 , . . . , Xn ]. Thus, the composite g ◦ f : K n → K k is a polynomial mapping which clearly restricts on X to the composite function (g ◦ f ) : X → Z. Since the restriction to X of any element of K[X1 , . . . , Xn ] determines an element of O(X), the the polynomial mappings are closed under composites. Let ξi be the image in O(X) of the polynomial Xi . Then Xi : K n → K restricts to ξi on X. But Xi (a1 , . . . , an ) = ai , and the result follows.


361

In particular, Corollary 9.3.31 shows that the affine varieties and polynomial mappings form a category. Definition 9.3.32. Let K be an algebraically closed field. We write VarK for the category whose objects and are the affine varieties over K and whose morphisms are given as follows. If X and Y are affine varieties over K, then the set of morphisms, MorVarK (X, Y), from X to Y in VarK is the set of polynomial mappings from X to Y. We now show that polynomial mappings induce homomorphisms of coordinate rings. Proposition 9.3.33. Let X ⊂ K n and Y ⊂ K m be affine varieties over the algebraically closed field K and let f : X → Y be a polynomial mapping, given by f (x) = (f1 (x), . . . , fm (x)) for x ∈ X, where fi ∈ O(X) for i = 1, . . . , m. Then there is a unique K-algebra homomorphism O(f ) : O(Y) → O(X) with the property that O(f )(ξi ) = fi for i = 1, . . . , m, where ξi is the image of Xi under the canonical map π : K[X1 , . . . , Xm ] → O(Y). If g : Y → Z is another polynomial mapping, then O(f ) ◦ O(g) = O(g ◦ f ). Also, O(1X ) = 1O(X) . Thus, O is a contravariant functor from the category, VarK , of affine varieties and polynomial mappings over K to the category of commutative K-algebras of finite type and K-algebra homomorphisms. Proof Since ξ1 , . . . , ξm generate O(Y) as a K-algebra, there is at most one K-algebra homomorphism carrying ξi to fi for all i. We must show that such a homomorphism exists. For each i, let fi be a lift of fi under the canonical map π : K[X1 , . . . , Xn ] → O(X). Then there is a K-algebra homomorphism εf1 ,...,fm : K[X1 , . . . , Xm ] → K[X1 , . . . , Xn ] obtained by evaluating Xi at fi for i = 1, . . . , m. By the universal property of quotient rings, it suffices to show that εf1 ,...,fm (IY ) ⊂ IX : We may then define O(f ) to be the unique homomorphism making the following diagram commute. εf1 ,...,fm / K[X1 , . . . , Xn ] K[X1 , . . . , Xm ] π O(Y)

π O(f )

/ O(X)

Thus, let h(X1 , . . . , Xm ) ∈ IY . We wish to show that εf1 ,...,fm (h) ∈ IX , meaning that it vanishes on each element of X. By Lemma 9.3.30, ε (h) = h ◦ f as a f1 ,...,fm

function on K n , where f(x) = (f1 (x), . . . , fm (x)) for x ∈ K n . Since fi is a lift of fi to K[X1 , . . . , Xn ], fi (x) = fi (x) for all x ∈ X. Thus, as a function on X, f = f . Since f (X) ⊂ Y and since h vanishes on Y, εf1 ,...,fm (h) ∈ IX , as desired. If g : Y → Z ⊂ K k is a polynomial map, write g(y) = (g1 (y), . . . , gk (y)) for y ∈ Y, where gi ∈ O(Y) for all i. Let gi be a lift of gi to K[X1 , . . . , Xm ] for i = 1, . . . , k. Then the following diagram commutes. ε εg1 ,...,gk / K[X1 , . . . , Xm ] f1 ,...,fm/ K[X1 , . . . , Xn ] K[X1 , . . . , Xk ] π O(Z)

π O(g)

/ O(Y)

π O(f )

/ O(X)


362

Thus, O(f ) ◦ O(g)(ξi ) is the image in O(X) of εf1 ,...,fm ( gi ) = gi (f1 , . . . , fm ). Now, an element of O(X) is determined by its value as a function on X. But for x ∈ X, we have gi (f1 (x), . . . , fm (x))

= gi (f (x)) = gi (f (x)),

since f (x) ∈ Y. Since they have the same effect on elements of X, O(f ) ◦ O(g)(ξi ) must be equal to the i-th component function of g ◦ f . Thus, the uniqueness statement used to define O(g ◦ f ) gives O(f ) ◦ O(g) = O(g ◦ f ). Finally, Corollary 9.3.31 shows that the identity map of X is given by 1X (x) = (ξ1 (x), . . . , ξn (x)) for all x ∈ X, so O(1X ) is characterized by O(1X )(ξi ) = ξi for i = 1, . . . , n. This forces O(1X ) = 1O(X) . Proposition 9.3.34. Let X ⊂ K n and Y ⊂ K m be affine varieties over the algebraically closed field K. Then the passage from f : X → Y to O(f ) : O(Y) → O(X) gives a oneto-one correspondence between the polynomial mappings from X to Y and the K-algebra homomorphisms from O(Y) to O(X). Proof By Proposition 9.3.33, it suffices to show that every K-algebra homomorphism γ : O(Y) → O(X) is equal to O(f ) for some polynomial mapping f : X → Y. Thus, let γ : O(Y) → O(X) be a K-algebra homomorphism, and let fi = γ(ξi ) ∈ O(X) for i = 1, . . . , m. (As above, ξi is the image of Xi under the canonical map π : K[X1 , . . . , Xm ] → O(Y).) Let f : X → K m be the polynomial mapping given by f (x) = (f1 (x), . . . , fm (x)) for x ∈ X. By the characterization of the maps O(f ) given in Proposition 9.3.33, it suffices to show that f (X) ⊂ Y. Now, Y = V(IY ) is the set of elements of K m which vanish under every element of the ideal of Y, so it suffices to show that if g ∈ IY and x ∈ X, then g vanishes on f (x). Let fi be a lift of fi to K[X1 , . . . , Xn ] for i = 1, . . . , m and let f : K n → K m be the polynomial mapping whose i-th component function is fi for i = 1, . . . , m. Then f(x) = f (x) for all x ∈ X, so it suffices to show that the polynomial inducing g ◦ f lies in IX . Lemma 9.3.30 shows that g ◦ f is induced by εf1 ,...,fm (g). Note that the fi were chosen to make the following diagram commute. εf1 ,...,fm / K[X1 , . . . , Xn ] K[X1 , . . . , Xm ] π O(Y)

π γ

/ O(X)

The kernels of the vertical maps π are precisely IY and IX . Since g ∈ IY , its image under εf1 ,...,fm must lie in IX . Thus, the relationships between affine varieties given by polynomial mappings and the relationships between their corrdinate rings given by K-algebra homomorphisms are precise mirrors of one another.


363

Corollary 9.3.35. Let X and Y be affine varieties over the algebraically closed field K. Then there is a polynomial isomorphism between X and Y (i.e., a bijective polynomial mapping f : X → Y whose inverse function is polynomial) if and only if the coordinate rings O(X) and O(Y) are isomorphic as K-algebras. Proof If f : X → Y is a polynomial isomorphism, then O(f ) is an isomorphism by the functoriality of the coordinate ring: If f −1 : Y → X is the inverse function of f , then O(f ) ◦ O(f −1 ) = O(1X ) = 1O(X) and O(f −1 ) ◦ O(f ) = O(1Y ) = 1O(Y) . −1 If O(f ) is an isomorphism, then Proposition 9.3.34 shows that O(f ) = O(g) for some polynomial mapping g : Y → X. But then O(f ◦ g) = 1O(Y) , so Proposition 9.3.34 shows that f ◦ g = 1Y . Similarly, g ◦ f = 1X . We next show that the polynomial functions between affine varieties may be identified with the algebraic maps between the maximal ideal spectra of their coordinate rings. Proposition 9.3.36. Let X ⊂ K n and Y ⊂ K m be affine varieties over the algebraically closed field K. Let f : X → Y be a polynomial mapping. Then there is a commutative diagram ∗

Max(O(X))

O(f )

∼ =

/ Max(O(Y)) ∼ =

X

/ Y

f

where the vertical maps are the standard isomorphisms. Thus, the algebraic maps from Max(O(X)) to Max(O(Y)) correspond precisely under the standard isomorphisms to the polynomial mappings from X to Y. Proof The standard isomorphism from Max(O(X)) to X is the composite π∗

ν −1

∼ =

∼ =

Max(O(X)) −→ V (IX ) −−→ X, where π : K[X1 , . . . , Xn ] → O(X) is the canonical map. The analogous result holds for Y. Let f : K n → K m be a polynomial mapping which restricts to f on X. Then the top and bottom squares of the following diagram commute. ∗

Max(O(X))

O(f )

π∗

/ Max(O(Y)) π∗

O(f)∗ / Max(K[X1 , . . . , Xm ]) Max(K[X1 , . . . , Xn ]) O O ν ∼ = KO n

ν ∼ = f

∪ X

/ Km O ∪

f

/ Y


364

It suffices to show that the middle square commutes. For x = (a1 , . . . , an ) ∈ K n , ν(x) is the kernel of the evaluation map εa1 ,...,an , and hence O(f)∗ (ν(x)) is the kernel of εa1 ,...,an ◦ O(f). Let fi be the i-th component function of f. Then O(f) is precisely the evaluation map εf1 ,...,fm : K[X1 , . . . , Xm ] → K[X1 , . . . , Xn ]. Thus, εa1 ,...,an ◦ O(f) is the evaluation map from K[X1 , . . . , Xm ] → K which evaluates Xi at fi (a1 , . . . , an ). The kernel of this evaluation map is precisely ν(f(x)), so the middle square commutes. Corollary 9.3.37. Polynomial mappings are continuous in the Zariski topology. The polynomial mappings give a much more rigid notion of equivalence of varieties than is provided by the Zariski topology alone. Indeed, it is quite possible for varieties to be homeomorphic as spaces but inequivalent as varieties. Consider the following example. Example 9.3.38. Let K be an algebraically closed field. Let κ : K[X, Y ] → K[X] be the K-algebra homomorphism taking X to X 2 and Y to X 3 , and let p = ker κ. Since im κ is a domain, p is prime, so O(V(p)) = K[X, Y ]/p by Corollary 9.3.23. (We leave it as an exercise to show that p = (Y 2 − X 3 ).) Write κ : O(V(p)) = K[X, Y ]/p → K[X] = O(K) for the homomorphism induced by κ. Then Proposition 9.3.34 shows that κ = O(f ), where f : K → V(p) is given by f (x) = (x2 , x3 ) for all x ∈ K. By inspection, f : K → V(p) is injective. We claim it is a homeomorphism. But the coordinate ring of V(p) is isomorphic to im κ = K[X 2 , X 3 ], which cannot be generated as a K-algebra by a single element. Thus, Corollary 9.3.35 shows that V(p) is not isomorphic to K as a variety (i.e., through polynomial mappings). The coordinate ring O(K) = K[X] is a P.I.D. Thus, as shown in Example 1 of Examples 9.3.17, the proper closed subsets of K are precisely the finite subsets of K. Since affine varieties are T1 spaces, f (S) is closed in V(p) for every proper closed subspace S of K. Thus, to show that f is a homeomorphism, it suffices to show it is onto. Thus, by Proposition 9.3.36, it suffices to show that i∗ : Max(K[X]) → Max(K[X 2 , X 3 ]) is onto, where i : K[X 2 , X 3 ] ⊂ K[X] is the inclusion. For this, we shall make use of the theory of integral dependence developed in Section 12.7. Note first that since K[X] is generated by 1, X as a K[X 2 , X 3 ]-module, it is integral over K[X 2 , X 3 ] by Proposition 12.7.2. Now Proposition 12.7.22 shows that for each maximal ideal m of K[X 2 , X 3 ] there is a prime ideal p of K[X] with m = p ∩ K[X 2 , X 3 ]. But every nonzero prime ideal of K[X] is maximal, so m lies in the image of i∗ : Max(K[X]) → Max(K[X 2 , X 3 ]). Exercises 9.3.39. 1. Give the proof of Lemma 9.3.14. 2. Let A be a commutative ring. Show that A[X 2 , X 3 ] is free as an A[X 2 ]-module, with basis 1, X 3 . Deduce that the kernel of the evaluation map εX 2 ,X 3 : A[X, Y ] → A[X 2 , X 3 ] is the principal ideal generated by Y 2 − X 3 . 3. Let K be a field and fix n > 0. Suppose the map ν : K n → Max(K[X1 , . . . , Xn ]) is onto. Show that K is algebraically closed. 4. Let K be any field. Show that a maximal ideal m lies in the image of ν : K n → Max(K[X1 , . . . , Xn ]) if and only if there is an element (a1 , . . . , an ) ∈ K n such that f (a1 , . . . , an ) = 0 for all f ∈ m.


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5. Show that we may identify Max(R[X]) with the upper half plane in C. What is the map Max(C[X]) → Max(R[X]) induced by the natural inclusion? 6. What is Spec(A × B)? 7. Let A be a commutative ring. Show that the topology on Spec(A) is T0 , i.e., that for any pair of points p, q, at least one of the two points is open in the subspace topology on {p, q}. 8. Let A be a commutative ring. Show that the sets U ({a}) for a ∈ A form a basis for the Zariski topology on Spec(A). It is customary to write X = Spec(A) and Xa = U ({a}). 9. Let A be a commutative ring. Show that Spec(A) is compact in the sense that any open cover of Spec(A) has a finite subcover. (Many writers call a non-Hausdorff space with this property quasi-compact.)

9.4

Tensor Products

The theory of tensor products is one of the most important tools in module theory. We have already seen it used implicitly in more than one argument. One especially important application, which we shall emphasize here, is extension of rings, whereby, if M is an A-module and f : A → B is a ring homomorphism, we create an associated B-module from the tensor product B ⊗A M . This construction is functorial, and has important applications to exactness arguments. Tensor products are a good example of an important concept that really cannot be adequately understood except via universal mapping properties. We shall use some somewhat nonstandard terminology in the next definition, as there are two definitions of bilinearity common in the literature, and we wish to distinguish between them. Definitions 9.4.1. Let A be a ring and suppose given a right A-module M and a left A-module N . Let G be an abelian group. A function f : M × N → G is said to be weakly A-bilinear if f (m + m , n)

= f (m, n) + f (m , n)

f (m, n + n ) = f (m, n) + f (m, n ) f (ma, n) = f (m, an)

and

for all choices of m, m ∈ M , n, n ∈ N , and a ∈ A. If A is commutative, then we need not distinguish between left and right modules. Let M1 , M2 , and N be A-modules. We say that f : M1 × M2 → N is A-bilinear if for each m1 ∈ M1 and each m2 ∈ M2 , the maps f (m1 , −) : M2 → N

and

f (−, m2 ) : M1 → N

are A-module homomorphisms. If A is commutative, then an A-bilinear map is easily seen to be weakly A-bilinear. Note also that for A-arbitrary, any weakly A-bilinear map is Z-bilinear. We shall give a formal construction of the tensor product, but its purpose is that it is the universal group for bilinear functions. Thus, we shall have an abelian group


366

M ⊗A N and a weakly A-bilinear map ι : M × N → M ⊗A N with the following universal property: For each abelian group G and each weakly A-bilinear map f : M × N → G, there is a unique group homomorphism f : M ⊗A N → G such that the following diagram commutes. f / G M ×N : JJ tt JJ t t J t tt ι JJJ$ tt f M ⊗A N Surprisingly, it turns out that if A is commutative, then M ⊗A N is an A-module that is universal in exactly the same manner for A-bilinear mappings. Thus, the two different notions of bilinearity are “solved” by exactly the same construction. Indeed, almost everything we know about tensor products will be derived from this universal property, rather than the actual construction of the tensor product. Probably the most important fact coming out of the construction itself is that M ⊗A N is generated as an abelian group by the image of ι. We now construct the tensor product. First, let F (M, N ) be the free abelian group (i.e., free Z-module) on the set M × N . Thus, using the notation of Section 7.7, , F (M, N ) = Z. (m,n)∈M ×N

Continuing this notation, we write e(m,n) for the canonical basis element corresponding to the (m, n)-th summand. We then have a function i : M × N → F (M, N ) with i(m, n) = e(m,n) . Now let H ⊂ F (M, N ) be the subgroup generated by {e(m+m ,n) − e(m,n) − e(m ,n) | m, m ∈ M, n ∈ N } ∪{e(m,n+n ) − e(m,n) − e(m,n ) | m ∈ M, n, n ∈ N } ∪{e(ma,n) − e(m,an) | m ∈ M, n ∈ N, a ∈ A} Definition 9.4.2. With the conventions above, we set the tensor product M ⊗A N i equal to F (M, N )/H, and set ι : M × N → M ⊗A N equal to the composite M × N − → π → M ⊗A N , where π is the canonical map onto F (M, N )/H. F (M, N ) − We write m ⊗ n for the element ι(m, n) ∈ M ⊗A N . The map ι : M × N → M ⊗A N is easily seen to be weakly A-bilinear. That is important in showing that the following proposition gives universality for weakly bilinear maps. Proposition 9.4.3. Suppose given a right A-module M , a left A-module N , and a weakly A-bilinear map f : M ×N → G. Then there is a unique group homomorphism f : M ⊗A N such that the following diagram commutes. f / G M ×N : JJ tt JJ t t J t tt ι JJJ$ tt f M ⊗A N


367

Proof The function f : M × N → G is defined on the generators of the free abelian group F (M, N ). By the universal property of free abelian groups (Lemma 7.7.32), there is a unique group homomorphism f : F (M, N ) → G such that f(e(m,n) ) = f (m, n) for all (m, n) ∈ M × N . Since f is weakly A-bilinear, f is easily seen to vanish on the generators of H ⊂ F (M, N ). So the Noether Isomorphism Theorem, which gives the universal property of factor groups, shows that there is a unique group homomorphism f : M ⊗A N → G with f ◦ π = f. Here, π : F (M, N ) → M ⊗A N is the canonical map. The composite of the uniqueness of f with respect to f and the uniqueness of f with respect to f gives the desired result. Warning: Not every element of M ⊗A N has the form m ⊗ n. The generic element has the form m1 ⊗ n1 + · · · + mk ⊗ nk for k ≥ 0 (the empty sum being 0), with mi ∈ M and ni ∈ N for 1 ≤ i ≤ k. Remarks 9.4.4. In practice, in defining homomorphisms out of tensor products, one often neglects to explicitly write down a weakly A-bilinear function from M × N to G, but does so implicitly, by a statement such as “setting f (m ⊗ n) = expression defines a homomorphism f : M ⊗A N → G.” Of course, f ◦ ι is the desired weakly A-bilinear function, and is specified in a statement such as the above, since m ⊗ n = ι(m, n). It is left to the reader to verify that the expression that is supplied in such a statement does in fact define a weakly A-bilinear function. We shall follow this convention of implicitly defined bilinear functions from now on, other than in explicit statements about universal properties of the tensor product. We reiterate, as pointed out in the above warning, that specifying f on terms of the form m⊗n only gives the value of f on a generating set for M ⊗A N , and not on every element. The following fundamental observation is immediate from the fact that 0 + 0 = 0 in either M or N . We shall use it without explicit mention. Lemma 9.4.5. Let M and N be right and left A-modules, respectively. Then for m ∈ M and n ∈ N , the elements m ⊗ 0 and 0 ⊗ n are 0 in M ⊗A N . The next lemma says that the tensor product is a functor of two variables. Lemma 9.4.6. Let f : M → M and g : N → N be left and right A-module homomorphisms, respectively. Then there is a homomorphism, which we shall denote by f ⊗ g : M ⊗A N → M ⊗A N whose effect on generators is given by (f ⊗ g)(m ⊗ n) = f (m) ⊗ g(n). If f : M → M and g : N → N are left and right A-module homomorphisms, respectively, then (f ⊗ g ) ◦ (f ⊗ g) = (f ◦ f ) ⊗ (g ◦ g). Finally, 1M ⊗ 1N is the identity map of M ⊗ N . Proof That setting (f ⊗ g)(m ⊗ n) = f (m) ⊗ g(n) gives rise to a well defined homomorphism follows precisely as in Remarks 9.4.4. The rest follows since any two homomorphisms that agree on the generators of a group must be equal. The functoriality of tensor products is all that we need to establish their universality for A-bilinear mappings when A is commutative.


368

Proposition 9.4.7. Let A be a commutative ring and let M and N be A-modules. Then there is a natural A-module structure on M ⊗A N which is defined on generators by a · (m ⊗ n) = (am) ⊗ n. Under this convention, ι : M × N is A-bilinear, and is universal with respect to that property: For any A-bilinear map f : M × N → N , there is a unique A-module homomorphism f : M ⊗A N → N such that the following diagram commutes. f / N M ×N : JJ tt JJ t t JJ t tt ι JJ$ tt f M ⊗A N Proof For a ∈ A, write μa : M → M for the map induced by multiplication by a. Because A is commutative, μa is an A-module homomorphism for each a ∈ A. Thus, μa ⊗ 1N : M ⊗A N → M ⊗A N is a group homomorphism. We use this to define multiplication by a on M ⊗A N . That this gives an A-module structure on M ⊗A N may be deduced from the basic properties of tensor products and of A-module structures, together with the fact that (μa ⊗ 1N ) ◦ (μb ⊗ 1N ) = (μa ◦ μb ) ⊗ 1N = μab ⊗ 1N . We leave the rest as an exercise for the reader. There is a very useful generalization of Proposition 9.4.7 in which M has module structures coming from two different (and possibly non-commutative) rings. It will give us a way to turn A-modules into B-modules, induced by any ring homomorphism from A to B. Definition 9.4.8. Let A and B be rings. We say that M is B-A-bimodule if the following hold. 1. M is a left B-module. 2. M is a right A-module. 3. These two module structures interact as follows. For m ∈ M , b ∈ B, and a ∈ A, we have b(ma) = (bm)a. Examples 9.4.9. 1. A itself is an A-A-bimodule via ordinary multiplication. Its A-A-submodules are its two-sided ideals. 2. If f : A → B is a ring homomorphism, then any B-B-bimodule may be pulled back to an A-B, a B-A, or an A-A-bimodule, via f . 3. If A is commutative, then any A-module M is an A-A-bimodule, where both actions are just the usual action on M . 4. For any ring A, identify An with the space of 1 × n row matrices. Then An is a right module over Mn (A) by setting x · M equal to the matrix product xM for x ∈ An and M ∈ Mn (A). This gives An the structure of an A-Mn (A)-bimodule.


369

If M is a B-A-bimodule and N is a left A-module, we shall put a left B-module structure on M ⊗A N which will be universal for weakly A-bilinear maps with the following additional property. Definition 9.4.10. Let M be a left B-module and let N be any set. We say that a function f : M × N → N is B-linear in M if N is a left B-module and for each n ∈ N the function f (−, n) : M → N is a B-module homomorphism. Let M be a B-A-bimodule and let b ∈ B. Then multiplication by b, μb : M → M is easily seen to be a homomorphism of right A-modules. Thus, the proof of the next proposition proceeds along the lines of that of Proposition 9.4.7. Proposition 9.4.11. Let M be a B-A-bimodule and let N be a left A-module. Then there is a natural B-module structure on M ⊗A N which is defined on generators by b · (m ⊗ n) = (bm) ⊗ n. Under this convention, ι : M × N is B-bilinear in M and satisfies the following universal property: if N is a left B-module and if f : M × N → N is weakly A-bilinear as well as being B-linear in M , then there is a unique B-module homomorphism f : M ⊗A N → N such that the following diagram commutes. f / N M ×N : JJ tt JJ t t JJ t tt ι JJ$ tt f M ⊗A N

Similarly, if M is a right A-module and N is an A-B-bimodule, then the tensor product M ⊗A N inherits a right B-module structure and satisfies a similar universal property. An important application of the role of bimodule structures in tensor products comes from the B-A-bimodule structure on B that arises from a ring homomorphism f : A → B. Given a left A-module M , we obtain a left B-module B ⊗A M . The passage from M to B ⊗A M is functorial, and is known as “extension of rings” or “base change,” among other names. It satisfies an important universal property: Proposition 9.4.12. Let f : A → B be a ring homomorphism. Let M be a left Amodule and let N be a left B-module. Regarding N as an A-module via f , let g : M → N be an A-module homomorphism. Then there is a unique B-module homomorphism g : B ⊗A M → N such that the following diagram commutes. g / N M J t: JJ t J tt αM JJ$ tt g B ⊗A M Here, αM : M → B ⊗A M is the A-module homomorphism defined by αM (m) = 1 ⊗ m for m ∈ M . Explicitly, g is defined on generators by setting g(b ⊗ m) = bg(m).


370

Proof Setting g(b⊗m) = bg(m) is easily seen to induce a left B-module homomorphism from B ⊗A M to N . Uniqueness follows from the fact that 1 ⊗ m = αM (m), and hence the commutativity of the diagram requires that g(1 ⊗ m) = g(m). Recall that if M and N are A-modules, then HomA (M, N ) is the group (under addition of homomorphisms) of A-module homomorphisms. The following may be expressed in category theoretic language by saying that passage from M to B ⊗A M is a left adjoint to the forgetful functor from B-modules to A-modules induced by f : A → B. Corollary 9.4.13. Let f : A → B be a ring homomorphism. Let M be a left A-module and N a left B-module. Let αM : M → B ⊗A M be the A-module homomorphism defined by αM (m) = 1 ⊗ m for all m ∈ M . Then there is an isomorphism of abelian groups α∗

→ HomA (M, N ) HomB (B ⊗A M , N ) −−M ∼ =

∗ defined by αM (h) = h ◦ αM . Additionally, the isomorphism is natural in the sense that if g : M → M is an A-module homomorphism and h : N → N is a B-module homomorphism, then the following diagram commutes.

HomB (B ⊗A M , N )

∗ αM

/ HomA (M, N )

γ ∗ αM / HomA (M , N ) HomB (B ⊗A M , N ) β

Here, β and γ are defined by β(k) = h ◦ k ◦ (1B ⊗ g)

and

γ(l) = h ◦ l ◦ g

for k ∈ HomB (B ⊗A M , N ) and l ∈ HomA (M, N ). ∗ is well defined, and is clearly Proof Since αM is an A-module homomorphism, αM a homomorphism between the Hom groups. Let h ∈ HomB (B ⊗A M , N ). Then the following diagram commutes. ∗ αM (h) / N M J t: JJ t J tt αM JJ$ tt h B ⊗A M ∗ So αM is an isomorphism by Proposition 9.4.12. The reader may verify the naturality statement.

The simplest type of base change is the trivial one: B = A and f is the identity map. Here, Corollary 9.4.13 gives us an isomorphism α∗

HomA (A ⊗A M , N ) −−M → HomA (M, N ) ∼ =

∼ =

for any pair of A-modules M and N . For formal reasons, this implies that αM : M −→ A ⊗A M for all A-modules M . We can also prove it directly, with a direct construction of the inverse homomorphism:


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Lemma 9.4.14. Let M be a left A-module and let αM : M → A ⊗A M be the A-module homomorphism given by αM (m) = 1 ⊗ m for all m ∈ M . Then αM is an isomorphism whose inverse, εM : A ⊗A M → M , is defined on generators by εM (a ⊗ m) = am. Additionally, the isomorphisms εM and their inverses are natural in the sense that if f : M → M is an A-module homomorphism, then the following diagram commutes. εM / A ⊗A M M 1A ⊗ f f εM / A ⊗A M M Proof It is easy to see that εM is well defined and that εM ◦ αM = 1M . So εM and αM are inverse isomorphisms, providing that αM is onto. But each generator a ⊗ m is equal to αM (am), so the image of αM contains a generating set for A ⊗A M . The rest is left to the reader. We’ve already seen a couple of examples of extension of rings without knowing it. Corollary 9.4.15. Let S be a multiplicative subset of the commutative ring A. Then for any A-module M there is an isomorphism of S −1 A-modules ϕM : S −1 A ⊗A M → S −1 M whose effect on generators is given by ϕM ((a/s) ⊗ m) = am/s for a ∈ A, s ∈ S, and m ∈ M. This isomorphism is natural in the sense that if f : M → N is an A-module homomorphism, then the following diagram commutes, S −1 A ⊗A M 1⊗f S −1 A ⊗A N

ϕM / S −1 M S −1 (f ) ϕN / −1 S N

where 1 is the identity of S −1 A. Proof The point is that Lemma 7.11.14 shows that the canonical map η : M → S −1 M satisfies exactly the same universal property in terms of A-module maps into S −1 Amodules that Proposition 9.4.12 shows for the mapping αM : M → S −1 A ⊗A M . In particular, the universal property for η shows that there exists a unique S −1 A-module homomorphism ψ : S −1 M → S −1 A ⊗A M such that the following diagram commutes. αM / S −1 A ⊗ M M OO A OOO 7 OOO ooo o o O' η oo ψ S −1 M And ϕM : S −1 A ⊗A M → S −1 M , if placed in the diagram in the opposite direction from that of ψ, would make it commute as well.


372

Thus, the composite S −1 A-module homomorphisms ϕM ◦ ψ and ψ ◦ ϕM restrict to the identity maps on the images of αM and η, respectively. Since these images generate S −1 A ⊗A M and S −1 M , respectively, as modules over S −1 A, ϕM and ψ must be inverse isomorphisms. The naturality statement holds by an easy diagram chase. Similarly, Lemma 7.7.19 shows that if a is a two-sided ideal in the ring A, then the passage from a left A-module M to the A/a-module M/aM satisfies the same universal property as the extension of rings. We leave the proof of the following to the reader. Corollary 9.4.16. Let a be a two-sided ideal in the ring A. Then for any left A-module M , there is an isomorphism of A/a-modules βM : A/a ⊗A M → M/aM whose effect on generators is given by βM (a ⊗ m) = am for a ∈ A and m ∈ M . This isomorphism is natural in the sense that if f : M → N is an A-module homomorphism, then the following diagram commutes, A/a ⊗A M 1⊗f

A/a ⊗A N

βM / M/aM f βN / N/aN

where 1 is the identity map of A/a, and f is the map induced by f . We may now reinterpret Nakayama’s Lemma in terms of extension of rings. Corollary 9.4.17. (Nakayama’s Lemma, second form) Let a be a two-sided ideal which is contained in the Jacobson radical of the ring A and let M be a finitely generated A-module. Then A/a ⊗A M = 0 if and only if M = 0. Proof M/aM = 0 if and only if M = aM . Now apply Nakayama’s Lemma. We’d like to know that extension of rings takes free modules to free modules. For this purpose, it suffices to understand what tensoring does to a direct sum. *k Proposition 9.4.18. Let N1 , . . . , Nk be left A-modules and let ιi : Ni → i=1 Ni be the canonical inclusion for i = 1, . . . , k. Then for any right A-module M , there is an isomorphism " k # k , , ω: (M ⊗A Ni ) −→ M ⊗A Ni i=1

i=1

whose restriction to M ⊗A Ni is 1M ⊗ ιi for i = 1, . . . , k. If M is a B-A-bimodule, then ω is an isomorphism of left B-modules; if each Ni is an A-B-bimodule, then ω is an isomorphism of right B-modules. Similarly, given A-modules M* 1 , . . . , Mk and a left A-module N , there is an *right k k isomorphism ω : i=1 (Mi ⊗A N ) → ( i=1 Mi ) ⊗A N whose restriction to Mi ⊗A N is ιi ⊗ 1N for i = 1, . . . , k. Once again, appropriate bimodule structures give module isomorphisms. These isomorphisms are natural in all variables.


373

*k *k Proof Let β : M ⊗A ( i=1 Ni ) → i=1 (M ⊗A Ni ) be the homomorphism whose effect on generators is given by β(m ⊗ (n1 , . . . , nk )) = (m ⊗ n1 , . . . , m ⊗ nk ). Then the composites ω * ◦ β and β ◦ ω are *k easily seen to give the identity maps on genk erating sets for M ⊗A ( i=1 Ni ) and i=1 (M ⊗A Ni ), and hence ω and β are inverse isomorphisms. We leave the rest to the reader. Corollary 9.4.19. Let f : A → B be a ring homomorphism. Then B ⊗A An is a free B-module of rank n with basis 1 ⊗ e1 , . . . , 1 ⊗ en , where e1 , . . . , en is the canonical basis of An . Similarly, if we regard An as a right A-module, then An ⊗A B is a free right B-module with basis e1 ⊗ 1, . . . , en ⊗ 1. Proof We have B ⊗A An ∼ =B = (B ⊗A A)n . Now notice that the isomorphism B ⊗A A ∼ from Lemma 9.4.14 is one of left B-modules, and takes 1 ⊗ 1 to 1. The proof for right modules is identical. Recall that Mm,n (A) is the set of m×n matrices over A. Given a ring homomorphism f : A → B, we write f∗ : Mm,n (A) → Mm,n (B) for the function that takes a matrix (aij ) to the matrix whose ij-th coordinate is f (aij ). Recall from Section 7.10 that if g : An → Am is a homomorphism of right A-modules (or left ones, of course, if A is commutative), then g is induced by left-multiplying a column vector by the m × n matrix whose i-th column is g(ei ). Similarly, if g : An → Am is a left A-module map, then g is induced by rightmultiplying a row vector by the n × m matrix whose i-th row is g(ei ). In particular, there are two different conventions possible for the matrix of g if A is commutative. It is now an easy verification to show the following. Corollary 9.4.20. Let f : A → B be a ring homomorphism. Let g : An → Am be a homomorphism of right A-modules, represented by the matrix M . Then the matrix of g ⊗ 1B with respect to the bases coming from Corollary 9.4.19 is precisely f∗ (M ). Similarly, if the left A-module homomorphism g : An → Am is represented (under whichever convention one chooses to use in the case where both A and B are commutative) by the matrix M , then the matrix of 1B ⊗ g with respect to the bases of Corollary 9.4.19 is f∗ (M ). f

g

Suppose given ring homomorphisms A − → B − → C. We’d like to know that the extension of rings from A to B and then from B to C coincides with the extension in one step from A to C. The verification of this depends on an understanding of the associativity properties of iterated tensor product. In the study of cartesian products, the existence of a single n-fold product greatly simplifies the discussion of iterated products and their associativity properties. We shall make use of the same device in the study of iterated tensor products. Definitions 9.4.21. Let A1 , . . . , An−1 be rings. Suppose given a right A1 -module M1 , a left An−1 -module Mn , together with Ai−1 -Ai -bimodules, Mi for 2 ≤ i ≤ n − 1. Let G be an abelian group. Then a function f : M1 × · · · × Mn → G


374

is weakly A1 , . . . , An−1 -multilinear if the maps f (m1 , . . . , mi−1 , −, −, mi+2 , . . . , mn ) : Mi × Mi+1 → G are weakly Ai -bilinear for each i = 1, . . . , n − 1 and for all possible choices of elements mj ∈ Mj for j = i, i + 1. If A is commutative and if M1 , . . . , Mn are A-modules, then a function f : M1 × · · · × M n → N is said to be A-multilinear if N is an A-module and the maps f (m1 , . . . , mi−1 , −, mi+1 , . . . , mn ) : Mi → N are A-module homomorphisms for each i = 1, . . . , n and all possible choices of elements mj ∈ Mj for j = i. A-multilinear maps are easily seen to be weakly A, . . . , A-multilinear. Note that the word “weakly” may be safely omitted in the A1 , . . . , An -multilinear case if the rings A1 , . . . , An are not all equal, as there’s no stronger concept with which to confuse it. Note that if A is commutative, then a weakly A, . . . , A-multilinear map is A-multilinear if and only if it is A-linear on M1 in the sense of Definition 9.4.10. The construction of the n-fold tensor product has no surprises. Suppose given a right A1 -module M1 , a left An−1 -module Mn , and Ai−1 -Ai -bimodules Mi for i = 2, . . . , n − 1. Let F (M1 , . . . , Mn ) be the free abelian group on M1 × · · · × Mn , where e(m1 ,...,mn ) is the basis element corresponding to (m1 , . . . , mn ) ∈ M1 × · · · × Mn . Let H be the subgroup of F (M1 , . . . , Mn ) generated by all elements of the form e(m1 ,...,mi +mi ,...,mn ) − e(m1 ,...,mi ,...,mn ) −e(m1 ,...,mi ,...,mn ) , for i = 1, . . . , n and all possible choices of mj ∈ Mj for j = 1, . . . , n and mi ∈ Mi , together with all elements of the form e(m1 ,...,mi a,mi+1 ,...,mn ) − e(m1 ,...,mi ,ami+1 ,...,mn ) , for i = 1, . . . , n−1, and all possible choices of a ∈ Ai , and mj ∈ Mj for j = 1, . . . , n. Definition 9.4.22. Under the conventions above, set M1 ⊗A1 · · · ⊗An−1 Mn = F (M1 , . . . , Mn )/H and write ι : M1 × · · · × Mn → M1 ⊗A1 · · · ⊗An−1 Mn for the map that takes (m1 , . . . , mn ) to the image in M1 ⊗A1 · · · ⊗An−1 Mn of the basis element e(m1 ,...,mn ) . The proof of the next proposition is analogous to the proofs of Propositions 9.4.3, 9.4.7, and 9.4.11. Proposition 9.4.23. Suppose given a right A1 -module M1 , a left An−1 -module Mn , together with Ai−1 -Ai -bimodules, Mi for i = 2, . . . , n − 1. Let f : M1 × · · · × Mn → G be a weakly A1 , . . . , An−1 -multilinear function. Then there is a unique group homomorphism f : M1 ⊗A1 · · · ⊗An−1 Mn → G such that the following diagram commutes. f M1 × · · · × M n j/4 G TTTT jjjj j j TTTT j jjj ι TTT* jjjj f M1 ⊗A1 · · · ⊗An−1 Mn Suppose in addition M1 is an A0 -A1 -bimodule. Then M1 ⊗A1 · · · ⊗An−1 Mn is a left A0 -module by an action in which a · m1 ⊗ · · · ⊗ mn = (am1 ) ⊗ · · · ⊗ mn .


375

Moreover, if G is a left A0 -module and f is A0 -linear in M1 , then the induced map f : M1 ⊗A1 · · · ⊗An−1 Mn → G is a homomorphism of left A0 -modules. The analogous statement holds for the case where Mn is an An−1 -An -bimodule and f is An -linear in Mn . In particular, if A is commutative and M1 , . . . , Mn are A-modules, then there is an A-module structure on M1 ⊗A · · · ⊗A Mn induced by a · m1 ⊗ · · · ⊗ mn = (am1 ) ⊗ · · · ⊗ mn . If G is an A-module and f : M1 × · · · × Mn is an A-multilinear map, then the induced map f : M1 ⊗A · · · ⊗A Mn → G is an A-module homomorphism. As in the bilinear case, we shall generally define our multilinear maps implicitly, rather than explicitly. Corollary 9.4.24. Suppose given a right A-module M1 , a left B-module M3 , and an A-B-bimodule M2 . Then there are isomorphisms γ1

γ2

∼ =

∼ =

(M1 ⊗A M2 ) ⊗B M3 ←−− M1 ⊗A M2 ⊗B M3 −→ M1 ⊗A (M2 ⊗B M3 ) whose effects on generators are given by γ1 (m1 ⊗ m2 ⊗ m3 ) γ2 (m1 ⊗ m2 ⊗ m3 )

= (m1 ⊗ m2 ) ⊗ m3 = m1 ⊗ (m2 ⊗ m3 ).

and

Proof The formulæ above induce well defined homomorphisms γ1 and γ2 as stated, so it suffices to construct their inverse homomorphisms. We shall confine ourselves to the case of γ1 . Here, we note that for m3 ∈ M3 , there is a homomorphism from M1 ⊗A M2 to M1 ⊗A M2 ⊗B M3 that carries each generator m1 ⊗ m2 to m1 ⊗ m2 ⊗ m3 . This is enough to construct a homomorphism δ : (M1 ⊗A M2 ) ⊗B M3 → M1 ⊗A M2 ⊗B M3 whose effect k k on generators is given by δ(( i=1 mi ⊗ mi ) ⊗ m ) = i=1 (mi ⊗ mi ⊗ m ). And δ is easily seen to give an inverse to γ1 . Note that if M1 is a C-A-bimodule, then γ1 and γ2 are isomorphisms of left C-modules. Similarly, we get right module isomorphisms if M3 is a B-C-bimodule. f

g

→B− → C. Then there is a Corollary 9.4.25. Suppose given ring homomorphisms A − natural isomorphism of left C-modules C ⊗B (B ⊗A M ) ∼ = C ⊗A M for A-modules M . Proof The isomorphisms γi are clearly natural in all three variables. It suffices to note that (C ⊗B B) ⊗A M ∼ = C ⊗A M by Lemma 9.4.14. The theory of localization, together with Nakayama’s Lemma, now gives a valuable application of extension of rings.


376

Corollary 9.4.26. Let M be a finitely generated module over the commutative ring A. Then M = 0 if and only if A/m ⊗A M = 0 for every maximal ideal m of A. Proof Let m be a maximal ideal of A. Recall from Proposition 7.11.24 that A/m is isomorphic to the quotient ring Am /mm of the localization Am of A. Thus, A/m ⊗A M ∼ = Am /mm ⊗A M ∼ = Am /mm ⊗A (Am ⊗A M ) = Am /mm ⊗A Mm . m

m

Since M is finitely generated over A, there is a surjection A → M for some k ≥ 0. Since localization is an exact functor, we see that Mm is finitely generated over Am . Thus, if A/m ⊗A M = 0, then Mm = 0 by the second form of Nakayama’s Lemma. Proposition 7.11.25 shows that M = 0 if and only if Mm = 0 for every maximal ideal of M , so the result follows. k

Exercises 9.4.27. 1. Let a and b be two-sided ideals of A. Show that A/a ⊗A A/b ∼ = A/(a + b). 2. Let f : A → B be a ring homomorphism and let M and N be a right and a left Bmodule, respectively. Then we may consider them to be A-modules as well. Show that there is a natural surjection from M ⊗A N onto M ⊗B N . In particular, every tensor product is a quotient group of the appropriate tensor product over Z. † 3. Let a be a two-sided ideal of A, and let M and N be a right and a left A/a module, respectively. Show that the natural map M ⊗A N → M ⊗A/a N of the preceding problem is an isomorphism. † 4. Show that if A and B are commutative rings and f : A → B is a ring homomorphism, then there is a natural B-module isomorphism from (B ⊗A M ) ⊗B (B ⊗A N ) to B ⊗A (M ⊗A N ) for A-modules M, N . 5. Let A be a commutative ring and let m, n > 0. Show that Am ⊗A An is a free A-module of rank mn, with basis ei ⊗ ej for 1 ≤ i ≤ m and 1 ≤ j ≤ n. Here, e1 , . . . , em and e1 , . . . , en are the canonical bases of Am and An , respectively. † 6. Give the generalization of Proposition 9.4.18 to infinite direct sums, and deduce that extension of rings carries infinitely generated free modules to free modules. 7. Let f : A → B be a ring homomorphism. Show that passage from a left A-module M to the extended B-module B⊗A M provides a left adjoint to the forgetful functor from left B-modules to left A-modules (i.e., the functor that considers a B-module to be an A-module via f ). 8. Let A be a ring. Show that if M is a left A-module, then the action of A on M induces a homomorphism α : A ⊗Z M → M . Conversely, if M is an abelian group, then any homomorphism α : A⊗Z M → M defines the function (α◦ι) : A×M → M , which we may then investigate to see if it gives a module action. Show that α : A ⊗Z M → M induces an A-module structure on M if and only if the following diagrams commute: ν ⊗ 1M/ 1A ⊗ α/ A ⊗Z M Z ⊗Z M A ⊗Z M A ⊗Z A ⊗Z M JJ JJ JJ J μ ⊗ 1M α α εM JJJJ J $ α / M A ⊗Z M M


377

Here, μ is induced by the multiplication of A, ν is the unique ring homomorphism from Z to A, and εM is the isomorphism induced by the action of Z on M as in Lemma 9.4.14.

9.5

Tensor Products and Exactness

Here, we shall explore the relationships between tensor products and exact sequences, with particular attention paid to extension of rings. As an application, we shall study the A-module homomorphisms f : An → Am between finitely generated free modules, and extend, as much as we can, the sort of results that are known to hold for finitely generated vector spaces over a division ring. We’ve seen one example, tensoring with S −1 A, where S is a multiplicative subset of the commutative ring A, where extension of rings is an exact functor. However, extension of rings is often not an exact functor. For instance, as we shall see in Exercises 9.5.21, extension of rings from Z to Z2 fails to be exact. Thus, the next definition has significance. Definition 9.5.1. A right A-module M is flat if the passage from left A-modules N to M ⊗A N and from A-module homomorphisms f to 1M ⊗ f gives an exact functor. Similarly, a left A-module N is flat if − ⊗A N gives an exact functor. As we noted above, Corollary 9.4.15, together with the proof that passage to modules of fractions is exact, gives the following corollary. Corollary 9.5.2. Let S be a multiplicative subset of the commutative ring A. Then S −1 A is a flat A-module. Similarly, the next corollary follows from Lemma 9.4.14. Corollary 9.5.3. A itself is flat as either a left or a right A-module. While extension of rings is not in general an exact functor, it does have some exactness properties. Definitions 9.5.4. Let F be a covariant functor between two categories of modules. 1. We say that F is right exact if for each short exact sequence g

f

M − →M − → M → 0 in the domain category of F , the resulting sequence F (f )

F (g)

F (M ) −−−→ F (M ) −−−→ F (M ) → 0 is exact. 2. We say that F is left exact if for each short exact sequence f

g

0 → M − →M − → M in the domain category of F , the resulting sequence F (f )

F (g)

0 → F (M ) −−−→ F (M ) −−−→ F (M ) is exact.


378

One can also study partial exactness properties of contravariant functors. We shall do so (without giving formal definitions of what constitutes “left” and “right”) in our study of the Hom functors in Section 9.7. Proposition 9.5.5. Let M be a right A-module. Then the functor which takes N to M ⊗A N and takes f to 1M ⊗ f is a right exact functor from the category of left Amodules to abelian groups. Similarly, if N is a left A-module, then − ⊗A N is a right exact functor on right A-modules. Proof We treat the case of M ⊗A −. The other is similar. Suppose given a short exact sequence f

g

N − →N − → N → 0. We wish to show that 1M ⊗f

1M ⊗g

M ⊗A N −−−−→ M ⊗A N −−−−→ M ⊗A N → 0 is exact. Consider first the exactness at M ⊗A N . This amounts to showing that 1M ⊗ g is onto. But if n = g(n), then m ⊗ n = (1M ⊗ g)(m ⊗ n) for all m ∈ M , and hence the image of 1M ⊗ g contains a set of generators for M ⊗A N , so 1M ⊗ g is onto. Clearly, (1M ⊗ g) ◦ (1M ⊗ f ) = 0. Thus, if B ⊂ M ⊗A N is the image of 1M ⊗ f , then B is contained in the kernel of 1M ⊗ g. We obtain an induced map h : (M ⊗A N )/B → M ⊗A N . It suffices to show that h is an isomorphism. To show this, we construct a map the other way. As usual, we use the universal mapping property of the tensor product. This time, we shall explicitly define a weakly A-bilinear function k : M ×N → (M ⊗A N )/B, as follows. For a pair (m, n ) ∈ M ×N , choose n ∈ N with g(n) = n , and set k(m, n ) = m ⊗ n, the element of (M ⊗A N )/B represented by m ⊗ n. To see that k(m, n ) does not depend on which n ∈ g −1 (n ) we choose, suppose that g(n) = g(n1 ) = n . By the exactness of the original sequence, n − n1 = f (n ) for some n ∈ N . But then (m ⊗ n) − (m ⊗ n1 ) = m ⊗ (n − n1 ) = (1M ⊗ f )(m ⊗ n ) lies in B. Thus, k is a well defined function, and is easily seen to be weakly A-bilinear. Thus, there is a homomorphism k : M ⊗A N → (M ⊗A N )/B extending k. Now if g(n) = n and m ∈ M , then (h ◦ k)(m ⊗ n ) = h(m ⊗ n) = (1M ⊗ g)(m ⊗ n) = m ⊗ n . Since h ◦ k is the identity on a set of generators for M ⊗A N , it must be the identity map everywhere. For m ⊗ n ∈ (M ⊗A N )/B, we have (k ◦ h)(m ⊗ n) = k(m ⊗ g(n)) = m ⊗ n, since any lift to N of g(n) may be used in defining k. So k ◦ h is the identity on a set of generators, and hence k and h are inverse to one another. The right exactness of extension of rings has some immediate applications, but let us first develop some tools for recognizing flat modules.


379

Proposition 9.5.6. Let F be a covariant functor from A-modules to abelian groups. Then F is exact if and only if it takes short exact sequences to short exact sequences. Proof To see that an exact functor, F , preserves short exact sequences, it suffices to 0 0 show that F (0) = 0. Note that 0 − →0− → 0 is an exact sequence in which the morphisms are both isomorphisms. Since any functor preserves isomorphisms, the same is true of F (0)

F (0)

F (0) −−−→ F (0) −−−→ F (0). Thus, F (0) = 0, as claimed. f

g

For the converse, suppose given an exact sequence M − →M − → M of A-modules. Then we have a commutative diagram 0 JJ t: 0 JJ tt JJ t t J$ tt M/ ker g JJ g : JJ πtttt JJ t t $ t f g / / M M M J t: JJ JJ ttt t J t t f π $ M / ker f JJ : JJ tt JJ tt t J$ tt 0 0 in which all the straight lines are exact. Suppose then that F takes short exact sequences to short exact sequences. Then it is easy to see that it must take injections to injections and take surjections to surjections. Thus, if we apply F to the above diagram, the diagonal straight lines will remain exact. Thus, im F (f ) = im F (f ) and ker F (g) = ker F (π), and hence the exactness F (f )

F (g)

of F (M ) −−−→ F (M ) −−−→ F (M ) follows from the exactness of five-term diagonal sequence. Corollary 9.5.7. Let F be a right exact functor from A-modules to abelian groups. Then F is exact if and only if for any injective A-module homomorphism f : M → M , the induced map F (f ) is also injective. The following consequence of Corollary 9.5.7 has an easy direct proof, as well. Proposition 9.5.8. Let N1 and N2 be left A-modules. Then N1 ⊕ N2 is a flat module if and only if both N1 and N2 are flat. The analogous statement holds for right modules as well. Proof Suppose given an injective right A-module homomorphism f : M → M . The naturality of the isomorphism of Proposition 9.4.18 shows that we may identify f ⊗A 1(N1 ⊕N2 ) with (f ⊗A 1N1 ) ⊕ (f ⊗A 1N2 ). But clearly, a direct sum of homomorphisms is injective if and only if each one of them is injective, and hence the result follows. The next corollary now follows from Lemma 9.4.14 and induction on n. Corollary 9.5.9. For n ≥ 1, the free module An is flat as either a right or a left Amodule. The next lemma can be useful in recognizing flatness.


380

Lemma 9.5.10. Let M and N be left and right A-modules, respectively, and suppose that k m ⊗n i i = 0 in M ⊗A N . Then there are finitely generated submodules M0 ⊂ M and i=1 k N0 ⊂ N such that mi ∈ M0 and ni ∈ N0 for all i, and i=1 mi ⊗ ni = 0 in M0 ⊗A N0 . Proof As in Section 9.4, we write F (M, N ) for the free abelian group on M × N and write e(m,n) for the basis element corresponding to the ordered pair (m, n) ∈ M × N . k Then our construction of tensor products shows that i=1 e(mi ,ni ) must be equal to a sum of integral multiples of terms of the form e(xi +xi ,yi ) − e(xi ,yi ) − e(xi ,yi ) , e(xi ,yi +yi ) − e(xi ,yi ) −e(xi ,yi ) , and e(xi ai ,yi ) −e(xi ,ai yi ) . So take M0 to be the submodule of M generated by m1 , . . . , mk together with the x’s and x ’s and take N0 to be the submodule of N generated by n1 , . . . , nk together with the y’s and y ’s. Corollary 9.5.11. Let N be a left A-module with the property that every finitely generated submodule of N is flat. Then N is flat. Proof Let f : M → M be an injective homomorphism of right A-modules and let k 9.5.10 provides a i=1 mi ⊗ ni be an element of the kernel of f ⊗ 1N . Then Lemma k finitely generated submodule N0 containing n1 , . . . , nk such that i=1 f (mi ) ⊗ ni = 0 in M ⊗A N0 . k By our hypothesis, N0 is flat, and hence i=1 mi ⊗ ni = 0 in M ⊗A N0 , and hence also in M ⊗A N . We can now characterize the flat modules over a P.I.D. Corollary 9.5.12. Let A be a P.I.D. Then an A-module is flat if and only if it is torsionfree. Proof Let M be a torsion-free A-module. Then every finitely generated submodule of M is also torsion-free, and hence free by Proposition 8.9.6. By Corollary 9.5.9, the finitely generated submodules of M are flat, so M is flat by Corollary 9.5.11. 1⊗η Conversely, suppose that M is a flat A-module. Then M ⊗A A −−→ M ⊗A K is injective, where η : A → K is the canonical inclusion of A in its field of fractions. By Corollary 9.4.15, we may identify 1 ⊗ η with the canonical map η : M → M(0) from M to its localization at (0). But the kernel of η is easily seen to be the torsion submodule of M , and hence M is torsion-free. There are times when tensoring with even a non-flat module will preserve short exactness. Recall that a short exact sequence g

f

0 → N − →N − → N → 0 splits if f admits a retraction, meaning an A-module homomorphism r : N → N such that r ◦ f = 1N . Corollary 9.5.13. Suppose given a split short exact sequence of left A-modules: f

g

→N − → N → 0. 0 → N − Then for any right A-module M , the sequence 1⊗f

1⊗g

0 → M ⊗A N −−→ M ⊗A N −−→ M ⊗A N → 0 is exact and splits.


381

Proof Note that any map that admits a retraction is injective. Thus, since tensoring with M is right exact, it suffices to show that 1 ⊗ f admits a retraction. But if r is a retraction for f , then 1 ⊗ r is a retraction for 1 ⊗ f . We now consider the applications of this material to homomorphisms of free modules. We’ve already seen, in Theorem 7.9.8, that the rank of a free module is well defined over a commutative ring. The argument there implicitly used extension of rings from A to the field A/m, where m is a maximal ideal of A. Now, we’re given a general extension of rings functor, so we can give a more general result: Proposition 9.5.14. Let A be any ring that admits a ring homomorphism f : A → D, with D a division ring. Then the free left A-modules Am and An are isomorphic if and only if m = n. Proof Let g : Am → An be an A-module isomorphism and let g −1 be the inverse isomorphism. Since extension of rings is a functor, 1D ⊗g is a D-module isomorphism from Dm to Dn , with inverse 1D ⊗ g −1 . Since dimension is well defined for finite dimensional vector spaces over a division ring (Corollary 7.9.5), m = n. This does indeed cover the case of commutative rings by precisely the same argument as that given earlier, as the commutative ring A maps to the field A/m if m is a maximal ideal of A. Here is another result about vector spaces that extends to a more general context. Proposition 9.5.15. Let A be a commutative ring, or more generally, any ring that admits a ring homomorphism f : A → D, with D a division ring. Let g : Am → An be a surjective homomorphism. Then m ≥ n. Proof By the right exactness of the tensor product, 1D ⊗ g : Dm → Dn is surjective. Now apply Proposition 7.9.2. Of course, if D is a division ring and if f : Dm → Dn is surjective, then this determines the isomorphism type of ker f . In particular, any surjection from Dn to itself must be an isomorphism. We shall extend this latter result to commutative rings: Proposition 9.5.16. Let A be a commutative ring. Then any surjective homomorphism f : An → An is an isomorphism. Proof Let K = ker f . Then the exact sequence i

f

0→K− → An − → An → 0 splits by Proposition 7.7.51. Thus, for each maximal ideal m of A, the sequence 1⊗i

1⊗f

0 → A/m ⊗A K −−→ A/m ⊗A An −−→ A/m ⊗A An → 0 is exact by Corollary 9.5.13. In particular, the dimension of the kernel of 1 ⊗ f must be 0, and hence A/m ⊗A K = 0 for each maximal ideal m of A. Note that if r : An → K is a retraction for i, then r is surjective, and hence K is finitely generated. Thus, K = 0 by Corollary 9.4.26.


382

We shall now show that if A is a commutative ring, and if f : Am → An is injective, then m ≤ n. If A is an integral domain, then this is almost immediate, because tensoring with its field of fractions gives an exact functor to a category of vector spaces. If A is not a domain, we shall need some sort of substitute for the field of fractions. The thing to notice is that in an integral domain, 0 is minimal in the partially ordered set of all prime ideals under inclusion. So the technique we shall use in general is to localize at a minimal prime ideal. Before we can do that we need to know that minimal primes exist. If A has finite Krull dimension, this is obvious. For the general case, we need further argument. Lemma 9.5.17. A commutative ring A has minimal prime ideals. Proof A minimal prime is a maximal element in the partially ordered set of prime ideals under reverse inclusion. We apply Zorn’s Lemma. Given a totally ordered collection of prime ideals, an “upper bound” in this ordering would be given by their intersection, provided it is prime. Normally, intersections of prime ideals are not prime.(Look at Z, for instance.) But if we’re intersecting a totally ordered collection, say p = i∈I pi , suppose that ab is in p. Then if a is not in p, it must not be in pi for some i. But then b ∈ pj whenever pj ⊂ pi . But since the collection {pj | j ∈ I} is totally ordered, for j ∈ I we either have pj ⊂ pi or pi ⊂ pj , so in either case b ∈ pj . So b must be in p, the intersection of the pj . So p is prime, and the hypotheses of Zorn’s Lemma are satisfied. If f : Am → An is injective, so is its localization at a minimal prime ideal. Thus, we may assume that A has exactly one prime ideal, p, which therefore is its nil radical (Proposition 9.1.10). In other words, every element of p is nilpotent. The following will be useful. Lemma 9.5.18. Let A be a commutative ring with only one prime ideal, p. Let p1 , . . . , pk be nonzero elements of p. Then there is a nonzero element p ∈ p which annihilates each of the pi . Proof We claim that p may be taken to have the form pr11 . . . prkk , with ri ≥ 0 for , where s is the 1 ≤ i ≤ k. We show this by induction on k. If k = 1, just take p = ps−1 1 smallest positive integer for which ps1 = 0. For k > 1, assume by induction that we’re rk−1 given an element p = pr11 . . . pk−1 , such that p = 0, and p annihilates p1 , . . . , pk−1 . Let t be the smallest integer such that ptk = 0. If p pt−1 = 0, then that’s our value k of p. Otherwise, let rk be the largest non-negative integer such that p prkk = 0, and set p = p prkk . Proposition 9.5.19. Let A be a commutative ring with only one prime ideal. Let f : Am → M be an injective A-module homomorphism for some A-module M . Then 1 ⊗ f : A/p ⊗A Am → A/p ⊗ M is also injective. Proof We may identify 1 ⊗ f with the unique A/p-module homomorphism that makes the following diagram commutative. f / M Am πm

(A/p)m

π f

/ M/pM,


383

Where π : A → A/p and π : M → M/pM are the canonical maps. Suppose that (a1 , . . . , am ) ∈ ker f , where ai = π(ai ) for 1 ≤ i ≤ m. Then f (a1 , . . . , am ) ∈ pM . Thus, we can write f (a1 , . . . , am ) = p1 m1 +· · ·+pk mk , where p1 , . . . , pk are nonzero elements of p, and m1 , . . . , mk ∈ M . By Lemma 9.5.18, there is a nonzero element p ∈ p which annihilates p1 , . . . , pk . But then pf (a1 , . . . , am ) = 0. Since f is injective, this says p annihilates (a1 , . . . , am ). But since A is local, with maximal ideal p, every zero divisor of A lies in p. In particular, we must have ai ∈ p for all i. But then π(ai ) = 0 for all i, and hence (a1 , . . . , am ) = π n (a1 , . . . , am ) = 0. Thus, the kernel of f is trivial. We can now put it all together. Theorem 9.5.20. Let A be a commutative ring and let f : Am → An be an injective A-module homomorphism. Then m ≤ n. Proof Since localization is exact, we may localize at a minimal prime ideal. Thus, we may assume that A is a local ring with a unique prime ideal p. But now, Proposition 9.5.19 shows that the homomorphism obtained by extension of rings to A/p is also injective. Since A/p is a field, the result follows. Exercises 9.5.21. 1. Deduce that Z2 is not a flat Z-module by tensoring it with the exact sequence f2

π

0 → Z −→ Z − → Z2 → 0 from Problem 3 of Exercises 7.7.52. Is Zn a flat Z-module for n > 2? 2. Show that Z2 is not a flat Z2r -module for r > 1. Note that Z2r is a local ring with only one prime ideal. 3. Let A be a domain. For which ideals a is A/a flat as an A-module? † 4. Show that an infinite direct sum of flat modules is flat. Deduce that all free modules are flat. f0

f1

5. Suppose given a direct system M0 −→ M1 −→ · · · of right A-modules, together with a left A-module M . (a) Show that − lim →(Mi ⊗A N ) is naturally isomorphic to (lim −→ Mi ) ⊗A N . (b) Deduce that if each Mi is flat, so is lim −→ Mi .

9.6

Tensor Products of Algebras

Let A be a commutative ring and let B and C be A-algebras, via homomorphisms f : A → B and g : A → C. We shall show that B ⊗A C is then an A-algebra, and in fact satisfies an important universal mapping property with respect to A-algebra homomorphisms out of B and C. A ring called A, in this section, will be assumed to be commutative.


384

Proposition 9.6.1. Let B and C be A-algebras, via f : A → B and g : A → C. Then there is an A-algebra structure on B ⊗A C whose product is given on generators by (b ⊗ c) · (b ⊗ c ) = bb ⊗ cc . In particular, B ⊗A C is commutative if and only if both B and C are commutative. In addition, if we define ι1 : B → B ⊗A C and ι2 : C → B ⊗A C by ι1 (b) = b ⊗ 1 and ι2 (c) = 1 ⊗ c, then the ιi are A-algebra homomorphisms, and satisfy the following universal property. 1. Every element of the image of ι1 commutes with every element of the image of ι2 . 2. Suppose given an A-algebra D, together with A-algebra homomorphisms h1 : B → D and h2 : C → D such that every element of the image of h1 commutes with every element of the image of h2 . Then there is a unique A-algebra homomorphism h : B ⊗A C → D such that the following diagram commutes. ι1 / ι2 B ⊗A C o B ? C ?? ?? ?? ?? h h1 ?? h2 ? D If B, C, and D are commutative, this says simply that B ⊗A C, together with the maps ιi , is the coproduct of B and C in the category of commutative A-algebras. Proof First note that the A-algebra structures on B and C imply that there is an A-module homomorphism μ : B ⊗A C ⊗A B ⊗A C → B ⊗A C whose effect on generators is given by μ(b ⊗ c ⊗ b ⊗ c ) = bb ⊗ cc . By the associativity of the tensor product (e.g., Corollary 9.4.24), the iterated tensor product B ⊗A C ⊗A B ⊗A C is naturally isomorphic to (B ⊗A C) ⊗A (B ⊗A C), so that μ induces a product operation on B ⊗A C via x · y = μ(x ⊗ y) for x, y ∈ B ⊗A C. By the laws of the tensor product (B ⊗A C) ⊗A (B ⊗A C), this product satisfies both distributive laws. We claim that this product makes B ⊗A C a ring, with multiplicative identity 1 ⊗ 1. On generators, we have (1 ⊗ 1) · (b ⊗ c) = b ⊗ c = (b ⊗ c) · (1 ⊗ 1). Since 1 ⊗ 1 acts as the identity on a set of generators, it must act as the identity everywhere. Similarly, it suffices to show associativity of products of generators, where it follows from the associativity of the products in B and C. The homomorphisms ιi defined above are easily seen to be ring homomorphisms. Note that if ν and ν give the A-algebra structures of B and C, respectively, then ιi ◦ ν(a) = a ⊗ 1 = 1 ⊗ a = ι2 ◦ ν (a) for all a ∈ A, and that this element lies in the center of B ⊗A C. Since the ιi , ν, and ν are ring homomorphisms, this composite is also, and gives B ⊗A C an A-algebra structure such that the ιi are A-algebra homomorphisms. We have (b ⊗ 1) · (1 ⊗ c) = b ⊗ c = (1 ⊗ c) · (b ⊗ 1), and hence the elements of the image of ι1 commute with the elements of the image of ι2 . Let h1 : B → D and h2 : C → D be A-algebra homomorphisms with the property that every element of the image of h1 commutes with every element of the image of h2 .


385

Then, as the reader may check, there is an A-module homomorphism h : B ⊗A C → D whose effect on generators is given by h(b ⊗ c) = h1 (b)h2 (c). The fact that each h1 (b) commutes with each h2 (c) shows that h(x · y) = h(x)h(y) as x and y range over a set of generators of B ⊗A C. Since h is a homomorphism of abelian groups, this is enough to show that h(xy) = h(x)h(y) for all x, y ∈ B ⊗A C. Since h(1 ⊗ 1) = 1, h is a ring homomorphism, which is easily seen to be an A-algebra homomorphism as well. It is easy to see that h is the unique A-algebra homomorphism that makes the diagram commute. We’ve already seen some examples of tensor products of algebras. Proposition 9.6.2. Let A be a commutative ring and let B be an A-algebra via f : A → B. Then there is an isomorphism of A-algebras η : B ⊗A Mn (A) → Mn (B) which makes the following diagram commute. ι1 / ι2 B ⊗A Mn (A) o Mn (A) B J JJ t JJ t t JJ t J tt tt ι JJJJ η t JJ tt f∗ $ ztt Mn (B) Here ι : B → Mn (B) is given by ι(b) = bIn , while f∗ takes (aij ) to the matrix whose ij-th entry is f (aij ). The maps ιi are the structure maps of the tensor product of algebras. Proof Because the square A ι

Mn (A)

f

/ B ι

f∗ / Mn (B)

commutes, Mn (B) has an A-algebra structure such that both f∗ and ι : B → Mn (B) are A-algebra homomorphisms. It is easy to see that every element in ι(B) commutes with every element in the image of f∗ . So the universal property of the tensor product of algebras provides an A-algebra homomorphism η : B ⊗A Mn (A) → Mn (B) that makes the above diagram commute. It suffices to show that η is an isomorphism. Write eij for the matrix whose ij-th coordinate is 1 and whose other coordinates are all 0. Recall from Lemma 7.10.5 that if C is any ring, then {eij | 1 ≤ i, j ≤ n}, in any order, gives a basis for Mn (C) as either a left or a right C-module. By extension of rings, B ⊗A Mn (A) is free as a left B-module, with basis {1 ⊗ eij | 1 ≤ i, j ≤ n}. But η(1 ⊗ eij ) = eij , so, viewing η as a left B-module homomorphism, we see that it carries a basis of B ⊗A Mn (A) onto a basis of Mn (B). By Lemma 7.7.36, this is sufficient to conclude that η is an isomorphism. There is an analogous result for monoid rings, which we shall treat as an exercise. We shall see other examples later.


386

Exercises 9.6.3. ‡ 1. Let B be an A-algebra, via f : A → B, and let M be a monoid. Show that there is a natural A-algebra isomorphism from B ⊗A A[M ] to B[M ]. 2. Combining Proposition 9.6.2 with the preceding problem, we see that if A is commutative and M is a monoid, then there is an isomorphism between Mn (A[M ]) and Mn (A)[M ]. Write down the isomorphism explicitly and give a proof of isomorphism that doesn’t use tensor products. 3. Let M and M be monoids. Show that there is a natural A-algebra isomorphism from A[M ] ⊗A A[M ] to A[M × M ]. 4. Let K be a subfield of the fields L and L . Is L ⊗K L a field? 5. Let B be an A-algebra with structure map ν : A → B. Show that there is a homomorphism of A-modules (but not necessarily of A-algebras), μ : B ⊗A B → B induced by μ(b ⊗ b ) = bb . Show also that the following diagrams commute. B ⊗A B ⊗A B 1⊗μ

μ ⊗ 1/

B ⊗A B μ

μ / B B ⊗A B ν ⊗ 1/ 1⊗ν A ⊗A B B ⊗A B o B ⊗A A JJ t JJ t JJ tt t J μ tt εB JJJJ tt εB J$ zttt B Here, the maps εB are the natural isomorphisms of Lemma 9.4.14. Conversely, suppose given an A-module B, together with A-module homomorphisms ν : A → B and μ : B ⊗A B → B such that the above diagrams commute. Show that B is an A-algebra with structure map ν and with ring product given by bb = μ(b ⊗ b ). 6. Let B be an A-algebra and let μ : B ⊗A B → B be induced by the ring product of B. Show that B is a commutative ring if and only if μ is an A-algebra homomorphism.

9.7

The Hom Functors

Let M and N be A-modules. Recall that HomA (M, N ) is the abelian group of A-module homomorphisms from M to N . Here, the group structure on HomA (M, N ) is given by (f +g)(m) = f (m)+g(m). Without extra hypotheses (e.g., A being commutative), there is no module structure on HomA (M, N ). We shall use the same notation HomA (M, N ) whether we are discussing the homomorphisms of left modules or of right modules, and the theory proceeds identically. Thus, in general, we shall not make reference to the sidedness of the module structures. In applications, it must either be made clear by the context or be explicitly stated.


387

We should immediately discuss the functoriality of Hom, as it is good to keep it in mind as we go. Here, for an A-module M , HomA (M, −) gives a covariant functor from A-modules to abelian groups. For f : N → N , the induced map HomA (M, f ) : HomA (M, N ) → HomA (M, N ) is generally written as f∗ , and is given by f∗ (g) = f ◦ g. Similarly, for an A-module N , HomA (−, N ) is a contravariant functor. Here, if f : M → M , the induced map HomA (f, N ) : HomA (M, N ) → HomA (M , N ) is generally written as f ∗ , and is given by f ∗ (g) = g ◦ f . We shall first establish the exactness properties of the Hom functors. Lemma 9.7.1. For any A-module M , the functor HomA (M, −) is left exact; i.e., if we’re given an exact sequence g

f

0 → N − →N − → N , then the induced sequence f∗

g∗

0 → HomA (M, N ) −→ HomA (M, N ) −→ HomA (M, N ) is exact. Proof Clearly, f∗ is injective because f is, and g∗ ◦ f∗ = (g ◦ f )∗ = 0. Let h : M → N be in the kernel of g∗ . Then im h ⊂ ker g. But the kernel of g is equal to the image of f . Thus, since f is injective, the induced map f : N → ker g is an isomorphism. Thus, h is f −1

h

→ ker g −−→ N . the image under f∗ of the composite M − The contravariant direction of Hom also has exactness properties. Lemma 9.7.2. Let N be an A-module and let f

g

M − →M − → M → 0 be an exact sequence of A-modules. Then the following sequence is exact. g∗

f∗

0 → HomA (M , N ) −→ HomA (M, N ) −→ HomA (M , N ) Proof The injectivity of g ∗ follows directly from the surjectivity of g. It is also immediate that f ∗ ◦ g ∗ = 0. Let h : M → N be in the kernel of f ∗ . Thus, h ◦ f = 0. Thus, h vanishes on the image of f . By the exactness of the original sequence, the image of f is the kernel of g. But then h factors through the canonical map from M to M/(ker g). But the Noether Theorem says we may identify this canonical map with g itself. In other words, there is a homomorphism h : M → N such that h ◦ g = h. But this is just another way of saying that h = g ∗ h for some h ∈ HomA (M , N ). We have so far avoided a discussion of module structures on the Hom groups. The proof of the next lemma is easy, and is left to the reader. Lemma 9.7.3. Bimodule structures may be used to put module structures on Hom groups in any of the following ways. 1. If M is an A-B-bimodule and N is a left A-module, then there is a left B-module structure on HomA (M, N ) via (b · f )(m) = f (mb). 2. If M is a B-A-bimodule and N is a right A-module, then there is a right B-module structure on HomA (M, N ) via (f · b)(m) = f (bm).


388

3. If N is an A-B-bimodule and M is a left A-module, then there is a right B-module structure on HomA (M, N ) via (f · b)(m) = f (m)b. 4. If N is a B-A-bimodule and M is a right A-module, then there is a left B-module structure on HomA (M, N ) via (b · f )(m) = bf (m). If A is commutative, then any A-module is an A-A-bimodule. So each of the procedures above puts an A-module structure on HomA (M, N ). In fact, each of the above procedures puts the same A-module structure on HomA (M, N ). In particular, for any ring A, if M is a left A-module, then so is HomA (A, M ), by the first item above. Similarly, if M is a right A-module, then the second item puts a right A-module structure on HomA (A, M ). Recall from Corollary 7.7.9 that the evaluation map ε : HomA (A, M ) → M is an isomorphism of abelian groups, where ε(f ) = f (1). In fact, it is more: Proposition 9.7.4. The evaluation map ε : HomA (A, M ) → M gives a natural isomorphism of A-modules. The various module structures on Hom allow us to demonstrate the relationship between Hom and the tensor products. Proposition 9.7.5. Suppose given a right B-module M1 , a B-A-bimodule M2 , and a right A-module M3 . Then there is an isomorphism of abelian groups, natural in all three variables: ϕ : HomB (M1 , HomA (M2 , M3 )) → HomA (M1 ⊗B M2 , M3 ). Explicitly, if f ∈ HomB (M1 , HomA (M2 , M3 )), then the effect of ϕ(f ) on generators is given by ϕ(f )(m1 ⊗ m2 ) = f (m1 )(m2 ). If M1 is a C-B-bimodule, then ϕ is an isomorphism of right C-modules. Proof For f ∈ HomB (M1 , HomA (M2 , M3 )), it is easy to see that the displayed formula for ϕ(f ) gives rise to a well defined homomorphism. And ϕ itself is then easily seen to be a homomorphism. To see that ϕ is an isomorphism, we construct its inverse. Define ψ : HomA (M1 ⊗B M2 , M3 ) → HomA (M1 , HomA (M2 , M3 )) as follows. For g ∈ HomA (M1 ⊗B M2 , M3 ), we set ((ψ(g))(m1 ))(m2 ) = g(m1 ⊗ m2 ). We leave it to the reader to check that ϕ and ψ are inverse isomorphisms, and to verify the claims regarding naturality and C-module structures. Corollary 9.7.6. Let A be a commutative ring and let Mi be an A-module for i = 1, 2, 3. Then there is a natural isomorphism ϕ : HomA (M1 , HomA (M2 , M3 )) → HomA (M1 ⊗A M2 , M3 ) of A-modules.


389

As in the case of tensor products, it is useful to understand the behavior of Hom groups with respect to direct sums. Proposition 9.7.7. Let M, M1 , . . . , Mk and N, N1 , . . . , Nk be left A-modules. Then there are natural isomorphisms k k , , α : HomA ( Mi , N ) → HomA (Mi , N ), i=1

i=1

whose i-th coordinate map is ι∗i , where ιi is the natural inclusion of Mi in β : HomA (M,

k , i=1

Ni ) →

k ,

*k i=1

Mi , and

HomA (M, Ni ),

i=1

*k whose i-th coordinate map is (πi )∗ , where πi is the natural projection map from i=1 Ni onto Ni . These isomorphisms are natural in all variables. If each Mi is an A-B-bimodule or if N is an A-B-bimodule, then α is an isomorphism with respect to the appropriate Bmodule structure. Similarly, if each Ni is an A-B-bimodule or if M is an A-B-bimodule, then β is an isomorphism with respect to the appropriate B-module structure. Proof In fact, these follow from the universal properties of the coproduct and product in the category of A-modules. Thus, α is an isomorphism because a homomorphism out of a direct sum is determined uniquely by its restrictions to the individual summands. Similarly, β is an isomorphism because finite direct sums coincide with finite direct products. Thus, a homomorphism into a finite direct sum is determined uniquely by its component functions. We leave the statements about naturality and module structures to the reader. Definitions 9.7.8. Let M be an A-module. Define the dual module, M ∗ , by M ∗ = HomA (M, A). Since A is an A-A-bimodule, M ∗ is a right A-module when M is a left A-module, and M ∗ is a left A-module when M is a right A-module. In either case, the dual of M ∗ , M ∗∗ = HomA (HomA (M, A), A) is defined, and is an A-module from the same side that M is. The next proposition is fundamental in our understanding of the duality of free modules. Proposition 9.7.9. Let M be a finitely generated free left A-module, with basis m1 , . . . , mn . Then the dual module M ∗ is a free right A-module, with basis m∗1 , . . . , m∗n , where m∗i (a1 m1 + · · · + an mn ) = ai . We call m∗1 , . . . , m∗n the dual basis of m1 , . . . , mn . Proof Let g : An → M be the isomorphism defined by the basis m1 , . . . , mn : g(a1 , . . . , an ) = a1 m1 + · · · + an mn . Then g ∗ : M ∗ → (An )∗ is an isomorphism (because (g −1 )∗ is its inverse). But Proposition 9.7.7 provides an isomorphism α : (An )∗ → (A∗ )n . And A∗ = HomA (A, A) is isomorphic to A (Proposition 9.7.4) by the correspondence, ε : HomA (A, A) → A, which takes a left A-module homomorphism f : A → A to f (1). In Proposition 9.7.4 it was shown that ε is a left A-module map with respect to the left module structure on A∗ induced by the A-A-bimodule structure on the A serving as the domain. Here, we are using the right module structure on A∗ which comes from the bimodule structure of the A that receives the maps f ∈ HomA (A, A). But we see that


390

ε is a module map with respect to this structure also: ε(f a) = (f a)(1) = f (1)a by the definition of f a. But this last is (ε(f ))a. Thus, we have a composite isomorphism of right A-modules: g∗

εn

→ (A∗ )n −→ An . M ∗ −→ (An )∗ − α

But this composite is easily seen to carry the dual basis m∗1 , . . . , m∗n to the canonical basis of An . Note that the homomorphism m∗i depends on the entire basis m1 , . . . , mn , and not just on the element mi . A different basis whose i-th term is also mi could give rise to a different homomorphism m∗i . Explicitly, m∗i is the composite g −1

π

i M −−→ An −→ A

∼ =

where g is the isomorphism induced by the basis m1 , . . . , mn , and πi is the projection onto the i-th factor. We see from Proposition 9.7.9 that if A is commutative and if M is a finitely generated free A-module, then M and M ∗ are isomorphic. However, even in the case where A is a field, there is no natural isomorphism between M and M ∗ : if we’re given two different bases for M and map each one of them to its dual basis, we will get two different isomorphisms from M to M ∗ . But if we dualize twice, some naturality returns. The reader may verify the following lemma. Lemma 9.7.10. Let M be an A-module for any ring A. Then there is an A-module homomorphism κM : M → M ∗∗ defined as follows. For m ∈ M , κM (m) is the homomorphism from M ∗ to A defined by evaluating at m: κM (m)(f ) = f (m). In addition, κM is natural in M in the sense that if g : M → N is an A-module homomorphism, then the following diagram commutes. M κM M ∗∗

g

g ∗∗

/ N κN / N ∗∗

Here, for h ∈ M ∗∗ and f ∈ N ∗ , g ∗∗ (h)(f ) = h(f ◦ g). We shall borrow a definition from functional analysis. Definition 9.7.11. An A-module M is reflexive if κM : M → M ∗∗ is an isomorphism. Proposition 9.7.12. Finitely generated free modules are reflexive. Proof Let M be a finitely generated free module with basis m1 , . . . , mn . Then the dual module M ∗ is free, with basis m∗1 , . . . , m∗n , by Proposition 9.7.9. ∗∗ ∗ ∗ ∗∗ But then M ∗∗ is free on the dual basis, m∗∗ 1 , . . . , mn , to m1 , . . . , mn . Here, mi : ∗ ∗∗ ∗ ∗ M → A is given by mi (m1 a1 + · · · + mn an ) = ai . Thus, M and M ∗∗ are both free modules of the same rank, and hence are abstractly isomorphic. But as the reader may check, κM (mi ) = m∗∗ i . Thus, κM carries a basis of M to a basis of M ∗∗ , and hence is an isomorphism.


391

The fact that M is finitely generated here is essential. Infinitely generated free modules are not reflexive, because their duals are infinite direct products of copies of A, rather than infinite direct sums. The proof of Proposition 9.7.12 brings up a useful notation that originated in exactly this context, called the Kronecker delta. Definition 9.7.13. The Kronecker delta, δij , is defined to be 0 or 1, depending on the values of i and j. Explicitly, 1 if i = j δij = 0 otherwise. Thus, if m1 , . . . , mn is the basis of a free module and if m∗1 , . . . , m∗n is its dual basis, then m∗j (mi ) = δij . Indeed, this fact uniquely determines the dual basis m∗1 , . . . , m∗n . Exercises 9.7.14. 1. Let a be a two-sided ideal of A and let M be a left A-module. Show that there is a natural isomorphism of left A-modules ε : HomA (A/a, M ) → {m ∈ M | a ⊂ Ann(m)} given by ε(f ) = f (1). Note the use of the modules HomA (A/(a), M ) in the proof of the Fundamental Theorem of Finitely Generated Modules Over a P.I.D. 2. Let M, M and N be A-modules, with M a submodule of M . Show that HomA (M/M , N ) is isomorphic to the subgroup of HomA (M, N ) consisting of those homomorphisms that factor through the canonical map π : M → M/M . 3. Give an example of a ring A and an A-module M for which HomA (M, −) is not an exact functor. 4. Give an example of a ring A and an A-module N for which HomA (−, N ) is not an exact functor. 5. Let A be an integral domain and let M be an A-module. Let π : M → M/Tors(M ) be the canonical map. Show that the induced map of dual modules, π ∗ : (M/Tors(M ))∗ → M ∗ is an isomorphism. 6. Let M be a finitely generated module over the principal ideal domain A. Show that the double dual module M ∗∗ is naturally isomorphic to M/Tors(M ). 7. Suppose given A-modules M , {Mi | i ∈ I}, N and {Ni | i ∈ I}. (a) Show that there is a natural isomorphism , . α HomA ( Mi , N ) − → HomA (Mi , N ). i∈I

i∈I

Here, * the i-th factor of α(f ) is f ◦ ιi , where ιi is the canonical inclusion of Mi in i∈I Mi .


392

(b) Show that there is a natural isomorphism .

HomA (M,

β

Ni ) − →

i∈I

.

HomA (M, Ni ).

i∈I

Here, the i-th factor of β(f ) is πi ◦ f , where πi is the canonical projection of ! N i onto its i-th factor. i∈I (c) Suppose that M is finitely generated. Show that the preceding map β restricts to an isomorphism , β , Ni ) − → HomA (M, Ni ). HomA (M, i∈I

i∈I

8. Interpret the relationship between Hom and tensor in terms of adjoint functors. Which of the results on these functors would then have automatic proofs?

9.8

Projective Modules

Unless it’s explicitly stated to the contrary, all modules in this section are left A-modules, and all homomorphisms are A-module homomorphisms. Recall that a short exact sequence g

f

→M − → M → 0 0 → M − splits if g admits a section, meaning an A-module homomorphism s : M → M such that g ◦ s = 1M . Recall from Proposition 7.7.49 that if the sequence above splits, we obtain an A-module isomorphism from M ⊕ M to M . Definition 9.8.1. An A-module P is projective if every short exact sequence g

f

0 → M − →M − →P →0 of A-modules splits. Note that this is equivalent to the statement that every surjective A-module homomorphism g : M → P admits a section. There is another characterization of projectives that is often used. Proposition 9.8.2. Let P be an A-module. Then the following are equivalent: 1. P is projective. 2. Given a diagram P h M

g

/ M

/ 0


393

where the horizontal row is exact, there is a lift of h to M , i.e., a map h : P → M making the following diagram commute. P h h g / M M

/ 0

3. HomA (P, −) is an exact functor. Proof The second property says that if g : M → M is onto, then g∗ : HomA (P, M ) → HomA (P, M ) is onto. But since HomA (N, −) is left exact for any A-module N , the second property is equivalent to the third by Proposition 9.5.6. The second property also easily implies the first. Given a surjection g : M → P , a section of g is a lift of the identity map of P to M . Thus, it suffices to show that the first property implies the second. Suppose given a surjective homomorphism g : M → M and a homomorphism h : P → M . Recall that the pullback of g and h is the subgroup M ×M P ⊂ M × P specified by M ×M P = {(m, p) | g(m) = h(p)}. Since h and g are A-module homomorphisms, M ×M P is an A-submodule of M × P . The projections of M × P onto its factors provide A-module homomorphisms h : M ×M P → M and g : M ×M P → P making the following diagram commute. g / M ×M P P g h h / M / 0 M We claim that g is surjective. To see this, let p ∈ P . Since g is surjective, h(p) = g(m) for some m ∈ M . But this says that (m, p) ∈ M ×M P , and hence p = g(m, p). Since P is projective and g is surjective, there is a section s for g. But then h ◦ s gives the desired lift of h. We’ve already seen some projective modules in action. Indeed, Proposition 7.7.51 translates precisely to the following statement. Corollary 9.8.3. Free modules are projective. The reader may now ask whether all projective modules are free. A quick answer is “no,” but there is then a rich field of study in determining what all the projective modules are. In a certain stabilized sense, this question is answered for the finitely generated projectives by the study of the group K0 (A) introduced in Section 9.9. Definition 9.8.4. We say that a submodule N ⊂ M is a direct summand of M if M is the internal direct sum of N and some other submodule N ⊂ M . As was the case for groups, this means that the homomorphism μ : N ⊕ N → M is an isomorphism.

given by

μ(n, n ) = n + n


394

It is important to note the distinction between saying that N is isomorphic to a direct summand of M and that it in fact is a direct summand. For instance, the ideal (2) ⊂ Z is isomorphic to Z itself, but is not a direct summand of Z, as may be seen by the next lemma: the sequence ⊂

0 → (2) −→ Z → Z2 → 0 fails to split. Lemma 9.8.5. Suppose given a submodule N ⊂ M . Then the following are equivalent. 1. N is a direct summand of M . 2. The inclusion i : N ⊂ M admits a retraction. 3. The sequence ⊂

0 → N −→ M → M/N → 0 splits. Proof Suppose that N is a direct summand of M with complementary summand N , and let μ : N ⊕ N → M be the isomorphism given by μ(n, n ) = n + n . Then π1 ◦ μ−1 gives a retraction for i, where π1 : N ⊕ N → N is the projection map. Thus, the first condition implies the second. The second and third conditions are equivalent by Proposition 7.7.49, which also ∼ = shows that these two conditions imply the existence of an isomorphism h : N ⊕ M/N −→ M such that h(n, 0) = i(n) for all n ∈ N . But then M is easily seen to be the internal direct sum of N and h(0 ⊕ M/N ). We obtain an important characterization of projective modules. Proposition 9.8.6. A module is projective if and only if it is a direct summand of a free module. In particular, a direct summand of a projective module is projective. An A-module is a finitely generated projective module if and only if it is a direct summand of a finitely generated free module. Proof Suppose P is projective, and construct a short exact sequence 0 → Q → F → P → 0, where F is free, and is finitely generated if P is finitely generated. (We may do this in the finitely generated case by Corollary 7.7.41, and in the infinitely generated case by Problem 2 of Exercises 7.7.42.) In either case, since P is projective, the exact sequence splits, giving an isomorphism P ⊕Q ∼ = F by Proposition 7.7.49. Since any module isomorphic to a free module is free, P ⊕ Q is free, and is finitely generated if P is finitely generated. And P is a direct summand of P ⊕ Q. Notice that a direct summand N of a finitely generated module M is always finitely generated: If r : M → N is a retraction for N ⊂ M , then r is surjective.


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Thus, it suffices to show that a direct summand of a projective module is projective. Suppose that N ⊕ N is projective and that we’re given a commutative diagram N h M

g

/ M

/ 0

where the bottom row is exact. Since N ⊕N is projective, there is a map k : N ⊕N → M that makes the following diagram commute. N ⊕ N k M

π / N h g / M

/ 0

Here, π is the projection onto the first factor. But then k ◦ ι1 : N → M gives a lift of h, where ι1 is the inclusion of the first factor in N ⊕ N . We can now give an example of a non-free projective. Example 9.8.7. Let A and B be nonzero rings. Then the ideals A × 0 and 0 × B are direct summands of the ring A × B, and hence are projective A × B-modules. Note that even if A is isomorphic to A × B as a ring, A × 0 is not free as an A × B-module, as AnnA×B (A × 0) = 0 × B, while AnnA×B (A × B) = 0. And more applications of Proposition 9.8.6 follow: Corollary 9.8.8. Every finitely generated projective module over a P.I.D. is free. Proof If A is a P.I.D., then any submodule of a free A-module is torsion-free. But the Fundamental Theorem of Finitely Generated Modules over a P.I.D. shows that any finitely generated torsion-free A-module is free. We shall show in Section 12.4 that every infinitely generated projective module over a P.I.D. is free. We return to giving corollaries of Proposition 9.8.6. Corollary 9.8.9. Projective modules are flat. Proof By Proposition 9.5.8, a direct summand of a flat module is flat. By Corollary 9.5.9 in the finitely generated case, or Problem 4 of Exercises 9.5.21 in the general case, free modules are flat. Corollary 9.8.10. Let f : A → B be a ring homomorphism and let P be a projective left A-module. Then B ⊗A P is a projective B-module. Also, if P is finitely generated as an A-module, then B ⊗A P is finitely generated as a B-module. Proof Because P is projective, there is an A-module Q such that P ⊕ Q is free over A. But then B ⊗A (P ⊕ Q), which is isomorphic to (B ⊗A P ) ⊕ (B ⊗A Q), is a free B-module, via Corollary 9.4.19 in the finitely generated case, or Problem 6 of Exercises 9.4.27 in


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the general case. Thus, B ⊗A P is a direct summand of a free B-module, and hence is projective. As for finite generation, if P is finitely generated, there is a surjective A-module homomorphism f : An → P . But then 1B ⊗ f gives a surjection from B n ∼ = B ⊗A An onto B ⊗A P , since tensoring is right exact. The next result should, by now, be expected. Proposition 9.8.11. Any direct sum of projective modules is projective. * Proof The simplest argument is just to apply Proposition 9.8.2. Let P = i∈I Pi , where each Pi is projective. Suppose given a surjective A-module homomorphism g : M → M and an A-module map h : P → M . Let ιi : Pi → P be the canonical inclusion and let hi = (h◦ιi ) : Pi → M . Since Pi is projective, there is a homomorphism hi : Pi → M with g ◦ hi = hi . By the universal property of the direct sum, there is a unique homomorphism h : P → M such that h ◦ ιi = hi for each i ∈ I. But then g ◦ h agrees with h on each summand Pi , and hence g ◦ h = h. We are primarily interested in finitely generated projective modules. There are a number of reasons for this, including the fact that many of the interesting applications of projectives are restricted to finitely generated phenomena. It is also the case that if we allow the consideration of infinitely generated phenomena, then the behavior of projectives becomes in a sense too simple. For instance, consider the following. Proposition 9.8.12. (Eilenberg Swindle) Let P be a projective module. Then there is a free module F such that P ⊕ F is free. Proof Since P is projective, it is isomorphic to a direct summand of a free module. Thus, there is a module Q (which must also be projective) such that P ⊕ Q is free. Let F be a direct sum of infinitely many copies of P ⊕ Q: F = (P ⊕ Q) ⊕ (P ⊕ Q) ⊕ · · · . Since P ⊕Q∼ = Q ⊕ P , we have P ⊕ F ∼ = P ⊕ (Q ⊕ P ) ⊕ (Q ⊕ P ) ⊕ · · · . But if we reassociate this sum, we get (P ⊕ Q) ⊕ (P ⊕ Q) ⊕ · · · , which is just F . Explicitly, P ⊕ F ∼ = F. We wish to be able to describe the projective modules over certain types of rings. Here, we consider local rings. Lemma 9.8.13. Let A be a local ring with maximal ideal m. Let f : M → N be an A-module homomorphism, with N finitely generated. Then f is onto if and only if 1 ⊗ f : A/m ⊗A M → A/m ⊗A N is onto. Proof We have an exact sequence f

π

M− →N − → C → 0, where π is the canonical map onto the cokernel of f . Since the sequence remains exact when tensored with A/m, it suffices that C = 0 if and only if A/m ⊗A C = 0. But this follows from Nakayama’s Lemma, since the Jacobson radical of a local ring is its maximal ideal. Projective modules over local rings are well behaved: Proposition 9.8.14. Let A be a local ring. Then every finitely generated projective A-module is free.


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Proof Let P be a finitely generated projective module over A and let x1 , . . . , xn be a minimal generating set for P in the sense that no proper subset of it generates P . Let f : An → P be the homomorphism that takes the canonical basis element ei to xi for i = 1, . . . , n. Then f is surjective. Let K be its kernel. We obtain an exact sequence i

π

0→K− → An − → P → 0, which splits, because P is projective. Thus, by Corollary 9.5.13, the sequence 1⊗i

1⊗π

0 → A/m ⊗A K −−→ A/m ⊗A An −−→ A/m ⊗A P → 0 is also split exact, where m is the maximal ideal of A. If a proper subset of 1 ⊗ x1 , . . . , 1 ⊗ xn were to generate A/m ⊗A P , then a proper subset of x1 , . . . , xn would generate P , by Lemma 9.8.13. Thus, 1 ⊗ x1 , . . . , 1 ⊗ xn must be a basis for the vector space A/m ⊗A P . But this says that 1 ⊗ f is an isomorphism, and hence A/m ⊗A K = 0. But then K = 0 by Nakayama’s Lemma, and hence P is the free module with basis x1 , . . . , xn . In fact, it is also true that every infinitely generated projective module over a local ring is free, but we shall not prove it here. Note that in the infinitely generated case, we would not be able to use Nakayama’s Lemma in our argument. Recall that if p is a prime ideal of a commutative ring A and if P is a finitely generated projective module over A, then Pp = Ap ⊗A P is a finitely generated projective Ap -module by Corollary 9.8.10. Thus, we may make the following. Definitions 9.8.15. Let P be a finitely generated projective module over the commutative ring A, and let p be a prime ideal of A. Then the p-rank of P is the rank of the finitely generated free module Pp over Ap . We say that P has constant rank if the p-rank of P is the same for each prime ideal p of A. The relationship between the ranks at various primes depends in part on the way that the prime ideals are nested. Proposition 9.8.16. Let A be an integral domain. Then every finitely generated projective A-module has constant rank. Proof Let P be a finitely generated A-module and let p be a prime ideal of A. Let K be the field of fractions of A. Suppose that Pp is free of rank r over Ap . Then K ⊗Ap Pp is an r-dimensional vector space over K. But K ⊗Ap Pp = K ⊗Ap Ap ⊗A P = K ⊗A P = P(0) , so the rank of P at (0) is r, also. Exercises 9.8.17. 1. Let A be a local ring with only one prime ideal, and let x1 , . . . , xm be a generating set for An . Suppose also that x1 , . . . , xk are linearly independent. Show that there is a subset of x1 , . . . , xm that contains x1 , . . . , xk and is a basis for An . † 2. Let A and B be rings. Show that every submodule of (A×B)n has the form M ×N , where M and N are submodules of An and B n , respectively. Show also that every finitely generated projective module of A × B has the form P × P , where P and P are finitely generated projective over A and B, respectively.


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3. What are all the finitely generated projective modules over Zn ? 4. Give an example of a commutative ring A and a finitely generated projective module P over A whose rank is not constant. 5. Let P be a finitely generated projective left A-module. Show that the dual module P ∗ is a finitely generated projective right A-module. 6. Let M = M1 ⊕ M2 be a direct sum of two A-modules. Show that M is reflexive if and only if M1 and M2 are reflexive. Deduce that finitely generated projective modules are reflexive.

9.9

The Grothendieck Construction: K 0

We show here how the finitely generated projective modules can be used to construct a functor, K0 , from rings to abelian groups, which detects, in a stable sense, the deviation of a finitely generated projective module from being free. Consider first the set F P (A) of isomorphism classes of finitely generated left projective modules. One may construct this set by first taking the set consisting of the direct summands of An for all n ≥ 0 (which is a set because the collection of all subsets of An is a set), and identifying any two such direct summands that are isomorphic. If P is a finitely generated projective over A, we denote its isomorphism class by [P ] ∈ F P (A). Notice that F P (A) is an abelian monoid under the operation [P ] + [Q] = [P ⊕ Q]. The identity element is the 0 module. The group we shall define here, called K0 (A), is an abelian group which satisfies a universal mapping property with respect to the abelian monoid F P (A): There is a monoid homomorphism η : F P (A) → K0 (A) with the property that if f : F P (A) → G is a monoid homomorphism, with G an abelian group, then there is a unique group homomorphism f : K0 (A) → G such that the following diagram commutes. f / G F P (A) t: JJ t t JJ tt η JJ$ tt f K0 (A) The universal mapping property is solved by a general construction called the Grothendieck construction. There are two different ways to build it. The one we give first is similar to the construction of rings of fractions. Suppose given an abelian monoid M , written in additive notation. The elements of the Grothendieck construction G(M ) are formal differences m − n of elements m, n ∈ M , in the same way that a ring of fractions consists of formal quotients a/s. Here, the element m − n represents the equivalence class of the ordered pair (m, n) under the equivalence relation that sets (m, n) ∼ (m , n ) if there is an r ∈ M such that m + n + r = m + n + r. We leave it to the reader to show that this is, in fact, an equivalence relation. Given that it is, we see that m − n = m − n in G(M ) if and only if m + n + r = m + n + r in M for some r ∈ M . As the reader may verify, setting (m − n) + (m − n ) = (m + m ) − (n + n ) gives a well defined binary operation on G(M ). Under this operation, G(M ) is an abelian monoid with identity element 0 = 0 − 0. But then n − m is an inverse for m − n, so that G(M ) is an abelian group. We also have a monoid homomorphism η : M → G(M ) defined by η(m) = m − 0.


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It is common to refer to G(M ) as the Grothendieck group of M . Note that m − 0 = m − 0 in G(M ) if and only if there is an m ∈ M such that m + m = m + m in M . This sets up the following definition and lemma. Definition 9.9.1. Let M be an abelian monoid. We say that elements m, m are stably equivalent in M if there is an element m ∈ M such that m + m = m + m in M . Lemma 9.9.2. Let M be an abelian monoid and let η : M → G(M ) be the canonical map to its Grothendieck group. Then elements m, m ∈ M become equal in G(M ) if and only if they are stably equivalent in M . The Grothendieck construction satisfies the following universal property. Proposition 9.9.3. Let M be an abelian monoid. Then for any monoid homomorphism f : M → G with G an abelian group, there is a unique group homomorphism f : G(M ) → G such that the following diagram commutes. f / G M? ? ?? ?? η ?? f G(M ) Proof Note that m − n = η(m) − η(n), so if f : G(M ) → G is a group homomorphism such that f ◦ η = f , then we must have f (m − n) = f (m) − f (n). Thus, it suffices to show that if f : M → G is a monoid homomorphism, then f (m−n) = f (m)−f (n) gives a well defined group homomorphism, as then automatically, (f ◦ η)(m) = f (m) − 0 = f (m) for all m ∈ M . But this is an easy check and is left to the reader. In fact, the group G in the last proposition need not be abelian. There is an alternative construction of the Grothendieck group which has a certain appeal, so we give it. Thus, for an abelian monoid M , let F (M ) be the free abelian group generated by M , with canonical basis {em | m ∈ M }. Let H ⊂ F (M ) be the subgroup generated by the elements {em1 +m2 − em1 − em2 | m1 , m2 ∈ M }. We define G (M ) = F (M )/H, and define η : M → G (M ) by η (m) = em . Note the fact that the generators of H have been set equal to 0 implies that η is a homomorphism. Proposition 9.9.4. The homomorphism η : M → G (M ) satisfies the same universal property as η: For any monoid homomorphism f : M → G with G an abelian group, there is a unique group homomorphism f : G (M ) → G such that the following diagram commutes. f / G M? ?? ? ?? ? η ?? f G (M ) In consequence, there is an isomorphism η : G (M ) → G(M ) with η ◦ η = η. Proof Since f : M → G is a function and G is abelian, there is a unique group homomorphism, f , from the free abelian group F (M ) to G such that f (em ) = f (m)


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for all m ∈ M . The fact that f is a monoid homomorphism shows that f vanishes on the generators of H, and hence H ⊂ ker f . Thus, there is a unique homomorphism f : G (M ) → G such that f ◦ π = f , where π : F (M ) → G (M ) is the canonical map. But now f ◦ η = f as desired. The uniqueness of such an f follows from the universal properties of free abelian groups and of factor groups. Since η : M → G(M ) and η : M → G (M ) are monoid homomorphisms, the universal properties of η and η give us unique homomorphisms η : G (M ) → G(M ) and η : G(M ) → G (M ) such that η ◦ η = η and η ◦ η = η , respectively. By the uniqueness statements for the universal properties, η ◦ η = 1G(M ) and η ◦ η = 1G (M ) . Definitions 9.9.5. Let A be a ring. Then K0 (A) is the Grothendieck group of the monoid F P (A) of isomorphism classes of finitely generated projective left A-modules. ) 0 (A) for K0 (A)/H, where H is the subgroup generated by [A] = [A] − 0. We write K ) 0 (A) is sometimes known as the projective class group of A, and sometimes as the K reduced projective class group, as some authors use the former term for K0 (A). Without ) 0 (A) as the reduced K0 group of A. ambiguity, we can refer to K We should immediately establish the functoriality of these groups. Proposition 9.9.6. Extension of rings makes K0 a functor from rings to abelian groups. Here, if f : A → B is a ring homomorphism, K0 (f )([P ]) = [B ⊗A P ]. ) 0 (f ) : K ) 0 (A) → K ) 0 (B). Since K0 (f )([A]) = [B], there is an induced map K Proof Let P be a finitely generated projective A-module. Then Corollary 9.8.10 shows us that B ⊗A P is finitely generated and projective over B. By the commutativity of direct sums with tensor products (Proposition 9.4.18), the passage from [P ] to [B ⊗A P ] induces a monoid homomorphism from F P (A) to F P (B). The universal property of the Grothendieck construction then gives a unique homomorphism K0 (f ), which has the stated effect on the generators, [P ]. Lemma 9.4.14 shows that K0 (1A ) is the identity map of K0 (A), so it suffices to show that K0 (g) ◦ K0 (f ) = K0 (g ◦ f ) for any ring homomorphism g : B → C. But this is immediate from Corollary 9.4.25. K0 (A) fits into a general theory called algebraic K-theory, in which functors Ki (A) are defined for each i ∈ Z. We shall study K1 in Section 10.9. ) 0 occur in many situations of importance in algebra, topology, and analysis. K0 and K For instance, if G is a finite group, and K is a field whose characteristic does not divide the order of G, then, using the theory of semisimple rings, one may show that K0 (K[G]) is isomorphic (as an abelian group) to the representation ring RK (G) of G over K. ) 0 (A) is the And we shall show in Chapter 12 that if A is a Dedekind domain, then K class group, Cl(A), which plays an important role in number theory. ) 0 (Z[G]) have numerous applications in topology. For instance, Wall The groups K has shown that for appropriate spaces X, there is a finiteness obstruction, σ(X) ∈ ) 0 (Z[π1 (X)]), which vanishes if and only if X is homotopy equivalent to a finite simplicial K ) 0 (Z[G]) play an important role in the calculation of complex. As a result, the groups K the surgery obstruction groups, which are used to classify manifolds. These last two examples have a useful connection. We shall see in Chapter 12 that Z[ζp ] is a Dedekind domain if p is prime. We shall not prove the following theorem here.


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Theorem (Rim’s Theorem) Let p be a prime and write Zp = T . Let f : Z[Zp ] → Z[ζp ] ) 0 (f ) is an isomorphism. be the ring homomorphism specified by f (T ) = ζp . Then K Note that every finitely generated projective module P determines an element [P ] = [P ]−0 in K0 (A). Since K0 (A) is an abelian group, it may actually be possible to compute with it, and detect whether [P ] = [Q] in K0 (A) for the finitely generated projective modules P and Q. In the best of all possible worlds, this would imply that P and Q were isomorphic (as would be the case if [P ] were equal to [Q] in F P (A), rather than K0 (A)). For instance, we shall see in Chapter 12 that if A is a semisimple ring, then [P ] = [Q] in K0 (A) if and only if P ∼ = Q. However, for general rings, we must fall back on Lemma 9.9.2, which tells us that [P ] = [Q] in K0 (A) if and only if P and Q are stably equivalent in F P (A), meaning that P ⊕ P ∼ = Q ⊕ P for some finitely generated projective module P of A. We shall make a formal definition for this relationship. Definitions 9.9.7. We say that the finitely generated projective modules P and Q are stably isomorphic if there is a finitely generated projective module P such that P ⊕ P ∼ = Q ⊕ P . We say that a finitely generated projective module P is stably free if P ⊕ An ∼ = Am for some m, n ≥ 0. Notice that P can be stably free without being stably equivalent to a free module, if it should happen that P ⊕ An ∼ = Am with n > m. But extension of rings tells us that this cannot happen if A admits a ring homomorphism to a division ring D, as then dim(D ⊗A P ) + n = m. The following lemma gives a characterization of stable isomorphism and discusses its ) 0 (A). We have already verified its initial statement. role in K0 (A) and K Lemma 9.9.8. Let P and Q be finitely generated projective A-modules. Then [P ] = [Q] in K0 (A) if and only if P and Q are stably isomorphic. Moreover, P and Q are stably isomorphic if and only if there is an isomorphism P ⊕ An ∼ = Q ⊕ An for some n ≥ 0. ) 0 (A) if and only if In consequence, the classes of P and Q become equivalent in K m ∼ n ) 0 (A) if P ⊕ A = Q ⊕ A for some nonnegative integers m and n. Thus, [P ] = 0 in K and only if P is stably free. Proof Suppose that P and Q are stably isomorphic, say P ⊕ P ∼ = Q ⊕ P for a finitely ∼ generated projective module P over A. But then P ⊕ P ⊕ Q = Q ⊕ P ⊕ Q for any Q . Since P is a direct summand of a finitely generated free module, we must have P ⊕ An ∼ = Q ⊕ An for some n. ) 0 (A) if and only if [P ] − [Q] = [Ar ] − [As ] in Now P and Q become equivalent in K K0 (A) for some r and s, meaning that P ⊕ As and Q ⊕ Ar are stably isomorphic. It is hard to conceive of the fact that P and Q may be stably isomorphic but not isomorphic. But there do exist examples where this is so. And it is often very difficult to verify that stable isomorphism implies isomorphism for a given ring. For instance, the following theorem, proven independently by Quillen and Suslin, was a longstanding conjecture, special cases of which were solved by some eminent mathematicians. It is still known as the Serre Conjecture, despite the fact that it has been proven.


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Theorem (Serre Conjecture) Let K be a field. Then every finitely generated projective module over a polynomial ring K[X1 , . . . , Xn ] is free. In this case, using the theory of K0 -groups, it is not hard to show that every finitely generated module over K[X1 , . . . , Xn ] is stably free. The hard part is showing that the stably free modules are all free. We shall not prove it here. Note, in light of the Eilenberg Swindle of Section 9.8, how important it is that P and Q are being summed with An , rather than an infinitely generated free module in the definition of stable isomorphism. So far we haven’t made any computations of K0 . First, consider the following. Lemma 9.9.9. Let N be the monoid of non-negative integers under addition. Then there is a commutative diagram i / Z N? ? ?? ı ?? = η ?? ∼ G(N) where η : N → G(N) is the canonical map to the Grothendieck construction, and i is the usual inclusion of N in Z. Proof Since i : N ⊂ Z is a monoid homomorphism, the universal property of the Grothendieck construction produces a group homomorphism ı : G(N) → Z which makes the diagram commute. Since Z is the free abelian group on the generator 1, there is a group homomorphism j : Z → G(N) such that j(1) = η(1) = 1 − 0. Then ı ◦ j(1) = 1, so that ı ◦ j = 1Z . But since the elements of N generate G(N), and since 1, in turn, generates N, j must be onto. Thus, ı and j are inverse isomorphisms. A slicker proof follows from the fact that N is the free monoid on 1, while Z is the free group on 1. In consequence i : N ⊂ Z satisfies the same universal property as η : N → G(N). Recall that the rank of a free module is not well defined for all rings (i.e., there are rings for which Am ∼ = An for some m = n), but that it is well defined, for instance, if A admits a ring homomorphism to a nonzero division ring. Corollary 9.9.10. Suppose that every finitely generated projective left A module is free, or, more generally, that every finitely generated projective left A module is stably free. Suppose in addition that the rank of a free A-module is well defined. Then K0 (A) is ) 0 (A) = 0. isomorphic to Z, with generator [A], and hence K Proof We assume throughout the argument that the rank of a free A-module is well defined. If every finitely generated projective module is free, then F P (A) is isomorphic to the monoid N of non-negative integers, and the result follows from Lemma 9.9.9. If the finitely generated projectives are only stably free, the generic element of K0 (A) has the form [P ] − [Q] = η([P ]) − η([Q]). But since P and Q are stably isomorphic to free modules, this element may be written in the form [An ] − [Am ] for some m, n ≥ 0. (Here, A0 is the 0 module, whose class gives the identity element of F P (A), and hence also of K0 (A).) Thus, [A] generates K0 (A) as a group. It suffices to show that the homomorphism Z → K0 (A) which takes 1 to [A] is injective. Since every nonzero


403

subgroup of Z is cyclic, generated by a positive element, it suffices to show that [An ] = 0 in K0 (A) for n > 0. Lemma 9.9.8 shows that [An ] = 0 in K0 (A) if and only if An+m ∼ = Am for some m ≥ 0. But this would contradict the statement that rank is well defined. Corollary 9.8.8 and Proposition 9.8.14 show that every finitely generated projective module over a P.I.D. or a local ring, respectively, is free. Corollary 9.9.11. If A is a P.I.D., a local ring, or a division ring, K0 (A) is the infinite ) 0 (A) = 0. cyclic group generated by [A] and K We shall make some additional computations of K0 in Chapter 12. We get some additional structure on K0 (A) if A is commutative. Proposition 9.9.12. Let A be a commutative ring. Then K0 (A) is also a commutative ring, with the multiplication given by [P ] · [Q] = [P ⊗A Q]. In addition, if f : A → B is a homomorphism of commutative rings, then K0 (f ) is a ring homomorphism as well. Thus, K0 gives a functor from commutative rings to commutative rings. Proof The key is to show that if P and Q are finitely generated projectives, then so is P ⊗A Q. To see this, suppose that P ⊕ P ∼ = An and Q ⊕ Q ∼ = Am . Then (P ⊕ P ) ⊗A mn (Q ⊕ Q ) is isomorphic to A by a couple of applications of the commutativity of tensor products and direct sum (Proposition 9.4.18). But the commutativity of tensor products with direct sums also shows that P ⊗A Q is a direct summand of (P ⊕ P ) ⊗A (Q ⊕ Q ). Proposition 9.4.18 also shows that for fixed Q, the map which sends [P ] to [P ⊗A Q] gives a monoid homomorphism from F P (A) to itself. By the universal property of the Grothendieck construction, this extends to a group homomorphism r[P ] : K0 (A) → K0 (A). Since P ⊗A Q is isomorphic to Q ⊗A P , we see that if [P ] − [Q] = [P ] − [Q ] in K0 (A), then r[P ] − r[Q] agrees with r[P ] − r[Q ] on the image of η : F P (A) → K0 (A). Since K0 (A) is generated by the image of η, we see that the tensor product induces a commutative binary operation on K0 (A) that satisfies the distributive law. Associativity of this multiplication follows from Corollary 9.4.24, and [A] is a multiplicative identity by Lemma 9.4.14. Finally, if f : A → B is a homomorphism of commutative rings, then K0 (f ) is a ring homomorphism by Problem 4 of Exercises 9.4.27. Corollary 9.9.13. Let A be a P.I.D. or a local ring. Then K0 (A) is isomorphic to Z as a ring. We mentioned earlier that if G is a finite group and K is a field whose characteristic does not divide the order of G, then K0 (K[G]) is naturally isomorphic as an abelian group to the representation ring RK (G) of G over K. Thus, if G is abelian, we have ring structures on K0 (K[G]) coming from either its structure as a representation ring or as K0 of a commutative ring. These ring structures are quite different. For instance, the multiplicative identity of the representation ring is K (with the trivial G-action), while the multiplicative identity of K0 (K[G]) is K[G]. Proposition 9.9.14. Let A be a ring that admits a homomorphism to a division ring. Then the sequence ∗ ) 0 (A) → 0 0 → K0 (Z) −→ K0 (A) → K

i

) 0 (A). is split short exact, and hence K0 (A) ∼ =Z⊕K If A is commutative and f : A → K is a homomorphism to a field, then we may ) 0 (A) with the kernel of f∗ , which is an ideal of K0 (A). identify K


404

Proof Let f : A → D be a homomorphism to a division ring. Then f ◦ i : Z → D induces an isomorphism of K0 , so f∗ may be used to construct a retraction for i∗ . Now apply Proposition 7.7.49. ) 0 (A) → K0 (A) induced by this retraction has image equal to Note that the section K the kernel of f∗ . The result follows. We shall see below that if A is a product of fields, say A = K1 × K2 , then the projection maps A → Ki have different kernels. Thus, there is no preferred splitting of ) 0 (A) with an the exact sequence above, nor is there an unambiguous identification of K ideal of K0 (A) for general commutative rings. If A is an integral domain, however, this problem does not occur. The next result can be useful. Proposition 9.9.15. Let A and B be rings. Then K0 (A × B) is naturally isomorphic to K0 (A) × K0 (B). If A and B are commutative, then this isomorphism is a ring isomorphism. Proof By Problem 2 of Exercises 9.8.17, every finitely generated projective module over A × B has the form P × P , where P is a finitely generated projective over A and P is a finitely generated projective over B. But this says that F P (A × B) is isomorphic to F P (A) × F P (B) as a monoid. Since finite products and coproducts agree in the categories of abelian groups and abelian monoids, it is easy to see that the Grothendieck construction on a product M ×M of abelian monoids is naturally isomorphic to G(M )× G(M ). If A and B are commutative, then the natural map (K0 (π1 ),K0 (π2 ))

K0 (A × B) −−−−−−−−−−→ K0 (A) × K0 (B) is a ring homomorphism, since K0 (f ) is a ring homomorphism for any homomorphism f of commutative rings. Thus, it suffices to show that (K0 (π1 ), K0 (π2 )) coincides with the isomorphism constructed above, i.e., that K0 (π1 )([P ×P ]) = [P ] and K0 (π2 )([P ×P ]) = [P ] if P and P are finitely generated projectives over A and B, respectively. We have K0 (π1 )([P × P ]) = [A ⊗A×B (P × P )]. A ⊗A×B (P × P ) ∼ = (A × B/0 × B) ⊗A×B (P × P ) ∼ = (P × P )/(0 × B)(P × P ) ∼ = P Here, the second isomorphism is from Corollary 9.4.16. The case of π2 is analogous. ) 0 (A × B) does not split as a product the way that K0 (A × B) does. Notice that K Thus, even if one is primarily interested in phenomena regarding the reduced K0 groups, it is sometimes useful to consider the unreduced groups as well. ) 0 (A) which can be useful. We define an There is an alternative description of K equivalence relation ≈ on F P (A) by setting [P ] ≈ [Q] if P ⊕ An is isomorphic to Q ⊕ Am / for some m and n. Write F P (A) for the set of equivalence classes on F P (A) given by / this relation, and write [[P ]] ∈ F P (A) for the equivalence class of [P ]. / Lemma 9.9.16. F P (A) is an abelian group under the operation [[P ]]+[[Q]] = [[P ⊕Q]].


405

Proof The direct sum operation in F P (A) is easily seen to respect equivalence classes / with respect to the relation ≈, and therefore induces a binary operation on F P (A). It inherits commutativity, associativity, and an identity element from F P (A), so it suffices to show that every element has an inverse. But if [P ] ∈ F P (A), then there is a finitely generated projective module Q such that P ⊕ Q is isomorphic to An for some n. Since / [[An ]] = [[0]] = 0 in F P (A), [[Q]] is the inverse of [[P ]]. A nice application of this is the next corollary, which only requires the uniqueness of inverses in a group. Corollary 9.9.17. Let P be a finitely generated projective module over A, and let Q and Q be finitely generated projective modules such that P ⊕ Q and P ⊕ Q are both free. Then Q ⊕ An ∼ = Q ⊕ Am for some m, n ≥ 0. / Note that F P is a functor via extension of rings. ) 0 and F / Proposition 9.9.18. There is a natural isomorphism between the functors K P. ) 0 (A), there is a homomorphism ν : F / ) 0 (A) defined Proof Since [An ] = 0 in K P (A) → K by ν([[P ]]) = [P ]. Clearly, this is natural in A. Since K0 (A) is generated by the elements of F P (A), ν is onto. Let [[P ]] be in the kernel of ν. Then Lemma 9.9.8 shows that P is / stably free. But that says that [P ] ≈ [0], and hence that [[P ]] = 0 in F P (A). Exercises 9.9.19. 1. Calculate K0 (Zn ). 2. Let f : A → B be a ring homomorphism, where A is commutative. Show that K0 (B) is a module over K0 (A). 3. Let A be a commutative ring and let p be a prime ideal of A. Show that we may ) 0 (A) with the kernel of K0 (η), where η : A → Ap is the canonical map. identify K 4. Let A be a commutative ring and let P be a finitely generated projective A-module. Show that [P ] = 0 in K0 (A) if and only if P is the 0-module. (Hint: Show that Pp = 0 for each prime ideal p.) Note that this gives us no information about the properties of stable isomorphism between nonzero modules. 5. Verify, using the first construction of the Grothendieck group, that if f : M → G is a monoid homomorphism, where M is abelian, but G is not, then there is a unique group homomorphism f : G(M ) → G such that f ◦ η = f . 6. Show that the second construction of the Grothendieck group may be (drastically) modified to work in the nonabelian world. Thus, if M is a possibly nonabelian monoid, we can construct a group G (M ) and a monoid homomorphism η : M → G (M ) which is universal in the sense that if f : M → G is a monoid homomorphism into a group G, then there is a unique group homomorphism f : G (M ) → G such that f ◦ η = f . Now using the preceding exercise, show that if M is abelian, there is a group isomorphism η : G(M ) → G (M ) such that η ◦ η = η .


9.10

406

Tensor Algebras and Their Relatives

Throughout this section, A is a commutative ring. Here, we describe the tensor algebra of an A-module M , together with some algebras derived from it. The tensor algebra is the free A-algebra on an A-module M . Definition 9.10.1. Let M be an A-module. Then the tensor algebra, TA (M ), of M over A is given by , M ⊗k , TA (M ) = k≥0

where

M ⊗k

⎧ ⎨ =

A M ⊗A · · · ⊗A M ⎩

if k = 0 if k > 0.

k times

The product in TA (M ) is induced by setting (m1 ⊗ · · · ⊗ mk ) · (m1 ⊗ · · · ⊗ ml ) = m1 ⊗ · · · ⊗ mk ⊗ m1 ⊗ · · · ⊗ ml and by letting A act via its usual module structure on M ⊗k . One way of interpreting this product is by noting that it may be obtained from the natural isomorphisms M ⊗k ⊗A M ⊗l ∼ = M ⊗k+l . In particular, it is well defined, and satisfies the distributive laws. It is easy to see that TA (M ) is an A-algebra via this product. The easiest example to analyze is the tensor algebra of A itself. Notice that if we think of A as the free A-module on a basis element e1 , then A⊗k is the free A-module on ek1 = e1 ⊗ · · · ⊗ e1 . k times

We obtain the following lemma. Lemma 9.10.2. The A-algebra homomorphism εe1 : A[X] → TA (A) obtained by evaluating X at e1 is an isomorphism. To develop the universal properties of tensor algebras more generally, let i : M → TA (M ) be the canonical inclusion of M = M ⊗1 . Then the following proposition shows that TA (M ) is the free A-algebra on M . Proposition 9.10.3. Let B be an A-algebra and let f : M → B be an A-module homomorphism. Then there is a unique A-algebra homomorphism f : TA (M ) → B that makes the following diagram commute. f / B M? ?? ? ?? ? i ?? f TA (M )


407

Proof Just set f (m1 ⊗ · · · ⊗ mk ) = f (m1 ) . . . f (mk ). As the right-hand side is Amultilinear in m1 , . . . , mk , this specifies an A-module homomorphism, which is then easily seen to be a ring homomorphism. In particular, note that the elements of M = M ⊗1 generate TA (M ) as an A-algebra. We shall be able to construct A-algebras with other kinds of universal properties as quotient rings of TA (M ). Indeed, every A-algebra is a quotient of a tensor algebra. Definition 9.10.4. Let M be an A-module and let a be the two-sided ideal of TA (M ) generated by all elements of the form m1 ⊗ m2 − m2 ⊗ m1 , with m1 , m2 ∈ M . Then the symmetric algebra, SA (M ), of M over A is the quotient ring TA (M )/a. Example 9.10.5. Since TA (A) is commutative, it is easy to see that TA (A) ∼ = SA (A) ∼ = A[X]. Let ı : M → SA (M ) be the composite of i : M → TA (M ) with the canonical map of the tensor algebra onto the symmetric algebra. Then the following proposition shows that SA (M ) is the free commutative A-algebra on M . Proposition 9.10.6. Let B be a commutative A-algebra and let f : M → B be an Amodule homomorphism. Then there is a unique A-algebra homomorphism f : SA (M ) → B that makes the following diagram commute. f / B M? ?? ? ?? ? ı ?? f SA (M ) Proof By Proposition 9.10.3, there is a unique A-algebra homomorphism f : TA (M ) → B such that f ◦ i = f . Since B is commutative, f (m1 ⊗ m2 − m2 ⊗ m1 ) = 0 for all m1 , m2 ∈ M . Thus, the ideal a is contained in the kernel of f , and hence f factors uniquely through the canonical map from TA (M ) to SA (M ). The next example, the exterior algebra, is easiest to understand in the context of graded algebras, so we shall develop these first. Definitions 9.10.7. A graded A-module is an A-module M with an explicit direct sum decomposition M=

∞ ,

Mi

i=0

as an A-module. The elements of the submodule Mi are called the homogeneous elements of degree i. A graded homomorphism f : M → N between graded modules is one with the property that f (Mi ) ⊂ Ni for all i ≥ 0. A graded A-algebra, B, is a graded A-module which is an A-algebra (compatible with the A-module structure) under a multiplication satisfying 1. 1 ∈ B0 . 2. If b1 ∈ Bi and b2 ∈ Bj , then b1 b2 ∈ Bi+j .


408

A graded A-algebra B is called skew commutative (or skew symmetric, or commutative in the graded sense) if for b1 ∈ Bi and b2 ∈ Bj , we have b1 b2 = (−1)ij b2 b1 Skew commutative algebras are important in algebraic topology and in homological algebra, as the cohomology ring of a topological space is skew commutative, as are most of the cohomology rings that arise purely algebraically. We shall see in Exercises 9.10.28 that if 2 is a unit in A, then the exterior algebra on M is the free skew commutative algebra on M , if we take M to be a graded module whose elements are all homogeneous of degree 1. Examples 9.10.8. 1. The tensor algebra TA (M ) may be considered as a graded A-algebra with (TA (M ))i = M ⊗i . 2. Specializing the preceding example to M = A gives an isomorph of the following: The polynomial ring A[X] may be considered as a graded ring in which (A[X])i = {aX i | a ∈ A}. Thus, the homogeneous elements are the monomials. 3. Let X1 , . . . , Xn be variables, and choose arbitrary non-negative integers ki , 1 ≤ i ≤ n, for the degree of Xi . Then the polynomial ring A[X1 , . . . , Xn ] has a graded algebra structure in which the monomial aX1r1 . . . Xnrn is a homogeneous element of degree r1 k1 + · · · + rn kn . We would like to show that the symmetric algebras are graded algebras as well. To see this, we should define graded ideals. Definitions 9.10.9. Let M be a graded A-module. Then a graded submodule, N , of M is a direct sum N=

∞ ,

Ni

i=0

where Ni is a submodule of Mi . If B is a graded A-algebra, then a graded ideal, a, of B is an ideal of B which is graded as a submodule of B. Let M be a graded A-module. Notice that a submodule N of M is a graded submodule if and only if for each n ∈ N , the component of n lying in each* Mi is also in N . (Here, ∞ we mean the component in the direct sum decomposition M = i=0 Mi .) Lemma 9.10.10. Let B be a graded A-algebra and let S ⊂ B be any subset consisting of homogeneous elements of B. Then the ideal of B generated by S is a homogeneous ideal. *∞ i=0 ai is a graded ideal of B, then the graded module B/a = *∞Of course, if a = B /a is a graded A-algebra. Since the elements m1 ⊗ m2 − m1 ⊗ m1 of the tensor i i=0 i algebra are homogeneous of degree 2 in the standard grading, we obtain the following corollary. Corollary 9.10.11. The symmetric algebra SA (M ) has a natural grading in which the elements of M are homogeneous of degree 1.


409

Notice that if all the nonzero homogeneous elements of a graded algebra have even degree, then the concepts of commutativity and skew commutativity coincide. Thus, if we set M ⊗k to have homogeneous degree 2k in TA (M ) for all k ≥ 0, then the resultant grading on SA (M ) gives an algebra that is skew commutative as well as commutative. Definition 9.10.12. Let M be an A-module and let b ⊂ TA (M ) be the two-sided ideal generated by all elements of the form m ⊗ m, with m ∈ M . Then the exterior algebra ΛA (M ) is the quotient ring TA (M )/b. We write ΛkA (M ) for the image in ΛA (M ) of M ⊗k . We write m1 ∧ · · · ∧ mk for the image in ΛkA (M ) of the element m1 ⊗ · · · ⊗ mk of TA (M ). If we grade TA (M ) by setting M ⊗i to be the homogeneous elements of degree i, then b is homogeneous, inducing a grading on ΛA (M ) in which ΛiA (M ) is the module of homogeneous elements of degree i. We shall take this as the standard grading on ΛA (M ). As above, the simplest example to work out is ΛA (A). Here, ae1 ⊗ ae1 is a multiple of e1 ⊗ e1 , so b is the ideal generated by e1 ⊗ e1 . Utilizing the isomorphism of Lemma 9.10.2, which preserves grading, we see that ΛA (A) is isomorphic as a graded algebra to A[X]/(X 2 ), which has basis 1, X as an A-module, with X 2 = 0. Lemma 9.10.13. Think of A as the free module on e1 . Then Λ0A (A) and Λ1A (A) are free on 1 and e1 , respectively, while ΛkA (A) = 0 for k > 1. For general A-modules M , we have the following facts about ΛA (M ). Lemma 9.10.14. Let M be an A-module. Then the exterior algebra ΛA (M ) is skew commutative. Moreover, the square of any element of Λ2k+1 (M ) is zero, for all k ≥ 0. A Proof For m1 , m2 ∈ M , we have (m1 + m2 )2 = m21 + m1 ∧ m2 + m2 ∧ m1 + m22 . But the squares of elements of M are in b and hence are trivial in ΛA (M ), and hence m1 ∧ m2 = −m2 ∧ m1 for all m1 , m2 ∈ M . But an easy induction now shows that (m1 ∧ · · · ∧ mk ) ∧ (m1 ∧ · · · ∧ ml ) = (−1)kl (m1 ∧ · · · ∧ ml ) ∧ (m1 ∧ · · · ∧ mk ). The distributive law now shows ΛA (M ) to be skew commutative. Finally, if α, β ∈ Λ2k+1 (M ), then the fact that they anticommute shows that (α + A 2 β) = α2 + β 2 . Clearly, any element of the form m1 ∧ · · · ∧ m2k+1 squares to 0. Since the elements of this form generate Λ2k+1 (M ), the squaring operation on Λ2k+1 (M ) must be A A trivial. Notice that there are no hypotheses about A in the preceding result. If 2 were invertible in A, then the statement regarding the squares of odd-degree elements would follow from the skew commutativity. The point is that skew commutativity implies that if x is homogeneous of odd degree, then x2 = −x2 . Lemma 9.10.15. Let A be a commutative ring in which 2 is invertible, and let B be a skew commutative graded A-algebra. Then the square of any homogeneous element of odd degree is trivial.


410

Note that all we really need is that 2 fails to be a zero-divisor in B. Thus, this sort of analysis applies to situations where A is a domain of characteristic = 2 and B is free as an A-module. We apply the results above to the study of universal properties of exterior algebras. Let M be an A-module. We may consider it to be a graded A-module in which all elements are homogeneous of degree 1. We call this the graded A-module in which M is concentrated in degree 1. We write ı : M → ΛA (M ) for the composite of the natural inclusion of M = M ⊗1 in TA (M ) with the canonical map onto the quotient ring ΛA (M ). Note that ı is a graded map from the above grading on M to the standard grading on ΛA (M ). Also, since the ideal b of TA (M ) is generated by elements of degree 2, ı is an isomorphism of M onto Λ1A (M ). As in the proof of Proposition 9.10.6, the following proposition is immediate from the universal property of the tensor algebra. Proposition 9.10.16. Let M be an A-module, graded by concentration in degree 1 and let B be a graded A-algebra. Let f : M → B be a graded A-module map with the property that f (m)2 = 0 for each m ∈ M . Then there is a unique graded A-algebra homomorphism f : ΛA (M ) → B that makes the following diagram commute. f / B M? ? ?? ?? ? ı ?? f ΛA (M ) Here, ΛA (M ) is given the standard grading. If 2 is invertible in A and if B is skew commutative, then every graded A-algebra map f : M → B has the above property. Thus, ΛA (M ) is the free skew commutative A-algebra on M when 2 is a unit in A. We wish to determine the structure of ΛA (An ). One way to do this is via determinant theory, as exterior algebras turn out to be a generalization of determinants. But the argumentation turns out to be simpler if we develop the theory of skew commutative algebras a bit more. By this method, we shall develop some more powerful results, as well as shed some light on determinant theory. We first give an alternative formulation of the tensor product of graded algebras. Definitions 9.10.17. Let M and N be graded A-modules. Then the preferred grading on M ⊗A N is obtained by setting (M ⊗A N )k =

k ,

Mi ⊗A Nk−i .

i=0

A C for the algebra structure on Let B and C be graded A-algebras. Write B ⊗ B ⊗A C obtained as follows: If b1 , b2 , c1 , c2 are homogeneous elements, then b1 ⊗ c1 · b2 ⊗ c2 = (−1)|b2 |·|c1 | b1 b2 ⊗ c1 c2 , where |b2 | and |c1 | stand for the degrees of b2 and c1 , respectively. A C as the graded The standard terminology in homological algebra is to refer to B ⊗ tensor product of B and C. We shall follow this convention.


411

Warning: The graded tensor product is generally not isomorphic as an algebra to the usual algebra structure on B ⊗A C (the one from Proposition 9.6.1). The latter, which is always commutative if B and C are commutative, can be thought of as the graded tensor product if B and C are both concentrated in degree 0 (or in even degrees). If we restrict attention to skew commutative A-algebras, the graded tensor product has an important universal property. Note first that it is easy to see that if B and C are A C. skew commutative, then so is B ⊗ Proposition 9.10.18. The graded tensor product is the coproduct in the category of skew commutative A-algebras. In other words, if B, C, and D are skew commutative A-algebras and if f : B → D and g : C → D are graded A-algebra homomorphisms, A C → D such that the then there is a unique graded A-algebra homomorphism h : B ⊗ following diagram commutes. ι1 / ι2 A C o B ? C B⊗ ?? ?? ?? ?? h f ?? g ? D Here, ι1 (b) = b ⊗ 1 and ι2 (c) = 1 ⊗ c. Proof Since b ⊗ 1 · 1 ⊗ c = b ⊗ c, if there is a graded A-algebra homomorphism h making A C is generated by the diagram commute, we must have h(b ⊗ c) = f (b)g(c). Since B ⊗ the elements of the form b ⊗ c, the homomorphism h is uniquely defined by the diagram. For the existence of h, note that setting h(b ⊗ c) = f (b)g(c) specifies a graded A A C. And if b1 , b2 , c1 , and c2 are homogeneous, then the module homomorphism on B ⊗ skew commutativity of D shows that h(b1 ⊗ c1 · b2 ⊗ c2 ) = h(b1 ⊗ c1 ) · h(b2 ⊗ c2 ). A C is generated by elements b ⊗ c with b and c homogeneous, h is a homoSince B ⊗ morphism of graded algebras. For an A-module M , we have Λ0A (M ) = A and Λ1A (M ) = M . Thus, if M and N are A ΛA (N ) are given A-modules, then the homogeneous elements in degree 1 in ΛA (M ) ⊗ by (M ⊗A A) ⊕ (A ⊗A N ) ∼ = M ⊕ N . This isomorphism gives the degree 1 part of the A-algebra isomorphism in the following proposition. Proposition 9.10.19. Let M and N be A-modules. Then there is a natural isomorphism of graded rings: A ΛA (N ) ∼ ΛA (M ) ⊗ = ΛA (M ⊕ N ). ∼

= A ΛA (N ))1 −→ Proof Write α : (ΛA (M ) ⊗ M ⊕N for the isomorphism described above. Then Propositions 9.10.16 and 9.10.18 provide a unique graded A-algebra homomorphism

A ΛA (N ) → ΛA (M ⊕ N ) α : ΛA (M ) ⊗ which restricts to α in degree 1, while Proposition 9.10.16 provides a unique graded A-algebra homomorphism, β, in the opposite direction, restricting to α−1 in degree 1. Thus, both α ◦ β and β ◦ α restrict to the identity in degree 1. Since both algebras are generated by their elements of degree 1, β and α must be inverse isomorphisms.


412

Thus, given our analysis of ΛA (A), we are now able to compute ΛA (An ). Proposition 9.10.20. Let n ≥ 1. Then for 1 ≤ k ≤ n, ΛkA (An ) is a free A-module of n rank k and with basis given by any ordering of {ei1 ∧ · · · ∧ eik | i1 < · · · < ik }. Here, e1 , . . . , en is the canonical basis for An . For k > n, ΛkA (An ) = 0, while Λ0A (An ) is, of course, A. Proof Write An = Ae1 ⊕ · · · ⊕ Aen . Then Proposition 9.10.19 gives an isomorphism of graded A-algebras A · · · ⊗ A ΛA (Aen ) → ΛA (An ) α : ΛA (Ae1 ) ⊗ which carries ei1 ⊗ · · · ⊗ eik onto ei1 ∧ · · · ∧ eik . The result now follows easily from Lemma 9.10.13: Since Λ0A (A) = A and ΛkA (A) = 0 for k > 1, the homogeneous terms of A · · · ⊗ A ΛA (Aen ) are given by the terms degree k in ΛA (Ae1 ) ⊗ , Λ1A (Aei1 ) ⊗A · · · ⊗A Λ1A (Aeik ). i1 1, the first row of M is (a11 , . . . , a1m , 0, . . . , 0), where (a11 , . . . , a1m ) is the first row of M . Thus, det M = a11 det M11 − a12 det M12 + · · · + (−1)1+m a1m det M1m . But for j ≤ m, we can write M1j as a block sum: M1j = M1j ⊕ M . Since M1j is (m − 1) × (m − 1), our induction hypothesis gives det M1j = (det M1j )(det M ), and hence m (−1)1+j a1j (det M1j )(det M ) det M = j=1

= (det M )

m

(−1)1+j a1j (det M1j )

j=1

= (det M )(det M ) The fact that matrices whose determinants are units are always invertible, together with the product rule for determinants, shows that the set of n×n matrices of determinant 1 forms a subgroup of Gln (A). Definition 10.3.8. We write Sln (A) for the group of n × n matrices over A of determinant 1. We call it the n-th special linear group of A. Special linear groups tend to contain the hard part in terms of our understandings of general linear groups: Proposition 10.3.9. For any commutative ring A and any n > 0, we have a split short exact sequence of groups ⊂

0 → Sln (A) −→ Gln (A) −−→ A× → 0. det

Proof It suffices to construct a section for det : Gln (A) → A× . We define s : A× → Gln (A) by s(a) = (a) ⊕ In−1 . Proposition 10.3.7 shows that det s(a) = a, and the result follows. Recall that if N is a finitely generated free A-module and if f ∈ EndA (N ), then the matrices of f with respect to any two bases are similar. Definition 10.3.10. Let N be a finitely generated free A-module and let f ∈ EndA (N ). Then the determinant of f , det f , is the determinant of the matrix of f with respect to any basis of N . Let N be a free module of rank n. Then a choice of basis for n gives a ring isomorphism from EndA (N ) to Mn (A). Recall that the unit group of EndA (N ) is called AutA (N ). Theorem 10.3.11. Let A be a commutative ring and let N be a finitely generated free A-module. Then the determinant gives a function det : EndA (N ) → A with the following properties.

CHAPTER 10. LINEAR ALGEBRA

427

(1) det(f ◦ g) = (det f )(det g). (2) An endomorphism f ∈ EndA (N ) is an isomorphism if and only if det f ∈ A× . (3) Restriction gives a group homomorphism det : AutA (N ) → A× . (4) If f ∈ EndA (N ) and g ∈ AutA (N ), then det(gf g −1 ) = det f . (5) If N = N1 ⊕ N2 , where N1 and N2 are finitely generated free modules, and if fi ∈ EndA (Ni ) for i = 1, 2, then det(f1 ⊕ f2 ) = (det f1 )(det f2 ). (6) Let f : A → B be a ring homomorphism, with B commutative. Then for g ∈ EndA (N ), the endomorphism 1B ⊗ g of B ⊗A N satisfies det(1B ⊗ g) = f (det g). Proof All these assertions but the last should be clear from the results in this section. The last follows from Corollary 9.4.20. Exercises 10.3.12. 1. Let K be a field and let V be a finite dimensional vector space over K. Let f ∈ EndK (V ). Show that det f = 0 if and only if ker f = 0. 2. Using rings of fractions, show that the preceding problem shows that if A is an integral domain and if N is a finitely generated free A-module, then f ∈ EndA (N ) has determinant 0 if and only if ker f = 0. 3. Show that if A is not an integral domain, then there are endomorphisms of free modules whose determinant is nonzero, but whose kernel is also nonzero. 4. Cramer’s Rule Let A be a commutative ring and let M ∈ Mn (A). Show that for y ∈ An , the i-th coordinate of M adj y is equal to the determinant of the matrix obtained by replacing the i-th column of M with y. Deduce that if det M ∈ A× , then the i-th coordinate of the solution of M x = y is equal to (det M )−1 times the determinant of the matrix obtained by replacing the i-th column of M with y. 5. Show that if M is a diagonal matrix (i.e., the off-diagonal entries are all 0), then the determinant of M is the product of the diagonal entries. M X † 6. Show that if M = , then det M = (det M )(det M ). Show the same 0 M M 0 for M = . Y M 7. Write fM : An → An for the linear transformation by the matrix M ∈ induced M X with M ∈ Mk (A) Mn (A). Show that the matrix M has the form 0 M and M ∈ Mn−k (A) if and only if fM (Ak ) ⊂ Ak , where Ak is the submodule of An generated by e1 , . . . , ek . 8. Let N, N1 , and N2 be free modules of finite rank, and suppose that the rows of the following commutative diagram are exact. 0

/ N1

⊂

f1

0

/ N1

/N

π

f ⊂

/N

/ N2

/0

f2 π

/ N2

/0

Show that tr(f ) = tr(f1 ) + tr(f2 ) and det f = det f1 · det f2 .


428

9. Permutation Matrices. For σ ∈ Sn , let Mσ ∈ Mn (A) be the matrix whose i-th column is eσ(i) . (a) Show that the passage from σ to Mσ gives an injective group homomorphism η : Sn → Gln (A): η(σ) = Mσ . (b) Show that η induces an action of Sn on An under which each σ carries ei to eσ(i) for 1 ≤ i ≤ n. (c) Show that det Mσ = ε(σ). (d) Let xi be the i-th column of M ∈ Mn (A) for i = 1, . . . , n. Show that the i-th column of M · Mσ is xσ(i) for all σ ∈ Sn and all i = 1, . . . , n. Compare this to the action of Sn on (An )n (i.e., the matrices, when considered as n-tuples of columns) given by Problem 23 of Exercises 3.3.23. (e) Let yi be the i-th row of M ∈ Mn (A) for i = 1, . . . , n. Show that the i-th row of Mσ · M is yσ−1 (i) . 10. The Vandermonde Determinant. Define Mn ⎛ 1 X1 X12 . . . ⎜ 1 X2 X22 . . . ⎜ Mn = ⎜ .. ⎝ . 1 Xn Xn2 . . .

∈ Mn (Z[X1 , . . . , Xn ]) by ⎞ X1n−1 X2n−1 ⎟ ⎟ ⎟. ⎠ Xnn−1

The element det Mn ∈ Z[X1 , . . . , Xn ] is called the Vandermonde Determinant. As shown in Proposition 7.3.23, we have ring isomorphisms Z[X1 , . . . , Xn ] ∼ = (Z[X1 , . . . , Xj−1 , Xj+1 , . . . , Xn ]) [Xj ] for j = 1, . . . , n, where the isomorphism takes each Xi to Xi . (a) Show that if i < j, then Xj − Xi divides det Mn in Z[X1 , . . . , Xn ]. ! (b) Show that det Mn is divisible by 1≤i<j≤n (Xj − Xi ). (Here, if n = 1, the latter is just the empty product, which we set equal to 1.) (c) Show by induction on n that the Vandermonde Determinant is given by . (Xj − Xi ). det Mn = 1≤i<j≤n

! For the inductive step, view both det Mn and 1≤i<j≤n (Xj − Xi ) as polynomials in Xn with coefficients in Z[X1 , . . . , Xn−1 ]. Show that each one is a polynomial of degree n − 1 in Xn with leading coefficient det Mn−1 . (d) Let K be a field and let x1 , . . . , xn be distinct elements of K. Let y1 , . . . , yn be any collection of (not necessarily distinct) elements of K. Show that there is a unique polynomial f (X) ∈ K[X] of degree n − 1 such that f (xi ) = yi for all i. ∼ =

11. Let P be a finitely generated projective module over A, and let α : P ⊕ Q −→ An . Define ι : EndA (P ) → EndA (P ⊕ Q) by ι(f ) = f ⊕ 1Q and define α∗ : EndA (P ⊕ Q) → EndA (An ) by α∗ (g) = α ◦ g ◦ α−1 . Define det : EndA (P ) → A to be the composite ι

α

det

∗ EndA (P ) − → EndA (P ⊕ Q) −→ EndA (An ) −−→ A.


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(a) Show that det : EndA (P ) → A is independent of the choice of isomorphism from P ⊕ Q to An . (b) Show that if we replace Q by Q⊕A and replace α by α ⊕1A , then the function det : EndA (P ) → A doesn’t change. (c) Deduce from Corollary 9.9.17 that if Q is any other finitely generated projective such that P ⊕ Q is free, then the above procedure applied to an isomorphism P ⊕ Q ∼ = Am will produce exactly the same function det : EndA (P ) → A. (d) Show that det : EndA (P ) → A satisfies all the properties listed in Theorem 10.3.11. 12. Let f ∈ EndA (N ), and let Λk (f ) be the endomorphism of ΛkA (N ) which restricts on generators to take each x1 ∧ · · · ∧ xk to f (x1 ) ∧ · · · ∧ f (xk ). Suppose that N is a free module of rank n. Thus, by Proposition 9.10.20, ΛnA (N ) is a free module of rank 1. (a) Deduce that Λn (f ) must be induced by multiplication by some a ∈ A. (b) Let x1 , . . . , xn be a basis for N and let M be the matrix of f with respect to this basis. Show, via Proposition 10.2.3, that f (x1 ) ∧ · · · ∧ f (xn ) = det M · x1 ∧ · · · ∧ xn . (c) Use this to give a new proof that if M and M are matrices for f with respect to two different bases, then det M = det M . In particular, note that we could have defined the determinant of an endomorphism f , simply by stipulating that det f is the a of part (a) above. Note that if x1 , . . . , xn is a basis for N , then det f is characterized by the equality f (x1 ) ∧ · · · ∧ f (xn ) = det f · x1 ∧ · · · ∧ xn . (d) Use the properties of Λn (f ) to give a quick proof that det(f ◦ g) = det f · det g. (e) Use the relationship between exterior algebras and direct sums to show that det(f ⊕ g) = det f · det g, where f and g are endomorphisms of free modules of rank n and k, respectively.

10.4

The characteristic polynomial

Let A be a commutative ring and let ι : A → A[X] be the natural inclusion. As is customary, we write ι(a) = a, and identify A as a subring of A[X]. Similarly, we consider Mn (A) to be a subring of Mn (A[X]) via ι∗ : Mn (A) → Mn (A[X]), and write M , rather than ι∗ (M ), for the image of M ∈ Mn (A) under this inclusion. Thus, if M ∈ Mn (A), we may consider the element XIn − M ∈ Mn (A[X]). We have the determinant map det : Mn (A[X]) → A[X], and hence det(XIn − M ) is a polynomial. Definition 10.4.1. Let M ∈ Mn (A). Then the characteristic polynomial of M is the polynomial chM (X) = det(XIn − M ). The characteristic polynomial turns out to be of greatest use in the analysis of matrices over a field. Note, however, that even if A is a field, the study of characteristic polynomials will involve the use of determinants over the commutative ring A[X]. Thus, it is not enough for the purposes of field theory to study determinants over fields alone.


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Lemma 10.4.2. Let M ∈ Mn (A). Then chM (X) is a monic polynomial of degree n. Proof Write M = (aij ). Then (XIn − M ) = (fij (X)), where if i = j −aij fij (X) = X − aii if i = j. Thus, fij (X) has degree ≤ 0 if i = j and has degree 1 if i = j. Now consider the expansion of det(XIn − M ) in terms of permutations. For every permutation other than the identity, at least two of the fσ(i),i must have degree ≤ 0. Thus, the product of terms associated with each nonidentity permutation is a polynomial of degree < n − 1. But the term corresponding to the identity permutation is (X − a11 ) . . . (X − ann ), a monic polynomial of degree n. Thus, det(XIn − M ) is also a monic polynomial of degree n. We shall summarize some of the most basic properties of characteristic polynomials, and then specialize to some of the applications for matrices over a field. Lemma 10.4.3. Let M and M be similar matrices in Mn (A). Then chM (X) = chM (X). Proof Suppose that M M (M )−1 = M . Then in Mn (A[X]), we have M (XIn − M )(M )−1 = M XIn (M )−1 − M M (M )−1 = XIn − M , since XIn is in the center of Mn (A[X]). Thus, XIn − M and XIn − M are similar in Mn (A[X]), and hence have the same determinant. Thus, we may define the characteristic polynomial of an endomorphism. Definition 10.4.4. Let N be a finitely generated free A-module and let f ∈ EndA (N ). Then the characteristic polynomial chf (X) of f is equal to the characteristic polynomial of the matrix of f with respect to any chosen basis of N . Let M ∈ Mn (A) and M ∈ Mk (A). Since XIn is a diagonal matrix, XIn −(M ⊕M ) is the block sum of XIn − M and XIn − M . We may now apply Proposition 10.3.7: Lemma 10.4.5. chM ⊕M (X) = chM (X) · chM (X). The roots of the characteristic polynomial have an important property. Lemma 10.4.6. Let M ∈ Mn (A). Then if we evaluate the characteristic polynomial of M at a ∈ A, we get chM (a) = det(aIn − M ), the determinant of the matrix (aIn − M ) ∈ Mn (A). In particular, a is a root of chM (X) if and only if det(aIn − M ) = 0. Proof Let ε : A[X] → A be the A-algebra homomorphism obtained by evaluating X at a. Then Proposition 10.2.13 gives a commutative diagram Mn (A[X])

ε∗ / Mn (A)

det

A[X]

det ε

The result follows, as ε∗ (XIn − M ) = aIn − M .

/ A


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An important result about the characteristic polynomial is what’s known as the Cayley–Hamilton Theorem. We give a generalization of it here. Theorem 10.4.7 (Generalized Cayley–Hamilton Theorem). Let N be a finitely generated module over the commutative ring A and let f ∈ End nA (N ). Let x1 , . . . , xn be a set of generators for N over A, and suppose that f (xj ) = i=1 aij xi . Write M = (aij ) ∈ Mn (A). Then chM (X) is in the kernel of the evaluation map εf : A[X] → EndA (N ) obtained by evaluating X at f . In other words, if chM (X) = X n + an−1 X n−1 + · · · + a0 , then chM (f ) = f n + an−1 f n−1 + · · · + a0 = 0 as an endomorphism of N . Proof First note that since XIn −M t is the transpose of XIn −M , chM (X) = chM t (X). Thus, we may use M t in place of M . The subalgebra A[f ] of EndA (N ) is the image of εf , and hence is commutative. Thus, Proposition 10.2.13 gives us a commutative diagram Mn (A[X])

εf ∗

det A[X]

/ Mn (A[f ]) det

.

/ A[f ]

εf

Since chM t (f ) = εf (chM t (X)), we see that chM t (f ) is the determinant of the matrix (f In − M t ) ∈ Mn (A[f ]). (This is, of course an abuse of notation, as the M t in f In − M t should be replaced by its image under the natural map Mn (A) → Mn (EndA (N )), which has not been assumed to be an embedding.) Let N n be the direct sum of n copies of N , thought of as column matrices with entries in N . Then the A[f ]-module structure on N induces an Mn (A[f ])-module structure on N n by the usual formula for the product of an n × n matrix with a column vector. Under this module structure, the formula used to define M shows that ⎞ ⎛ x1 ⎟ ⎜ (f In − M t ) · ⎝ ... ⎠ = 0, xn where x1 , . . . , xn is the set of generators in the statement. Multiplying both sides on the adj left by the adjoint matrix (f In − M t ) , we see that the diagonal matrix (chM t (f ))In annihilates the column vector with entries x1 , . . . , xn . But this says that the endomorphism chM t (f ) carries each of x1 , . . . , xn to 0. Since x1 , . . . , xn generate N , chM t (f ) = chM (f ) is the trivial endomorphism of N . Notice that if N is a free module with basis x1 , . . . , xn then the matrix M in the statement of the Generalized Cayley–Hamilton Theorem is just the matrix of f with respect to the basis x1 , . . . , xn . Thus, chM (X) is by definition equal to chf (X). Corollary 10.4.8. Let N be a finitely generated free module over the commutative ring A and let f ∈ EndA (N ). Then chf (f ) = 0 in EndA (N ).


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Of course, if we start with a matrix M and let fM be the endomorphism of An induced by M , then M is the matrix of fM with respect to the canonical basis. Moreover, the standard A-algebra isomorphism from Mn (A) to EndA (An ) carries M onto fM , so the classical formulation of the Cayley–Hamilton Theorem is an immediate consequence: Theorem 10.4.9 (Cayley–Hamilton Theorem). Let A be a commutative ring and let M ∈ Mn (A). Let α : A[X] → Mn (A) be the A-algebra homomorphism with α(X) = M . Then the characteristic polynomial chM (X) is in the kernel of α. Thus, chM (M ) = 0, meaning that if chM (X) = X n + an−1 X n−1 + · · · + a0 , then n M + (an−1 In )M n−1 + · · · + (a0 In ) = 0 in Mn (A). We now give a useful consequence of the Generalized Cayley–Hamilton Theorem. Corollary 10.4.10. Let N be a finitely generated module over the commutative ring A, and let a be an ideal of A such that aN = N . Then there is an element a ∈ a such that 1 − a ∈ AnnA (N ). Proof Let x1 , . . . , xn be a generating set for N . The elements of aN all have the form a1 n1 + · · · + ak nk for some k, where ai ∈ a and ni ∈ N for all i. Since the xi generate N and since a is an ideal, distributivity allows us to write any such element as a1 x1 + · · · + an xn with ai ∈ a for all i. n Since aN = N , we may write xj = i=1 aij xi for all j, where aij ∈ a for all i, j. Thus, the matrix M = (aij ) is a matrix for the identity map, 1N , of N in the sense of the statement of the Generalized Cayley–Hamilton Theorem. As a result, we see that chM (1N ) is the trivial endomorphism of N . But chM (1N ) is just multiplication by the element chM (1) ∈ A. Thus, chM (1) lies in the annihilator of N . n−1 Write chM (X) = X n + i=0 ai X i . Since the entries of M are all in a, Corollary 10.2.6 n−1 shows that the coefficients ai all lie in a. So N is annihilated by 1 − (− i=0 ai ). Exercises 10.4.11. 1. Let M ∈ Mn (A) and let chM (X) = X n + an−1 X n−1 + · · · + a0 . Show that an−1 = −tr(M ). † 2. Let M ∈ Mn (A) and let chM (X) = X n + an−1 X n−1 + · · · + a0 . Show that a0 = (−1)n det M . 3. Use Corollary 10.4.10 to give a new proof of Nakayama’s Lemma when the ring A is commutative. † 4. Let a and b ideals in the integral domain A such that ab = b. Suppose that b is nonzero and finitely generated. Deduce that a = A.

10.5

Eigenvalues and eigenvectors

Here, we study matrices over a field, K. Definitions 10.5.1. Let M ∈ Mn (K). We say that v ∈ K n is an eigenvector of M if M v = av for some a ∈ K. If v = 0, this determines a uniquely, and we say that a is the eigenvalue, or characteristic root, associated to v. The eigenspace for M of an element a ∈ K is the set of eigenvectors of M for which M v = av.


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More generally, if V is a vector space over K and if f ∈ EndK (V ), then we define the eigenvalues and eigenvectors of f analogously. The fact that eigenvalues are also called characteristic roots is suggestive. Lemma 10.5.2. Let M ∈ Mn (K). Then a ∈ K is root of the characteristic polynomial chM (X) if and only if it is the eigenvalue associated to a nonzero eigenvector of M . Moreover, the eigenspace of any a ∈ K with respect to M is a vector subspace of K n . Specifically, the eigenspace of a is the kernel of the K-linear map obtained by multiplication by the matrix aIn − M . Proof By Lemma 10.4.6, a is a root of chM (X) if and only if det(aIn − M ) = 0. But since K is a field, det(aIn − M ) = 0 if and only if the K-linear map obtained from multiplication by aIn − M fails to be invertible. However, a dimension count (Corollary 7.9.12) shows that every injective endomorphism of a finitely generated vector space is invertible. So det(aIn − M ) = 0 if and only if the K-linear map obtained from multiplication by aIn − M has a nontrivial kernel. But (aIn − M )v = aIn v − M v = av − M v, and hence (aIn − M )v = 0 if and only if M v = av. Recall that a diagonal matrix is a matrix whose off-diagonal entries are all 0. Definition 10.5.3. A matrix M ∈ Mn (K) is diagonalizable if M is similar to a diagonal matrix. It is valuable to be able to detect when a matrix is diagonalizable. Proposition 10.5.4. A matrix M ∈ Mn (K) is diagonalizable if and only if there is a basis of K n consisting of eigenvectors of M . Proof Suppose first that M is diagonalizable, so that M M (M )−1 is a diagonal matrix, for some M ∈ Gln (K). Recall from Corollary 7.10.10 that a matrix is invertible if and only if its columns form a basis of K n . Thus, the columns, (M )−1 e1 , . . . , (M )−1 en form a basis of K n . We claim that (M )−1 ei is an eigenvector of M for each i. To see this, note that if M = (aij ) is a diagonal matrix, then M ei = aii ei . Thus, the canonical basis vectors are eigenvectors for any diagonal matrix. In particular, for each i, there is an ai ∈ K such that (M M (M )−1 )ei = ai ei . Multiplying both sides by (M )−1 , we get M (M )−1 ei = ai (M )−1 ei , and hence (M )−1 ei is an eigenvector for M as claimed. Conversely, suppose that K n has a basis v1 , . . . , vn consisting of eigenvectors of M . Thus, for each i there is an ai ∈ K such that M vi = ai vi . Let M be the matrix whose i-th column is vi for 1 ≤ i ≤ n. Since its columns form a basis of K n , M is invertible. We claim that (M )−1 M M is a diagonal matrix. To see this, note that (M )−1 M M ei = (M )−1 M vi = (M )−1 ai vi = ai ei , and hence the canonical basis vectors are eigenvectors for (M )−1 M M . But this says that the i-th column of (M )−1 M M is ai ei , and hence the off-diagonal entries of (M )−1 M M are all 0. The reader who is approaching this material from a theoretical point of view may not appreciate the full value of this last result and those related to it. The point is that characteristic polynomials are reasonably easy to calculate. If we can then factor them,


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we can use Gauss elimination to find bases for the kernels of the transformations induced by the (aIn − M ) as a ranges over the roots of the characteristic polynomial. If the dimensions of these eigenspaces add up to the size of the matrix, then one can show that the bases of these eigenspaces fit together to form a basis of K n . Thus, modulo factoring the characteristic polynomial, the determination of whether a matrix is diagonalizable, and the actual construction of a diagonalization, are totally algorithmic. And indeed, this sort of hands-on calculation can arise in actual research problems. Exercises 10.5.5. 1. Let M ∈ Mn (K) and suppose that chM (X) = (X−a1 ) . . . (X−an ), where a1 , . . . , an are n distinct elements of K. Show that M is diagonalizable. 2. What is the relationship between the eigenspaces of M and the eigenspaces of M M (M )−1 for M ∈ Gln (K)? 3. Show that if θ ∈ R is not a multiple of π, then the matrix cos θ − sin θ Rθ = sin θ cos θ is not diagonalizable as an element of M2 (R), but is diagonalizable when regarded as an element of M2 (C). What are its complex eigenvalues? 4. Show that a block sum M1 ⊕ M2 is diagonalizable if and only if each of M1 and M2 is diagonalizable. 5. Let M1 and M2 be elements of Mn (K) which commute with each other, and let V be the eigenspace of a ∈ K with respect to M1 . Show that M2 preserves V (i.e., M2 v ∈ V for all v ∈ V ). 6. Suppose given a family {Mi | i ∈ I} of diagonalizable matrices in Mn (K) which commute with each other. Show that the Mi may be simultaneously diagonalized, in the sense that there is a single matrix M such that M Mi M −1 is a diagonal matrix for each i ∈ I. Conversely, show that any family of matrices that is simultaneously diagonalizable must commute with one another.

10.6

The classification of matrices

Let K be a field. We shall classify the elements of Mn (K) up to similarity. The key is the Fundamental Theorem of Finitely Generated Modules over the P.I.D. K[X]. The connection between matrices and K[X]-modules is as follows. First off, K n is an Mn (K)-module, where we identify K n with the space of n × 1 column vectors, and Mn (K) acts on K n by matrix multiplication from the left. Note that if a ∈ K, then the matrix aIn acts on K n as multiplication by a. In other words, the K-module structure on K n induced by the Mn (K)-module structure together with the usual inclusion of K in Mn (K) agrees with the original K-module structure on K n . Now let M ∈ Mn (K). Then there is a K-algebra homomorphism α : K[X] → Mn (K) which takes X to M . And there is a K[X]-module structure on K n obtained via α from the usual Mn (K)-module structure on K n : for v ∈ K n , Xv = M v, and the action of K ⊂ K[X] on K n agrees with the usual action of K on K n . Conversely, let N be a K[X]-module such that the induced K-module structure on N is n-dimensional. Because the elements a ∈ K commute with X in K[X], multiplication


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by X induces a K-linear map from N to N . Thus, a choice of basis for N will provide a matrix M ∈ Mn (K) corresponding to multiplication by X. Thus, we can pass back and forth between n × n matrices and K[X] modules whose underlying K-vector space has dimension n. Of course, the passage from modules to matrices involves a choice of basis. By varying this choice, we may vary the matrix up to similarity. Note that similarity is an equivalence relation on Mn (K). We call the resulting equivalence classes the similarity classes of n × n matrices. Proposition 10.6.1. Let K be a field. Then the passage between matrices and K[X]modules described above gives a one-to-one correspondence between the similarity classes of matrices in Mn (K) and the isomorphism classes of those K[X]-modules whose underlying K-vector space has dimension n. Proof Suppose that M, M ∈ Mn (K) are similar: say M = M M (M )−1 for some M ∈ Gln (K). Let N1 be the K[X]-module structure on K n induced by M , and let N2 be the K[X]-module structure on K n induced by M . Let fM : N1 → N2 be the K-linear map induced by M (i.e., fM (v) = M v for v ∈ N1 = K n ). Since M is invertible, fM is bijective. We claim that fM is a K[X]-module isomorphism from N1 to N2 . To see this, note that K[X] is generated as a ring by X and the elements of K. Thus, it suffices to show that fM (Xv) = XfM (v) for all v ∈ N1 and that fM (av) = afM (v) for all v ∈ N1 and a ∈ K. The latter statement is immediate, as fM is a K-linear map. For the former, write fM , fM for the linear transformations induced by multiplication by M and M , respectively. Then the equation M = M M (M )−1 implies that the following diagram commutes. N1

fM / N2

fM

fM f M / N1 N2

Now on N1 , multiplication by X is given by fM , and on N2 it is given by fM . So the commutativity of the diagram says precisely that fM (Xv) = XfM (v) for all v ∈ N1 . The passage from isomorphism classes of K[X]-modules whose underlying K-vector space has dimension n to similarity classes of matrices in Mn (K) is well defined because if f : N → N is a K[X]-module isomorphism and if x1 , . . . , xn is a basis for N , then the matrix for multiplication by X with respect to this basis is precisely equal to the matrix for multiplication by X in N with respect to the basis f (x1 ), . . . , f (xn ). If N is a K[X]-module whose underlying K-vector space has dimension n, then the matrix of X with respect to a basis of N produces a K[X]-module clearly isomorphic to N . Thus, to complete the proof, it suffices to show that if M, M ∈ Mn (K) such that the K[X] module structures on K n induced by M and M are isomorphic, then M and M are similar. To see this, let N1 and N2 be the K[X] module structures on K n induced by M and M , respectively, and let f : N1 → N2 be a K[X]-module isomorphism. Then f = fM for some M ∈ Mn (K). Moreover, since f is an isomorphism, M ∈ Gln (K). Reversing the steps in the earlier argument, the equation fM (Xv) = XfM (v) produces the commutative diagram displayed above, which then gives the matrix equation M = M M (M )−1 .


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Thus, we may apply the Fundamental Theorem of Finitely Generated Modules over a P.I.D. to classify matrices up to similarity. Note that any K[X] module N whose underlying K-vector space is finite dimensional must be a torsion module: Any torsion-free summands would be isomorphic to K[X], and hence infinite dimensional over K. Thus, we may use the second form of the Fundamental Theorem (Theorem 8.9.20), which says that every torsion module over K[X] may be written uniquely as a direct sum K[X]/((f1 (X))r1 ) ⊕ · · · ⊕ K[X]/((fk (X))rk ), where f1 (X), . . . , fk (X) are (not necessarily distinct) monic irreducible polynomials in K[X]. Next, recall from Proposition 7.7.35 that if f (X) ∈ K[X] has degree k, then K[X]/(f (X)) is a k-dimensional vector space over K, with a basis given by the images under the canonical map of 1, X, . . . , X k−1 . We obtain the following proposition. Proposition 10.6.2. Let M be a module over K[X] which has dimension n as a vector space over K. Then M has a unique (up to a reordering of the summands) decomposition M∼ = K[X]/((f1 (X))r1 ) ⊕ · · · ⊕ K[X]/((fk (X))rk ), where f1 (X), . . . , fk (X) are (not necessarily distinct) monic irreducible polynomials in K[X], ri > 0 for all i, and r1 deg f1 + · · · + rk deg fk = n.

The translation of this classification into matrices is called rational canonical form. To make this translation, we must study the matrices associated with the cyclic modules K[X]/(f (X)). Thus, suppose that f (X) is any monic polynomial of degree m over K and write x1 , . . . , xm for the images (in order) of 1, X, . . . , X m−1 under the canonical map π : K[X] → K[X]/(f (X)). We call this the standard basis of K[X]/(f (X)). Also, write f (X) = X m + am−1 X m−1 + · · · + a0 . Note that in the K[X]-module structure on K[X]/(f (X)), we have Xxi = xi+1 for 1 ≤ i < m. And Xxm is the image under the canonical map of X m . Since f (X) is in the kernel of the canonical map, this gives Xxm = −a0 x1 − · · · − am−1 xm . The resulting matrix has a name. Definition 10.6.3. Let f (X) = X m + am−1 X m−1 + · · · + a0 be a monic polynomial of degree m in K[X]. Then the rational companion matrix C(f ) of f is the m × m matrix whose i-th column is ei+1 for 1 ≤ i < m, and whose m-th column is ⎞ ⎛ −a0 ⎟ ⎜ .. ⎠. ⎝ . −am−1 We have shown the following. Lemma 10.6.4. Let f (X) = X m + am−1 X m−1 + · · · + a0 be a monic polynomial of degree m in K[X]. Then the matrix obtained from the standard basis of K[X]/(f (X)) of the transformation induced by multiplication by X is precisely the rational companion matrix C(f ).


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Definition 10.6.5. We say that a matrix M is in rational canonical form if it is a block sum M = C(f1r1 ) ⊕ · · · ⊕ C(fkrk ) of rational companion matrices, where f1 (X), . . . , fk (X) are (not necessarily distinct) monic irreducible polynomials in K[X], and the exponents ri are all positive. Propositions 10.6.1 and 10.6.2 now combine to give the classification of matrices up to similarity. Theorem 10.6.6. (Classification of Matrices) Every similarity class of n × n matrices over a field K contains a unique (up to permutation of the blocks) matrix in rational canonical form. The replacement of M by a matrix in its similarity class that’s in rational canonical form is called “putting M in rational canonical form.” We call the resulting matrix the rational canonical form of M . We next consider the connection between rational canonical form and the characteristic polynomial. Proposition 10.6.7. Let f (X) = X m + am−1 X m−1 + · · · + a0 be a monic polynomial of degree m in K[X], and let C(f ) be its rational companion matrix. Then the characteristic polynomial of C(f ) is precisely f (X). Proof Write XIn − C(f ) = (fij (X)) and consider the decomposition of its determinant in terms of permutations: det(XIn − C(f )) = ε(σ)fσ(1),1 (X) . . . fσ(m),m (X). σ∈Sm

Now for i < m, the only nonzero terms of the form fji (X) are fii (X) = X and fi+1,i (X) = −1. So for i < m, the term corresponding to σ ∈ Sm vanishes unless σ(i) = i or σ(i) = i + 1. Note that since σ is a permutation, if σ(i) = i + 1, then σ(i + 1) = i + 1. In particular, if i < m is the smallest index for which σ(i) = i, then we must have σ(j) = j + 1 for i ≤ j < m. Since σ fixes the indices less than i, this forces σ(m) = i. Thus, for the term fσ(1),1 (X) . . . fσ(m),m (X) to be nonzero, with σ not the identity permutation, we must have σ equal to one of the cycles (i i + 1 . . . m). Now the term corresponding to the identity permutation is X m−1 (X + am−1 ) = X m + am−1 X m−1 , which are the leading two terms of f (X). And for σ = (i i + 1 . . . m), we have ε(σ) = (−1)m−i , while ⎧ ⎪ if j < i ⎨X fσ(j),j (X) = −1 if i ≤ j < m ⎪ ⎩ ai−1 if j = m. Putting this together, we see that the signs from the −1’s in the matrix cancel against the sign of σ, and we have ε(σ)fσ(1),1 (X) . . . fσ(m),m (X) = ai−1 X i−1 , the i − 1-st term in f . Adding this up over the displayed cycles σ, we get chC(f ) (X) = f (X), as desired.


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Corollary 10.6.8. Suppose given a matrix M = C(f1r1 ) ⊕ · · · ⊕ C(fkrk ) in rational canonical form. Then the characteristic polynomial of M is given by chM (X) = (f1 (X))r1 . . . (fk (X))rk .

There’s one situation in which the characteristic polynomial forces the rational canonical form. Corollary 10.6.9. Let M be a matrix whose characteristic polynomial is square free, in the sense that chM (X) = f1 (X) . . . fk (X) where f1 (X), . . . , fk (X) are distinct monic irreducible polynomials over K. Then the rational canonical form for M is given by C(f1 ) ⊕ · · · ⊕ C(fk ).

However, nonsimilar matrices can have the same characteristic polynomial. For instance, C((X − a)3 ), C((X − a)2 ) ⊕ C(X − a) and C(X − a) ⊕ C(X − a) ⊕ C(X − a) are all 3 × 3 matrices whose characteristic polynomial is (X − a)3 . But all three lie in distinct similarity classes. There is another invariant, which, together with the characteristic polynomial, will characterize these particular examples. Recall that the minimal polynomial, minM (X), of a matrix M ∈ Mn (K) is the monic polynomial of least degree in the kernel of the evaluation map εM : K[X] → Mn (K) which evaluates X at M . In particular, minM (X) generates the kernel of εM , and hence the Cayley–Hamilton Theorem shows that minM (X) divides chM (X). We can relate the minimal polynomial to the Fundamental Theorem of Finitely Generated Modules over a P.I.D. as follows. Lemma 10.6.10. Let M ∈ Mn (K), and let N be K n with the K[X]-module structure induced by M . Then minM (X) generates the annihilator of N over K[X]. Proof A polynomial f (X) acts on N by f (X)v = f (M )v. So f (X) annihilates N if and only if the matrix f (M ) induces the 0 transformation. But this occurs only when f (M ) is the 0 matrix, meaning that f (X) is in the kernel of εM . But we may now apply our understanding of finitely generated torsion modules over K[X] in order to calculate the minimal polynomial of a matrix with a given rational canonical form. Corollary 10.6.11. Suppose given a matrix M = C(f1r1 ) ⊕ · · · ⊕ C(fkrk ) in rational canonical form. Then the minimal polynomial of M is the least common multiple of the polynomials firi .


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Proof The annihilator of a direct sum of modules is the intersection of the annihilators of the individual modules. In a P.I.D., the generator of an intersection of ideals is the least common multiple of the generators of the individual ideals (Proposition 8.1.32). Thus, if f1 (X), . . . , fk (X) are the monic irreducible polynomials that divide the characteristic polynomial of M , and if si is the largest exponent such that the companion matrix C(fisi ) appears in the rational canonical form of M for 1 ≤ i ≤ k, then minM (X) = (f1 (X))s1 . . . (fk (X))sk . In particular, the minimal polynomials of C((X − a)3 ), C((X − a)2 ) ⊕ C(X − a), and C(X − a) ⊕ C(X − a) ⊕ C(X − a) are (X − a)3 , (X − a)2 , and X − a, respectively. So the minimal polynomial tells these three rational canonical forms apart. Minimal polynomials also allow us to recognize diagonalizable matrices. Corollary 10.6.12. Let M ∈ Mn (K). Then M is diagonalizable if and only if minM (X) = (X − a1 ) . . . (X − ak ) for a1 , . . . , ak distinct elements of K. Proof If minM (X) has the stated form, then every rational companion block in the rational canonical form of M has the form C(X − ai ) = (ai ) for some i. In particular, the rational canonical form for M is a block sum of 1 × 1 matrices, hence a diagonal matrix. Conversely, a diagonal matrix is already in rational canonical form, with companion blocks C(X − a) = (a) for each diagonal entry a of the matrix. The least common multiple of the annihilators of the blocks then has the desired form. But the minimal polynomial is primarily useful as a theoretical tool, as it’s difficult to calculate in practice. It remains to discuss what happens to the rational canonical form when we pass from a given field to an extension field. The following can also be seen directly, using the structure of the rational companion blocks. Proposition 10.6.13. Let L be an extension field of K and let M ∈ Mn (K). Then the L[X]-module structure on Ln induced by the matrix M is isomorphic to the extended module L[X] ⊗K[X] K n obtained from the K[X]-module structure on K n induced by M . In particular, if K n ∼ = K[X]/(f1 (X)) ⊕ · · · ⊕ K[X]/(fk (X)), then Ln ∼ = L[X]/(f1 (X)) ⊕ · · · ⊕ L[X]/(fk (X)) for the same polynomials f1 (X), . . . , fk (X) ∈ K[X]. Thus, the rational canonical form for M in Mn (L) may be found by determining the prime factorizations of the fi (X) in L[X] and applying the Chinese Remainder Theorem. Proof Let f : K n → K n be the linear map induced by M . Then, as noted in Corol1⊗f lary 9.4.20, the effect of M on Ln may be identified with L ⊗K K n −−→ L ⊗K K n . Thus, n n if i : K → L ⊗K K is the natural inclusion, then it is a K[X]-module homomorphism, where the K[X]-module structure on L ⊗K K n is the restriction of the L[X]-module structure induced by M . The universal property of extension of rings (Proposition 9.4.12) provides a unique L[X]-module homomorphism ı : L[X] ⊗K[X] K n → L ⊗K K n whose restriction to K n is i. We claim that ı is an isomorphism.


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To see this, let j : L → L[X] be the natural inclusion. Then there is a well defined homomorphism j ⊗ 1 : L ⊗K K n →L[X] ⊗K[X] K n , which is easily seen to satisfy m ı ◦ (j ⊗ 1) = 1L⊗K K n . But if f (X) = i=0 αi X i ∈ L[X], then for each v ∈ K n , we have f (X) ⊗ v =

m

αi ⊗ M i v

i=0

in L[X] ⊗K[X] K n . Thus, j ⊗ 1 is onto, and ı is an isomorphism. We may now obtain the direct sum decomposition of Ln by applying the natural isomorphisms B ⊗A A/a ∼ = B/Ba for B a commutative A-algebra and a an ideal of A. Exercises 10.6.14. 1. What is the rational canonical form of the rotation matrix cos θ − sin θ Rθ = ? sin θ cos θ Deduce from the classification theorem that if θ, φ ∈ [−π, π], then Rθ and Rφ are similar in M2 (R) if and only if θ = ±φ. 2. What is the smallest n for which there exist two matrices M, M ∈ Mn (K) such that chM (X) = chM (X) and minM (X) = minM (X) but M and M are not similar? 3. Write Z3 = T . What is the rational canonical form of the transformation of Q[Z3 ] induced by multiplication by T ? 4. Write Z4 = T . What is the rational canonical form of the transformation of Q[Z4 ] induced by multiplication by T ? What do you get if you replace Q by C? 5. Write Z8 = T . What is the rational canonical form of the transformation of Q[Z8 ] induced by multiplication by T ? What do you get if you replace Q by R? What if you replace Q by C? 6. Let K be any field and let M ∈ Mn (K). Show that M is similar to its transpose. 7. Let L be an extension field of K and let f1 (X) and f2 (X) be two monic irreducible elements of K[X]. Show that f1 and f2 have no common prime factors in L[X]. (Hint: How do f1 and f2 factor as polynomials over an algebraic closure of L?) Deduce that if M, M ∈ Mn (K), then M and M are similar in Mn (K) if and only if they are similar in Mn (L).

10.7

Jordan canonical form

Let K be a field. Suppose that the characteristic polynomial of M ∈ Mn (K) factors as a product of degree 1 polynomials in K[X]: chM (X) = (X − a1 )r1 . . . (X − ak )rk where a1 , . . . , ak are distinct elements of K. (This will always happen, for instance, if the field K is algebraically closed.) Then we shall give an alternative way of representing the similarity class of M in Mn (K) by a canonical form.


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In fact, we simply replace each rational companion matrix in the rational canonical form of M by the matrix obtained for an alternate choice of basis for the summand in question. The matrix obtained from this alternative basis is called a Jordan block. The key is the following. Lemma 10.7.1. Let a ∈ K and write v1 , . . . , vm for the images under the canonical map K[X] → K[X]/((X − a)m ) of 1, X − a, . . . , (X − a)m−1 . Then v1 , . . . , vm is a basis for K[X]/((X − a)m ) over K. Proof The elements v1 , . . . , vm are all nonzero, as the polynomials (X − a)k with 0 ≤ k < m have degree less than that of (X − a)m . Notice that for 2 ≤ i ≤ m, vi = (X − a)i−1 v1 , and that (X − a)m v1 = 0. This is enough to show the linear independence of v1 , . . . , vm : Let a1 v1 + · · · + am vm = 0. If the coefficients ai are not all 0, let i be the smallest index for which ai = 0. Then 0 = (X − a)m−i (a1 v1 + · · · + am vm ) = ai vm . But vm = 0, contradicting the assumption that the coefficients were not all 0. Since K[X]/((X − a)m ) has dimension m over K, the result follows. For 1 ≤ i < m, we have vi+1 = (X − a)vi , and hence Xvi = vi+1 + avi . And (X − a)vm = 0, so that Xvm = avm , and vm is an eigenvector for the transformation induced by X, with eigenvalue a. Definition 10.7.2. Let a ∈ K and let m ≥ 1. The m × m Jordan block with eigenvalue m is the matrix, J(a, m), whose i-th column is aei + ei+1 if 1 ≤ i < m and whose m-th column is aem . Thus, the diagonal entries of J(a, m) are all a’s, and the off-diagonal entries are all 0 except for the entries just below the diagonal, which are all 1’s. Thus, ⎛

⎞ a 0 0 J(a, 3) = ⎝ 1 a 0 ⎠ . 0 1 a From the discussion above, we see that J(a, m) is the matrix for the transformation of K[X]/((X − a)m ) induced by X, with respect to the basis v1 , . . . , vm . The next lemma is immediate from Lemma 10.6.4. Lemma 10.7.3. Let a ∈ K and let m ≥ 1. Then the Jordan block J(a, m) is similar to the rational companion matrix C((X − a)m ) in Mm (K). In particular, for matrices whose characteristic polynomials factor as products of degree 1 terms, we’ll be able to translate back and forth between rational canonical form and a canonical form using Jordan blocks. Definition 10.7.4. A matrix M ∈ Mn (K) is in Jordan canonical form if it is a block sum of Jordan blocks: M = J(a1 , r1 ) ⊕ · · · ⊕ J(ak , rk ) where a1 , . . . , ak are (not necessarily distinct) elements of K and the exponents ri are positive for 1 ≤ i ≤ k.


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Theorem 10.6.6 and Lemma 10.7.3 now combine to give the following theorem. Theorem 10.7.5. Let M ∈ Mn (K) be a matrix whose characteristic polynomial factors as a product of degree 1 polynomials over K. Then the similarity class of M in Mn (K) contains a unique (up to permutation of blocks) matrix in Jordan canonical form. The replacement of M by a matrix in its similarity class that’s in Jordan canonical form is called “putting M in Jordan canonical form.” We call the resulting matrix the Jordan canonical form of M . The Jordan blocks have certain features that make them useful. For simplicity of notation, we write ker M for the kernel of the transformation induced by a matrix M . Lemma 10.7.6. Let a ∈ K and let M = J(a, m) for some m > 0. Then for 1 ≤ k ≤ m, a basis for ker(aIm − M )k is given by the canonical basis vectors em−k+1 , . . . , em . In particular, the eigenspace of a with respect to M has dimension 1. For b = a, however, (bIm − M )k is invertible for all k. Proof A glance at the matrix aIm − M shows that 2 −ei+1 if i < m (aIm − M )ei = 0 if i = m. Induction on k then shows that

2 (−1)k ei+k (aIm − M ) ei = 0 k

if i ≤ m − k if i > m − k.

The kernel of (aIm − M )k is now easily seen to be as stated. Since chM (X) = chC((X−a)k ) (X) = (X − a)k , a is the only eigenvalue of M . So if b = a, then bIm − M invertible, and hence its powers are invertible. (Alternatively, it is easy to see that det(bIm − M ) = (b − a)m .) This, in turn, allows us to calculate the Jordan canonical form of any matrix whose characteristic polynomial is a product of degree 1 polynomials. The next proposition is immediate by adding up the dimensions of the kernels in the different Jordan blocks. Proposition 10.7.7. Let M ∈ Mn (K) such that chM (X) is a product of degree 1 polynomials in K[X]. Then for each eigenvalue a of M , the number of Jordan blocks with eigenvalue a occurring in the Jordan canonical form of M is equal to the dimension of the eigenspace of a with respect to M . More generally, in the Jordan canonical form of M , the number of Jordan blocks J(a, m) occurring with m ≥ k is equal to dim ker (aIn − M )k − dim ker (aIn − M )k−1 .

But this now says that if we can factor the characteristic polynomial as a product of degree 1 terms, then we may calculate the Jordan canonical form of a matrix algorithmically. The point is that Gauss elimination gives an algorithm for calculating a basis for the kernel of each (aIn − M )k . Exercises 10.7.8.


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1. A matrix M ∈ Gln (K) is said to be periodic if it has finite order in the group Gln (K). The period of M is its order. Let K be an algebraically closed field and let M be a periodic matrix over K. Suppose that M has order m in Gln (K), where m is not divisible by the characteristic of K. Show that M is diagonalizable. Deduce that such an M is similar to a matrix of the form (a1 ) ⊕ · · · ⊕ (an ), where ai is an m-th root of unity in K for each i, and the least common multiple of the orders of a1 , . . . , an in K × is m. 2. Give an example of a periodic matrix over the algebraic closure of Zp that is not diagonalizable. ‡ 3. Let G be a finite abelian group and let K be an algebraically closed field whose characteristic does not divide the order of G. Let ρ : G → Gln (K) be a group homomorphism. Show that there is a matrix M such that M ρ(g)M −1 is a diagonal matrix for each g ∈ G. Deduce that if |G| = m and if ζ is a primitive m-th root of unity in K, then there are homomorphisms ρi : G → ζ ⊂ K × for i = 1 . . . , n, such that ρ is conjugate as a homomorphism into Gln (K) to the homomorphism that carries each g ∈ G to the block sum (ρ1 (g)) ⊕ · · · ⊕ (ρn (g)). (In the language of representation theory, this says that any finite dimensional representation of G over K is a direct sum of one-dimensional representations.) 4. Let K be any field. Construct a homomorphism ρ : K → Gl2 (K), from the additive group of K into Gl2 (K), such that ρ(a) not diagonalizable if a = 0. 5. Let M be a periodic matrix over a field K. Show that the minimal polynomial of M over K divides X m − 1, where m is the period of M . Deduce that if the characteristic of K does not divide m and if K contains a primitive m-th root of unity (i.e., an element with order m in K × ), then M is diagonalizable over K. (The reader who has not read Chapter 11 may assume that K has characteristic 0.)

10.8

Generators for matrix groups

Here, A is permitted to be any ring. A common technique for studying matrix groups is Gauss elimination. It is based on row and column operations obtained by multiplying by the elementary matrices: Definitions 10.8.1. Let x ∈ A and let i, j be distinct elements of {1, . . . , n}. We write Eij (x) for the n × n matrix whose diagonal entries are all 1 and whose only off-diagonal entry that is nonzero is the ij-th entry, which has value x. We shall refer to the matrices Eij (x) as the elementary matrices. If M is an n×k matrix, then the passage from M to Eij (x)·M is called an elementary row operation on M . If M is a k × n matrix, then the passage from M to M · Eij (x) is called an elementary column operation on M . The elementary row and column operations behave as follows. Proposition 10.8.2. Let x ∈ A and let i, j be distinct elements of {1, . . . , n}. Let M ∈ Mn (A). Then the product Eij (x) · M coincides with M in all rows but the i-th. The


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i-th row of Eij (x) · M is equal to the sum yi + xyj , where yi and yj are the i-th and j-th rows of M , respectively. Here, we treat row vectors as left A-modules. The product M ·Eij (x) coincides with M in all columns but the j-th. The j-th column of M · Eij (x) is the sum xj + xi x, where xj and xi are the j-th and i-th columns of M , respectively. Here, we treat column vectors as right A-modules. Proof Recall that the k-th row of a matrix M is the product ek ·M , where the canonical basis vector ek is regarded as a row vector. The result for Eij (x) · M now follows from the fact that the k-th row of Eij (x) is ek if k = i, and is ei + xej if k = i. The case for M · Eij (x) is similar. A special case of row operations allows us to determine a subgroup of Gln (A). Corollary 10.8.3. For fixed i and j, we have Eij (x) · Eij (y) = Eij (x + y) for all x, y ∈ A. Thus, each Eij (x) is invertible, with inverse Eij (−x), and the set {Eij (x) | x ∈ A} forms a subgroup of Gln (A) isomorphic to the additive group of A. Recall that the commutator, [x, y], of two group elements is defined by [x, y] = xyx−1 y −1 , and that [x, y] = e if and only if x and y commute. The reader may verify the next lemma by testing the transformations’ effect on the canonical basis vectors. Lemma 10.8.4. If i = j and i = j , then [Eij (x), Ei j (y)] = e. Here, x, y ∈ A, and, of course, i = j and i = j . If i, j, and k are all distinct, then [Eij (x), Ejk (y)] = Eik (xy).

The commutators [Eij (x), Eji (y)] are computable, but are less useful than the ones displayed above. Definition 10.8.5. A square matrix M = (aij ) is upper triangular if all the entries below the main diagonal are 0 (i.e., if aij = 0 for i > j). A square matrix M = (aij ) is lower triangular if all the entries above the main diagonal are 0 (i.e., if aij = 0 for i < j). Elementary matrices are useful in understanding certain triangular matrices. Lemma 10.8.6. Any upper or lower triangular matrix whose diagonal entries are all 1’s is a product of elementary matrices. Proof We shall treat the lower triangular case and leave the rest to the reader. Thus, suppose that M = (aij ) is a lower triangular matrix in Mn (A) whose diagonal entries are all 1’s. Let Mj =

n .

Eij (aij ).

i=j+1

Then the reader may verify that M = M1 . . . Mn−1 .


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In 0 In Y or are products of eleX Ik 0 Ik mentary matrices, a fact that will allow us to show the following. In particular, matrices of the form

Lemma 10.8.7. Let M1 , M2 ∈ Gln (A). Then there is a product, E, of elementary matrices over A with the property that E · (M1 M2 ⊕ In ) = M2 ⊕ M1 . In 0 M1 M2 0 Proof Multiplication by takes M . Now M ⊕ I to 1 2 n M2 In M −1 In 1 In In − M1 M2 In − M1 multiply by , and we get . 0 In M2 In In M2 In − M1 0 Multiplying this by gives . Finally, we obtain M2 ⊕ −In In 0 M1 −1 In In − M1 . M1 by multiplying this last matrix by 0 In Since Eij (x)−1 = Eij (−x), the inverse of a product of elementary matrices is also a product of elementary matrices. Thus, induction gives the following. Corollary 10.8.8. Let M1 , . . . , Mk ∈ Gln (A). Then there is a product, E, of elementary matrices with the property that E · (M1 ⊕ · · · ⊕ Mk ) = (Mk . . . M1 ) ⊕ I(k−1)n .

We shall make use of this for block sums of 1 × 1 matrices in the following. The technique we shall use in the next proof is known as Gauss elimination. Proposition 10.8.9. Let D be a division ring and let M ∈ Gln (D). Then there is a product, E, of elementary matrices over D, and an element a ∈ D such that M = E · ((a) ⊕ In−1 ) . In particular, Gln (D) is generated by the elementary matrices, together with the matrices of the form (a) ⊕ In−1 , with a ∈ D× . Proof By Corollary 10.8.8, it suffices to show that M may be reduced to a diagonal matrix by a sequence of elementary row operations. First, if the 11 entry of M is 0, choose any row whose first column entry is nonzero, and add it to the first (i.e., multiply M by E1i (1), where the i-th row has a nonzero entry in the first column). Thus, after applying an elementary row operation, if necessary, we may assume that the 11 entry!of our matrix is nonzero. If the resulting matrix is (aij ), n we may now multiply it by i=2 Ei1 (−ai1 a−1 11 ), obtaining a matrix with no nonzero off-diagonal entries in the first column. Assume, inductively, that we’ve altered M by a sequence of elementary row operations so that the first k − 1 columns have no nonzero off-diagonal entries. The diagonal entries in these k − 1 columns must be nonzero, as otherwise the linear transformation induced by our matrix will have a nontrivial kernel. Thus, the first k − 1 columns form a basis for the (right) vector subspace of Dn generated by e1 , . . . , ek−1 . Thus, the k-th column


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must have a nonzero entry below the (k − 1)-st row; otherwise, the first k columns of our matrix will be linearly dependent. Note that for i ≥ k, the first k − 1 entries of the i-th row are empty at this point. So adding a multiple of the i-th row to any other row will not change the first k − 1 columns. If the kk-th entry is 0, then multiply the matrix by Eki (1), where i > k is chosen so that the ik-th entry is nonzero. Thus, without altering the first k − 1 columns, we may assume that the kk-th entry is nonzero. Suppose that our matrix is now (bij ). Multiply it by . Eik (−bik b−1 kk ), 1≤i≤n i =k

and we obtain a matrix whose off-diagonal entries in the first k columns are all 0. The result now follows by induction on n. The analogous result fails if D is replaced by some quite reasonable looking commutative rings. A method of analyzing the failure of this sort of result in a stable sense (i.e., as n → ∞) will be studied in Section 10.9. Proposition 10.8.9 has a useful interrelation with determinant theory if we are working over a field. Recall that the n-th special linear group, Sln (A), of a commutative ring A is the group of matrices of determinant 1. Proposition 10.8.10. Let K be a field and let n > 0. Then Sln (K) is the subgroup of Gln (K) generated by the elementary matrices. Proof It is easy to see (e.g., by Problem 6 of Exercises 10.3.12) that the determinant of either an upper triangular matrix or a lower triangular matrix is given by the product of the diagonal terms. Thus, elementary matrices lie in Sln (K). But then, if M = E · ((a) ⊕ In−1 ), with E a product of elementary matrices and a ∈ K, then det M = a. We may now make use of our calculations of commutators in Lemma 10.8.4. We first need a definition. Definition 10.8.11. A group G is perfect if the commutator subgroup [G, G] = G. Proposition 10.8.12. Let K be a field. Then for n ≥ 3, the special linear group Sln (K) is perfect, and is the commutator subgroup of Gln (K). Proof By Lemma 10.8.4, every elementary matrix is a commutator of elementary matrices, so Sln (K) is generated by commutators of elements of Sln (K). Thus, Sln (K) is perfect and is contained in the commutator subgroup of Gln (K). But Sln (K) Gln (K), and the quotient group Gln (K)/Sln (K) is the abelian group K × . So Sln (K) contains the commutator subgroup of Gln (K) by Corollary 5.4.8. Exercises 10.8.13. 1. Let A be a Euclidean domain. Show that every element of Gln (A) may be written as a product E · ((a) ⊕ In−1 ), where E is a product of elementary matrices and a ∈ A× . In particular, Sln (A) is generated by elementary matrices. and b = E21 (1). Show that x = † 2. Show that Sl2 (Z)

is generated by a = E212 (1) −1 has order 4, and y = b order 3. Deduce that ba−1 b = 01 −1 a b = 01 −1 −1 0 Sl2 (Z) is generated by torsion elements, despite the fact that it contains elements of infinite order.


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3. Show that Sln (Z) is generated by torsion elements for all n. 4. Let f : Zn → Zn be an injective homomorphism. Show that the cokernel of f is a finite group of order |det f |. 5. Let A be a commutative ring. Find the center of Sln (A).

10.9

K1

Once again, A can be any ring. Here, we study in a stable sense the failure of Gauss elimination to solve matrix problems over general rings. Stability here means allowing the size of our matrices to grow. Since we shall limit our interest to invertible matrices, we shall use the stability map i : Gln (A) → Gln+1 (A) given by i(M ) = M ⊕ (1). Clearly, i is a group homomorphism. We can think of each Gln (A) as embedded in the set of infinite matrices over A by the embedding that takes M ∈ Gln (A) to M ⊕ I∞ . Under this identification, i : Gln (A) → Gln+1 (A) is an inclusion of subsets. (Alternatively, we can just work abstractly in the direct limit of the inclusions i.) Definition 10.9.1. We set Gl(A) = Gl∞ (A) to be the ascending union of the Gln (A): Gl(A) =

∞

Gln (A).

n=1

We call it the infinite general linear group of A. The group structure we place on Gl(A) is, of course, the unique group structure that agrees with the usual group structure on each Gln (A) ⊂ Gl(A). Notice that if i = j and if i and j are both ≤ n, then i : Gln (A) → Gln+1 (A) carries the n × n elementary matrix Eij (x) to the (n + 1) × (n + 1) elementary matrix Eij (x) for all x ∈ A. Thus, for each pair i, j of distinct positive integers, there is a uniquely defined elementary matrix Eij (x) ∈ Gl(A). Definition 10.9.2. We write E(A) ⊂ Gl(A) for the subgroup of Gl(A) generated by the elementary matrices. The next lemma is a key in understanding E(A). Lemma 10.9.3. Let M ∈ Gln (A). Then M ⊕ M −1 is a product of elementary matrices in Gl2n (A). Proof By Lemma 10.8.7, there is a product of elementary matrices, E, in Gl2n (A), such that M ⊕ M −1 = E · (M −1 M ⊕ In ) = E.

Recall that a perfect group is one which is its own commutator subgroup. Proposition 10.9.4. E(A) is a perfect group and is the commutator subgroup of Gl(A).


448

Proof Lemma 10.8.4 shows that E(A) is perfect and is contained in the commutator subgroup of Gl(A). It suffices to show that if M1 , M2 ∈ Gl(A), then the commutator [M1 , M2 ] = M1 M2 M1−1 M2−1 is in E(A). Choose n large enough that M1 and M2 both lie in Gln (A). Then in Gl2n (A),

[M1 , M2 ] = (M1 ⊕ M1−1 )(M2 ⊕ M2−1 ) · (M2 M1 )−1 ⊕ (M2 M1 ) . The three matrices on the right are in E(A), by Lemma 10.9.3. Elementary row operations, if conducted stably, may be used to replace a matrix by any element of its right coset under E(A) — so stably, Gauss elimination picks matrix representatives for the elements of the quotient group Gl(A)/E(A). Thus, a calculation of this group will give us an indication of the inherent limitations on Gauss elimination for the study of matrices over A. Definition 10.9.5. Let A be a ring. Then K1 (A) = Gl(A)/E(A). The groups K1 (A) are sometimes called Whitehead groups, as they were first studied by J.H.C. Whitehead. However, we shall reserve that term for a related group, to be defined in Exercises 10.9.10, that was the primary motivation for Whitehead in developing this theory. The Whitehead groups in the latter sense are functors on groups, denoted W h(G) for a group G. They arise in the s-Cobordism Theorem, which detects when an inclusion M ⊂ W of manifolds is the inclusion of M in its product with I. They also arise in surgery theory, which is used to classify manifolds up to homeomorphism, diffeomorphism, etc. The study of K1 continues the study of algebraic K-theory that we began with K0 . We can think of K1 as giving a generalized notion of determinant theory, which works even over noncommutative rings. The point is that K1 (A) is the abelianization of Gl(A), so that similar matrices have the same value when viewed in K1 (A). Note that if f : A → B is a ring homomorphism, then the induced homomorphism f∗ : Gln (A) → Gln (B) carries elementary matrices to elementary matrices. Thus, there is an induced map K1 (f ) : K1 (A) → K1 (B) making K1 a functor from rings to abelian groups. If A is commutative, note that i : Gln (A) → Gln+1 (A) preserves determinants. Thus, there is a homomorphism det : Gl(A) → A× that restricts on each Gln (A) to the usual determinant map. Lemma 10.9.6. Let A be a commutative ring. Then there is a split short exact sequence ⊂

0 → Sl(A) −→ Gl(A) −−→ A× → 0 where Sl(A) =

∞ n=1

det

Sln (A) ⊂ Gl(A).

Proof The splitting comes from the fact that det : Gl1 (A) → A× is an isomorphism. Since elementary matrices have determinant 1, E(A) ⊂ Sl(A). Definition 10.9.7. Let A be a commutative ring. We write SK1 (A) for the quotient group Sl(A)/E(A). The following is then immediate from the above.


449

Proposition 10.9.8. Let A be a commutative ring. Then there is a split short exact sequence of abelian groups ⊂

0 → SK1 (A) −→ K1 (A) −−→ A× → 0. det

Thus, since the groups are abelian, there is an isomorphism K1 (A) ∼ = SK1 (A) ⊕ A× .

Calculations of K1 tend to be difficult. We shall not present any nontrivial calculations here. The next result is immediate from Proposition 10.8.10. Corollary 10.9.9. Let K be a field. Then SK1 (K) = 0 and K1 (K) = K× . Exercises 10.9.10. 1. Show that K1 (A × B) ∼ = K1 (A) ⊕ K1 (B). 2. Let G be a group. The 1 × 1 matrices {(±g) | g ∈ G} form a subgroup of Gl1 (Z[G]), which we denote by ±G. We write π(±G) for the image of ±G in K1 (Z[G]), and write W h(G) = K1 (Z[G])/π(±G), the Whitehead group of G. Show that if G is abelian, there is a split short exact sequence 0 → SK1 (Z[G]) → W h(G) → Z[G]× / ± G → 0.

Chapter 11

Galois Theory Galois Theory studies the algebraic extensions of a field k. One wishes to determine when two algebraic extensions are isomorphic by an isomorphism that is the identity on k, and to determine all such isomorphisms between them. In particular, for a given extension field K of k, we would like to be able to calculate autk (K), the group of automorphisms (as a ring) of K that are the identity on k. Notice that our interest here is in the ring homomorphisms between fields. It is customary to refer to these as homomorphisms of fields, and to implicitly assume, unless specifically stated to the contrary, that all mappings under discussion are homomorphisms of fields. Thus, in this chapter, when we speak of embeddings between fields or automorphisms of a field, we shall assume that the mappings are homomorphisms of fields. Our primary interest will be in the finite extensions of k, as these are the extensions about which we can say the most. However, it is important to note that the study of infinite algebraic extensions is an area of ongoing research. In Section 11.1, we show that if L is an algebraic closure of k, then every algebraic extension of k embeds in it. (We also show that any two algebraic closures of k are isomorphic by an isomorphism which restricts to the identity map on k.) Thus, our focus of interest is on the relationships between the subfields of the algebraic closure of L. If an extension of k is obtained by adjoining all roots of a given collection of polynomials, it is what’s known as a normal extension. Normal extensions K of k have the nice property that any two embeddings over k of K in an algebraic closure, L, of k have the same image. Thus, there is a unique subfield of L that is isomorphic to K, and any two embeddings of K in L differ by an automorphism of K. Thus, in an essential way, the study of the homomorphisms out of K is determined by the calculation of autk (K). We develop the theory of normal extensions in Section 11.2, and then apply it to classify the finite fields in Section 11.3. The separable extensions are the easiest to analyze. These are the extensions K of k with the property that the minimal polynomial over k of any element of K has no repeated roots. In particular, every algebraic extension is separable in characteristic 0. For a finite, separable extension, K, of k, the number of embeddings of K over k in an algebraic closure of k is equal to the degree of the extension. We develop the theory of separable extensions in Section 11.4. Extensions that are normal and separable are known as Galois extensions. If K is a Galois extension of k, we shall write Gal(K/k) for the automorphism group autk (K). 450

CHAPTER 11. GALOIS THEORY

451

We call it the Galois group of K over k. In Section 11.5, we prove the Fundamental Theorem of Galois Theory, which studies the Galois group of a finite Galois extension. Among other things, it shows that the fixed points in K of the action of Gal(K/k) are precisely k, and that the subfields of K containing k are precisely the fixed fields of the subgroups of Gal(K/k). Using the Fundamental Theorem of Galois Theory, we prove the Primitive Element Theorem, showing that every finite separable extension is simple. Then, in Section 11.6, we prove the Fundamental Theorem of Algebra, that the complex numbers, C, are algebraically closed. In Section 11.7, we study cyclotomic extensions and characterize their Galois groups. In particular, we obtain a calculation of Gal(Q(ζn )/Q). We use this in Section 11.8 to study the splitting fields of polynomials of the form X n − a. Galois groups for examples of such extensions are computed in the text and exercises. Sections 11.9 and 11.10 are concerned with Galois extensions with abelian Galois groups. We obtain a complete classification of the finite extensions of this sort, provided that the base field has a primitive n-th root of unity, where n is an exponent for the Galois group. In Section 11.11, we use our analysis of abelian extensions and of cyclotomic extensions to characterize the finite Galois extensions whose Galois group is solvable. As a corollary, we give a proof of Galois’ famous theorem that there is no formula for finding roots of polynomials of degree ≥ 5. We continue the discussion of such formulæ in Section 11.12. Section 11.13 gives the Normal Basis Theorem, which shows that if K is a finite Galois extension of k with Galois group G, then there is a basis for K as a k-vector space consisting of the elements of a single orbit of the action of G on K. This shows that K is free on one generator as a module over k[G]. Section 11.14 defines and studies the norm and trace functions, NK/k : K → k and trK/k : K → k for a finite extension K of k. These are extremely useful in studying the ring of integers of a number field. We shall make use of them in Chapter 12.

11.1

Embeddings of Fields

Definition 11.1.1. Let K be an extension field of k. Then an intermediate field between k and K is a subfield of K containing k. It is customary to indicate field extensions by vertical or upward slanting lines. Thus, the diagrams K

K1

k

K1

L ? ?? ?? ?? ?

?? ?? ?? ??

K2

k

may be read as follows: The left-hand diagram indicates that K is an extension of k and that K1 is an intermediate field between them. The right-hand diagram gives an extension, L, of k, together with two intermediate fields, K1 and K2 .


452

Definition 11.1.2. Let K and L be extensions of the field k. An embedding of K in L over k is a homomorphism ν : K → L of fields such that ν restricts on k to its original inclusion map into L. Pictorially, this gives ν

K? ?? ??

/ L

k If K is an extension of k and ν : k → L is any homomorphism of fields, then an extension of ν to K is a homomorphism ν : K → L whose restriction to k is ν. Here, we could make a diagram as follows. K

k

ν

ν

/ L

/ ν(k)

Recall that if f : A → B is a ring induced homomorphism

n then the n homomorphism, i f∗ : A[X] → B[X] is given by f∗ = i=0 f (ai )X i . i=0 ai X Lemma 11.1.3. Let ν : k → L be an embedding of fields and let K = k(α) be a finite simple extension of k. Let f (X) be the minimal polynomial of α over k. Then for any root β of ν∗ (f ) in L, there is a unique extension of ν to K that carries α to β. Indeed, the extension factors as a composite k(α) ∼ = k1 (β) ⊂ L, where k1 = ν(k). Moreover, if ν is an extension of ν to K, then ν(α) must be a root of ν∗ (f ) in L. Thus, there is a one-to-one correspondence between the extensions of ν to K and the roots of ν∗ (f ) in L. Proof Let ν be an extension of ν to K. Then, applying ν to both sides of the equation f (α) = 0, we see that (ν∗ (f ))(ν(α)) = 0, so ν(α) is a root of ν∗ (f ), as claimed. Also, note that every element of K may be written as g(α) for some polynomial g(X) ∈ k[X]. As above, ν(g(α)) = (ν∗ (g))(ν(α)), and hence ν is determined by its effect on α. Finally, if β is a root of ν∗ (f ), we may construct a commutative diagram k[X]/(f (X)) εα ∼ = k(α)

ν∗ / k1 [X]/(ν∗ (f )) ∼ = εβ ∼ = ν / k1 (β)

with ν(α) = β. For algebraic extensions of the form k(α, β), the situation is more complicated, as the natural method of analysis proceeds by stages: First extend ν over k(α), and then extend further to k(α, β). To get control over this procedure, we need to know the minimal


453

polynomial of β over k(α). Finding this can be a nontrivial task, even if we know the minimal polynomial of β over k. The point is that if ν : k → L is a homomorphism, and if ν : k(α) → L is an extension, then it may be the case that not every root of ν∗ (minβ/k ) is a root of ν ∗ (minβ/k(α) ). And only the latter roots may be used to extend ν. In practice, we shall be forced to confront such issues head on. But let us build up some theory first. Proposition 11.1.4. Let ν : k → L be an embedding of fields, where L is algebraically closed, and let K be any algebraic extension of k. Then ν extends to an embedding of K in L. Proof If K is a finite extension of k, we may argue by induction on [K : k]. If α ∈ K is not in k, then ν∗ (minα/k ) has a root in L because L is algebraically closed. Thus, ν extends to ν : k(α) → L by Lemma 11.1.3. But [K : k(α)] < [K : k], so ν extends to an embedding of K by induction. If K is infinite over k, we use Zorn’s Lemma. We define a partially ordered set whose elements consist of pairs (K , ν ), where K is an intermediate field between k and K, and ν : K → L extends ν. Here, (K , ν ) ≤ (K , ν ) if K ⊂ K and ν extends ν . Here, if {(Ki , νi ) | i ∈ I} is a totally ordered subset of this partially ordered set, let K = i∈I Ki . Then K is a subfield of K, and there is an embedding ν : K → L defined by setting ν (α) = νi (α), for any i such that α ∈ Ki . Now (K , ν ) is an upper bound for the totally ordered subset in question, and hence the hypothesis of Zorn’s Lemma is satisfied. Thus, there is a maximal element (K , ν ) in this partially ordered set. But then K must equal K, as otherwise, we could extend ν over a simple extension K (α) of K in K, in which case (K , ν ) would not be a maximal element in this set of extensions. Corollary 11.1.5. Let ν : k1 → k2 be an isomorphism of fields and let Li be an algebraic closure of ki for i = 1, 2. Then there is an embedding ν : L1 → L2 that extends ν: L1

k1

ν

/ L2

ν

/ k2

Moreover, any such extension of ν gives an isomorphism from L1 to L2 . In particular, if L1 and L2 are algebraic closures of a field k, then L1 and L2 are isomorphic over k. Proof Considering ν as an embedding into L2 , Proposition 11.1.4 provides an extension to an embedding ν : L1 → L2 , as claimed. Similarly, if μ : k2 → k1 is the inverse of the isomorphism ν, then μ provides an embedding of k2 in L1 . Let μ : L2 → L1 extend μ. Then (ν ◦ μ) : L2 → L2 is an embedding over k2 . But since L2 is algebraic over k2 , Proposition 8.2.19 shows that any embedding of L2 in itself over k2 is an automorphism of L2 . So ν ◦ μ is surjective, and hence ν : L1 → L2 is surjective as well.


11.2

454

Normal Extensions

Definitions 11.2.1. Let f (X) be a polynomial over the field k and let K be an extension field of k. We say that f splits in K[X] (or splits in K) if f (X) = a(X − α1 )r1 . . . (X − αk )rk in K[X], with a, α1 , . . . , αk ∈ K and ri > 0 for i = 1, . . . , k. We say that K is a splitting field for f (X) if f splits as above in K[X] and K = k(α1 , . . . , αk ). Notice that since a is the leading coefficient of the right-hand side of the displayed equation, the fact that f (X) ∈ k[X] implies that a ∈ k. Of course, the elements α1 , . . . , αk are algebraic over k as they are roots of f (X). Examples 11.2.2. 1. We’ve seen in Section 8.8 that X n − 1 splits in Q(ζn )[X]. Since ζn is a root of X n − 1, Q(ζn ) is a splitting field for X n − 1 over Q. √ 2. Let a > 0 in Q. Then {ζnk n a | 0 ≤ k < n} gives n distinct roots of X n − a. Thus, √ Q( n a, ζn ) is a splitting field of X n − a over Q. 3. The complex numbers C is a splitting field for X 2 + 1 over R. The next lemma is basic. Lemma 11.2.3. Let K be an extension of k and suppose that f (X) ∈ k[X] splits in K[X] via f (X) = a(X − α1 )r1 . . . (X − αk )rk with a, α1 , . . . , αk ∈ K and with ri > 0 for i = 1, . . . , k. Let K1 be an intermediate field between k and K such that each of the roots αi lies in K1 . Then f (X) splits as above in K1 [X] as well. In particular, the subfield k(α1 , . . . , αk ) of K is a splitting field of f . Proof Both sides of the displayed equation lie in K1 [X]. Since the natural map from K1 [X] to K[X] is injective, the result follows. In an algebraically closed field, every polynomial of positive degree splits. Thus, Lemma 11.2.3 implies that splitting fields exist. Corollary 11.2.4. Let L be an algebraically closed field containing k and suppose that f (X) ∈ k[X] splits as f (X) = a(X − α1 )r1 . . . (X − αk )rk in L[X], with a, αi ∈ L and ri > 0 for i = 1, . . . , k. Then K = k(α1 , . . . , αk ) is a splitting field for f . We shall treat the uniqueness of splitting fields in Proposition 11.2.8. The splitting field of a single polynomial is all we need think about if we’re concerned only with finite extensions. But for infinite extensions, we need a generalization.


455

Definition 11.2.5. Let {fi (X) | i ∈ I} be any set of polynomials over the field k. Then K is a splitting field for {fi (X) | i ∈ I} if 1. Each polynomial fi (X) with i ∈ I splits in K[X]. 2. If {αj | j ∈ J} is the set of all roots in K of polynomials in {fi (X) | i ∈ I}, then K = k(αj | j ∈ J). Note that the splitting field for a finite set {f1 (X), . . . , fk (X)} is just the splitting field for the product f1 (X) . . . fk (X). The property of being a splitting field is one of the fundamental properties we will use in our analysis of Galois theory. We abstract it as follows. Definition 11.2.6. An extension K of k is a normal extension1 if K is the splitting field for some family of polynomials in k[X]. Note that if K is a splitting field for the polynomials {fi (X) | i ∈ I} over k and if K1 is any intermediate field between k and K, then K is also the splitting field for the polynomials {fi (X) | i ∈ I} over K1 . Lemma 11.2.7. Let K be a normal extension of k and let K1 be an intermediate field between k and K. Then K is a normal extension of K1 . However, we shall see in Example 11.2.10 that an intermediate field between k and a normal extension K of k need not be normal over k. As was done in Corollary 11.2.4 for a single polynomial, we may construct a splitting field for any set of polynomials in k[X] as a subfield of any algebraic closure, L, of k: If {fi (X) | i ∈ I} is any set of polynomials over k, then each fi (X) splits in L[X], since L is algebraically closed. Thus, if {αj | j ∈ J} is the set of all roots in L of the polynomials in {fi (X) | i ∈ I}, then k(αj | j ∈ J) is a splitting field for {fi (X) | i ∈ I} over k. We now show that splitting fields are unique. Proposition 11.2.8. Let {fi (X) | i ∈ I} be a set of polynomials over the field k, and let K be a splitting field for {fi (X) | i ∈ I}. Let L be any extension field of k with the property that each fi (X) splits in L[X] and let {αj | j ∈ J} be the set of all roots in L of the polynomials in {fi (X) | i ∈ I}. Then every embedding of K in L over k has image equal to k(αj | j ∈ J), and hence gives an isomorphism from K onto k(αj | j ∈ J). Note that if L is an algebraically closed field containing k, then K embeds in L over k by Proposition 11.1.4. As a result, we see that any two splitting fields for {fi (X) | i ∈ I} are isomorphic over k. Proof Let ν : K → L be an embedding over k. Let {βj | j ∈ J } be the set of all roots in K of polynomials in {fi (X) | i ∈ I}. Then each βj is a root of one of the irreducible factors, say p(X), of one of the fi in k[X]. By Lemma 11.1.3, ν(βj ) must also be a root of p(X), and hence ν(βj ) ∈ {αj | j ∈ J}. Thus, ν restricts to a function ν : {βj | j ∈ J } → {αj | j ∈ J}. Since K = k(βj | j ∈ J ), it suffices to show that this function is onto. 1 Some authors, e.g., Emil Artin, have used the expression “normal extension” to describe what we shall call a Galois extension.


456

Note that for each irreducible factor, p(X), of one of the fi (X), ν carries the roots of p in K into the set of roots of p in L. Since ν is injective, it suffices to show that p has the same number of roots in K as in L. But if p(X) = a(X − βj1 )r1 . . . (X − βjk )rk in K[X], with a ∈ k, with βj1 , . . . βjk distinct and with ri > 0 for all i, then p(X) = a(X − ν(βj1 ))r1 . . . (X − ν(βjk ))rk in L[X]. Since ν is injective, p has k distinct roots in L, just as it did in K. We can now characterize normal extensions. Proposition 11.2.9. Let K be an algebraic extension of k. Then the following conditions are equivalent. 1. K is a normal extension of k. 2. If L is an algebraic closure of K, then every embedding of K in L over k has image K, and hence gives an automorphism of K. 3. For every element α ∈ K, the minimal polynomial of α over k splits in K[X]. Proof Let K be a normal extension of k. Since K is algebraic over k, so is any algebraic closure, L, of K (Corollary 8.2.18), and hence L is an algebraic closure for k as well. Thus, if K is the splitting field for {fi (X) | i ∈ I}, Then Proposition 11.2.8 shows that K is the subfield of L obtained by adjoining all the roots in L of the polynomials in {fi (X) | i ∈ I}, and that any embedding of K in L over k must have image K. Thus, the first condition implies the second. Suppose that the second condition holds. Let L be an algebraic closure for K and let α ∈ K. Then the minimal polynomial, f (X), of α over k splits in L[X]. To show that it splits in K[X], it suffices to show that every root of f (X) in L must lie in K. But if β is a root of f (X) in L, Lemma 11.1.3 provides an isomorphism ν : k(α) → k(β) over k. By Proposition 11.1.4, ν extends to an embedding ν : K → L. But the second condition says that the image of ν must be K, and hence β must lie in K. So the second condition implies the third. Now suppose that the third condition holds. Since K is algebraic over k, K = k(αi | i ∈ I) for some collection of elements αi ∈ K that are algebraic over k (Corollary 8.2.17). But the minimal polynomials of the αi over k must split in K[X], and hence K is the splitting field for the set of minimal polynomials of the αi over K. So the third condition implies the first. We may use Proposition 11.2.9 to show that if K is a normal extension of k, then an intermediate field between k and K need not be normal over k. √ √ Example 11.2.10. √ Q( 4 2) is an intermediate field between Q and Q( 4 2, i). The minimal polynomial of 4 2 over Q is X 4 − 2, √ which is irreducible over Q by Eisenstein’s √ 4 as Q( 2) is contained in the real criterion. But X 4 − 2 fails to split in Q( 4 2)[X], √ √ 4 4 4 numbers, and hence does not contain the roots ±i 2 of X − 2. So Q( 2) is not normal √ over Q. But Q( 4 2, i) is normal over Q, as it is the splitting field of X 4 − 2. We now give another application of Proposition 11.2.9.


457

Corollary 11.2.11. Let K be an extension of k and let {Ki | i ∈ I} be a collection of intermediate fields that are normal over k. Then ∩i∈I Ki is a normal extension of k. Proof Let α ∈ ∩i∈I Ki and let f (X) be the minimal polynomial of α over k. Then f splits in each Ki [X], and hence also in K[X]. In particular, the set of roots of f in each Ki must coincide, and hence each root of f in K must lie in ∩i∈I Ki . Since f splits in K[X], Lemma 11.2.3 shows that it splits in (∩i∈I Ki )[X]. The result now follows from Proposition 11.2.9. In particular, let K be a normal extension of k and let K1 be an intermediate field between k and K. Then the collection of intermediate fields between K1 and K that are normal over k is nonempty. Taking their intersection, we see that the following definition makes sense. Definition 11.2.12. Let K be a normal extension of k and let K1 be an intermediate field between k and K. Then the normal closure of K1 over k in K is the smallest intermediate field between K1 and K that is normal over k. And normal closures may be constructed as follows. Proposition 11.2.13. Let K be a normal extension of k and let α1 , . . . , αn ∈ K. Let fi (X) be the minimal polynomial of αi for i = 1, . . . , n and let β1 , . . . , βk be the collection of all roots in K the polynomials f1 , . . . , fn . Then k(β1 , . . . , βk ) is the normal closure of k(α1 , . . . , αn ) over k in K. Also, if f (X) = f1 (X) . . . fn (X), then k(β1 , . . . , βk ) is a splitting field for f (X) over k. Proof Write K1 = k(α1 , . . . , αn ) and K2 = k(β1 , . . . , βk ). Since K is a normal extension of k and since fi (X) is the minimal polynomial of αi ∈ K over k, Proposition 11.2.9 shows that each fi (X) splits in K[X], and hence f (X) does also. Since β1 , . . . , βk is the set of all roots of f (X) in K, Lemma 11.2.3 shows that K2 is the splitting field for f over k, and hence is normal over k. But any normal extension of k in K containing α1 , . . . , αn must contain all the roots of the fi (X), and hence must contain K2 . Thus, K2 is the smallest normal extension of k in K containing K1 , and the result follows. Since any finite extension of k may be written as k(α1 , . . . , αn ) for some collection of elements α1 , . . . , αn , and since the splitting field of a single polynomial is finite, we obtain the following corollary. Corollary 11.2.14. Let K be a normal extension of k and let K1 an intermediate field that is finite over k. Then the normal closure of K1 over k in K is a finite extension of k. The proof of Proposition 11.2.13 generalizes to give the following proposition. Proposition 11.2.15. Let K be a normal extension of k and let {αi | i ∈ I} ⊂ K. Let fi (X) be the minimal polynomial of αi for i ∈ I. Then the normal closure of k(αi | i ∈ I) over k in K is a splitting field for {fi (X) | i ∈ I}. Exercises 11.2.16. 1. Show that every extension of degree 2 is normal. 2. Show that for all n > 2, there are extensions of Q of degree n that are not normal.


458

3. Show that normality of extensions is not transitive: Give an example of extensions k ⊂ K1 ⊂ K such that K1 is normal over k, K is normal over K1 , but K is not normal over k. 4. Let K be a normal extension of k and let K1 be an intermediate field between k and K. Let L be any extension field of K. Show that the image of any embedding of K1 in L over k must lie in K. 5. Let K be a normal extension of k and let K1 be an intermediate field between k and K. Show that every embedding of K1 in K extends to an automorphism of K. 6. Let K be an algebraic extension of k. Let L1 and L2 be two different extensions of K that are both normal over k. Show that the normal closures of K over k in L1 and L2 are isomorphic over K.

11.3

Finite Fields

Let k be a finite field. Then k must have characteristic p for some prime p, and hence k is a finite extension of Zp . We shall show that for each r > 0 there is a unique field of order pr . Recall that for any commutative ring A of characteristic p, the Frobenius map ϕ : A → A, defined by ϕ(a) = ap for all a ∈ A, is a ring homomorphism. We shall write ϕn for the n-fold composite of ϕ with itself. Proposition 11.3.1. Let k be a finite field of order q = pr and let K be the splitting n field of X q − X over k. Then [K : k] = n. n

Proof Since q n ≡ 0 mod p, the formal derivative of X q − X is just 1. Thus, by n n Lemma 8.7.7, X q − X has no repeated roots in K. Since X q − X splits in K, this says n n that X q − X has q n distinct roots in K. Write S ⊂ K for the set of roots of X q − X in K. We claim that S is a subfield of K. To see this, note that since q n = prn , we have α ∈ S if and only if ϕrn (α) = α, with ϕ the Frobenius, as above. Since ϕ is a ring homomorphism, the elements fixed by ϕrn are easily seen to form a subfield of K. n Thus, S is a field containing all the roots of X q − X. Since K is obtained from k by adjoining these roots, we must have K = S, and hence K has q n elements. But then [K : k] = n, as claimed. Thus, for any positive n, a finite field has an extension of degree n. Our next result shows that this extension is unique. Proposition 11.3.2. For any finite field k and any positive integer n, any two extensions of k of degree n are isomorphic over k. Thus, k has a unique extension of degree n. n

Proof We have already shown that if k has order q, then the splitting field of X q − X is an extension of k of degree n. Thus, by our uniqueness result for splitting fields (Proposition 11.2.8), it suffices to show that any extension of k of degree n is a splitting n field for X q − X. If K is an extension of degree n over k, then it has q n elements. Thus, the unit group, × K , of K has order q n − 1. So every element α ∈ K× has exponent q n − 1 there, and n hence is a root of X q −1 − 1. But then α is also a root of X · (X q

n

−1

n

− 1) = X q − X.

CHAPTER 11. GALOIS THEORY n

459 n

Since 0 is also a root of X q − X, X q − X has q n distinct roots in K, and hence splits in K[X]. Since K is generated by the roots of this polynomial, it must be its splitting field over k. So, applying the above with k = Zp , we obtain the following corollary. Corollary 11.3.3. For each prime p and each positive integer n, there is a unique isomorphism class of fields of order pn . We next analyze the uniqueness properties of subfields. Proposition 11.3.4. Let K be the field of order pn . Then if k is a subfield of K, k must have order pr for some r dividing n. Conversely, if r divides n, then there is a unique subfield of K of order pr , given by r

k = {α ∈ K | αp = α}. Proof Since we’re in characteristic p, any subfield, k, of K has order pr for some r. And, if [K : k] = s, then (pr )s = pn , so r divides n. If r divides n, then Proposition 11.3.1 shows that the field of order pr has an extension of order pn . Since there’s only one field, here denoted K, of order pn , K must have a subfield of order pr . Thus, suppose that k is any subfield of K of order pr . It suffices to show that r

k = {α ∈ K | αp = α}. r

r

Of course, {α ∈ K | αp = α} is the set of roots of X p − X in K. And the proof r of Proposition 11.3.2 shows that every element of k must be a root of X p − X, simply r because k is a field with pr elements. Since X p − X may have at most pr roots in K, k must be the entire collection of such roots in K, and hence must coincide with the displayed subset. Of course, there is much more of interest in the study of finite fields. For instance, if K is the field of order pn , and if α ∈ K is not contained in a proper subfield, then K = Zp (α). We may then ask about the minimal polynomial of α over Zp . Of course, every irreducible polynomial of degree n over Zp occurs in this manner. One source of elements α with K = Zp (α) is as follows. Since K× is finite, it is a cyclic group by Corollary 7.3.14. We shall refer to the generators of K× as the primitive (pn − 1)-st roots of unity over Zp . If α is one such, then no proper subfield of K may contain α, and hence K = Zp (α). Exercises 11.3.5. 1. Find the smallest number pn such that the field, K, of order pn has an element α that is not a primitive (pn − 1)-st root of unity, but K = Zp (α), anyhow. 2. Show that X 4 + X + 1 is irreducible over Z2 . Find a primitive 15-th root of unity in the field K = Z2 [X]/(X 4 + X + 1). 3. Let p be a prime congruent to −1 mod 4. Show that X 2 + 1 is irreducible in Zp [X], and hence K = Zp [X]/(X 2 + 1) is the field of order p2 . Note that K has a multiplication similar to that of the complex numbers. 4. Find a primitive 24-th root of unity in K = Z5 [X]/(X 2 − 2). 5. What are the infinite subfields of the algebraic closure of Zp ?


11.4

460

Separable Extensions

Definition 11.4.1. Let K be an extension of k. We say that α ∈ K is separable over k if α is algebraic over k and if α is not a repeated root of its minimal polynomial over k. We say that an extension K of k is separable if every element of K is separable over k. Proposition 8.7.8, together with our understanding of splitting fields, will give us a better ideal of what separability means. Proposition 11.4.2. Let K be an algebraic extension of k and let α ∈ K. Let f (X) be the minimal polynomial of α over k. Then α is separable over k if and only if f (X) has no repeated roots in its splitting field over k. In this case, if L is any extension of k in which f splits, then f (X) factors as f (X) = (X − α1 ) . . . (X − αn ) in L[X], where α1 , . . . , αn are distinct elements of L. All algebraic extensions in characteristic 0 are separable. In characteristic p = 0, if α is not separable over k, then, if L is any extension of k in which the minimal polynomial f (X) splits, we have a factorization k

k

f (X) = (X − α1 )p . . . (X − αn )p

in L[X], where α1 , . . . , αn are distinct elements of L and k > 0. Proof Proposition 8.7.8 tells us that α is separable over k if and only if the formal derivative f (X) = 0, in which case f (X) has no repeated roots. Thus, if f splits in L[X], it must split as stated. Proposition 8.7.8 goes on to say that in characteristic 0, no irreducible polynomial has repeated roots, while in characteristic p, any irreducible polynomial with repeated k roots has the form f (X) = h(X p ), where k > 0 and h(X) is an irreducible polynomial in k[X] with h (X) = 0. Thus, h(X) is separable. Let L be an algebraically closed field containing k. Then in L[X], we have h(X) = (X − β1 ) . . . (X − βn ) for distinct elements β1 , . . . , βn ∈ L. k Since L is algebraically closed, we can find roots αi of X p −βi for i = 1, . . . , n. Thus, in L[X], we have k

k

k

k

X p − βi = X p − αip = (X − αi )p , where the last equality comes from the fact that the Frobenius map is a homomorphism. Putting this together, we see that k

k

k

f (X) = h(X p ) = (X − α1 )p . . . (X − αn )p

in L[X], and hence also in k(α1 , . . . , αn )[X]. The result now follows from the uniqueness of splitting fields (Proposition 11.2.8).


461

Since all algebraic extensions in characteristic 0 are separable, it might appear at first that the study of separability is mainly of interest in characteristic p. However, separability has important consequences whenever it holds. We shall explore these consequences below. Proposition 11.4.2 allows us to recast separability in terms of polynomials. Definition 11.4.3. An irreducible polynomial over a field k is separable if it has no repeated roots in its splitting field. More generally, we say that a polynomial f (X) ∈ k[X] is separable if each of its irreducible factors in k[X] is separable.2 The next corollary is now immediate from Proposition 11.4.2. Corollary 11.4.4. Let L be an extension of k and let α ∈ L be algebraic over k. Then α is separable over k if and only if the minimal polynomial, f (X), of α over k is separable. Proposition 11.4.2 also gives the following. Corollary 11.4.5. Let L be an extension field of k and let α ∈ L be separable over k. Let K be any intermediate field between k and L. Then α is separable over K. Proof The minimal polynomial of α over K divides the minimal polynomial of α over k in K[X]. Since minα/k (X) has no repeated roots in its splitting field, the same must hold for minα/K . Corollary 11.4.6. Let K be a separable extension of k and let K1 be an intermediate field between k and K. Then K is separable over K1 , and K1 is separable over k. Proof K is separable over K1 by Corollary 11.4.5. K1 is separable over k because every element of K is separable over k. Here is another case in which separability is automatic. Proposition 11.4.7. Every algebraic extension of a finite field is separable. Proof We first show that every algebraic extension of Zp is separable. Thus, suppose that K is an algebraic extension of Zp , and let f (X) be the minimal polynomial of α ∈ K over Zp . Then if α is not separable over Zp , Proposition 8.7.8 shows that f (X) = g(X p ) for some g(X) ∈ Zp [X]. But in Zp [X], Lemma 8.6.6 shows that g(X p ) = (g(X))p , which is not irreducible. So α must have been separable. Now suppose given a finite field k of characteristic p and an algebraic extension K of k. Then k is a finite extension of Zp , and hence K is algebraic over Zp by Corollary 8.2.18. But we’ve just shown that K must be separable over Zp . Since k is an intermediate field between Zp and K, the result follows from Corollary 11.4.6. In the case of finite extensions, there is another very useful way to look at separability. Definition 11.4.8. Let K be a finite extension of k and let L be an algebraic closure for k. Then the separability degree of K over k, written [K : k]s , is the number of distinct embeddings of K over k into L. 2 Warning: Some authors call a polynomial separable if it has no repeated roots in its splitting field. We have chosen the present definition for its role in Corollary 11.4.16.


462

Note that since any two algebraic closures of k are isomorphic over k, the calculation of [K : k]s is independent of the choice of algebraic closure. Proposition 11.4.2 shows that if α is algebraic over k and f is its minimal polynomial, then there is a factorization of the form f (X) = (X − α1 )r . . . (X − αn )r in any field in which f (X) splits, where α1 , . . . , αn are distinct. Here, r = 1 if k has characteristic 0. In characteristic p, we have r = pk , where k = 0 if and only if f is separable. Lemma 11.4.9. Let K be a simple extension K = k(α) of k, with α algebraic over k, and suppose that the minimal polynomial f (X) of α over k splits as f (X) = (X − α1 )r . . . (X − αn )r in its splitting field. Then [K : k]s = n, and hence [K : k] = r · [K : k]s . Thus, [K : k] = [K : k]s if and only if α is separable over k, and if α is inseparable over k, then [K : k] = pk · [K : k]s , where p is the characteristic of k and k > 0. Proof Let L be an algebraic closure of K, and suppose that the factorization of f (X) above takes place in a splitting field that’s contained in L. Then Lemma 11.1.3 shows there are exactly n embeddings of K into L over k, obtained by mapping α to α1 , . . . , αn . Thus, [K : k]s = n. The rest follows from the above summary of the results in Proposition 11.4.2. Notice that since the embeddings of an extension k(α1 , . . . , αn ) over k are determined by their effect on α1 , . . . , αn , [K : k]s is finite for any finite extension K of k. The next lemma will be useful for dealing with non-simple extensions. Lemma 11.4.10. Let K be a finite extension of k and let ν : k → k1 be an isomorphism. Let L1 be an algebraic closure for k1 . Then [K : k]s is equal to the number of distinct extensions ν : K → L1 of ν to K. Proof Let L be an algebraic closure for k. Let ν : L → L1 be an embedding of L in L1 that extends ν: L

k

ν / L1

ν

/ k1

Then ν is an isomorphism by Corollary 11.1.5. Let S be the set of embeddings of K into L1 extending ν and let T be the set of embeddings of K into L over k. Then there’s a bijection ν∗ : T → S given by ν∗ (μ) = ν ◦ μ.


463

Corollary 11.4.11. Let K be a finite extension of k and let K1 be an intermediate field between k and K. Then [K : k]s = [K : K1 ]s [K1 : k]s . Proof Let L be an algebraic closure of k. Then each of the [K1 : k]s distinct embeddings of K1 in L over k has [K : K1 ]s distinct extensions to K. Now we can put things together. Proposition 11.4.12. Let K be a finite extension of k. Then [K : k]s divides [K : k], and the two are equal if and only if K is a separable extension of k. If K is inseparable over k, we have [K : k] = pr [K : k]s , where p is the characteristic of k and r > 0. If K = k(α1 , . . . , αk ), then K is a separable extension of k if and only if α1 , . . . , αk are separable over k. Proof Write K = k(α1 , . . . , αk ). Let K0 = k and let Ki = k(α1 , . . . , αi ) for 1 ≤ i ≤ k. Then [K : k]s

=

k .

[Ki−1 (αi ) : Ki−1 ]s ,

while

i=1

[K : k]

=

k .

[Ki−1 (αi ) : Ki−1 ].

i=1

For i = 1, . . . , k, Lemma 11.4.9 tells us that [Ki : Ki−1 ]s divides [Ki : Ki−1 ], with the two being equal if and only if αi is separable over Ki−1 . If αi is inseparable over Ki−1 , then [Ki : Ki−1 ] = pri · [Ki : Ki−1 ]s , where p is the characteristic of k and ri > 0. By passage to products, we see that if αi is separable over Ki−1 for i = 1, . . . , k, then [K : k]s = [K : k], and that otherwise [K : k] = pr · [K : k]s , where p is the characteristic of k and r > 0. Suppose that αi is separable over k for i = 1, . . . , n. Then Corollary 11.4.5 shows that each αi is separable over Ki−1 , and hence [K : k]s = [K : k] by the argument above. Thus, it suffices to show that if [K : k]s = [K : k], then K is a separable extension of k. But if [K : k]s = [K : k] and if α ∈ K, then the multiplicativity formulæ for degrees and separability degrees show that [k(α) : k]s = [k(α) : k]

and

[K : k(α)]s = [K : k(α)].

And the former equality shows that α is separable over k by Lemma 11.4.9. Corollary 11.4.13. Let K be a finite extension of k and let K1 be an intermediate field between k and K. Then K is separable over k if and only if both K1 is separable over k and K is separable over K1 .


464

Proof The multiplicativity formulæ for degrees and for separability degrees show that [K : k]s = [K : k] if and only if both [K : K1 ]s = [K : K1 ] and [K1 : k]s = [K1 : k]. We may now generalize this to infinite algebraic extensions. Proposition 11.4.14. Let K be an algebraic extension of k and let K1 be an intermediate field between k and K. Then K is separable over k if and only if both K1 is separable over k and K is separable over K1 . Proof By Corollary 11.4.6, it suffices to show that if K is separable over K1 and K1 is separable over k, then K is separable over k.Thus, let α ∈ K and let f (X) be the n minimal polynomial of α over K1 . Say f (X) = i=0 αi X i with αi ∈ K1 for i = 0, . . . , n and with αn = 1. Then f (X) is also the minimal polynomial of α over the finite extension k(α0 , . . . , αn−1 ) of k. Since α is not a repeated root of f (X), α is separable over k(α1 , . . . , αn ). By Proposition 11.4.12, k(α1 , . . . , αn )(α) is separable over k(α1 , . . . , αn ). And any intermediate field between k and K is separable over k. Since the extensions are now finite over k, the result follows from Corollary 11.4.13. Corollary 11.4.15. Let K = k(αi | i ∈ I), where αi is separable over k for each i ∈ I. Then K is a separable extension of k. Proof Each element α ∈ K is contained in a finite extension k(αi1 , . . . , αik ) of K. The result now follows from Proposition 11.4.12. The next corollary is now immediate from Corollary 11.4.4. Corollary 11.4.16. A polynomial f (X) ∈ k[X] is separable if and only if the splitting field of f is a separable extension of k. More generally, if {fi (X) | i ∈ I} is any collection of separable elements of k[X], then the splitting field of {fi (X) | i ∈ I} is a separable extension of k. Corollary 11.4.17. Let K be a normal extension of k and let K1 be an intermediate field between k and K such that K1 is separable over k. Then the normal closure of K1 over k in K is a separable extension of k as well. Proof Write K1 = k(αi | i ∈ I). Since αi is separable over k, its minimal polynomial, fi (X), over k is separable. By Proposition 11.2.15, the normal closure of K1 over k in K is a splitting field for {fi (X) | i ∈ I}, so the result follows from Corollary 11.4.16. Proposition 11.4.12 shows that the separability degree of a finite extension divides its degree. Definitions 11.4.18. Let K be a finite extension of k. Then the inseparability degree, [K : k]i , of K over k is defined by [K : k]i = [K : k]/[K : k]s . If [K : k]i = [K : k], we say that K is a purely inseparable extension of k. We may now apply Proposition 11.4.12. Corollary 11.4.19. Let k be a field of characteristic p = 0 and let K be a finite extension of k. Then [K : k]i is a power of p.


465

Exercises 11.4.20. 1. Show that every algebraic extension of a perfect field is separable. † 2. Let k be a field of characteristic p = 0. Let K be an extension of k and let K1 be the collection of elements of K that are separable over k. Show that K1 is an intermediate field between k and K, and hence a separable extension of k. We shall refer to K1 as the separable closure of k in K. (a) Show that the separable closure of K1 in K is just K1 . (b) Let α ∈ K. Show that the minimal polynomial of α over K1 has the form r X p − β for some r ≥ 0 and some β ∈ K1 . (c) Show that K is a purely inseparable extension of K1 and that [K1 : k] = [K : k]s , while [K : K1 ] = [K : k]i .

11.5

Galois Theory

We shall be able to treat finite extensions of the following sort. Definitions 11.5.1. An extension field K of k is a Galois extension if it is normal and separable. If K is a Galois extension of k, then the Galois group, Gal(K/k), of K over k is the group of automorphisms (as a field) of K over k. The group structure on the Galois group, of course, is given by composition of automorphisms. We’d like to be able to compute the Galois groups of finite Galois extensions. This can require some detailed study of the extension in question. Indeed, it can be difficult to compute the degree of a particular Galois extension. Look at the work, for instance, that went into the calculation of [Q(ζn ) : Q] in Section 8.8. However, we shall see that this calculation is the most difficult part of computing Gal(Q(ζn )/Q). Here, we shall develop the general theory of Galois groups. Note that if K is a Galois extension of k, then the Galois group, G = Gal(K/k), acts on K, making K a G-set. We wish to determine the orbit of an element α ∈ K under this action. Note that since K is a normal extension of k, the minimal polynomial of α over K splits in K[X]. Lemma 11.5.2. Let K be a Galois extension of k and let α ∈ K. Suppose that the minimal polynomial, f (X), of α over k splits as f (X) = (X − α1 ) . . . (X − αk ) in K[X]. Then the orbit of α under the action of G = Gal(K/k) is precisely {α1 , . . . , αk }. Proof Let σ ∈ G. Since σ is an automorphism over k, σ∗ : K[X] → K[X] carries f (X) to itself. Thus, by Lemma 11.1.3, σ(α) must be a root of f (X). Thus, it suffices to show that for any root, αi , of f (X) there is an automorphism, σ, of K over k, with σ(α) = αi . First, Lemma 11.1.3 provides an isomorphism σ0 : k(α) → k(αi ) over k with σ0 (α) = αi . Regarding σ0 as an embedding into an algebraic closure, L, of K, we may extend it to an embedding σ : K → L over k by Proposition 11.1.4. But K is a normal extension of k, so any embedding of K in L has image K by Proposition 11.2.9. Thus, σ may be regarded as an automorphism of K.


466

In particular, if K is a Galois extension of k and if f (X) ∈ k[X] is a polynomial that splits in K[X], then the set of roots of f in K form a sub-G-set of K. Note that if K is the splitting field of f over k and if α1 , . . . , αn are the roots of f in K, then K = k(α1 , . . . , αn ), and hence any automorphism of K over k is determined by its effect on α1 , . . . , αn . Corollary 11.5.3. Let K be the splitting field over k of the separable polynomial f (X) ∈ k[X] and let α1 , . . . , αn be the roots of f in K. Then every element σ ∈ Gal(K/k) restricts to give a permutation of {α1 , . . . , αn }. Moreover, the passage from elements of the Galois group to their restriction to {α1 , . . . , αn } induces an injective group homomorphism ρ : Gal(K/k) → Sn .

We shall make extensive use of the following notions. Definitions 11.5.4. Let K be a Galois extension of k and let G = Gal(K/k). For each subgroup H ⊂ G, the fixed field of H, KH , is defined by KH = {α ∈ K | σ(α) = α for all σ ∈ H}. For an intermediate field, K1 , between k and K, we define the subgroup GK1 associated to K1 by GK1 = {σ ∈ G | σ(α) = α for all α ∈ K1 }. Lemma 11.5.5. Let K be a Galois extension of k and let G = Gal(K/k). Then for each subgroup H ⊂ G, KH is an intermediate field between K and k, and for each intermediate field K1 between K and k, GK1 is a subgroup of G. Lemma 11.5.6. Let K be a Galois extension of k and let G = Gal(K/k). Let K1 be an intermediate field between k and K. Then K is a Galois extension of K1 and the subgroup GK1 of G associated to K1 is the Galois group Gal(K/K1 ) of K over K1 . Proof K is normal over K1 by Lemma 11.2.7, and is separable over K1 by Corollary 11.4.6. So K is a Galois extension of K1 as claimed. Since k is a subfield of K1 , every automorphism of K over K1 also fixes k. So we may regard Gal(K/K1 ) as a subgroup of Gal(K/k): It is the subgroup consisting of those elements that fix K1 . By definition, this subgroup is GK1 . Note that Example 11.2.10 shows that in a Galois extension K of k, there may be intermediate fields not normal over k. We may now apply Corollary 11.4.6. Lemma 11.5.7. Let K be a Galois extension of k, and let K1 be an intermediate field between K and k. Then K1 is a Galois extension of k if and only if K1 is normal over k. Thus, if an intermediate field K1 between k and K is normal over k, then we can discuss the relationship between Gal(K/k) and Gal(K1 /k).


467

Lemma 11.5.8. Let K be a Galois extension of k and let K1 be an intermediate field between k and K such that K1 is normal over k. Then every element of Gal(K/k) carries K1 onto itself, and hence restricts to an automorphism of K1 over k. Passage from automorphisms of K over k to their restrictions to K1 gives a group homomorphism, ρ : Gal(K/k) → Gal(K1 /k). Proof Let L be an algebraic closure of K. Since K is algebraic over K1 , L is also an algebraic closure of K1 . The restriction of any automorphism of K over k to K1 gives an embedding of K1 over k in K, and hence also in L. By Proposition 11.2.9, this embedding must have image K1 . The rest follows easily. We shall refer to ρ : Gal(K/k) → Gal(K1 /k) as the restriction homomorphism. Lemma 11.5.9. Let K be a finite Galois extension of k and let G = Gal(K/k). Then the order of G is given by |G| = [K : k]. Proof Let L be an algebraic closure for K. Since K is a normal extension, every embedding of K in L over k has image K, and hence gives an automorphism of K (Proposition 11.2.9). Thus, the order of G is equal to the number of distinct embeddings of K in L over k, which is by definition the separability degree, [K : k]s of K over k. But since K is separable over k, [K : k]s = [K : k] (Proposition 11.4.12), and the result follows. Lemma 11.5.10. Let G be a finite group of automorphisms of a field K and let k = KG , the fixed field of G. Then the degree of K over k is less than or equal to the order of G: [K : k] ≤ |G|. Proof Let |G| = n and write G = {σ1 , . . . , σn }. Let α1 , . . . , αm be a basis for K as a vector space over k. Supposing that m > n, we shall derive a contradiction. We have n linear equations in m unknowns in K given by σ1 (α1 )x1 + · · · + σ1 (αm )xm .. . σn (α1 )x1 + · · · + σn (αm )xm

=

0

=

0.

Since m > n, this system has a nontrivial solution. Let β1 , . . . , βm be a nontrivial solution with the smallest possible number of nonzero entries. By a permutation of the indices of the α’s and β’s, if necessary, we may arrange that for some r > 0, we have βi = 0 for 1 ≤ i ≤ r, but βi = 0 for i > r. Thus, we obtain equations σ1 (α1 )β1 + · · · + σ1 (αr )βr .. . σn (α1 )β1 + · · · + σn (αr )βr

=

0

=

0.

Note that since the αi are all nonzero, we must have r > 1. Multiplying everything by βr−1 , we may assume that βr = 1. Also, since σ1 is an automorphism of K over k and since the αi are linearly independent over k, σ1 (α1 ), . . . , σ1 (αr )


468

are linearly independent over k, and hence at least one of the βi must lie outside of k. Permuting indices if necessary, we may assume that β1 is not in k. Since k is the fixed field of the action of G on K, there must be an element τ ∈ G for which τ (β1 ) = β1 . Now apply τ to both sides of each of the above equations. Since G is a group, the elements τ σ1 , . . . , τ σn form a permutation of σ1 , . . . , σn . So, permuting rows, we see that τ (β1 ), . . . , τ (βr ) is a solution to the last displayed set of linear equations. But then τ (β1 ) − β1 , . . . , τ (βr ) − βr is also a solution for these equations. Since τ (β1 ) − β1 = 0, this is a nontrivial solution. But since βr = 1 and since τ is an automorphism of K, the r-th term is 0. Thus, we obtain a nontrivial solution to our original set of equations which has less than r nonzero entries. This contradicts the minimality of r, and hence contradicts the assumption that n < m. We may now prove the Fundamental Theorem of Galois Theory. Theorem 11.5.11. (Fundamental Theorem of Galois Theory) Let K be a finite Galois extension of k and let G = Gal(K/k). Then there is a one-to-one correspondence between the set of subgroups of G and the set of intermediate fields between k and K that associates to each subgroup H of G the fixed field KH and associates to each intermediate field K1 between k and K the group GK1 = Gal(K/K1 ). A subgroup H of G is normal in G if and only if KH is a normal extension of k. Moreover, if H G, then the restriction map ρ : G → Gal(KH /k) is surjective, with kernel H, and hence Gal(KH /k) ∼ = G/H. Thus, if K1 is an intermediate field that is normal over k, we have a short exact sequence ⊂

ρ

1 → Gal(K/K1 ) −→ Gal(K/k) − → Gal(K1 /k) → 1 of groups. Proof We first show that for a subgroup H ⊂ G, we have H = Gal(K/KH ). Since H fixes KH , we have H ⊂ Gal(K/KH ), so it suffices to show that |Gal(K/KH )| ≤ |H|. But Lemma 11.5.9 gives |Gal(K/KH )| = [K : KH ] and Lemma 11.5.10 shows that [K : KH ] ≤ |H|, so H = Gal(K/KH ) as claimed. Thus, the one-to-one correspondence will follow if we show that K1 = KGal(K/K1 ) for each intermediate field K1 . Clearly, K1 ⊂ KGal(K/K1 ) , so it suffices to show that if α ∈ K does not lie in K1 , then there is an element σ ∈ Gal(K/K1 ) for which σ(α) = α. Now the minimal polynomial, minα/K1 (X), of α over K1 has degree > 1, as α ∈ K1 . Moreover, since K is a normal extension of K1 , minα/K1 (X) splits in K[X], by Proposition 11.2.9. Since K is separable over K1 , minα/K1 (X) has no repeated roots in K, by Proposition 11.4.2. Thus, the number of roots of minα/K1 (X) in K is equal to its degree. In particular, minα/K1 (X) has a root β ∈ K, with β = α. Let ν : K1 (α) → K1 (β) be the isomorphism over K1 that carries α to β. Let L be an algebraic closure of K. Then Proposition 11.1.4 provides an extension of ν


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to an embedding ν : K → L. Since K is a normal extension of K1 , any embedding of K in L over K1 has image K (Proposition 11.2.9). Thus, ν may be viewed as an automorphism of K, and hence an element of Gal(K/K1 ). Since ν(α) = α, we have shown that KGal(K/K1 ) = K1 , as claimed. Let H be a subgroup of G = Gal(K/k) and let σ ∈ G. Then it’s easy to see that the fixed field of σHσ −1 is given by KσHσ

−1

= σ(KH ).

By the one-to-one correspondence between the subgroups of G and the intermediate fields between k and K, we see that σ normalizes H if and only if σ(KH ) = KH . Thus, H is normal in G if and only if every element of G carries KH onto itself. If KH is a normal extension of k, then every element of G carries KH onto itself by Lemma 11.5.8. Conversely, suppose that every element of G carries KH onto itself and let L be an algebraic closure of K. Since K is algebraic over KH , so is L, and hence L is an algebraic closure of KH also. Thus, Proposition 11.2.9 shows that KH is a normal extension of k if every embedding of KH in L over k has image KH . Thus, let ν : KH → L be an embedding over k. Then ν extends to an embedding ν : K → L. But K is normal over k, so ν gives an automorphism of K, and hence an element of G. But our assumption on KH is that every element of G carries KH onto itself, so the image of ν must be KH . Thus, KH is normal over k. In particular, we see that H G if and only if KH is a normal extension of k. Suppose, now, that H G. Let L be an algebraic closure of K. Then any automorphism, σ, of KH over k may be considered to be an embedding of KH in L. Since K is algebraic over KH , σ extends to an embedding of K in L, which must have image K, since K is a normal extension of k. In particular, σ extends to an automorphism of K, and hence σ is in the image of the restriction homomorphism ρ : G → Gal(KH /k). Thus, ρ is onto. But its kernel is the subgroup of G that fixes KH , which, by the one-to-one correspondence above, is precisely H. Thus, if we’re given a finite Galois extension K of k, we can find all the intermediate fields by finding the subgroups of Gal(K/k) and determining their fixed fields. We shall carry this out in some exercises in later sections. Meanwhile, here’s an example in which we already know the intermediate fields. Example 11.5.12. Finite fields: Let K be the unique (Corollary 11.3.3) field with pn n elements. Then K is the splitting field of X p − X over Zp by Proposition 11.3.1, and hence is a normal extension of Zp . Proposition 11.4.7 shows K to be separable over Zp , so K is a Galois extension of Zp . Now let ϕ : K → K be the Frobenius homomorphism. Since K is finite, ϕ is an automorphism of K. As a ring homomorphism, it fixes 1, and hence fixes Zp , so ϕ ∈ Gal(K/Zp ). Since K is finite, K× is a cyclic group of order pn − 1, by Corollary 7.3.14. Let α be a generator of K× . Then for k < n, k

ϕk (α) = αp = α, so the order of ϕ in Gal(K/Zp ) is at least n. But ϕn takes each element of K to its pn -th n power, and hence is the identity map, as each element of K is a root of X p − X. Thus, ϕ has order n in Gal(K/Zp ). But the order of Gal(K/Zp ) is equal to [K : Zp ], which is n, so Gal(K/Zp ) is the cyclic group generated by ϕ. Note that the subfields of K, as enumerated in Proposition 11.3.4, are given as the fixed fields of ϕr for integers r dividing n.


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We may also make some immediate calculations using the discriminant. Recall from Corollary 11.5.3 that the Galois group of the splitting field of a separable polynomial f embeds as a subgroup of the permutation group on the roots of f . Corollary 11.5.13. Let K be the splitting field of the separable polynomial f (X) over k and let α1 , . . . , αn be the roots of f in K. Let ρ : Gal(K/k) → Sn be the inclusion induced by restricting an automorphism of K over k to its action on {α1 , . . . , αn }. Let . δ= (αj − αi ) ∈ K. i<j

Then an element, σ, of Gal(K/k) gives an even permutation of {α1 , . . . , αn } if and only if it fixes δ. Thus, ρ gives an embedding of Gal(K/k) into An if and only if the discriminant Δ(f ), which is the square of δ, has a square root in k. Proof The use of δ to test whether ρ(σ) is even follows from Proposition 7.4.8. In particular, ρ(Gal(K/k)) is contained in An if and only if Gal(K/k) fixes δ. But the Fundamental Theorem of Galois Theory says this happens if and only if δ ∈ k. That δ is a square root of the discriminant Δ(f ) follows from Corollary 7.4.13. Since Δ(f ) ∈ k, the result follows. Note that the value of δ depends on the ordering of the roots of f . In particular, δ is only determined up to sign. The behavior of the discriminant is enough to characterize the behavior of splitting fields of separable cubics. Corollary 11.5.14. Let K be the splitting field over k of the separable, irreducible cubic f (X). Then 1. If Δ(f ) is a square in k, then Gal(K/k) = Z3 . 2. If Δ(f ) is not a square in k, then Gal(K/k) = S3 . Proof Let α be a root of f in K. Then f is the minimal polynomial of α, so k(α) has degree 3 over k. Thus, [K : k], which is the order of Gal(K/k), is divisible by 3. Since f is irreducible and separable, it has three distinct roots, so Gal(K/k) embeds in S3 . The only subgroups of S3 whose order is divisible by 3 are A3 = Z3 and S3 itself. The result now follows from Corollary 11.5.13. The case of reducible cubics, of course, depends only on finding the number of roots of the cubic that lie in k. We shall treat the issue of finding the roots of a cubic in Section 11.11. Remarks 11.5.15. We’ve only given the explicit formula for the discriminant of a cubic of the form X 3 + pX + q: In Problem 5 of Exercises 7.4.14, it is shown that Δ(X 3 + pX + q) = −4p3 − 27q 2 . But the general case may be recovered from this by the method of completing the cube. Note that if f (X) = X 3 + bX 2 + cX + d, then setting g(X) = f (X − (b/3)) gives us a cubic of the desired form. But if α1 , α2 , and α3 are the roots of f , then α1 +b/3, α2 +b/3, and α3 + b/3 are the roots of g. Thus, f and g have the same value for δ, and hence for Δ, as well.


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Note that in all of the above, we started out with a Galois extension and studied the behavior of its Galois group. Suppose, on the other hand, that we start out with a group G of automorphisms of a field K and set k equal to the fixed field KG . What can we say about K as an extension of k? First, we consider the induced action on polynomials. Suppose that G is a group of automorphisms of the ring A and that B = AG , the fixed ring of the action of G. For each σ ∈ G, we have an induced ring homomorphism σ∗ : A[X] → A[X], obtained by applying σ to the coefficients of the polynomials. Passage from σ to σ∗ gives an action of G on A[X]. The following lemma is immediate. Lemma 11.5.16. Let G be a group of automorphisms of the ring A and let B = AG . Then the fixed ring of the induced G-action on A[X] is given by A[X]G = B[X].

We now give a theorem of Artin. Theorem 11.5.17. Let G be a finite group of automorphisms of the field K and let k = KG . Then K is a Galois extension of k, and Gal(K/k) = G. Proof Let α ∈ K and let α1 , . . . , αk be the orbit of α under the action of G, with, say, α = α1 . Thus, {α1 , . . . , αk } is the set of all distinct elements of K of the form σ(α) with σ ∈ G. In particular, {α1 , . . . , αk } is the smallest sub-G-set of K containing α. Write f (X) =

k .

(X − αi ) ∈ K[X].

i=1

Since {α1 , . . . , αk } is preserved by the action of G, we have σ∗ (f (X)) = f (X) for all σ ∈ G, and hence f (X) ∈ K[X]G = k[X]. Thus, since α is a root of f , the minimal polynomial of α over k must divide f (X). It’s not hard to show that f is the minimal polynomial of α over k, but it’s unnecessary for our argument. Since f has no repeated roots, neither does the minimal polynomial, so α is separable over k. Also, the minimal polynomial of α splits in K[X]. Since α was chosen arbitrarily, K is a normal extension of k by Proposition 11.2.9. Thus, K is a Galois extension of k, and G embeds in Gal(K/k). In particular, we have k = KG = KGal(K/k) . Since passage from subgroups of G to their fixed fields gives a one-to-one correspondence between the subgroups of Gal(K/k) and the intermediate fields between k and K, we must have G = Gal(K/k). The following application of this result plays an important role in understanding the unsolvability of polynomials of degree > 4. We shall come back to it later. Example 11.5.18. Let k be a field. Then the symmetric group Sn acts on the field of rational functions k(X1 , . . . , Xn ) by permuting the variables. In particular, Sn is the Galois group of k(X1 , . . . , Xn ) over the fixed field k(X1 , . . . , Xn )Sn .


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Thus, every finite group is the Galois group of some field extension. Next, we give an application of the Fundamental Theorem. We shall prove the Primitive Element Theorem: that every finite separable extension is simple. Recall, here, that an extension K of k is simple if K = k(α) for some α ∈ K that is algebraic over k. We first give a characterization of simple extensions. Proposition 11.5.19. A finite extension K of k is simple if and only if there are only finitely many intermediate fields between K and k. Proof Suppose that K is simple, with K = k(α). Let K1 be an intermediate field between k and K and let g(X) be the minimal polynomial of α over K1 , say g(X) = X k + αk−1 X k−1 + · · · + α0 . Let K2 = k(α0 , . . . , αk−1 ) be the field obtained by adjoining the roots of g(X) to k. Then g(X) lies in K2 [X] and has α as a root. Since K2 is a subfield of K1 and since g(X) is irreducible in K1 [X], it must be irreducible in K2 [X]. Thus, g(X) is the minimal polynomial of α over K2 . Since K is the result of adjoining α to any intermediate field between k and K, we obtain [K : K2 ] = deg g(X) = [K : K1 ], and hence K2 = K1 . In particular, this says that K1 is obtained by adjoining the coefficients of g(X) to k. Note that if f (X) is the minimal polynomial of α over k, then g(X) is a monic factor of f (X) in K[X]. Since f (X) has only finitely many monic factors in K[X], there can be only finitely many intermediate fields. For the converse, let K be a finite extension of k such that there are only finitely many intermediate fields between k and K. If K is a finite field, then K× is cyclic, and hence K = k(α) for any generator α of K× . Thus, we shall assume that K, and hence also k, is infinite. Since [K : k] is finite, we have K = k(α1 , . . . , αn ) for some α1 , . . . , αn ∈ K. By induction on n, it suffices to show that if K = k(α, β), then K = k(γ) for some γ. Consider the subfields of the form k(α + aβ) for a ∈ k. Since there are only finitely many intermediate fields between k and K and since k is infinite, there must be elements a = b ∈ k such that k(α + aβ) = k(α + bβ) = K1 ⊂ K. But then (α + aβ) − (α + bβ) = (a − b)β is in K1 , and hence β and α are in K1 . Thus, K1 = K, and we may take γ = α + aβ. The Primitive Element Theorem is now an easy application of the Fundamental Theorem. Theorem 11.5.20. (Primitive Element Theorem) Every finite separable extension is simple. Proof Let K be a finite separable extension of k and let L be the normal closure of K over k in an algebraic closure of K. Then Corollary 11.4.17 shows L to be a Galois extension of k. By Corollary 11.2.14, L is finite over k, and hence is a finite Galois extension of k.


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Since K is an intermediate field between k and L, it suffices to show that there are only finitely many intermediate fields between k and L. But this is immediate from the one-to-one correspondence between the intermediate fields between k and L and the subgroups of the Galois group Gal(L/k): Since the Galois group is finite, it has only finitely many subgroups. In computing Galois groups, it is often important to see how an extension is built up out of smaller extensions. Definition 11.5.21. Suppose given a diagram

K1

L ? ?? ??

?? ?? ?

K2

k of field extensions. Then the compositum, K1 K2 , of K1 and K2 in L is the smallest subfield of L containing both K1 and K2 . If L = K1 K2 , we say that L is a compositum of K1 and K2 . However, it should be noted that extension fields K1 and K2 of k may have composita that are not isomorphic over k: The specific embeddings of K1 and K2 in the larger field are needed to determine the compositum. We can often deduce information about the compositum from information about K1 and K2 . We shall discuss this further in the exercises. Note that in the abstract, neither K1 nor K2 is assumed to be algebraic over k. Exercises 11.5.22. 1. Compute the Galois groups over Q of the splitting fields of the following polynomials. (a) X 3 − 3X + 1 (b) X 3 − 4X + 2 (c) X 3 + X 2 − 2X − 1 2. Let K be a Galois extension of k. Show that an element α ∈ K is a primitive element for K over k (i.e., that K = k(α)) if and only if no non-identity element σ ∈ Gal(K/k) satisfies σ(α) = α. 3. Let L be a compositum of K1 and K2 over k. Suppose that K1 is algebraic over k, with K1 = k(αi | i ∈ I). Show that L is algebraic over K2 , with L = K2 (αi | i ∈ I). 4. Let L be a compositum of K1 and K2 over k. Show that if any of the following are descriptors of K1 as an extension of k, then the same descriptor holds for L as an extension of K2 . (a) normal (b) separable (c) Galois


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(d) finite (e) simple † 5. Let L be a compositum of K1 and K2 over k, and suppose that K1 is a Galois extension of k. Show that each element of Gal(L/K2 ) restricts on K1 to an element of Gal(K1 /k). Show that this restriction map provides an injective group homomorphism ρ : Gal(L/K2 ) → Gal(K1 /k). Deduce that if K1 is a finite Galois extension of k, then [L : K2 ] divides [K1 : k]. 6. Let L be a compositum of K1 and K2 over k, and suppose that K1 is a finite Galois extension of k. (a) Suppose K1 ∩ K2 = k, where the intersection is as subfields of L. Show that [L : K2 ] = [K1 : k]. (Hint: Let α be a primitive element for K1 over k and let f (X) be the minimal polynomial of α over K2 . Show that the coefficients of f (X) must lie in K1 ∩ K2 , and hence f is also the minimal polynomial of α over k.) (b) Conversely, suppose that K1 ∩K2 properly contains k. Show that [L : K2 ] is a proper divisor of [K1 : k]. Thus, K1 ∩K2 = k if and only if [L : K2 ] = [K1 : k]. (c) Show that L is a quotient ring of K1 ⊗k K2 , and that L ∼ = K1 ⊗k K2 if K1 ∩ K2 = k. 7. Let K1 be the splitting field over k of f (X) ∈ k[X]. Let K2 be any extension of k. Show that the splitting field of f (X) over K2 is a compositum of K1 and K2 . 8. Let K1 and K2 be extension fields of k, and suppose that K1 is a finite normal extension of k. Show that any two composita of K1 and K2 over k are isomorphic over K2 (and hence over k). 9. Find extension fields K1 and K2 of Q with at least two different composita. (Hint: Look for an example where K1 = K2 .) 10. Let K be the algebraic closure of Zp . Calculate Gal(K/Zp ). 11. Give an example of a finite extension that is not simple.

11.6

The Fundamental Theorem of Algebra

We show that the complex numbers, C, are algebraically closed. We shall make use of the following application of the topology of the real line. Lemma 11.6.1. Every odd degree polynomial over the real numbers, R, has a real root. Proof Let f (X) ∈ R[X] be a polynomial of degree n, where n is odd. Dividing by the leading coefficient, if necessary, we may assume that f is monic, so that f (X) = X n + an−1 X n−1 + · · · + a0 . We claim that if |x| is large enough, then |xn | > |an−1 xn−1 + · · · + a0 |. To see this, note that if |x| > 1, then |an−1 xn−1 + · · · + a0 | < (|an−1 | + · · · + |a0 |)|xn−1 |.


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So if |x| > 1 is greater than |an−1 | + · · · + |a0 |, then |xn | is indeed greater than |an−1 xn−1 + · · · + a0 |. In particular, if |x| is large enough, then f (x) has the same sign as xn , which, since n is odd, has the same sign as x. Explicitly, if c is greater than both 1 and |an−1 |+· · ·+|a0 |, then f (−c) is negative and f (c) is positive, so the Intermediate Value Theorem shows that f must have a root in the open interval (−c, c). Corollary 11.6.2. The field of real numbers, R, has no nontrivial extensions of odd degree. Proof Let K be a nontrivial odd degree extension of R and let α be an element of K that’s not in R. Then [R(α) : R] divides [K : R], and hence is odd. But [R(α) : R] is the degree of the minimal polynomial of α over R. Since every odd degree polynomial over R has a root in R, there are no irreducible elements of R[X] whose degree is an odd number greater than 1. So this is impossible. Corollary 11.6.3. Let K be a finite extension of R. Then the degree of K over R must be a power of 2. Proof Suppose, to the contrary, that there is an extension K of R whose degree over R is divisible by an odd prime. Let L be the normal closure of K over R in an algebraic closure of K. Then L is finite over R by Corollary 11.2.14, and hence is a finite Galois extension of R. By the multiplicativity formula for degrees, we see that [L : R] is divisible by an odd prime. Let G be the Galois group of L over R and let G2 be a 2-Sylow subgroup of G. Let K1 = KG2 , the fixed field of G2 in K. Then the Fundamental Theorem of Galois Theory shows that G2 = Gal(K/K1 ), and hence the degree of K over K1 is |G2 | by Lemma 11.5.9. Thus, the multiplicativity formula for degrees gives [K1 : R] = [G : G2 ], the index of G2 in G. By the Second Sylow Theorem, [G : G2 ] is odd, and must be greater than 1, as |G| is divisible by an odd prime. In particular, K1 is a nontrivial odd degree extension of R, which is impossible by Corollary 11.6.2. Of course, every extension of C is an extension of R. Corollary 11.6.4. Let K be a finite extension of C. Then the degree of K over C must be a power of 2. There’s only one further result needed to prove that C is algebraically closed. Lemma 11.6.5. Every polynomial of degree 2 over C has a root in C. Proof Completing the square gives aX 2 + bX + c = a(X +

b 2 b2 ) + (c − ), 2a 4a

so a root is obtained by setting X + (b/2a) equal to any complex square root of (b2 − 4ac)/4a2 . In particular, it suffices to show that any complex number has a square root in C. The easiest way to see this is via the exponential function. The point is that for r, θ ∈ R with r ≥ 0, the complex number reiθ gives the point in the plane whose polar


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coordinates√are (r, θ). Thus, every complex number may be written in the form reiθ , and hence has reiθ/2 as a square root. Or, explicitly, if x, y ∈ R with y = 0, then the reader may check that a square root for x + yi is given by 3 r−x y +i 2 2(r − x) where r = x2 + y 2 . But every degree 2 extension is simple. Corollary 11.6.6. There are no degree 2 extensions of C. And now, we have the main theorem of this section. Theorem 11.6.7. (Fundamental Theorem of Algebra) The field of complex numbers, C, is algebraically closed. Proof It suffices to show that there are no irreducible elements of C[X] of degree greater than 1. To see this, note that if f (X) is an irreducible polynomial of degree > 1, then the splitting field, K, of f (X) is a nontrivial extension of C. By Corollary 11.6.4, [K : C] is a power of 2, and hence the Galois group Gal(K/C) is a 2-group. By Corollary 5.2.5, there is an index 2 subgroup, H, of Gal(K/C). But then KH is a degree 2 extension of C by the Fundamental Theorem of Galois Theory. Since C has no extensions of degree 2, the result follows. Exercises 11.6.8. 1. Show that every irreducible element of R[X] has either degree 1 or degree 2. 2. Show that the roots in C of an irreducible quadratic in R[X] are complex conjugates of one another.

11.7

Cyclotomic Extensions

Definitions 11.7.1. Let K be a field. Then the n-th roots of unity in K are the roots in K of the polynomial X n − 1. We say that ζ ∈ K is a primitive n-th root of unity if ζ has order n in K× . Lemma 11.7.2. Let K be a field. Then the set of n-th roots of unity in K forms a subgroup of K× . Moreover, this subgroup is cyclic and contains all the elments of K× with exponent n. In particular, K contains a primitive n-th root of unity if and only if there are n distinct roots of X n − 1 in K. In this case, any primitive n-th root of unity generates the group of n-th roots of unity in K. Proof An element ζ ∈ K is an n-th root of unity if and only if ζ n = 1. And this is equivalent to ζ having exponent n in K× . The elements of exponent n in an abelian group always form a subgroup. So the n-th roots of unity form a subgroup of K× . Since X n − 1 has at most n roots, this subgroup is finite. And Corollary 7.3.14 shows that any finite subgroup of K× is cyclic. Thus, the group of n-th roots of unity in K has order n if and only if K× has an element of order n.


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Examples 11.7.3. 1. Every element x ∈ R× with |x| = 1 has infinite order in R× . So the only roots of unity in R are ±1, the primitive first root and primitive second root, respectively. 2. Let K be the finite field of order pn . Then K× is a cyclic group of order pn − 1. So K has primitive r-th roots for all r dividing pn − 1. 3. We’ve seen that ζn = e2πi/n is a primitive n-th root of unity in C. So the group of n-th roots in C (or in any subfield of C containing Q(ζn )) is ζn = {ζnk | 0 ≤ k < n}. For many purposes, it is useful to adjoin a primitive n-th root of unity to a field k. Note that if ζ is a primitive n-th root of unity in an extension field K of k, then k(ζ) contains n distinct roots of X n − 1. Thus, X n − 1 splits in k(ζ), and hence k(ζ) is a splitting field for X n − 1 over k. Conversely, if K is a splitting field for X n − 1 over k, then K contains a primitive n-th root of unity if and only if X n − 1 has n distinct roots in K. If K does contain a primitive n-th root of unity, ζ, then the roots of X n − 1 in K are the distinct powers of ζ. Since K is the splitting field of X n − 1 over ζ, we see that K = k(ζ). Corollary 11.7.4. Let k be a field. Then we may adjoin a primitive n-th root of unity to k if and only if X n − 1 has n distinct roots in its splitting field over k. The result of adjoining a primitive n-th root of unity to k, when possible, is always a splitting field for X n − 1. The next thing to notice is that it isn’t always possible to adjoin a primitive n-th root to a field k. For instance, in Zp [X], we have X p − 1 = (X − 1)p , which has only the root 1 in any field of characteristic p. So it is impossible to adjoin a primitive p-th root of unity to a field of characteristic p. Proposition 11.7.5. We may adjoin a primitive n-th root to a field k if and only if n is not divisible by the characteristic of k. Proof The formal derivative of X n − 1 is nX n−1 . If n is divisible by the characteristic of k, then nX n−1 = 0, and hence Lemma 8.7.7 tells us that X n − 1 has repeated roots in its splitting field. In particular, there cannot be n distinct roots of X n − 1. If n is not divisible by the characteristic of k, 0 is the only root of nX n−1 in any extension of k. In particular, X n − 1 and nX n−1 have no common roots in the splitting field, K, of X n − 1 over k. Lemma 8.7.7 now shows that X n − 1 has no repeated roots in K, so there must be n distinct roots there. Definition 11.7.6. Let k be a field and let n be a positive integer not divisible by the characteristic of k. We shall refer to the splitting field of X n − 1 over k as the cyclotomic extension obtained by adjoining a primitive n-th root of unity to k. Since the roots of X n − 1 form a subgroup of K× , we may make an improvement on Corollary 11.5.3. Proposition 11.7.7. Let K be the cyclotomic extension of k obtained by adjoining a primitive n-th root of unity to K. Let ζ be a primitive n-th root of unity in K and write ζ = {ζ k | 0 ≤ k < n} for the group of n-th roots of unity in K. Write Aut(ζ) for the group of automorphisms of the cyclic group ζ. Then each σ ∈ Gal(K/k) restricts on ζ to an element of Aut(ζ). We obtain an injective group homomorphism ρ : Gal(K/k) → Aut(ζ).


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Recall from Proposition 4.5.4 that the automorphism group of a cyclic group of order n is isomorphic to Z× n , the group of units of the ring Zn . In multiplicative notation, the m automorphism of ζ induced by a unit k ∈ Z× to ζ mk . The group structure n takes ζ × on Zn is calculated in Section 4.5. Of crucial importance for our study here is the fact that Z× n is abelian. Recall from Proposition 8.8.7 that [Q(ζn ) : Q] = φ(n), where φ(n) is the order of Z× n. Since ρ : Gal(Q(ζn )/Q) → Aut(ζn ) is injective, we obtain the following corollary. Corollary 11.7.8. The Galois group of Q(ζn ) over Q restricts isomorphically to the automorphism group of the abelian group ζn . Recall from Corollary 7.11.10 that every field, K, of characteristic 0 has a unique, natural, Q-algebra structure. So it has a unique subfield isomorphic to Q, and every automorphism of K restricts to the identity on that subfield. Moreover, every subfield of K must contain Q, so the collection of intermediate fields between Q and K coincides with the collection of all subfields of K. Example 11.7.9. We have Z× 8 = {1, 3, 5, −1}, with a group structure isomorphic to Z2 × Z2 . Note that the automorphism of ζ8 induced by −1 is induced by Galois automorphism obtained by restricting the complex conjugation map to Q(ζ8 ). The Galois automorphisms corresponding to 3 and 5 are induced by the standard isomorphisms over Q from Q(ζ8 ) to Q(ζ83 ) and Q(ζ85 ). These isomorphisms exist because ζ8 , ζ83 , ζ85 , and ζ87 all have the same minimal polynomial, Φ8 (X), over Q by Proposition 8.8.7. The subgroups of Z× 8 form the following lattice, with the downward lines representing inclusions of subgroups.

3

?? ?? ?? ??

e ? ?? ?? ?? ??

5

Z× 8

{±1}

With this, we can find all the subfields of Q(ζ8 ). Since each nontrivial subgroup of Z× 8 has index 2, we can find its fixed field by finding a single element outside of Q that it fixes. In the case of the subgroup {±1}, we know that complex conjugation exchanges ζ8 and ζ 8 , and hence fixes √ ζ8 + ζ 8 = 2 re ζ8 = 2. (Here, the last equality comes from the fact that ζ8 = √12 + √12 i.) So the fixed field of √ {±1} is Q( 2). The remaining fixed fields may be calculated similarly, using the known effect of the


479

subgroups in question on ζ8 . We obtain the following lattice of subfields. Q(ζ8 ) ???? ?? ?? ??

√ √ Q(i) Q(i 2) Q( 2) ?? ?? ?? ?? ?? Q Exercises 11.7.10. 1. Let k be a field of characteristic p = 0 and suppose that n is divisible by p. How many roots does X n − 1 have in its splitting field over k? Is the splitting field in question a cyclotomic extension of k? 2. Let K be a cyclotomic extension of k. Show that every intermediate field between k and K is a Galois extension of k. 3. What are the orders of the cyclotomic extensions obtained by adjoining a primitive n-th root of unity to Zp for all primes p ≤ 11 and all n prime to p with n ≤ 11? 4. Show that the fixed field of the restriction to Q(ζn ) of complex conjugation is Q(ζn + ζ n ). Deduce that Q(ζn + ζ n ) = Q(ζn ) ∩ R. What are the roots of the minimal polynomial of ζn + ζ n over Q? 5. Find all the subfields of Q(ζ5 ). 6. What is the Galois group of Q(ζ7 + ζ 7 ) over Q? What are the other subfields of Q(ζ7 )? 7. What is the Galois group of Q(ζ9 + ζ 9 ) over Q? What are the other subfields of Q(ζ9 )? 8. Find all the subfields of Q(ζ12 ). 9. Find all the subfields of Q(ζ16 ). 10. Show that every cyclic group occurs as the Galois group of a Galois extension of Q. 11. Let k be a field and let n be an integer that is not divisible by the characteristic of k. Since the cyclotomic polynomial Φn (X) has integer coefficients, we may regard it as an element of k[X]. Let K be the cyclotomic extension obtained by adjoining a primitive n-th root of unity to k. Show that K is the splitting field of Φn (X) over k and that the roots of Φn (X) in K are precisely the primitive n-th roots of unity. for all r > 0. Deduce that if K is the 12. Show that Φpr (X) is irreducible over Q p , then cyclotomic extension obtained by adjoining a primitive pr -th root to Q p r r p ) = Z× Gal(K/Q − 1 = X p in Zp [X].) pr . (Hint: Note that (X + 1) p


480

13. Let n be an integer relatively prime to p and let K be the cyclotomic extension of Zp obtained by adjoining a primitive n-th root of unity. Show that Gal(K/Zp ) is the subgroup of Z× n generated by p. What, then, is the degree of K over Zp ? . 14. Suppose that (n, p − 1) > 2. Show that Φn (X) is not irreducible over Q p

11.8

n-th Roots

We shall study the Galois theory of the splitting fields of polynomials of the form X n − a. Lemma 11.8.1. Let k be a field and let 0 = a ∈ k. Let n be a positive integer not divisible by the characteristic of k. Then the splitting field of X n −a over k is a separable extension of k and has the form K = k(α, ζ), where α is a root of X n − a and ζ is a primitive n-th root of unity. The set of roots in K of X n − a is {αζ k | 0 ≤ k < n}, and hence Xn − a =

n−1 .

(X − αζ k )

k=0

in K[X]. Proof Let L be an algebraic closure of K. Choose any root α of X n − a in L and any primitive n-th root of unity, ζ ∈ L, and set K = k(α, ζ). Then {αζ k | 0 ≤ k < n} gives n distinct roots of X n − a in K, and hence X n − a factors as stated in K[X]. In particular, X n − a has no repeated roots in K, and hence is a separable polynomial over k. Since K is the splitting field of X n − a over k, it is separable over k by Corollary 11.4.16. The simplest special case of such extensions is when k already contains a primitive n-th root of unity. Proposition 11.8.2. Let k be a field containing a primitive n-th root of unity, ζ. Let 0 = a ∈ k and let K be a splitting field for X n − a over k. Then the Galois group of K over k is a subgroup of Zn . Explicitly, if α is any root of X n − a in K, then for each σ ∈ Gal(K/k) there is an integer s such that σ(αζ k ) = αζ k+s for 0 ≤ k < n. The passage from σ to s gives an injective group homomorphism from Gal(K/k) to Zn . Proof The roots of X n −a are given by {αζ k | 0 ≤ k < n}, so Corollary 11.5.3 shows that restricting elements of Gal(K/k) to their effect on {αζ k | 0 ≤ k < n} gives an injective group homomorphism from Gal(K/k) to the group S({αζ k | 0 ≤ k < n}) of permutations on {αζ k | 0 ≤ k < n}. For σ ∈ Gal(K/k), let σ(α) = αζ s . Since ζ ∈ k, it is fixed by σ, so σ(αζ k ) = σ(α)σ(ζ k ) = αζ s ζ k = αζ k+s for all values of k. In other words, the action of σ on the set {αζ k | 0 ≤ k < n} coincides with multiplication by ζ s . Define ϕ : Zn → S({αζ k | 0 ≤ k < n}) by ϕ(s)(αζ k ) = αζ k+s . Then ϕ is easily seen to be an injective group homomorphism. Since the image of the restriction map ρ : Gal(K/k) → S({αζ k | 0 ≤ k < n}) lies in the image of ϕ, the result follows.


481

Thus, if k contains a primitive n-th root of unity and 0 = a ∈ k, then the Galois group over k of the splitting field of X n − a is determined by the degree of this splitting field over k. As usual, the degree is not always easy to compute. For the general case of splitting fields of polynomials of the form X n − a with n not divisible by the characteristic of k, Lemma 11.8.1 shows that the splitting field, K, must contain a principal n-th root of unity, ζ. But then k(ζ) is a normal extension of k. We may apply Proposition 11.8.2 to describe the Galois group of K over k(ζ) and apply Proposition 11.7.7 to describe the Galois group of k(ζ) over k. We can then put these calculations together using the Fundamental Theorem of Galois Theory. Recall from Section 4.7 that if H and H are groups, then an extension of H by H is a group G, together with a surjective homomorphism f : G → H with kernel H. In particular, the Fundamental Theorem of Galois Theory shows that if K is a Galois extension of k and if K1 is an intermediate field normal over k, then the restriction map ρ : Gal(K/k) → Gal(K1 /k) presents Gal(K/k) as an extension of Gal(K/K1 ) by Gal(K1 /k). Recall from Lemma 4.7.9 that if H is abelian and if f : G → H is an extension of H by H , then there is an induced action of H on H via automorphisms, αf : H → Aut(H), defined as follows. For x ∈ H , write x = f (g). Then αf (x) acts on H by αf (x)(h) = ghg −1 for all h ∈ H. This action is an important invariant of the extension. Notice that if αf is not the trivial homomorphism, then G cannot be abelian. Proposition 11.8.3. Let k be a field and let 0 = a ∈ k. Let n be a positive integer not divisible by the characteristic of k and let K be the splitting field of X n − a over k. Let ζ be a primitive n-th root of unity in K and let K1 = k(ζ). Write H = Gal(K/K1 ), H = Gal(K1 /k) and G = Gal(K/k). Then the restriction map ρ : G → H presents G as an extension of H by H . We have inclusion maps H ⊂ Zn and H ⊂ Z× n , and the action of H on H induced by the extension ρ coincides with the composite ε ∼ → Aut(H) H ⊂ Z× n = Aut(Zn ) −

where ε is obtained by restricting an automorphism of Zn to its effect on H. Proof Recall that a cyclic group has a unique subgroup of order d for every d that divides is order. Thus, every subgroup of a cyclic group is characteristic, so ε is well defined (Lemma 3.7.8). The only other point that isn’t covered by the discussion prior to the statement of this proposition is the verification that the action of H on H is given by ε. Thus, let σ be k rk an automorphism of K1 over k. Then there is an element r ∈ Z× n such that σ(ζ ) = ζ for all k. An extension of σ to an automorphism of K is determined by its effect on α. For our purposes, all we need to know is that if σ extends σ, then σ(α) = αζ s for some value of s. Consequently, σ(αζ k ) = αζ rk+s . It’s now easy to see that σ −1 (αζ k ) = αζ t(k−s) , where t represents the inverse of r in Z× n. We now wish to calculate the effect of conjugating an automorphism τ ∈ H = Gal(K/K1 ) by σ. Here, Proposition 11.8.2 gives us an integer l such that τ (αζ k ) = αζ k+l for all k. And τ corresponds to the element l ∈ Zn under the identification of H with a subgroup of Zn . A direct verification now shows that στ σ −1 takes αζ k to αζ k+rl . This corresponds to the standard action of Z× n on Zn , as claimed. At this point, we need to consider some examples.


482

Example 11.8.4. We calculate the Galois group of the splitting field of X 4 − 2 over Q and determine the intermediate fields. Here, we have a diagram

√ 4 Q( 2)

√ 4 Q( 2, i) JJ t JJ4 2ttt JJ t t J t JJ JJ JJ 4 JJ Q

t ttt t t ttt 2

Q(i)

where the numbers on the extension lines represent the degrees of the extensions. These degrees are calculated as follows. 2 is irreducible over√Q by Eisenstein’s criterion, and hence is the minimal Here, X 4 −√ √ 4 4 polynomial of 4 2√over Q. So [Q( 2) : Q] = 4. Now Q( 2) ⊂ R, so it doesn’t contain √ i, and hence [Q( 4 2, i) : Q( 4 2)] = 2. The rest now follows from the multiplicativity of degrees, together with the √ fact that i has degree 2 over Q. √ Now the fact that [Q( 4 2, i) : Q(i)] = 4, shows that Gal(Q( 4 2, i)/Q(i)) = Z4 , by Proposition 11.8.2. And the non-trivial element of Gal(Q(i)/Q) extends to the restriction √ to Q( 4 2, i) of the complex conjugation √ map. Since complex conjugation has order 2, we see that the extension ρ : Gal(Q( 4 2, i)/Q) → Gal(Q(i)/Q) splits. It is easy to see that the resulting extension of Z4 by Z2 is precisely the dihedral group, D8 , of order 8. √ √ Here, we can take the generator, b, of√order 4 to√act via b( 4 2 ik ) = 4 2 ik+1 , and take the generator, a, of order 2 to act via a( 4 2 ik ) = 4 2 i−k . The inverted lattice of subgroups of D8 is given by o e ?OO ooo ??O?OOOO o o ?? OOO oo O ?? ooo o o ?? OOOOO o o O ooo a ab ab2 b2 ab3 ?? ? ?? ?? ?? ?? ? ?? ? ?? ? b a, b2 ab, b2 ?? ?? ?? ?? ? D8


483

The associated lattice of subfields is given by √ 4 Q( 2, i) ? O ooo ???OOOOO o o ?? OOO ooo OOO ?? ooo OOO o ?? o o OOO ?? oo o O o o √ √ √ √ 4 4 4 4 Q(ζ8 ) Q(ζ8−1 2) Q(ζ8 2) Q( 2) Q(i 2) ?? ?? ?? ?? ?? ?? ? ?? ?? ?? ?? ?? ?? √ √ Q(i) Q( 2) Q(i 2) ?? ?? ?? ?? ?? ?? ? Q √ √ To see this, note that ζ8 = √12 + √12 i is in Q( 4 2, i). Note also that both ζ8 4 2 and √ ζ8−1 4 2 are roots of X 4 + 2, and hence have degree 4 over Q by Eisenstein’s criterion. The rest follows by checking that the elements in question are left fixed by the appropriate subgroups of D8 . The preceding example, while straightforward in itself, points to a potential difficulty in calculating the Galois groups of splitting fields of polynomials of the form X n − a: It is√sometimes more difficult than in the preceding example to compute the degree of Q( n a, ζn ) over Q(ζn ). For instance, consider the following related example. √ Example 11.8.5. Consider the splitting field, Q( 8 2, ζ8 ), of X 8 −2 over Q. Once again, √ criterion, √ so Q( 8 2) has degree 8 over Q. But ζ8 satisfies X 8 − 2 satisfies Eisenstein’s √ the polynomial X 2 − 2X + 1 over Q( 8 2), so the degrees of our extensions are given as follows.

√ 8 Q( 2)

√ 8 Q( 2, ζ8 ) JJJ 2tttt JJ4J t JJ tt JJ JJ JJ J 8 JJ Q

tt tt t tt tt 4

Q(ζ8 )

√ ∼ Thus, Gal(Q( 8 2, ζ8 )/Q) is an extension of Z4 by Z× 8 = Z2 × Z2 . In this case, it turns out that the extension does not split. We shall leave the further analysis of this example to the exercises. Exercises 11.8.6. 1. Show that X 4 − 2 and X 4 + 2 have the same splitting field over Q. √ 2. Find a primitive element for Q(i, 4 2) over Q.


484

√ 3. Show that the Galois group, G, of Q( 8 2, ζ8 ) over Q contains an element of order 8, and hence is an extension of Z8 by Z2 . Compute the action of Z2 on Z8 induced by this extension and determine whether√G is a split extension of Z8 by Z2 . (Hint: Show that the minimal polynomial of 8 2 over Q(ζ8 ) is not invariant under the action of the Galois group of Q(ζ8 ) over Q.) √ 4. Show that Gal(Q( 8 2, ζ8 )/Q) cannot be a split extension of Z4 by Z2 × Z2 . √ 5. Find all the subfields of Q( 8 2, ζ8 ). 6. Show that the splitting field of X 8 + 2 over Q has degree 16 over Q. Calculate its Galois group over Q. Determine whether it is isomorphic to the Galois group of the splitting field of X 8 − 2 over Q. √ 7. Find all the subfields of Q( 3 2, ζ3 ). √ 8. Find all the subfields of Q( 5 2, ζ5 ). 9. Let 0 = a ∈ Q and let K be the splitting field of X n − a over Q. Suppose that K has degree n over Q(ζn ). Show that Gal(K/Q) is a split extension of Zn by Z× n. 10. Let a be an integer divisible by a prime √ number q, but not by q 2 . Let p be any prime. Show that the Galois group of Q( p a, ζp ) over Q is a split extension of Zp by Zp × . 11. Let a be an integer divisible by a prime number q, but not by q 2 . Let n = p1 , . . . , pk , where p1 , . . . , pk are distinct primes with √ the property that pi ≡ 1 mod pj for all i, j. Show that the Galois group of Q( n a, ζn ) over Q is a split extension of Zn by Zn × . 12. Let K be the splitting field over k of X n − a, where n is not divisible by the characteristic of k, and 0 = a ∈ k. Let ζ be a primitive n-th root of unity in K. Show that the extension ρ

1 → Gal(K/k(ζ)) → Gal(K/k) − → Gal(k(ζ)/k) → 1 splits whenever [K : k(ζ)] = [k(α) : k], where α is an arbitrarily chosen root of X n − a in K. Show that this latter condition holds if and only if k(ζ) ∩ k(α) = k. 13. Let K be a field containing a primitive n-th root of unity and let 0 = a ∈ K. Let α be a root of X n − a in some extension field of K. Show that the minimal polynomial of α over K has the form X d − αd , where d is the smallest divisor of n with the property that αd ∈ K. Show that in this case n/d is the largest divisor of n with the property that K contains an n/d-th root of a. 14. Let a be a positive integer divisible by a prime number q but not by q 2 . Show that the splitting field of√X n − a over Q is a split extension of Zn by Z× n if and only if the real k-th root, k a, of a does not lie in Q(ζn ) ∩ R for any k > 1 that divides n. 15. Give the analogue of the preceding problem for a negative integer a. 16. Let Ka be the splitting field of X 8 − a over Q. Calculate the degree of Ka over Q for all positive integers a. √ 17. Show that Q(ζ5 ) ∩ R = Q( 5). What does this say about the splitting fields of the polynomials X 10 − a for positive integers a? (Hint: Calculate the minimal polynomial of ζ5 + ζ 5 over Q.)


11.9

485

Cyclic Extensions

Definitions 11.9.1. We say that a Galois extension is cyclic if its Galois group is cyclic. Similarly, abelian extensions and solvable extensions will denote Galois extensions whose Galois groups are abelian and solvable, respectively. We say that a Galois extension has exponent n if its Galois group has exponent n in the sense of group theory (i.e., σ n = 1 for each σ ∈ G.) We shall first consider cyclic extensions whose degree is not divisible by the characteristic of k. Recall from Proposition 11.8.2 that if k contains a primitive n-th root of unity and if 0 = a ∈ k, then the splitting field of X n − a over k is a cyclic extension of exponent n. If k does not contain a primitive n-th root of unity, then there may be cyclic extensions of exponent n over k that are not splitting fields of polynomials of the form X n − a. For instance, the Galois group of Q(ζm ) over Q is Z× m , which has numerous cyclic quotient groups. And each one of these quotient groups is the Galois group over Q of an intermediate field between Q and Q(ζm ) by the Fundamental Theorem of Galois Theory. By varying m we may show that Q has cyclic extensions of every order. On the other hand, let K be the splitting field of X n − a over Q with 0 = a ∈ Q. Then a careful analysis of Proposition 11.8.3 shows that Gal(K/Q) cannot be cyclic unless either n = 2 or K = Q(ζn ) for n = 2, 4, pr , or 2pr for some odd prime p. This puts severe restrictions on the cyclic extensions of Q which arise as splitting fields of polynomials of the form X n − a, so we see that there are many cyclic extensions of Q that do not have this form. We shall not attempt a systematic study of cyclic extensions of degree n in the case that n is not divisible by the characteristic of k and yet k fails to contain a primitive n-th root of unity. But in the case that k does contain a primitive n-th root of unity, we shall give a converse to Proposition 11.8.2. Indeed, in this case, every cyclic extension of exponent n over k is the splitting field of a polynomial of the form X n − a. We shall make use of Artin’s Linear Independence of Characters theorem. Proposition 11.9.2. (Linear Independence of Characters) Let G be any group and K a field. Suppose given n distinct group homomorphisms, f1 , . . . , fn , from G to K× , where n ≥ 1. Then f1 , . . . , fn are linearly independent over K as functions from G to K. In other words, given any collection of elements a1 , . . . , an ∈ K that are not all 0, there is a g ∈ G such that a1 f1 (g) + · · · + an fn (g) = 0. Proof We argue by induction on n, with the result being trivial for n = 1. Assuming the result true for n − 1, suppose we’re given coefficients a1 , . . . , an ∈ K, not all 0, such that a1 f1 (g) + · · · + an fn (g) = 0 for all g ∈ G. By our induction hypothesis, we may assume that all the coefficients ai are nonzero. So by multiplying both sides of the equation by a−1 1 , if necessary, we may assume that a1 = 1. We obtain f1 (g) + a2 f2 (g) + · · · + an fn (g) = 0 for all g ∈ G. Since the homomorphisms f1 , . . . , fn are distinct, there is an x ∈ G such that f1 (x) = fn (x). Substituting xg for g in the above equation and applying the fact that the fi are homomorphisms, we obtain f1 (x)f1 (g) + (a2 f2 (x))f2 (g) + · · · + (an fn (x))fn (g) = 0


486

for all g ∈ G. Now multiply both sides of the equation by (f1 (x))−1 . Subtracting the result from the first displayed equation, the terms f1 (g) cancel and we are left with the following: a2 [1 − (f1 (x))−1 f2 (x)]f2 (g) + · · · + an [1 − (f1 (x))−1 fn (x)]fn (g) = 0 for all g ∈ G. Since f1 (x) = fn (x), this is a nontrivial dependence relation between f2 (x), . . . , fn (x), contradicting the inductive assumption. Since the embeddings from a field K into a field L restrict to homomorphisms from K× to L× , we obtain a proof of the following corollary, which was originally due to Dedekind. Corollary 11.9.3. Let σ1 , . . . , σn be distinct embeddings from the field K into a field L and let β1 , . . . , βn ∈ L be coefficients that are not all 0. Then there is an α ∈ K such that β1 σ1 (α) + · · · + βn σn (α) = 0.

Our most common use of this will be when σ1 , . . . , σn are automorphisms of K itself. Proposition 11.9.4. Let G be a group of automorphisms of K over k and let f : G → k× be a group homomorphism. Then there is an element α ∈ K such that f (σ) =

α σ(α)

for all σ ∈ G. Proof Consider the linear combination σ∈G f (σ) · σ. Since the coefficients f (σ) are all nonzero, we can find an a ∈ K such that σ∈G f (σ)σ(a) = 0. Set f (σ)σ(a). α= σ∈G

Note that each f (τ ) lies in k, and hence is fixed by each σ ∈ G. Thus, if we apply σ to each side of the equation, we get f (τ )(στ )(a) σ(α) = τ ∈G

=

(f (σ))−1 f (στ )(στ )(a)

στ ∈G

=

(f (σ))−1 · α,

where the second equality follows from the fact that f is a group homomorphism. The result now follows. We may now give the promised converse to Proposition 11.8.2. Proposition 11.9.5. Let k be a field containing a primitive n-th root of unity and let K be a cyclic extension of k of exponent n. Then K is the splitting field over k of a polynomial of the form X n − a for some a ∈ k.


487

Proof Since k has a primitive n-th root of unity, the splitting field for X d − a for any d dividing n and any element a ∈ K has the form k(α) for any given single root α of X d − a. In particular, if n = dk and if a ∈ k, then X d − a and X n − ak have the same splitting field over k. Thus, by induction, we may assume that K has degree n over k. Let σ be a generator of the Galois group of K over k and let ζ ∈ k be a primitive n-th root of unity. Then there is a homomorphism f : Gal(K/k) → k× such that f (σ) = ζ −1 . By Proposition 11.9.4, there is an element α ∈ K such that f (σ) = α/σ(α), and hence σ(α) = ζ · α. Thus, the orbit of α under the action of the Galois group is {ζ k α | 0 ≤ k < n}, and hence the minimal polynomial of α over k is g(X) =

n−1 .

(X − ζ k α)

k=0

by Lemma 11.5.2. The constant term of g(X) is, up to sign, a power of ζ times αn . Since ζ ∈ k, this implies that αn ∈ k. But then X n − αn has exactly the same roots in K as g(X) does, and hence g(X) = X n − αn . Since the minimal polynomial of α over k has degree n, k(α) = K, and the result follows. We now consider the case of cyclic extensions whose degree is divisible by the characteristic of k. We shall content ourselves with the study of the extensions whose degree is equal to the characteristic of k, as this will be sufficient for a description of solvable extensions. We will need a little preparation first. Definition 11.9.6. Let K be a finite Galois extension of k and let G = Gal(K/k). Let α ∈ K. Then TK/k (α) = σ(α). σ∈G

The function TK/k is a kind of a trace. We shall explore traces more generally in Section 11.14. Note that since we’re summing over all σ ∈ G, TK/k (α) is fixed by every element of G. Thus, TK/k (α) ∈ k. Note that as a linear combination of automorphisms of K, TK/k cannot be identically 0, by Corollary 11.9.3. The fact that elements of Gal(K/k) fix k now gives the next lemma. Lemma 11.9.7. Let K be a finite Galois extension of k. Then TK/k : K → k is a nontrivial homomorphism of k-vector spaces. With this, we can prove the following special case of an additive form of what’s known as Hilbert’s Theorem 90. Note that every Galois extension of prime degree is cyclic. Proposition 11.9.8. Let k be a field of characteristic p = 0 and let K be a Galois extension of k of degree p. Let σ be a generator of Gal(K/k). Then there is an α ∈ K such that α − σ(α) = 1. Proof Since TK/k : K → k is a nontrivial homomorphism of k-vector spaces, there is an element β ∈ K such that TK/k (β) = 1. Let α = β + 2σ(β) + 3σ 2 (β) + · · · + (p − 1)σ p−2 (β).


488

Then α − σ(α) = β + σ(β) + · · · + σ p−2 (β) − (p − 1)σ p−1 (β). Since −(p − 1) = 1, this is precisely TK/k (β) = 1, as required. This sets up a result of Artin and Schreier. Proposition 11.9.9. Let k be a field of characteristic p = 0. Then any Galois extension of degree p over k is the splitting field of a polynomial of the form X p − X − a with a ∈ k. Conversely, if a ∈ k and if X p − X − a has no roots in k, then its splitting field is a Galois extension of k of degree p. Proof Let K be a Galois extension of degree p over k and let σ generate the Galois group of K over k. Then Proposition 11.9.8 provides an element α ∈ K such that α − σ(α) = 1. Thus, σ(α) = α − 1, so α is not in k, but p

σ(αp − α) = (σ(α)) − σ(α) = (α − 1)p − (α − 1) = αp − α, since passage to p-th powers is a homomorphism. Since σ generates the Galois group of K over k, this implies that αp − α ∈ k. Thus, α is a root of X p − X − a, where a = αp − α ∈ k. In particular, the splitting field of X p − X − a is an intermediate field between k and K that properly contains k. Since K has degree p over k, this forces the splitting field of X p − X − a to equal K itself. For the converse, let a ∈ k and suppose that f (X) = X p − X − a has no roots in k. Since the formal derivative of X p − X − a has no roots, the splitting field, K, of f is a separable extension of k. Let α ∈ K be a root of f . Then (α + 1)p − (α + 1) = a, so α + 1 is also a root of f . By induction, the p distinct roots of f are α, α + 1, . . . , α + (p − 1). Thus, K = k(α). Let σ be any non-identity element of Gal(K/k). Then σ(α) = α + r for some r with 1 ≤ r ≤ p − 1. But then induction shows that σ k (α) = α + kr. Since r generates Zp additively, all the roots of f are in the orbit of α under the action of the Galois group. Thus, f is the minimal polynomial of α over k by Lemma 11.5.2, and hence K has degree p over k. Exercises 11.9.10. 1. Do X 2 − 2 and X 4 − 4 have the same splitting field over Q? 2. Let K be a finite Galois extension of k. A function f : Gal(K/k) → K× is called a solution to Noether’s equations if f (σ) · σ(f (τ )) = f (στ ) for all σ, τ ∈ Gal(K/k). (a) Show that f is a group homomorphism if and only if it takes value in k× . (b) Show that a function f : Gal(K/k) → K× is a solution to Noether’s equations if and only if there is a nonzero element α ∈ K such that f (σ) = α/σ(α) for all σ ∈ Gal(K/k).

11.10

Kummer Theory

Here, we generalize the preceding section to study abelian extensions of exponent n over a field containing a primitive n-th root of unity.


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Definition 11.10.1. Let k be a field containing a primitive n-th root of unity. A Kummer extension of k is the splitting field of a polynomial of the form f (X) = (X n − a1 ) . . . (X n − ak ), for elements a1 , . . . , ak ∈ k. One direction of our study is easy. Proposition 11.10.2. Let k be a field containing a primitive n-th root of unity and let K be a Kummer extension of k. In particular, let K be the splitting field over k of f (X) = (X n − a1 ) . . . (X n − ak ), where a1 , . . . , ak ∈ k. Then K is an abelian extension of k of exponent n. Proof Let ζ be a primitive n-th root of unity in k and let G = Gal(K/k). Let Ki ⊂ K be the splitting field of X n − ai over k and let Gi be the Galois group of Ki over k. Then we have restriction maps ρi : G → Gi , obtained by restricting the action of G to Ki . These assemble to give a homomorphism ρ:G→

k .

Gi

i=1

whose i-th component function is ρi . By Proposition 11.8.2, each Gi is a cyclic group of exponent n. Since abelian groups of exponent n are closed under products, it suffices to show that ρ is injective. By Corollary 11.5.3, an element of G is determined by its effect on the roots of f . But every root of f (X) is a root of X n − ai for some i, and hence lies in one of the Ki . Now any element of the kernel of ρ acts trivially on each Ki , and hence acts trivially on the set of roots of f (X). So it must be the identity element by Corollary 11.5.3. The converse here is very similar to the argument for the cyclic case. But we shall prove a little more. First, recall that if A and B are abelian groups, then HomZ (A, B) denotes the set of group homomorphisms from A to B, and that HomZ (A, B) itself has a natural abelian group structure given by addition of homomorphisms: If B is written in additive notation, then we have (f + g)(a) = f (a) + g(a) for all f, g ∈ HomZ (A, B) and a ∈ A. Lemma 11.10.3. Let k be a field containing a primitive n-th root of unity and let G be a finite abelian group of exponent n. Then there is an isomorphism of groups between G and HomZ (G, k× ). Proof Let ζ be a primitive n-th root of unity in k. Since any element of k× of exponent n is a root of X n − 1, and since the roots of X n − 1 in k× are precisely the powers of ζ, the image of any homomorphism from G to k× must lie in ζ, the subgroup of k× generated by ζ. Thus, we obtain an isomorphism ∗ HomZ (G, k× ). HomZ (G, ζ) −→

i

∼ =

Of course, ζ is isomorphic to Zn . For compatibility with earlier chapters, let us write G in additive notation. Thus, the Fundamental Theorem of Finite Abelian Groups


490

gives an isomorphism from G to a direct sum of cyclic groups (which were given to be of prime power order in the first proof of the Fundamental Theorem) whose order divides n: G∼ = Zm1 ⊕ · · · ⊕ Zmk , where mi divides n for i = 1, . . . , k. Proposition 9.7.7 now gives an isomorphism of groups HomZ (G, Zn ) ∼ =

k ,

HomZ (Zmi , Zn ),

i=1

so the result follows if we show that HomZ (Zm , Zn ) ∼ = Zm whenever m divides n. Now Proposition 2.5.17 gives a bijection from HomZ (Zm , Zn ) to the set of elements of Zn with exponent m. This bijection takes a homomorphism f : Zm → Zn to f (1), the result of evaluating f at the generator 1. Since Zn is abelian, this bijection is easily seen to be a group isomorphism from HomZ (Zm , Zn ) to the subgroup of Zn consisting of those elements of exponent m. Since m divides n, the elements of exponent m in Zn are precisely the elements of the cyclic subgroup of Zn of order m, so the result follows. The converse to Proposition 11.10.2 is included in the next proposition. Proposition 11.10.4. Let k be a field containing a primitive n-th root of unity and let K be an abelian extension of k of exponent n. Then K is a Kummer extension of k. We may also express the Galois group of K over k as follows. Let G = Gal(K/k) and let r be the smallest positive integer which is an exponent for G. Let A ⊂ K× be the collection of elements whose r-th powers lie in k. Write Ar and (k× )r for the sets of r-th powers of A and k× , respectively. Then A, Ar , and (k× )r are subgroups of K× , and there are group isomorphisms G∼ = A/k× ∼ = Ar /(k× )r . Proof This time, we shall apply the classification of finite abelian groups given by the Fundamental Theorem of Finitely Generated Modules over a P.I.D. Here, we obtain an isomorphism ∼ =

η : G −→ Zm1 ⊕ · · · ⊕ Zmk , where each mi divides mi−1 for i = 2, . . . , k. Thus, m1 is the smallest positive exponent for G. By our hypothesis, m1 divides n. We shall write G itself in multiplicative form. Write σi ∈ G for the element which corresponds under η to the image of the canonical generator of the summand Zmi under its standard inclusion in the direct sum. (Thus, each σi has order mi , and G is the internal direct product of the subgroups generated by the σi .) Let ζ be a primitive n-th root of unity in k, and write ζm for the principal m-th root of unity ζ n/m for all integers m dividing n. Let fi : G → k× be the homomorphism determined by 2 1 if i = j fi (σj ) = −1 ζmi if i = j. Notice that up to sign, fi is the image of σi under the isomorphism from G to HomZ (G, k× ) that is given in Lemma 11.10.3. Thus, HomZ (G, k× ) is generated by f1 , . . . , fk .


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By Proposition 11.9.4, we can find elements α1 , . . . , αk ∈ K such that fi (σ) = αi /σ(αi ) for all σ ∈ G and all i. Thus, 2 if i = j αj σi (αj ) = ζmi αi if i = j. j In particular, the orbit of αi under G is {ζm α | 0 ≤ j < mi }. Thus, as shown in i i mi the proof of Proposition 11.9.5, αi ∈ k, and the minimal polynomial of αi over k is X mi − ai , where ai = αimi . Since k contains a primitive n-th root of unity, k(αi ) contains all the roots of X n − n/mi ai . Thus, k(α1 , . . . , αk ) is a Kummer extension of k, being the splitting field over k of n/m1

f (X) = (X n − a1

n/mk

) . . . (X n − ak

).

Note that the displayed calculation of σi (αj ) shows that no non-identity element of G = Gal(K/k) acts as the identity on k(α1 , . . . , αk ). Thus, the Galois group of K over k(α1 , . . . , αk ) is the trivial group, and hence K = k(α1 , . . . , αk ). Thus, K is a Kummer extension of k, as claimed. We now consider the second part of the statement of this proposition. As observed above, the smallest positive exponent for G is r = m1 . Since passage to r-th powers is a homomorphism in an abelian group with multiplicative notation, the subsets A, Ar , and (k× )r are easily seen to be subgroups of K× . Moreover, passage to r-th powers induces a surjective homomorphism μ : A/k× → Ar /(k× )r . Now let α ∈ A and suppose that the class of α in A/k× lies in the kernel of μ. Thus, αr = ar for some a ∈ k× . But the r-th roots of ar all differ by multiplication by powers of ζr . Since ζr ∈ k, this forces α ∈ k× , and hence μ is an isomorphism. Thus, it suffices to provide an isomorphism from A/k× to HomZ (G, k× ). Let α ∈ A. Then αr ∈ k, and hence α is a root of the polynomial X r − αr ∈ k[X]. Since the roots of this polynomial all differ from α by multiplication by a power of ζr , we see that α/σ(α) must be a power of ζr for all σ ∈ G. In particular, if τ ∈ G, α/τ (α) is fixed by each element σ ∈ G. Thus, (α/σ(α)) · (α/τ (α)) = (α/σ(α)) · σ (α/τ (α)) = α/στ (α). Thus, there is a homomorphism fα : G → k× , defined by setting fα (σ) = α/σ(α). Since the elements of G are automorphisms of K, the passage from α to fα is easily seen to give a homomorphism γ : A → HomZ (G, k× ). And α ∈ A lies in the kernel of γ if and only if σ(α) = α for all σ ∈ G, so the kernel of γ is precisely k× . Thus, it suffices to show that γ is onto. But this is precisely the output of Proposition 11.9.4. To avoid making an elaboration on an earlier argument, simply consider the generators fi of HomZ (G, k× ) constructed above. We used Proposition 11.9.4 to construct elements αi ∈ K such that fi (σ) = αi /σ(αi ) for all σ ∈ G. And the αi were shown above to lie in A, and hence fi = γ(αi ). This shows that, with k, K, and A as in the statement of Proposition 11.10.4, then, with great redundancy, we may express K as the splitting field of {X r − a | a ∈ Ar }. Thus, K is determined uniquely, up to isomorphism over k, by the subset Ar ⊂ k× . This gives one direction of the next corollary.


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Corollary 11.10.5. Let k be a field containing a primitive r-th root of unity. Then there is a one-to-one correspondence between the finite abelian extensions of k whose smallest positive exponent is r and the subgroups B ⊂ k× such that (k× )r ⊂ B and B/(k× )r is a finite group whose smallest positive exponent is r. Proof It suffices to show that if B is such a subgroup, and if K is the splitting field of {X r − b | b ∈ B} over k, then there is a finite collection b1 , . . . , bk ∈ B such that K is the splitting field of {X r − bi | 1 ≤ i ≤ k}. To see this, suppose given a finite collection b1 , . . . , bs ∈ B, and let K1 ⊂ K be the splitting field of {X r − bi | 1 ≤ i ≤ s}. If K1 = K, then we’re done. Otherwise, let B1 be the set of all elements of k that are r-th powers of elements of K× 1 . Then Proposition 11.10.4 shows that the index of (k× )r in B1 is strictly less than the index of (k× )r in B. In particular, B1 = B, and hence we can find an element bs+1 that is in B but not in B1 . But then the splitting field of {X r − bi | 1 ≤ i ≤ s + 1} is strictly larger than K1 , so the result follows by induction on the index of (k× )r in B. Exercises 11.10.6. 1. Find fields k ⊂ K1 ⊂ K such that k contains a primitive n-th root of unity, K1 is a Kummer extension of k of exponent n, K is a Kummer extension of K1 of exponent n, but Gal(K/k) is not abelian. 2. Let k be a field of characteristic p and let a1 , . . . , ak ∈ k. Show that the splitting field of f (X) = (X p − X − a1 ) . . . (X p − X − ak ) over k is an abelian extension of exponent p. 3. Let k be a field of characteristic p and let K be a finite abelian extension of k of exponent p. Show that K is the splitting field of a polynomial of the form f (X) = (X p − X − a1 ) . . . (X p − X − ak ), with a1 , . . . , ak ∈ k. (Hint: Consider the subgroups of Gal(K/k) of index p.) 4. Let k be a field of characteristic p and let K be a finite abelian extension of k of exponent p. Let G = Gal(K/k), and write ψ : K → K for the homomorphism ψ(α) = αp − α. Show that G, HomZ (G, Zp ), (ψ −1 (k))/k, and (ψ(ψ −1 (k)))/(ψ(k)) are all isomorphic. 5. Let k be a field of characteristic p and let ψ : k → k be given by ψ(a) = ap − a. Show that there is a one-to-one correspondence between the isomorphism classes over k of finite abelian extensions of k of exponent p and the additive subgroups of k that contain ψ(k) as a subgroup of finite index. 6. Find extension fields Zp (X) ⊂ k ⊂ K such that k is an abelian extension of exponent p over Zp (X), K is an abelian extension of degree p over k, and Gal(K/Zp (X)) is nonabelian.


11.11

493

Solvable Extensions

We can now address the issue of finding finding roots of polynomials as functions of the coefficients. The starting point, of course, is the quadratic formula. Over any field k of characteristic not equal to 2, the roots of the quadratic aX 2 + bX + c have the form √ −b − b2 − 4ac . 2a √ Here, we allow b2 − 4ac to stand for either root of X 2 − (b2 − 4ac) in our preferred choice of splitting field for k. Indeed, in most fields (including C), there is no preferred choice for a square root, so the quadratic formula should be taken as specifying the two solutions in no particular order. Now, we consider the cubic. Example 11.11.1. Let k be a field whose characteristic is not equal to either 2 or 3. Consider the cubic g(X) = X 3 + bX 2 + cX + d. If b = 0, we “complete the cube” by making the substitution f (X) = g(X −(b/3)). This gives us a cubic that does not involve X 2 , which we write as f (X) = X 3 + pX + q. The solution to a cubic in this form is originally due to Cardano. Fix an algebraic closure, L, of k, in which all roots are to be taken, and let Δ = −4p3 − 27q 2 . As shown in in Problem 5 of Exercises 7.4.14, this is the discriminant of the cubic f (X). √ Choose a preferred square root, −3Δ, of −3Δ. We shall choose cube roots 4 4 √ √ −3Δ −3Δ q q 3 3 and β= − + α= − − 2 18 2 18 with some care. First, as the reader may verify, note that for any choices of α and β we have (3αβ)3 = −p3 . Now if ζ is a primitive third root of unity in L, the roots α and β are determined up to multiplication by a power of ζ. In particular, for any choice of α, there is a unique choice of β for which 3αβ = −p. Having thus chosen β, we set γ = α + β. Then γ3

= α3 + β 3 + 3αβ(α + β) = −q − pγ,

and hence γ is a root of f . Note that the relationship between α and β gives 4 γ=

3

q − − 2

√

⎛4 ⎞−1 √ −3Δ p ⎝ 3 q −3Δ ⎠ − − − . 18 3 2 18

Here, if we start by choosing a square root of −3Δ, then the three different choices for √ a cube root of −q/2 − −3Δ/18 will all give roots of f via this formula.


494

It’s not hard to see that the three roots thus constructed are all distinct unless the discriminant Δ = 0. But Corollary 7.4.13 shows that Δ(f ) = 0 if and only if f has repeated roots. We leave it to the reader to verify that in this degenerate case, that the procedure above produces all the roots of f in their correct multiplicities. Of course, the roots of the original polynomial g(X) may be obtained from those of f by subtracting b/3. The procedure for solving a quartic is more complicated, but the result is that the roots may be expressed as algebraic functions of the coefficients. Definition 11.11.2. An algebraic function of several variables is one obtained by composing the various operations of sums, differences, products, powers, quotients, and extraction of n-th roots. Note that the only coefficients introduced by these operations will be integers. We shall consider algebraic functions with more general coefficients in the next section. Due to the presence of n-th roots, of course, algebraic “functions” are not necessarily well defined as functions: Choices are made every time a root is extracted. But this is the best we can ask for, as far as solving for the roots of polynomials over arbitrary fields is concerned. Given that polynomials of degree ≤ 4 admit general solutions as algebraic functions (in characteristics not equal to 2 or 3), the thing that’s truly striking is that in degrees 5 and above there exist polynomials over Z whose roots cannot individually be expressed as algebraic functions of the coefficients. So not only is there no general formula for solving polynomials in degree ≥ 5, there are even examples (many of them, it turns out) which do not admit an individual formula of the desired sort for their solution. It was Galois who first discovered this, by developing a theory of solvable extensions. An extension is solvable, of course, if it is a Galois extension whose Galois group is a solvable group. A concept more closely related to solving polynomials is that of solvability by radicals. Definition 11.11.3. An extension k1 of k is said to be solvable by radicals if there is an extension K of k1 and a sequence of extensions k = K0 ⊂ K1 ⊂ · · · ⊂ Km = K such that each Ki is obtained from Ki−1 by adjoining a root of a polynomial X ni − ai for some ni > 0 and some ai ∈ Ki−1 . The next lemma is now clear. Lemma 11.11.4. The splitting field of a polynomial f (X) ∈ k[X] is solvable by radicals if and only if the roots of f may be expressed as algebraic functions on some collection of elements in k. In particular, if the roots of a polynomial f (X) ∈ k[X] may be written as algebraic functions of its coefficients, then the splitting field of f over k is solvable by radicals. In characteristic 0, we shall see that a finite Galois extension is solvable by radicals if and only if it is a solvable extension. In characteristic p = 0, the situation is more complicated: A Galois extension of degree p is certainly solvable, but not necessarily by radicals. Here, we shall relate solvable extensions to a different sequential solvability criterion, which agrees with solvability by radicals in characteristic 0. We shall continue the discussion of solvability by radicals in characteristic p in Exercises 11.11.10.


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Definition 11.11.5. We say that the extension k1 of k is solvable by cyclic extensions if there is an extension K of k1 and a sequence of extensions k = K0 ⊂ K1 ⊂ · · · ⊂ Km = K such that each extension Ki−1 ⊂ Ki has one of the following forms: 1. Ki is obtained from Ki−1 by adjoining a root of a polynomial of the form X ni − ai , where ni is not divisible by the characteristic of k and ai ∈ Ki−1 . 2. Ki is obtained from Ki−1 by adjoining a root of a polynomial of the form X p −X−ai for some ai ∈ Ki−1 . Here, p is the characteristic of k. This concept is indeed identical to solvability by radicals when the fields have characteristic 0. Lemma 11.11.6. Solvable extensions have the following closure properties. 1. Let K be a finite Galois extension of k. If K is solvable over k, then it is solvable over any intermediate field, and if K1 is an intermediate field that is normal over k, then Gal(K1 /k) must be solvable as well. Conversely, if K1 is an intermediate field such that K1 is solvable over k and K is solvable over K1 , then K must be solvable over k, as well. 2. Let K1 and K2 be finite solvable extensions of k and let L be a compositum of K1 and K2 over k. Then L is a solvable extension of k. Proof The first assertion is immediate from the Fundamental Theorem of Galois Theory and Corollary 5.5.10. For the second assertion, write Ki = k(αi ) and let fi (X) be the minimal polynomial of αi over k. Then L = k(α1 , α2 ) is the splitting field of f1 (X)f2 (X) over k, and hence is Galois over k. The restriction maps ρi : Gal(L/k) → Gal(Ki /k) provide a homomorphism (ρ1 ,ρ2 )

Gal(L/k) −−−−→ Gal(K1 /k) × Gal(K2 /k). Since an element of Gal(L/k) is determined by its effect on α1 and α2 , (ρ1 , ρ2 ) is injective. The result follows from Corollary 5.5.10. The next proposition should come as no surprise. Proposition 11.11.7. A finite Galois extension k1 of k is solvable by cyclic extensions if and only if it is a solvable extension. Proof Suppose that k1 is a finite solvable extension of k. Let n be the product of all the prime divisors of [k1 : k] not equal to the characteristic of k, and let ζ be a primitive n-th root of unity in the algebraic closure of k1 . Then k1 is contained in k1 (ζ), and we have inclusions k ⊂ k(ζ) ⊂ k1 (ζ). Since k(ζ) is one of the permissible types of extension over k in the definition of solvability by cyclic extensions, it suffices to show that k1 (ζ) is solvable by cyclic extensions over k(ζ).


496

Now k1 (ζ) is a compositum of k1 and k(ζ) over k. By Problem 5 of Exercises 11.5.22, k1 (ζ) is a Galois extension of k(ζ) and Gal(k1 (ζ)/k(ζ)) is isomorphic to a subgroup of Gal(k1 /k). By Corollary 5.5.7, this makes G = Gal(k1 (ζ)/k(ζ)) a solvable group. Also, k(ζ) has a primitive q-th root of unity for every prime q that divides |G| and is not equal to the characteristic of k. Write k1 (ζ) = K. Since G is solvable, Proposition 5.5.12 provides a subnormal series 1 = Gm Gm−1 . . . G0 = G, where the successive quotient groups Gi−1 /Gi are cyclic groups of prime order. Set Ki = KGi for i = 0, . . . , m. Applying the Fundamental Theorem of Galois Theory to the extensions Ki−1 ⊂ Ki ⊂ K, we see that Ki is a Galois extension of Ki−1 , with Galois group isomorphic to Gi−1 /Gi , a cyclic group of prime order, q. The extension Ki−1 ⊂ Ki is now easily seen to fall into one of the two types in Definition 11.11.5, using Proposition 11.9.5 when q is not equal to the characteristic of k, and using Proposition 11.9.9 otherwise. Thus, K = k1 (ζ) is solvable by cyclic extensions over K0 = k(ζ), as claimed. Conversely, suppose k1 is solvable by cyclic extensions over k, via a sequence k = K0 ⊂ K1 ⊂ · · · ⊂ Km = K. Here, Ki = Ki−1 (αi ), where αi ∈ Ki is a root of a polynomial fi (X) ∈ Ki−1 [X], and fi (X) may be written either as X ni − ai or as X p − X − ai , depending on whether the extension Ki−1 ⊂ Ki satisfies the first or the second condition of Definition 11.11.5. Since k1 ⊂ K, Lemma 11.11.6 shows that it suffices to show that K is contained in a solvable extension of k. We first claim that each Ki is contained in a solvable extension of Ki−1 : If fi (X) = X p − X − ai , this is immediate from Proposition 11.9.9. Otherwise let Ki be the splitting field of fi over Ki−1 . Then Proposition 11.8.3 shows that Gal(Ki /Ki−1 ) is an extension of an abelian group by an abelian group, and hence is solvable by Corollary 5.5.10. Thus, it suffices to show that if K0 ⊂ K1 ⊂ K2 such that K1 is contained in a finite solvable extension of K0 and K2 is contained in a finite solvable extension of K1 , then K2 is contained in a finite solvable extension of K0 . Of course, we may assume that K2 is solvable over K1 . Let L1 be a finite extension of K1 which is solvable over K0 and let L2 be the compositum of K2 and L1 over K1 in an algebraic closure L of K1 . By Problem 5 of Exercises 11.5.22, L2 is Galois over L1 , and Gal(L2 /L1 ) is isomorphic to a subgroup of Gal(K2 /K1 ), and hence is solvable. Let L2 be the normal closure of L2 in L over K0 . Then Gal(L2 /K0 ) is an extension of Gal(L2 /L1 ) by Gal(L1 /K0 ), so it suffices to show that Gal(L2 /L1 ) is solvable. Write L2 = L1 (α) and let σ1 , . . . , σk be the embeddings of L2 in L over K0 . Since L1 is normal over K0 , σi (L1 ) = L1 for all i, and hence σi (L2 ) = L1 (σi (α)). Thus L2 = L1 (σ1 (α), . . . , σk (α)), is the compositum of σ1 (L2 ), . . . , σk (L2 ) over L1 . Now σi (L2 ) is a solvable extension of σi (L1 ) = L1 . Thus, L2 is a compositum of finite solvable extensions of L1 , and hence is solvable by Lemma 11.11.6. We shall now show that there are splitting fields over Q of polynomials of degree 5 and higher that are not solvable by radicals. Lemma 11.11.8. Let k be any subfield of R and let f (X) be an irreducible polynomial of degree 5 over k with three real roots and two non-real complex roots. Let K be the splitting field of f over k. Then Gal(K/k) = S5 , and hence K is not a solvable extension of k.


497

Proof Let α be a root of f in K. Then f is its minimal polynomial, so k(α) has degree 5 over k. Thus, [K : k] is divisible by 5. We shall identify G = Gal(K/k) with its image in the permutation group of the set of roots of f . Since the only elements of order 5 in S5 are 5-cycles, we see that G contains a 5-cycle, σ. Since complex conjugation restricts to the identity on k ⊂ R, it must restrict to an automorphism of the normal extension K of k. Since conjugation fixes the three real roots of f and must move the two non-real roots, it gives a transposition, τ , as an element of G ⊂ S5 . Let us now number the roots of f to get an explicit isomorphism from the permutation group on the roots of f to S5 . First, let’s write τ = (1 2). Next, note that since the orbit under σ of any one root is the full set of roots of f , there is a power, say σ k , of σ such that σ k (1) = 2. Now, continue the numbering of the roots so that σ k = (1 2 3 4 5). Then conjugating τ by the powers of σ k , we see that (2 3), (3 4) and (4 5) lie in G. Since any transposition may be written as a composite of transpositions of adjacent numbers, G must contain all the transpositions, and hence all of S5 . Of course, A5 , as a nonabelian simple group (Theorem 4.2.7), is not solvable (Lemma 5.5.3). Since subgroups of solvable groups are solvable, K is not solvable over k. One of the interesting things about this study of solvability by radicals is that we can use the information about finite extensions of Q to deduce information about solutions of polynomials over algebraically closed fields, such as C. Theorem 11.11.9. Let k be any field of characteristic 0 and let k ≥ 5. Then there is no general formula that gives the roots of every polynomial of degree k over k as algebraic functions of its coefficients. Proof It is sufficient to show that there is a polynomial of degree k over Q ⊂ K whose roots cannot be expressed as algebraic functions of its coefficients. Note that if f is a polynomial of degree 5 over Q with this property, then f (X) · (X − 1)k−5 gives such a polynomial in degree k. Thus, it suffices to show that there is a polynomial of degree 5 over Q whose splitting field is not a solvable extension of Q. By Lemma 11.11.8, it suffices to find an irreducible polynomial of degree 5 over Q with exactly 3 real roots. We claim that f (X) = X 5 − 4X + 2 is such a polynomial. Clearly, f is irreducible by Eisenstein’s criterion. We shall use elementary calculus to show that it has the desired behavior with regard to roots. Note that the derivative f (X) = 5X 4 −4 has two roots, at ± 4 4/5. Between these two roots, f (X) is negative, and outside them, it is positive. Thus, f is strictly increasing on both (−∞, − 4 4/5] and [ 4 4/5, ∞), and is strictly decreasing on [− 4 4/5, 4 4/5]. Clearly, f (−2) is negative and suffices by the Intermediate Value f (2) is positive, so it 4 Theorem to show that f (− 4 4/5) is positive and f ( 4/5) is negative. But this follows easily from the fact that 4/5 ≤ 4 4/5 ≤ 1. Exercises 11.11.10. 1. Let p be a prime and let f be a polynomial of degree p over Q with exactly p − 2 real roots. Show that the Galois group of the splitting field of f over Q is Sp . 2. Let k be an algebraic extension of Zp . Show that every finite extension of k is both solvable and solvable by radicals. 3. Let K be a finite solvable extension of k in characteristic p = 0, such that [K : k] is relatively prime to p. Show that K is solvable by radicals over k.


498

† 4. We can recover an analogue of Theorem 11.11.9 for some infinite fields in characteristic p, provided that we first generalize the Fundamental Theorem of Galois Theory to the case of non-separable extensions. Here, if K is a normal extension of k, we write autk (K) for the group of automorphisms (as a field) of K which restrict to the identity on k. (a) Let K be a normal extension of k and let K1 be the separable closure of k in K. (See Problem 2 of Exercises 11.4.20.) Show that K1 is a Galois extension of k and that restriction of automorphisms to K1 gives an isomorphism of groups, ∼ =

ρ : autk (K) −→ Gal(K1 /k). (b) Suppose given extensions k ⊂ K1 ⊂ K2 , such that both K1 and K2 are normal over k. Show that restriction of automorphisms from K2 to K1 gives a well defined restriction homomorphism ρ : autk (K2 ) → autk (K1 ), inducing a short exact sequence ⊂

ρ

1 → autK1 (K2 ) −→ autk (K2 ) − → autk (K1 ) → 1. (c) Let n = pr s, where (p, s) = 1 and let a ∈ k. Let K be the splitting field of X n − a over k. Show that the separable closure of k in K is the splitting field of X s − a over k. (d) Let K be a normal extension of k which is solvable by radicals. Show that autk (K) is a solvable group. We shall see below that there is a polynomial over the field of rational functions Zp (X1 , . . . , Xn ) whose splitting field has Galois group Sn . Thus, polynomials of any degree ≥ 5 over a field whose transcendence degree over Zp is ≥ 5 do not admit a general formula to solve for their roots as algebraic functions of their coefficients.

11.12

The General Equation

We’ve seen that there can be no general formula for finding the roots of the polynomials of a given degree ≥ 5 as algebraic functions of the coefficients. Recall that the algebraic functions we’ve been considering can be thought of as having integer coefficients. The question can be generalized as follows. Definition 11.12.1. Let k be a field. Then a k-algebraic function is a function of several variables, defined on an extension of k, obtained by composing the various operations of sums, differences, products, powers, quotients, extraction of n-th roots, and addition or multiplication by arbitrary elements (to be thought of as constants) of k. Thus, a k-algebraic function is an algebraic function with coefficients in k. In a field of characteristic 0, Q-algebraic functions coincide with our earlier definition of algebraic functions with integer coefficients. Similarly, in characteristic p, the Zp algebraic functions coincide with the algebraic functions with integer coefficients. Thus, integer coefficients are the most general kind one could ask about for a formula that would apply to every field of a given characteristic.


499

But what about finding roots of polynomials over extensions of R? Every polynomial over R itself is solvable by radicals, because the roots lie in the radical extension C of R. So, abstractly, the roots are algebraic functions on elements of R. But this abstract fact tells us absolutely nothing about the way in which the roots depend on the coefficients of the polynomial. So we might ask if there is a general solution for polynomials over any extension field of R, where the roots are given as R-algebraic functions of the coefficients of the polynomial. Here, we cannot point to particular polynomials over R itself for counterexamples, as the splitting fields are always solvable by radicals. The counterexamples will be given by polynomials over extension fields of R. Nor is there anything special about R here. We may work over an arbitrary field k. For any field k, the symmetric group Sn acts on the field of rational functions k(X1 , . . . , Xn ) by permuting the variables. Write Si for the i-th elementary function si (X1 , . . . , Xn ) of the variables X1 , . . . , X1 . We’ve already seen (Proposition 7.4.7) that the fixed points of the restriction of the above action of Sn to the subring k[X1 , . . . , Xn ] is a polynomial algebra with generators S1 , . . . , Sn . Thus, the field of rational functions k(S1 , . . . , Sn ) is contained in the fixed field (k(X1 , . . . , Xn ))Sn . We shall show that the fixed field is equal to k(S1 , . . . , Sn ), which then shows, via Theorem 11.5.17, that k(X1 , . . . , Xn ) is a Galois extension of k(S1 , . . . , Sn ), with Galois group Sn . The relevance of this to solving polynomials comes from Lemma 7.4.4, which shows that in k(X1 , . . . , Xn )[X], we have the factorization X n − S1 X n−1 + S2 X n−2 + · · · + (−1)n Sn = (X − X1 ) . . . (X − Xn ). Thus, X1 , . . . , Xn are the roots of the separable polynomial fn (X) = X n − S1 X n−1 + S2 X n−2 + · · · + (−1)n Sn , and hence k(X1 , . . . , Xn ) is the splitting field for fn (X) over k(S1 , . . . , Sn ). Clearly, this extension is solvable by radicals if and only if there is a formula expressing the roots X1 , . . . , Xn as k-algebraic functions of S1 , . . . , Sn . For n ≤ 4, where such formulæ exist, they turn out to be universal, away from characteristics ≤ n, though that is not implied by the setup here: There cannot be a direct comparison of extensions between the one studied here and the splitting field of an arbitrary polynomial. Nevertheless, if there is a universal formula for finding the roots as k-algebraic functions of the coefficients of the polynomial, then it will show up in the study of this example. For n ≥ 5, we do not even need to verify that k(S1 , . . . , Sn ) is equal to the fixed field of the Sn action to draw conclusions regarding solvability. Since Sn is the Galois group of k(X1 , . . . , Xn ) over the fixed field of this action (Theorem 11.5.17), and since k(S1 , . . . , Sn ) is a subfield of this fixed field, Sn must be a subgroup of the Galois group of k(X1 , . . . , Xn ) over k(S1 , . . . , Sn ), and hence the Galois group cannot be solvable when n ≥ 5. Recall from Problem 4 of Exercises 11.11.10 that even in characteristic p = 0, a Galois extension which is solvable by radicals must be solvable. Therefore, no solution by radicals for the above polynomial fn (X) is possible for n ≥ 5. Note that since S1 , . . . , Sn are algebraically independent, k(S1 , . . . , Sn ) is isomorphic to k(X1 , . . . , Xn ). This shows that any field of transcendence degree ≥ 5 over Zp admits polynomials of degree ≥ 5 whose roots cannot be written as k-algebraic functions of the coefficients of the polynomial. We also obtain the next proposition.


500

Proposition 11.12.2. Let k be any field and let n ≥ 5. Then there is no general formula that expresses the roots of polynomials over extension fields of k as k-algebraic functions of the coefficients of the polynomials. Proposition 11.12.3. Let k be any field, and let Sn act on the field of rational functions k(X1 , . . . , Xn ) by permuting the variables. Then the fixed field of this action is k(S1 , . . . , Sn ), where Si is the i-th elementary symmetric function in the variables X1 , . . . , Xn . Thus, k(X1 , . . . , Xn ) is a Galois extension of k(S1 , . . . , Sn ) with Galois group Sn . Proof As noted above, since X1 , . . . , Xn are the roots of the separable polynomial fn (X), k(X1 , . . . , Xn ) is the splitting field of fn (X) over k(S1 , . . . , Sn ). Thus, if G is the Galois group of k(X1 , . . . , Xn ) over k(S1 , . . . , Sn ), then restricting the action of G to {X1 , . . . , Xn } gives in embedding of G in the group of permutations on {X1 , . . . , Xn }. Since the action of Sn on k(X1 , . . . , Xn ) fixes k(S1 , . . . , Sn ), Sn embeds in G, respecting the usual action of Sn on the variables. Thus, the inclusion of Sn in G is an isomorphism. The Fundamental Theorem of Galois Theory now shows that k(S1 , . . . , Sn ) must be the fixed field of the Galois group Sn .

11.13

Normal Bases

Definition 11.13.1. Let K be a finite Galois extension of k. Then a normal basis for K over k is an orbit G · α of the action of G = Gal(K/k) on K with the property that the elements of G · α form a basis for K as a vector space over k. Thus, if G = {σ1 , . . . , σn }, then a normal basis for K over k is a k-basis for K of the form σ1 (α), . . . , σn (α) for some α ∈ K. We shall show here that every finite Galois extension of an infinite field has a normal basis. So far, what we know is that every finite Galois extension has a primitive element. It is easy to see that an element α ∈ K is a primitive element for K over k if and only if the orbit of α under the action of G = Gal(K/k) has |G| elements, and hence is isomorphic to G as a G-set. Thus, if the orbit of α is a normal basis for K over k, then α is a primitive element. But the converse of this is false. For instance, i, −i fails to be a normal basis for Q(i) over Q. Clearly, it will be useful to derive a condition under which a collection of elements α1 , . . . , αn will be a basis for K over k. We generalize things slightly. Lemma 11.13.2. Let K be a finite separable extension of k and let L be the normal closure of K over k in an algebraic closure of K. Let σ1 , . . . , σn be the distinct embeddings of K over k into L, and let α1 , . . . , αn ∈ K. Then α1 , . . . , αn is a basis for K as a vector space over k if and only if the matrix (σi (αj )) ∈ Mn (L) is invertible. Proof Since any embedding, σ, of K over k into the algebraic closure of K extends to an embedding of the normal extension L, the image σ(K) must lie in L. Thus, by Proposition 11.4.12, n, which is the number of distinct embeddings of K in L over k, is equal to the degree of K over k. Let M = (σi (αj )). If M is not invertible, then its rows are linearly dependent over L. So we can find β1 , . . . , βn ∈ L such that β1 σ1 (αj ) + · · · + βn σn (αj ) = 0


501

n for all j. Thus, the linear combination i=1 βi σi vanishes on the k-linear subspace of n K generated by α1 , . . . , αn . But Corollary 11.9.3 shows that i=1 βi σi cannot be the trivial homomorphism, so α1 , . . . , αn cannot generate K as a vector space over k, and hence cannot form a basis for K over k. Conversely, nif α1 , . . . , αn are not linearly n independent, we can find coefficients a1 , . . . , an ∈ k such that i=1 ai αi = 0. But then j=1 aj σi (αj ) = 0 for all i, and hence the columns of M are not linearly independent. Lemma 11.13.2, which makes use of Corollary 11.9.3 in its proof, may be used to prove the following proposition, which generalizes Corollary 11.9.3 in the cases of greatest interest. We shall make use of Proposition 7.3.26, which shows that if K is an extension of the infinite field k, then any polynomial f (X1 , . . . , Xn ) over K which vanishes on every n-tuple (a1 , . . . , an ) ∈ kn must be the 0 polynomial. Proposition 11.13.3. (Algebraic Independence of Characters) Let K be a finite separable extension of the infinite field k, and let L1 be the normal closure of K over k in an algebraic closure of K. Let σ1 , . . . , σn be the distinct embeddings of K into L1 over k. Then σ1 , . . . , σn are algebraically independent over any extension field L of L1 , in the sense that if f (X1 , . . . , Xn ) is a polynomial over L such that f (σ1 (α), . . . , σn (α)) = 0 for all α ∈ K, then f = 0. Proof Let α1 , . . . , αn be a basis for K as a vector space over k and consider the matrix M = (σi (αj )). Then Lemma 11.13.2 shows M to be invertible. Write fM : Ln → Ln for the linear transformation induced by the matrix M . Then if we use the basis α1 , . . . , αn to identify K with kn , we see that the function which takes α ∈ kn to f (σ1 (α), . . . , σn (α)) is given by the restriction to kn of the composite f ◦ fM . Thus, our assumption on f is that f ◦ fM restricts to 0 on kn ⊂ Ln . But since fM is linear, its component functions are polynomials, so f ◦ fM = 0 by Proposition 7.3.26. But M is invertible, so f = f ◦ fM ◦ fM −1 = 0, as claimed. Finally, we can prove the Normal Basis Theorem. Theorem 11.13.4. (Normal Basis Theorem) Let K be a finite Galois extension of the infinite field k. Then K has a normal basis over k. Proof Let σ1 , . . . , σn be the distinct elements of G = Gal(K/k), with σ1 equal to the identity element. For each σ ∈ G, define a variable Xσ . To get an ordering of the variables, identify Xσi = Xi . Consider the matrix M over K[X1 , . . . , Xn ] whose ij-th entry is Xσi σ−1 . We claim j that M has nonzero determinant. To see this, make the substitution Xσ1 = 1 and Xσi = 0 for i > 1. Under this substitution, M evaluates to the identity matrix. By the naturality of determinants, det M must be nonzero. The determinant, det M , lies in the ground ring K[X1 , . . . , Xn ], so we may write det M = f (X1 , . . . , Xn ), where f (X1 , . . . , Xn ) is a nonzero polynomial in n variables. By Proposition 11.13.3, there is an α ∈ K such that f (σ1 (α), . . . , σn (α)) = 0.


502

By naturality of determinants, f (σ1 (α), . . . , σn (α)) is the determinant of the matrix obtained by substituting σi (α) for each occurrence of Xσi in the matrix M . Thus,

f (σ1 (α), . . . , σn (α)) = det σi σj−1 (α) . By Lemma 11.13.2, σ1−1 (α), . . . , σn−1 (α) forms a basis for K over k, which is clearly a normal basis, as desired. Exercises 11.13.5. 1. Find a normal basis for Q(ζ8 ) over Q. 2. Let K be a finite Galois extension of k and write G = Gal(K/k). Show that the action of G on K defines a k[G]-module structure on K that extends the usual k-module structure. Show that a normal basis of K over k induces an isomorphism of k[G]-modules from k[G] to K.

11.14

Norms and Traces

Let K be a finite extension of k and let α ∈ K. Write μα : K → K for the transformation induced by multiplication by α: μα (β) = αβ. Note that μα is transformation of vector spaces over K, and hence over k as well. We obtain a ring homomorphism, μ : K → Endk (K) by setting μ(α) = μα . The transformation μα contains quite a bit of information about α over k. For instance, if f (X) is a polynomial over k, then, as elements of Endk (K), we have μf (α) = f (μα ). In particular, the minimal polynomial of α over k is equal to the minimal polynomial of μα as a k-linear transformation of K. The easiest things to study about μα are its trace and determinant. Definitions 11.14.1. The norm of K over k, denoted by NK/k , is given by the composite μ

det

K− → Endk (K) −−→ k. The trace of K over k, denoted by trK/k , is given by the composite μ

tr

→ Endk (K) −→ k. K− Here, det and tr are defined as for any finite dimensional vector space over k. The standard properties of determinants and traces give us the following lemma. Lemma 11.14.2. Let K be a finite extension of k. Then trK/k : K → k is a homomorphism of vector spaces over k, while NK/k : K → k is a monoid homomorphism with the property that NK/k (α) = 0 if and only if α = 0. If a ∈ k, then trK/k (a) = [K : k] · a and NK/k (a) = a[K:k] . It can be useful to express the norm and trace in Galois-theoretic terms.


503

Proposition 11.14.3. Let K be a finite extension of k. Let L be an algebraic closure of K and let σ1 , . . . , σk be the collection of all distinct embeddings of K in L over k. Then for α ∈ K we have NK/k (α) trK/k (α)

[K:k]

i (σ1 (α) . . . σk (α)) and [K : k]i · (σ1 (α) + · · · + σk (α)) ,

= =

where [K : k]i is the inseparability degree of K over k. In particular, if K is separable over k, then NK/k (α)

= σ1 (α) . . . σk (α) and = σ1 (α) + · · · + σk (α).

trK/k (α)

Proof Let f (X) = X m + am−1 X m−1 + · · · + a0 be the minimal polynomial of α over k. Then 1, α, . . . , αm−1 is a basis for k(α) over k. Let β1 , . . . , βr be any basis for K over k(α). Then the matrix for multiplication by α with respect to the basis β1 , β1 α, . . . , β1 αm−1 , . . . , βr , βr α, . . . , βr αm−1 is the block sum of r = [K : k(α)] copies of the rational companion matrix, C(f ), of f (X): ⎞ ⎛ 0 0 ... 0 −a0 ⎜ 1 0 ... 0 −a1 ⎟ ⎟ ⎜ ⎟ ⎜ . .. C(f ) = ⎜ ⎟. ⎟ ⎜ ⎝ 0 0 . . . 0 −am−2 ⎠ 0 0 . . . 1 −am−1 It is easy to see, using the decomposition of the determinant in terms of permutations (or via Proposition 10.6.7 and Problem 2 of Exercises 10.4.11), that det(C(f )) = (−1)m a0 . Thus, [K:k(α)]

NK/k (α) = ((−1)m a0 )

.

Even more simply, we have trK/k (α) = [K : k(α)] · (−am−1 ). In Proposition 11.4.2, it is shown that f (X) = (X − α1 )[k(α):k]i . . . (X − αs )[k(α):k]i in L[X], where α1 , . . . , αs are all distinct, and s = [k(α) : k]s . Thus, there are s distinct embeddings, τ1 , . . . , τs , of k(α) in L, characterized by τi (α) = αi . Thus, am−1 a0

= −[k(α) : k]i (τ1 (α) + · · · + τs (α)) =

m

[k(α):k]i

(−1) (τ1 (α) . . . τs (α))

and

.

By Lemma 11.4.10, each τi has [K : k(α)]s distinct extensions to embeddings of K in L. Thus, [K:k(α)]

s and σ1 (α) . . . σk (α) = (τ1 (α) . . . τs (α)) σ1 (α) + · · · + σk (α) = [K : k(α)]s · (τ1 (α) + · · · + τs (α)) .

The result now follows from the fact that [K : k]i = [K : k(α)]i · [k(α) : k]i .


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The norm and trace satisfy an important transitivity property, a special case of which appeared in the preceding proof. Proposition 11.14.4. Let K be a finite extension of k and let L be a finite extension of K. Then NL/k trL/k

= =

NK/k ◦ NL/K trK/k ◦ trL/K .

and

More generally, if V is a finite dimensional vector space over K and if f ∈ EndK (V ), then detk (f ) trk (f )

= =

NK/k (detK (f )) trK/k (trK (f )).

and

Here, for F = K or k, detF (f ) and trF (f ) stand for the determinant and trace, respectively, of f when regarded as a transformation of a vector space over F . Proof Choose a basis v1 , . . . , vn for V over K and a basis α1 , . . . , αk for K over k. For f ∈ EndK (V ), let M = (βij ) be the matrix of f as a K-linear mapping, with respect to the basis v1 , . . . , vn . We may also consider the matrix of f as a k-linear mapping with respect to the kbasis α1 v1 , . . . , αk v1 , . . . , α1 vn , . . . , αk vn . The matrix of f with respect to this basis is a block matrix, where the blocks are the matrices of the k-linear transformations μβij of K, with respect to the basis α1 , . . . , αk . Thus, if φ : K → Mk (k) is the ring homomorphism that carries α ∈ K to the matrix of μα with respect to α1 , . . . , αk , then the matrix of f 5 = (φ(βij )). over k is the block matrix M Note that trK/k (α) = trk (φ(α)) and NK/k (α) = detk (φ(α)) for all α ∈ K. Thus, it 5) = trk (φ(trK (M ))), and detk (M 5) = detk (φ(detK (M ))). suffices to show that trk (M 5 For traces, this is obvious: The trace of M is the sum of the traces of the diagonal blocks φ(β11 ), . . . , φ(βnn ). 5 = (φ(βij )) gives a For determinants, note that the passage from M = (βij ) to M 5 ring homomorphism from Mn (K) to Mnk (k). Clearly, M is invertible if and only if M is invertible. In the non-invertible case, the determinants are 0, so it suffices to check the result on generators of Gln (K). By Proposition 10.8.9, Gln (K) is generated by elementary matrices and matrices of the form (α) ⊕ In−1 , with α ∈ K× . 5 is either upper or lower triangular with only If M is an elementary matrix, then M 5 1’s on the diagonal, so detk (M ) = 1. Since detK (M ) = 1, the desired equality holds. 5 is a block sum, and the result is easy to In the remaining case, M is diagonal and M check. Exercises 11.14.5. 1. Let K be a finite extension of k and let α ∈ K. Show that the matrix of μα (with respect to any k-basis of k) is diagonalizable over the algebraic closure of k if and only if α is separable over k. 2. Let K be a finite inseparable extension of k. Show that trK/k is the trivial homomorphism. 3. Calculate NC/R : C → R and trC/R : C → R explicitly.


505

4. Give √ explicit calculations of the norm and trace over Q of an extension of the form Q( m), where m ∈ Z is not a perfect square. 5. Let K be a finite extension of k and let f (X) be the minimal polynomial of α ∈ K over k. Let a ∈ k. Show that NK/k (a − α) = (f (a))[K:k(α)] .

Chapter 12

Hereditary and Semisimple Rings So far, we’ve been able to classify the finitely generated modules over P.I.D.s, and, via Galois theory, to develop some useful tools toward an understanding of the category of fields and the classification of particular types of extensions of a given field. In general, we’d like to be able to classify particular types of rings, and to classify the finitely generated modules, or even just the finitely generated projective modules, over a particular ring. Such results are hard to come by: Ring theory is incredibly rich and varied. Here, we shall look at generalizations of fields and P.I.D.s and obtain a few classification theorems. As a generalization of the concept of field, we shall study the semisimple rings: those rings with the property that every ideal of the ring is a direct summand of it. We shall develop a structure theory that will allow for a classification of these rings, modulo a classification of division rings. We shall also develop methods for classifying the modules over a semisimple ring. Given Maschke’s Theorem (Section 12.1) these give some of the fundamental techniques in the study of the representation theory of finite groups. The class of rings which simultaneously generalizes both P.I.D.s and semisimple rings is the class of hereditary rings: those with the property that every ideal is a projective module. The hereditary rings are much too broad a class of rings to permit any kind of useful classification results at this stage. Thus, we shall spend most of our time studying the hereditary integral domains, which, it turns out, are precisely the rings known as Dedekind domains (though we shall introduce the latter under a more classical definition). Dedekind domains first came to light in the study of number fields. Here, a number field is a finite extension field of the rational numbers. Number fields have subrings, called their rings of integers, that behave somewhat like the integers do as a subring of Q. Rings of integers are fundamental in the study of number theory. In fact, the false assumption that the ring of integers of a number field is a P.I.D. led to a false proof of Fermat’s Last Theorem. The circle of ideas surrounding this fact led to the first studies of ideal theory. The truth of the matter is that the ring of integers of a number field is a Dedekind domain. As a result, the study of the ideal and module theory of Dedekind domains is

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CHAPTER 12. HEREDITARY AND SEMISIMPLE RINGS

507

important for an understanding of number theory. In fact, the classification of the finitely generated projective modules over these rings of integers is fundamental in solving some of the basic questions in both algebra and topology. In Section 12.1, we prove Maschke’s Theorem, which states that if G is a finite group and K is a field whose characteristic does not divide the order of the group, then the group algebra K[G] is semisimple. We also characterize the finitely generated projective modules over a semisimple or hereditary ring. In Sections 12.2 and 12.3, we develop the theory of semisimple rings. We show that a semisimple ring is a product of matrix rings over division rings, where there is one factor in the product for each isomorphism class of simple left A-modules. We also show that every finitely generated left A-module may be written uniquely as a direct sum of simple modules. A complete determination of the ring and its finitely generated modules depends only on finding representatives for each isomorphism class of simple left A-modules. Applying this to a group algebra K[G] that satisfies the hypothesis of Maschke’s Theorem results in a calculation of the representations of G over K. In the exercises, we determine the simple K[G] modules and construct the associated splittings of K[G] for some familiar examples of groups and fields. We also show that left semisimplicity and right semisimplicity are equivalent. Then, in Section 12.4, we give characterizations of both hereditary and semisimple rings in terms of homological dimension. In Section 12.5, we give some basic material on hereditary rings. Then, in Section 12.6, we define and characterize the Dedekind domains. We give both unstable and stable classifications of the finitely generated projective modules over a Dedekind domain, modulo the calculation of the isomorphism classes of ideals of A. The isomorphism classes of nonzero ideals form a group under an operation induced by the product of ideals. This group is known as the class group, Cl(A). Calculations of class groups are important in ) 0 (A) ∼ numerous branches of mathematics. We show here that K = Cl(A). Section 12.7 develops the theory of integral extensions and defines and studies the ring of integers, O(K), of a number field. By means of a new characterization of Dedekind domains, we show that the rings of integers are Dedekind domains.

12.1

Maschke’s Theorem and Projectives

Definitions 12.1.1. A nonzero ring A is left hereditary if every left ideal is a projective A-module. A nonzero ring A is left semisimple if each left ideal a of A is a direct summand of A. Of course, semisimple rings are hereditary. It is important to underline the distinction between a submodule N ⊂ M being a direct summand and merely being isomorphic to one. (The reader may want to take another look at Definition 9.8.4 and Lemma 9.8.5.) For instance, in a P.I.D., a proper, nonzero ideal (a) is isomorphic to the full ring A, but is not a direct summand of it, as the sequence 0 → (a) → A → A/(a) → 0 fails to split. The point is that A/(a) is a torsion module, and hence cannot embed in A. Examples 12.1.2.


508

1. The simplest example of a hereditary ring is a P.I.D. Here, every ideal is free on one generator. However, as just noted, a P.I.D. is semisimple if and only if it is a field. 2. A product K × L of fields is easily seen to be semisimple: Its nonzero ideals are K × 0 and 0 × L. There are some interesting hereditary domains that are not P.I.D.s. Such domains are what’s known as Dedekind domains, and figure prominently in number theory. We shall study them in Sections 12.6 and 12.7. Since a direct summand of a finitely generated module is finitely generated, we obtain the following important observation. Lemma 12.1.3. Left semisimple rings are left Noetherian. In contrast, there are examples of hereditary rings that are not Noetherian. There is an alternative notion of semisimple rings that appears in the literature. We give the definition here, but defer most of the discussion until later. Definition 12.1.4. A nonzero ring is Jacobson semisimple if its Jacobson radical R(A) = 0. We shall see that a ring is left semisimple if and only if it is both Jacobson semisimple and left Artinian. We shall also see that every left semisimple ring is right semisimple. Thus, in the presence of Jacobson semisimplicity, the left Artinian property implies the right Artinian property. One important application of semisimple rings is to the study of the representations of finite groups. For a finite group G, representation theory concerns the classification of the finitely generated K[G]-modules, for a field K. The relevance of semisimple rings comes from Maschke’s Theorem, which says that if G is a finite group and if K is a field whose characteristic doesn’t divide the order of G, then the group ring K[G] is semisimple. A number of treatments of representation theory bypass this result in favor of a treatment depending on inner products in complex vector spaces. This approach only works for fields of characteristic 0, of course, and it gives some students the false impression that Maschke’s Theorem is something much deeper than it turns out to be. Thus, to illustrate how easy it is to prove Maschke’s Theorem, as well as to motivate the study of semisimple rings, we shall give it in this section. But first, we shall give the fundamental result regarding the finitely generated projective modules over hereditary and semisimple rings. Proposition 12.1.5. Let A be a left hereditary ring. Then every submodule of a finitely generated projective left A-module is projective. In fact, it is isomorphic to a finite direct sum of ideals.


509

If A is semisimple, then every submodule of a finitely generated projective module is a direct summand of it. Proof Since every finitely generated projective module is a direct summand of a free module, it suffices to show that every left submodule M ⊂ An is isomorphic to a finite direct sum of ideals, if A is hereditary, and is additionally a direct summand of An if A is semisimple. We argue by induction on n, with the case n = 1 being immediate from the definitions. Let π : An → A be the projection onto the last factor. Then π(M ) = a, a left ideal of A. Let K be the kernel of π : M → a and identify the kernel of π : An → A with An−1 . Then we have a commutative diagram

0

0

0 / K i / An−1

0 / M j / An

π

π

0 / a k / A

/ 0

/ 0

where the straight lines are all exact. By the inductive hypothesis, K is isomorphic to a finite direct sum of ideals. Since a is projective, the upper short exact sequence splits, and hence M ∼ = K ⊕ a is also isomorphic to a finite direct sum of ideals. Let P be projective and let N ⊂ P . By Proposition 9.8.6 and the definition, we see that ⊂

0 → N −→ P → P/N → 0 splits if and only if P/N is projective. So Lemma 9.8.5 shows that N is a direct summand of P if and only if P/N is projective. Thus, for the result at hand, if A is semisimple and M ⊂ An , it suffices to show that the cokernel of j : M ⊂ An is projective. Write C , C, and C , respectively, for the cokernels of i, j, and k in the above diagram. Then the Snake Lemma (Problem 2 of Exercises 7.7.52) shows that we may complete the above diagram to


0 / K i / An−1 / C 0

0

0

0

0 / M j / An / C 0

π

π

0 / a k / A / C 0

510

/ 0

/ 0

/ 0

so that all straight lines are exact. The semisimplicity of A shows k to be the inclusion of a direct summand, while i may be assumed to be so by induction. Thus, C and C are projective. But the short exact sequence of C’s then splits, and hence C is projective, too. We shall now give the proof of Maschke’s Theorem. We begin with two lemmas. Lemma 12.1.6. Let G be a finite group and let A be a commutative ring. Let M and N be A[G]-modules. Then there is an action of G on HomA (M, N ) given by (g · f )(m) = gf (g −1 m) for g ∈ G, f ∈ HomA (M, N ), and m ∈ M . Each g ∈ G acts via an A-module homomorphism on HomA (M, N ), and the fixed G points HomA (M, N ) are precisely the A[G]-module homomorphisms, HomA[G] (M, N ). Proof We treat the last assertion, leaving the rest to the reader. The point is that A[G] is generated by A and G. Thus, if f is in HomA (M, N ), then it already commutes with the actions of A on M and N , so it lies in HomA[G] (M, N ) if and only if f (gm) = gf (m) for all g ∈ G and m ∈ M . But multiplying both sides by g −1 , this says that (g −1 · f )(m) = f (m) for all g and m. Since passage from g to g −1 is a bijection on G, this is equivalent to f lying in G HomA (M, N ) . The hypothesis in Maschke’s Theorem that the characteristic of K does not divide the order of G will allow us to use the following technique. Lemma 12.1.7. Let A be a commutative ring and let G be a finite group whose order, |G|, maps to a unit under the unique ring homomorphism Z → A. Then for any two A[G]-modules, there is a retraction R : HomA (M, N ) → HomA[G] (M, N ) of A-modules given by 1 1 (g · f ), so that (R(f ))(m) = gf (g −1 m) R(f ) = |G| |G| g∈G

g∈G


511

for f ∈ HomA (M, N ) and m ∈ M . Proof Looked at in terms of group actions, the result is nearly immediate. It is also useful to compute it by hand. If f ∈ HomA[G] (M, N ), then 1 1 1 (R(f ))(m) = |G|f (m) gf (g −1 m) = f (m) = |G| |G| |G| g∈G

g∈G

for all m ∈ M , so R(f ) = f . For an arbitrary f ∈ HomA (M, N ) and x ∈ G, we have 1 (R(f ))(xm) = gf (g −1 xm) |G| g∈G

=

1 x(x−1 g)f ((x−1 g)−1 m) |G| g∈G

1 = x |G|

(x−1 g)f ((x−1 g)−1 m)

x−1 g∈G

= x(R(f ))(m). So R(f ) commutes with the action of G. As noted above, because it also commutes with the action of A, R(f ) is an A[G]-module homomorphism. The reader may easily verify that R is an A-module homomorphism between the Hom groups. The trick of adding things up over the elements of the group and then dividing by the order of G is an averaging technique akin to integration. In fact, the exact same argument applies, using the Haar integral for the averaging process, in the case where M and N are the spaces associated to representations of a compact Lie group. One obtains a result precisely analogous to the one we shall derive below for finitely generated modules over the group algebra of a finite group: Every finite dimensional representation over R or C of a compact lie group is a direct sum of irreducible representations. But we are getting ahead of the presentation here, so let us now prove Maschke’s Theorem. Theorem 12.1.8. (Maschke’s Theorem) Let G be a finite group and let K be a field whose characteristic does not divide the order of G. Then K[G] is semisimple. In fact, if N is a K[G]-submodule of any K[G]-module M , then N is a direct summand of M as a K[G] module. Proof The semisimplicity of K[G] follows from the statement of the second paragraph by taking M = K[G]. In fact, the second paragraph also follows from the semisimplicity of K[G], as we shall see in Proposition 12.4.1. Thus, suppose given an inclusion i : N ⊂ M of K[G]-modules. By Lemma 9.8.5, it suffices to produce a retraction for i, i.e., a K[G]-module homomorphism f : M → N that restricts to the identity map on N . Of course, any inclusion of vector spaces over K admits such a retraction as vector spaces, by Corollary 7.9.15. So let f : M → N be a homomorphism of K-vector spaces that restricts to the identity map on N . Since the characteristic of K does not divide the order of G, we may apply Lemma 12.1.7, producing a K[G]-module homomorphism R(f ) : M → N with 1 (R(f ))(m) = gf (g −1 m) |G| g∈G


512

for all m ∈ M . Applying this to n ∈ N , we see that f (g −1 n) = g −1 n since f is a retraction, so (R(f ))(n) = (1/|G|)|G|n = n, so R(f ) is a retraction of K[G]-modules. Exercises 12.1.9. 1. Show that the product of a finite set of hereditary rings is hereditary, and that the product of a finite set of semisimple rings is semisimple. 2. Write Z2 = T = {1, T } as a multiplicative group. Show that Z2 [Z2 ] = Z2 [T ] is not semisimple by showing that the ideal generated by 1 + T is not a direct summand. Deduce that the hypothesis that the characteristic of K does not divide the order of G is necessary in Maschke’s Theorem. 3. Write Z2 = T and let K be a field of characteristic = 2. Let a be the ideal of K[Z2 ] generated by 1 + T . Write down an explicit direct sum decomposition K[Z2 ] ∼ = a ⊕ b. Note that b is one-dimensional as a vector space over K. What effect does T have on a nonzero element of b? 4. A projection operator on an A-module M is an A-module homomorphism f : M → M such that f ◦ f = f . Show that a submodule N ⊂ M is a direct summand if and only if it is the image of a projection operator. Show that the complementary summand may be taken to be the image of 1M − f , which is also a projection operator. Deduce that a finitely generated module is projective if and only if it is the image of a projection operator on An for some n. 5. An element x ∈ A is idempotent if x2 = x. Show that a left ideal a ⊂ A is a direct summand of A if and only if it is the principal left ideal generated by an idempotent in A. 6. Show that A splits as a product of rings if and only if there is an idempotent element other than 0 or 1 in the center of A.

12.2

Semisimple Rings

For semisimple rings, Proposition 12.1.5 has strong implications. Corollary 12.2.1. Let A be a left semisimple ring. Then every finitely generated left A-module is projective. Proof Let M be a finitely generated left A-module. Then there is a surjection f : An → M , and hence an exact sequence ⊂

f

→ M → 0. 0 → K −→ An − Proposition 12.1.5 now shows the inclusion of K in An to be that of a direct summand, and hence the exact sequence splits. Thus, M is isomorphic to a direct summand of An , and hence is projective. Of course, we’d now like to develop techniques to classify the finitely generated modules over a semisimple ring. Indeed, by the end of this section, we shall obtain a complete classification, modulo the identification of the isomorphism classes of minimal nonzero left ideals. Note that A need not be semisimple in the next definition. Definition 12.2.2. An A-module M is semisimple if each of its submodules is a direct summand of it.


513

Clearly, a left semisimple ring is a ring that is semisimple as a left module over itself. Note that if N ⊂ N ⊂ M are inclusions of submodules and if N is a direct summand of M , then it must also be a direct summand of N : A retraction r : M → N of the inclusion of N restricts to a retraction r : N → N . Lemma 12.2.3. A submodule of a semisimple module is semisimple. Recall that a nonzero A-module M is simple, or irreducible, if it has no submodules other than 0 and itself. We shall use the next lemma as part of a characterization of semisimple modules. Lemma 12.2.4. Suppose that the A-module M is generated by simple submodules in the sense that there is a family {Mi | i ∈ I} of simple submodules of M such that i∈I Mi = M . Then there is a subset J ⊂ I such that M is the internal direct sum of the Mj for j ∈ J. Proof Consider the set, S, of all subsets * J ⊂ I with the property that the sum j∈J Mj is direct in the sense that the map j∈J Mj → M induced by the inclusions of the Mj is injective. We claim that Zorn’s Lemma shows that S has a* maximal element, J. To see this, note that for any J ⊂ I, if the natural map j∈J Mj → M fails to be injective, then there is a finite subset j1 , . . . , jk of J such that the natural map Mj1 ⊕ · · · ⊕ Mjk → M fails to inject. But if J is the union of a totally ordered subset F ⊂ S, then any finite subset j1 , . . . , jk of J must lie in one of the elements of F , so this cannot happen. Thus, F is bounded above in S, and the hypothesis of Zorn’s * Lemma is satisfied. Let J be a maximal element of S. We claim that j∈J Mj → M is onto. If not, then since M is generated by the submodules Mi , there must be an i such that Mi is not contained in j∈J Mj . But then, sinceMi is simple, we must have Mi ∩ j∈J Mj = 0. Thus, if we take J = J ∪ {i}, the sum j∈J Mj is easily seen to be direct, contradicting the maximality of J. Thus, M is the internal direct sum of the submodules Mj for j ∈ J. Lemma 12.2.5. A nonzero semisimple module, M , contains a simple module. Proof Let 0 = m ∈ M . If Am is contained in every nonzero submodule of M , then Am is simple, and we’re done. Otherwise, let N be maximal in the set of submodules of M that do not contain m. Since M is semisimple, it is an internal direct sum M = N + N . We claim that N is simple. Suppose it isn’t. By Lemma 12.2.3, it is semisimple, so it may be written as the internal direct sum of two nonzero submodules: N = P + Q. Thus, M is an internal direct sum M = N +P +Q. But the maximality of N shows that m must lie in both N +P and N + Q, contradicting the statement that the summation N + P + Q is direct. We can now characterize semisimple modules. Proposition 12.2.6. Let M be a nonzero A-module. Then the following conditions are equivalent. 1. M is a direct sum of simple modules. 2. M is a sum of simple submodules.


514

3. M is semisimple. Proof The first condition obviously implies the second, and the second implies the first by Lemma 12.2.4. We shall show that the second condition implies the third. Thus, suppose that M = i∈I Mi , where each Mi ⊂ M is simple. Let N ⊂ M be any submodule. We shall show that N is a direct summand of M . Note that Mi ∩ N is a submodule of the simple module Mi , and hence either Mi ⊂ N or Mi ∩ N = 0. Since the Mi generate M , if N = M , then there must be some indices i such that Mi ∩ N = 0. Thus, suppose N = M , and put a partial ordering by inclusion on the collection of subsets J ⊂ I with the property that N ∩ j∈J Mj = 0. Then Zorn’s Lemma shows that there is a maximal subset with this property. We claim that if J is a maximal such subset, then M is the internal direct sum of N and j∈J Mj . To see this, since N ∩ j∈J Mj = 0, it suffices to show that N + j∈J Mj = M . But the maximality of J shows that if i ∈ J, then N ∩ (Mi + j∈J Mj ) = 0, and hence Mi ∩ (N + j∈J Mj ) = 0. By the simplicity of Mi , we see that Mi ⊂ N + j∈J Mj . Since the Mi generate M , N + j∈J Mj = M , and hence the second condition implies the third. We now show that the third condition implies the second. Thus, suppose that M is semisimple. Let {Mi | i ∈ I} be the set of all simple submodules of M and let N = M . If N = M , then M is an internal direct sum M = N + N , with N = 0. By i i∈I Lemmas 12.2.3 and and 12.2.5, N contains a simple module M , which must be equal to one of the Mi , as those are all the simple submodules of M . But then, M ⊂ N ∩ N , which contradicts the directness of the summation M = N + N . Thus, M = N , and M is generated by simple modules. When we apply Proposition 12.2.6 to A itself, we get a nice bonus. Corollary 12.2.7. A left semisimple ring, A, is left Artinian, and is a finite direct sum of simple left ideals. In addition, the Jacobson radical, R(A) = 0, so that A is Jacobson semisimple. Proof Proposition 12.2.6 shows that A is a possibly infinite direct sum of simple left ideals. But a ring cannot be an infinite direct sum of ideals, as 1 will have only finitely many nonzero components with respect to this decomposition, and any multiple of 1 will involve only those summands. Thus, A is a finite direct sum of simple left ideals. Clearly, simple modules are Artinian, and by Lemma 7.8.7, any finite direct sum of Artinian modules is Artinian, so A is a left Artinian ring. Write A as an internal direct sum of simple left ideals: A = a1 + · · · + an . Let Ni = j =i aj . Then A/Ni is isomorphic to ai as an A-module. Since the submodules of A/Ni are in one-to-one correspondence with the submodules of A containing Ni , the Ni must be maximal left ideals. But ∩ni=1 Ni = 0, so R(A) = 0, as claimed. We can also use Proposition 12.2.6 to recognize the semisimplicity of a given ring. Corollary 12.2.8. Let D be a division ring and let n > 0. Then Mn (D) is semisimple. Proof Let ai ⊂ Mn (D) be the set of matrices whose nonzero entries all lie in the i-th column. Then D is the direct sum of the ideals ai , each of which is isomorphic to Dn , the space of column vectors.


515 ×

For any nonzero vector v ∈ Dn , there is a unit M ∈ Mn (D) with M e1 = v. So any Mn (D)-submodule of Dn that contains v must contain e1 , as well as any other nonzero element of Dn , so Dn is a simple module over Mn (D). We can now give a complete classification of the finitely generated modules over a semisimple ring, modulo the determination of the isomorphism classes as A-modules of the simple left ideals. The next lemma will play an important role. Lemma 12.2.9. (Schur’s Lemma) Let M be a simple module over the ring A. Then the endomorphism ring EndA (M ) is a division ring. If N is another simple module and if M and N are not isomorphic as A-modules, then HomA (M, N ) = 0. Proof It suffices in both cases to show that any nonzero homomorphism between simple modules M and N is an isomorphism (and hence a unit in EndA (M ) if M = N ). Thus, let f : M → N be a homomorphism of simple modules. If f = 0, then the kernel of f is not M . But since M is simple, the only other possibility is ker f = 0. Similarly, since N is simple, the image of f is either 0 or N , so if f is nontrivial, it is onto. Corollary 12.2.10. A left semisimple ring A has only finitely many isomorphism classes of simple modules, each of which is isomorphic to a simple left ideal. In consequence, every semisimple A-module is isomorphic to a direct sum of ideals, and hence is projective. Proof Let M be a simple A-module and let f : A → M be a nontrivial A-module homomorphism. By the simplicity of M , f is onto. Thus, M is finitely generated, and hence projective by Corollary 12.2.1. Thus, f admits a section, and hence M is isomorphic to a direct summand of A, and hence to a simple left ideal. By Proposition 12.2.6, we see that semisimple A-modules are isomorphic to direct sums of simple left ideals. Corollary 12.2.7 gives an A-module isomorphism A∼ = a1 ⊕ · · · ⊕ ak from A to a direct sum of simple left ideals. By Schur’s Lemma, if a is a simple left ideal of proj A, then at least one of the composites a ⊂ A −−→ ai must be an isomorphism: Otherwise, the inclusion a ⊂ A is the zero map. Thus, there are finitely many isomorphism classes of simple left ideals. In some cases, we obtain a decomposition A ∼ = a1 ⊕ · · · ⊕ ak of A as a direct sum of simple left ideals that are all isomorphic to one another as A-modules. For example, the proof of Corollary 12.2.8 shows that if D is a division ring, then Mn (D) is a direct sum of simple left ideals each of which is isomorphic to the space of column vectors, Dn . In this case, the above argument shows that every simple left ideal of Mn (D) is isomorphic to Dn . Thus, Mn (D) is an example of a simple ring: Definition 12.2.11. A simple ring is a semisimple ring with only one isomorphism class of simple left ideals. Summarizing the discussion of matrix rings, we have: Corollary 12.2.12. Let D be a division ring. Then Mn (D) is a simple ring for n ≥ 1.


516

Corollary 12.2.13. Let a and b be simple left ideals in the ring A and let M be a left A-module that is a sum of simple submodules all of which are isomorphic as modules to b. If a and b are not isomorphic as A-modules, then any A-module homomorphism f : a → M is trivial, a ⊂ Ann(M ), and hence aM = 0. Alternatively, if A is semisimple and a and b are isomorphic as A-modules, then aM = M . * Proof First note that by Lemma 12.2.4, M is a direct sum M = i∈I Mi , where each Mi is isomorphic to b. Suppose, then, that a and b are not isomorphic and let f : a → M be an A-module homomorphism. Then if πi : M → Mi is the projection onto the i-th summand, Schur’s Lemma shows that πi ◦ f = 0. Since a map into a direct sum is determined by its coordinate functions, f = 0. Now let m ∈ M . Then there is an A-module homomorphism f : a → M given by f (a) = am. In particular, if a is not isomorphic to b, f = 0, and hence a ⊂ Ann(m). Since this is true for all m ∈ M , a ⊂ Ann(M ). * Now suppose that a and b are isomorphic and that A is semisimple. We have aM = i∈I aMi , so it suffices to show that aMi = Mi for all i. The hypothesis now is that Mi is isomorphic to a, so let f : a → Mi be an isomorphism. Since A is semisimple, a is a direct summand of A, so f extends to an A-module homomorphism f : A → Mi . Since f is an A-module homomorphism out of A, there is an element m ∈ Mi such that f(x) = xm for all x ∈ A. Since f restricts to an isomorphism from a to Mi , am = Mi , so aMi = Mi , and hence aM = M , as claimed. In particular, if M is a sum of copies of a particular simple left ideal, a, then every simple submodule of M must be isomorphic to a. We can now split a semisimple ring into a product of simple rings, one for each isomorphism class of simple left ideals. We shall give a slicker proof of the following proposition in Section 12.3. We include it now, as a hands-on argument in instructive. Proposition 12.2.14. Let A be a semisimple ring and let a1 , . . . , ak be representatives of the distinct isomorphism classes of simple left ideals in A. Let bi ⊂ A be the sum of all the left ideals of A isomorphic to ai . Then bi is a two-sided ideal for each i, and A is the internal direct sum A = b1 ⊕ · · · ⊕ b k . Also, there are simple rings Bi for 1 ≤ i ≤ k and an isomorphism of rings ∼ =

f : A −→ B1 × · · · × Bk ∼ =

such that f restricts to an isomorphism fi : bi −→ Bi of A-modules for all i. In particular, fi (ai ) represents the single isomorphism class of simple left ideals of Bi . Proof By Proposition 12.2.6, A is a sum of simple left ideals, so A = b1 + · · · + bk . Since the simple left ideals generate A, bi will be a two-sided ideal if and only if bi c ⊂ bi for every simple left ideal c. If c ∼ = ai , then c ⊂ bi by the definition of the latter. Since c is a left ideal, bi c ⊂ c ⊂ bi , as desired. Suppose, then, that c ∼ ai . Since b is generated by ideals isomorphic to ai , bc = 0 by = Corollary 12.2.13. Thus bi is a two-sided ideal.


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We claim now by induction on i that each sum b1 + · · · + bi is direct. This is certainly true for i = 1. Assuming the result for i − 1, write a = (b1 + · · · + bi−1 ) ∩ bi . Then the result can only be false for i if a = 0. But a ⊂ bi , so if a = 0, then its simple left ideals must all be isomorphic to ai by Corollary 12.2.13. So another application of Corollary 12.2.13 shows that ai a = a. But a ⊂ b1 + · · · + bi−1 , so Corollary 12.2.13 also shows that a1 a = 0. Thus, A is the direct sum of the bi , as claimed. Clearly, bi bj = 0 for i = j. (This follows simply because bi and bj are two-sided ideals with bi ∩ bj = 0.) Thus, suppose that a = a1 + · · · + ak and b = b1 + · · · + bk are arbitrary elements of A, with ai , bi ∈ bi for all i. Then ab = a1 b1 + · · · + ak bk . Thus, if we write 1 = e1 + · · · + ek with ei ∈ bi for all i, then each ei acts as a twosided identity if we restrict the multiplication to the elements of bi alone. Thus, there is a ring Bi whose elements and whose operations are those of bi , and whose multiplicative identity element is ei . Clearly, A is isomorphic as a ring to the product of the Bi . Note that the A-module structure on bi pulls back from the obvious Bi -module structure, so ai is a simple Bi module, and Bi is a sum of copies of ai . Thus, Bi is a simple ring, as claimed. The next lemma is very easy, but may not be part of everyone’s developed intuition. Lemma 12.2.15. Let M be a left module over the ring A and let M n be the direct sum of n copies of M . Then EndA (M n ) is isomorphic to the ring of n × n matrices over EndA (M ). In particular, if M is a simple A-module and if D = EndA (M ), then EndA (M n ) is isomorphic to the ring of n × n matrices over the division ring D. Proof If f : M n → M n is an A-module homomorphism, then the matrix associated to f has ij-th entry given by the composite ιj

f

π

i M, → M n −→ M −→ M n −

where ιj is the inclusion of the j-th summand of M n and πi is the projection onto the i-th summand. The matrices then act on the left of M n in the usual manner, treating the n-tuples in M n as column vectors with entries in M . When A is commutative, the preceding argument is a straightforward generalization of the proof that the endomorphism ring of An is isomorphic to Mn (A). For noncommutative rings, the representation of EndA (An ) that’s given by Lemma 12.2.15 is different from the one we’re used to. The point is that the evaluation map ∼ =

ε : EndA (A) −→ A that takes f to f (1) is an anti-homomorphism: ε(f g) = ε(g) · ε(f ), so that EndA (A) is isomorphic to the opposite ring, Aop , of A. Thus, Lemma 12.2.15 gives an isomorphism from EndA (An ) to Mn (Aop ). However, the presence of opposite rings here should come as no surprise: The more customary method of representing the transformations of left free modules by right matrix multiplication induces an isomorphism from EndA (An ) to the opposite ring of Mn (A),


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as discussed in Proposition 7.10.22. (Indeed, some algebraists write homomorphisms of left modules on the right and use the opposite multiplication as the default in discussing composites and endomorphism rings.) The situation is resolved by the next lemma. Lemma 12.2.16. Passage from a matrix to its transpose induces an isomorphism op Mn (Aop ) ∼ = Mn (A) .

We shall now classify the finitely generated modules over a semisimple ring. Theorem 12.2.17. Let A be a semisimple ring and let a1 , . . . , ak represent the distinct isomorphism classes of simple left ideals of A (and hence of simple modules, by Corollary 12.2.10). Let M be a finitely generated module over A. Then there is a uniquely determined sequence r1 , . . . , rk of non-negative integers with the property that r M∼ = ar11 ⊕ · · · ⊕ akk .

Proof Corollary 12.2.1 shows that every finitely generated A-module is projective, so Proposition 12.1.5 shows that it is isomorphic to a finite direct sum of ideals. Lemma 12.2.3 shows that a nonzero ideal is semisimple as an A-module, and hence is a direct sum of simple ideals. Since semisimple rings are Noetherian, the direct sum must be finite. Thus, it suffices to show that if s ar11 ⊕ · · · ⊕ arkk ∼ = as11 ⊕ · · · ⊕ akk ,

then ri = si for all i. Now Corollary 12.2.13 shows that if we multiply the two modules on the left by the ∼ = ideal ai , we get an isomorphism ari i −→ asi i . Thus, we can argue one ideal at a time. In particular, it suffices to show that if a is any simple left ideal of A and if ∼ =

f : ar −→ as , then r = s. We shall apply the technique of proof of Lemma 12.2.15. Let D = EndA (a) and let Mf be the s × r matrix over D whose ij-th entry is the composite ιj

f

π

i a −→ ar − → as −→ a.

Then it is easy to see that f is induced by matrix multiplication by Mf . Similarly, f −1 is represented by an r × s matrix, Mf −1 , with entries in D. Since the passage from A-module homomorphisms to matrices is easily seen to be bijective, the fact that f and f −1 are inverse isomorphisms shows that Mf Mf −1 = Is and Mf −1 Mf = Ir . Since the rank of a D-module is well defined, this implies r = s. This implies the following, weaker result. Corollary 12.2.18. Let A be a semisimple ring and let a1 , . . . , ak be representatives of the isomorphism classes of the simple left ideals of A. Then K0 (A) is the free abelian group on the classes [a1 ], . . . , [ak ].


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Proof Theorem 12.2.17 shows that the monoid, F P (A), of isomorphism classes of finitely generated projectives over A is isomorphic to the submonoid of the free abelian group on [a1 ], . . . , [ak ] consisting of the sums r1 [a1 ] + · · · + rk [ak ] whose coefficients ri are all non-negative. (In other words, F P (A) is the free abelian monoid on [a1 ], . . . , [ak ].) The universal property of the Grothendieck construction gives a homomorphism from K0 (A) to the free abelian group on [a1 ], . . . , [ak ] that is easily seen to provide an inverse to the obvious map in the other direction. We can use Lemma 12.2.15 to give a version of what’s known as Wedderburn’s Theorem. We will give another version in Section 12.3. Theorem 12.2.19. Let A be a simple ring, with simple left ideal a. Let D = EndA (a). Then A is isomorphic as a ring to Mn (Dop ), where n is the number of copies of a that appear in an A-module isomorphism A ∼ = an . Thus, a ring A is simple if and only if it is isomorphic to Mn (D) for some n ≥ 1 and a division ring D. Proof If A ∼ = an , then Lemma 12.2.15 gives us an isomorphism Aop ∼ = EndA (an ) ∼ = Mn (D). = EndA (A) ∼ ∼ Mn (Dop ). Here, D is a division ring by Lemma 12.2.16 now gives an isomorphism A = op Schur’s Lemma, hence so is D . Conversely, Corollary 12.2.12 shows that Mn (D) is simple for any division ring D. Now let A be semisimple, and let a1 , . . . , ak represent the isomorphism classes of simple left ideals in A. Then Proposition 12.2.14 produces simple rings Bi for 1 ≤ i ≤ k such that A ∼ = B1 × · · · × Bk as a ring, and such that ai is isomorphic as an A-module to the simple left ideal ai of Bi . Note that since the map A → Bi induced by the product projection is surjective, pulling back the action gives an isomorphism ∼ =

EndBi (ai ) −→ EndA (ai ). Corollary 12.2.20. Let A be a semisimple ring and let a1 , . . . , ak represent the isomorphism classes of simple left ideals in A. Let Di = EndA (ai ) and let ni be the exponent of ai occurring in an A-module isomorphism n A∼ = an1 1 ⊕ · · · ⊕ ak k .

Then there is a ring isomorphism A∼ = Mn1 (D1 op ) × · · · × Mnk (Dk op ). Since finite products of semisimple rings are semisimple, we see that a ring is semisimple if and only if it is isomorphic to a finite product of matrix rings over division rings. Proof The only thing missing from the discussion prior to the statement is the identification of the integers ni . But this follows from the decomposition A ∼ = b 1 ⊕ · · · ⊕ bk of A as a direct sum of two-sided ideals that’s given in Proposition 12.2.14. Each bi is isomorphic to Bi as an A-module, and is a direct sum of ideals isomorphic to ai . So the number of copies of ai that appear in a direct sum decomposition as above for A is equal to the number of copies of ai that occur in such a decomposition of Bi .


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The entire theory of this section goes through, of course, for right modules. The end result is the same: products of matrix rings over division rings. Such matrix rings are both left and right simple. Corollary 12.2.21. A ring is left semisimple if and only if it is right semisimple and is left simple if and only if it is right simple.


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Exercises 12.2.22. 1. Show that a semisimple ring is commutative if and only if it is a product of fields. Show that the decomposition of a commutative semisimple ring A as a direct sum of simple ideals has the form A∼ = a1 ⊕ · · · ⊕ ak , where a1 , . . . , ak are simple left ideals all lying in distinct isomorphism classes. 2. Corollary 8.8.9 gives an explicit isomorphism from Q[Zn ] to a product of fields. Give analogous decompositions for R[Zn ] and C[Zn ]. 3. What are the idempotents in Q[Z2r ]? 4. Let A be any ring and let M be a finitely generated semisimple A-module. Suppose that r M∼ = M1r1 ⊕ · · · ⊕ Mk k ,

where M1 , . . . , Mk are mutually nonisomorphic simple A-modules. Show that EndA (M ) ∼ = Mr1 (D1 ) × · · · × Mrk (Dk ), where Di = EndA (Mi ). 5. Give an example of an Artinian ring A with a simple left ideal a such that aa = 0. Deduce from Corollary 12.2.13 that A cannot be semisimple.

12.3

Jacobson Semisimplicity

Recall that a nonzero ring is Jacobson semisimple if the Jacobson radical R(A) = 0 and is Jacobson simple if it has no two-sided ideals other than 0 and A. Recall from Corollary 12.2.7 that every semisimple ring is Jacobson semisimple, as well as being Artinian. Lemma 12.3.1. A simple ring A is Jacobson simple. Proof Let b be a nonzero two-sided ideal of A. By Lemmas 12.2.3 and 12.2.5, there is a simple left ideal a ⊂ b. As A is simple, it is a sum of simple left ideals isomorphic to a, and hence aA = A by Corollary 12.2.13. Since b is a two-sided ideal, aA ⊂ b, and the result follows. Jacobson semisimplicity is a very weak condition all by itself. The next example shows, among other things, that Jacobson semisimple rings need not be Artinian. Example 12.3.2. Let A be a P.I.D. with infinitely many prime ideals. Then unique factorization shows that the intersection of all the maximal ideals of A is 0, so A is Jacobson semisimple. Note that each of the quotient modules A/(p) with (p) prime is a simple module, so that A has infinitely many isomorphism classes of simple modules. We shall make use of the following concept, which could have been given much earlier. Definition 12.3.3. An A-module M is faithful if AnnA (M ) = 0.


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Since the annihilator of a module is a two-sided ideal, every nonzero module over a Jacobson simple ring is faithful. And every nonzero ring admits a simple module, as, if m is a maximal left ideal in the ring A, then Lemma 9.1.4 shows A/m to be a simple module. The next lemma now follows from Lemma 9.1.2, which shows every proper left ideal to be contained in a maximal one. Lemma 12.3.4. A Jacobson simple ring admits a faithful simple module. We’ve seen that an A-module structure on an abelian group M is equivalent to specifying a ring homomorphism ϕ : A → EndZ (M ). Here, for a ∈ A, the associated homomorphism ϕ(a) : M → M is given by ϕ(a)(m) = am. If we examine this more carefully, we can get additional information. First, note that if M is an A-module, then EndA (M ) is a subring of EndZ (M ), and that M is an EndA (M )-module via f · m = f (m) for f ∈ EndA (M ) and m ∈ M . Now write B = EndA (M ), and note that for a ∈ A, ϕ(a) is a B-module homomorphism. We obtain the following lemma. Lemma 12.3.5. Let M be an A-module and let B = EndA (M ). Then there is a canonical ring homomorphism ϕ : A → EndB (M ) given by ϕ(a)(m) = am for all a ∈ A and m ∈ M . In particular, the kernel of ϕ is AnnA (M ). Recall from Schur’s Lemma that if M is a simple module over the ring A, then EndA (M ) is a division ring. The next lemma now follows from Lemma 12.3.5. Lemma 12.3.6. Let M be a faithful simple module over the ring A, and let D = EndA (M ). Then the canonical map ϕ : A → EndD (M ) gives an embedding of A into the endomorphism ring of the vector space M over the division ring D. We shall show that if A is Artinian and M is a faithful simple A-module, then M is finite dimensional as a vector space over D and the map ϕ : A → EndD (M ) is an isomorphism. Since EndD (M ) is isomorphic to the ring of n × n matrices over a division ring, this will show that A is simple. We shall prove this using Jacobson’s Density Theorem (Theorem 12.3.8). The next lemma is a key step. Note that M here is semisimple, rather than simple. Lemma 12.3.7. Let M be a semisimple A-module and let B = EndA (M ). Then for any f ∈ EndB (M ) and any nonzero m ∈ M , there is an a ∈ A with f (m) = am. Proof This would be immediate if M were simple, as then Am = M . More generally, Am is a direct summand of M as an A-module, so there is an A-module homomorphism r : M → Am that is a retraction of M onto Am, i.e., if i : Am → M is the inclusion, then r ◦ i is the identity on Am. Now i ◦ r : M → M is an A-module homomorphism, hence an element of B = EndA (M ). Thus, f commutes with the action of i ◦ r, so f (am) = f ((i ◦ r)(am)) = (i ◦ r)f (am) for all a ∈ A, and hence f (Am) ⊂ Am. But this says f (m) ∈ Am, and the result follows. Theorem 12.3.8. (Jacobson Density Theorem) Let M be a simple A-module and let D = EndA (M ). Let f ∈ EndD (M ). Then for any finite subset m1 , . . . , mn of M , there is an a ∈ A such that f (mi ) = ami for all i.


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Proof Recall from Lemma 12.2.15 that EndA (M n ) is isomorphic to the ring of n × n matrices Mn (D), which acts by matrix multiplication on M n , where elements of M n are treated as n × 1 column vectors with entries in M . Write f n : M n → M n for the direct sum of n copies of f : Written horizontally, n f (x1 , . . . , xn ) = (f (x1 ), . . . , f (xn )) for each element (x1 , . . . , xn ) ∈ M n . Since the action of each d ∈ D commutes with that of f , it is easy to see that the action of a matrix (dij ) ∈ Mn (D) on M n commutes with the action of f n . In other words, if we set B = EndA (M n ), then f n ∈ EndB (M n ). Applying Lemma 12.3.7 to the element (m1 , . . . , mn ) ∈ M n , we see there’s an element a ∈ A such that a(m1 , . . . , mn ) = f n (m1 , . . . , mn ), i.e., (am1 , . . . , amn ) = (f (m1 ), . . . , f (mn )). Corollary 12.3.9. Suppose that the ring A has a faithful simple module M and let D = EndA (M ). Then A is left Artinian if and only if M is finite dimensional as a vector space over D. If M is an n-dimensional vector space over D, then the canonical map ϕ : A → EndD (M ) is an isomorphism. Since EndD (M ) is isomorphic to Mn (Dop ), we obtain that A is a simple ring. Moreover, A ∼ = M n as an A-module. Proof Lemma 12.3.6 shows that ϕ is injective. Suppose that M is a finite dimensional D-module, with basis m1 , . . . , mn . Then the Jacobson Density Theorem shows that for any f ∈ EndD (M ), there is an element a ∈ A such that f (mi ) = ami for all i. Since m1 , . . . , mn generate M over D, this says that f agrees everywhere with multiplication by a, and hence f = ϕ(a). Thus, ϕ : A → EndD (M ) is an isomorphism. op Now Proposition 7.10.22 gives an isomorphism from EndD (M ) to Mn (D) , and hence op to Mn (D ) by Lemma 12.2.16. Thus, A is left Artinian and simple. Now, Mn (Dop ) is a direct sum of n copies of its simple left ideal, so the same is true for A. The simple left ideal of A is isomorphic to M by Corollary 12.2.10, so A ∼ = M n. Conversely, suppose that M is an infinite dimensional vector space over D, and let m1 , m2 , . . . , mn , . . . be an infinite subset of M that is linearly independent over D. Then for each n > 1, there is an element of EndD (M ) that vanishes on m1 , . . . , mn−1 and is nonzero on mn . By the Jacobson Density Theorem, there is an a ∈ A that annihilates each of m1 , . . . , mn−1 , but does not annihilate mn . Thus, the inclusion AnnA (m1 , . . . , mn−1 ) ⊃ AnnA (m1 , . . . , mn ) is proper. Here, the annihilator of a finite set is taken to be the intersection of the annihilators of its elements. In particular, we have constructed an infinite descending chain of proper inclusions of left ideals of A, and hence A cannot be left Artinian if M is infinite dimensional over D. We obtain a characterization of simple rings. Theorem 12.3.10. (Wedderburn Theorem) Let A be a ring. Then the following conditions are equivalent. 1. A is Jacobson simple and left Artinian. 2. A is Jacobson simple and right Artinian. 3. A is left (resp. right) Artinian and has a faithful simple left (resp. right) module.


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4. A is simple. 5. A is isomorphic to the ring of n × n matrices over a division ring. Proof Of course, we’ve already seen in Theorem 12.2.19 that a ring is left simple if and only if it is right simple and that the last two conditions are equivalent. That will also come out of what we’ve done in this section, together with the fact, from Corollary 12.2.12, that the ring of n × n matrices over a division ring is both left and right simple. Now simple rings are Jacobson simple by Lemma 12.3.1 and Artinian by Corollary 12.2.7, so the fourth condition for left simple rings implies the first. Lemma 12.3.4 shows that the first condition implies the left version of the third, which implies the fifth by Corollary 12.3.9. The right side implications are analogous and the result follows. We now consider the Jacobson semisimple rings. Lemma 12.3.11. Let A be a left Artinian Jacobson semisimple ring. Then there is a finite collection M1 , . . . , Mk of simple left A-modules such that k

Ann(Mi ) = 0.

i=1

Proof Let {Mi | i ∈ I} be a set of representatives for the isomorphism classes of simple left A-modules. (The isomorphism classes form a set because every simple left A-module is isomorphic to A/m for some maximal left ideal m of A.) Then Proposition 9.1.7 shows that Ann(Mi ). R(A) = i∈I

Since A is left Artinian, we claim that R(A) is the intersection of a finite collection of the Ann(Mi ). To see this, note that the Artinian property shows the collection of finite k intersections of the ideals Ann(Mi ) has a minimal element. But if a = j=1 Ann(Mij ) is a minimal such intersection, then a ∩ Ann(Mi ) = a for all i. But then a ⊂ Ann(Mi ) for all i, and hence a = R(A). Since A is Jacobson semisimple, R(A) = 0, and the result follows. The key step is the following proposition. Proposition 12.3.12. Let A be a left Artinian Jacobson semisimple ring. Let M1 , . . . , Mk be pairwise non-isomorphic simple left A-modules such that k

Ann(Mi ) = 0.

i=1

Suppose also that this intersection is irredundant in the sense that for no proper subcollection of M1 , . . . , Mk will the intersection of the annihilators be 0. Let Di = EndA (Mi ). Then each Mi is a finite dimensional vector space over the division ring Di , and we have a ring isomorphism A∼ = Mn1 (D1 op ) × · · · × Mnk (Dk op ),


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where ni is the dimension of Mi as a vector space over Di . We obtain an A-module isomorphism n A∼ = M1n1 ⊕ · · · ⊕ Mk k .

In particular, every left Artinian Jacobson semisimple ring is semisimple. Proof Note that each Mi is a faithful simple module over A/Ann(Mi ). Since A/Ann(Mi ) is left Artinian, Theorem 12.3.10 shows it to be a simple ring. Note that EndA (Mi ) = EndA/Ann(Mi ) (Mi ), so Theorem 12.3.10 shows that Mi is finite dimensional over Di , and gives us a ring isomorphism op A/Ann(Mi ) ∼ = Mni (Di ).

Corollary 12.3.9 gives an A-module isomorphism A/Ann(Mi ) ∼ = Mini . Since A/Ann(Mi ) is a simple ring, it has no two-sided ideals other than 0 and itself. Thus, Ann(Mi ) is a maximal two-sided ideal of A. By the irredundancy of the intersection of the annihilators of the Mi , we see that if i = j, Ann(Mi ) = Ann(Mj ), and hence Ann(Mi ) + Ann(Mj ) = A. Thus, since the intersection of the Ann(Mi ) is 0, the Chinese Remainder Theorem gives us a ring isomorphism A∼ = A/Ann(M1 ) × · · · × A/Ann(Mk ).

The next proposition is now clear from Corollary 12.2.7, Lemma 12.3.11, Proposition 12.3.12, Corollary 12.2.8, and the fact that a finite product of semisimple rings is semisimple. Theorem 12.3.13. (Semisimple Wedderburn Theorem) Let A be a ring. Then the following conditions are equivalent. 1. A is Jacobson semisimple and left Artinian. 2. A is Jacobson semisimple and right Artinian. 3. A is semisimple. 4. A is isomorphic to a finite product of finite dimensional matrix rings over division rings. Let K be a field and let A be a K-algebra that is semisimple as a ring and finite dimensional as a vector space over K. Then any finitely generated A-module M is a finitely generated K-vector space, and there is a natural embedding of EndA (M ) as a K-subalgebra of EndK (M ). Thus, if M is simple, then D = EndA (M ) is a division algebra over K that is finite dimensional as a K-vector space. In practice, to obtain a decomposition of such an A as product of matrix rings, it is often convenient to hunt for simple modules Mi , calculate both Di and dimDi (Mi ), and continue this process until the resulting product of matrix rings has the same dimension as a K-vector space as A. Exercises 12.3.14. 1. Show that Jacobson semisimplicity is not preserved by localization.


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2. For any ring A, show that A/R(A) is Jacobson semisimple. 3. Let G be a finite group and let K be an algebraically closed field whose characteristic doesn’t divide |G|. Let M be a simple module over K[G]. Show that EndK[G] (M ) = K. Deduce that K[G] contains exactly dimK (M ) copies of M in its decomposition as a direct sum of simple modules. 4. Let G be a finite group and K a field. Then a K[G]-module structure whose underlying K-vector space is K n corresponds to a K-algebra homomorphism K[G] → Mn (K). Show that this induces a one-to-one correspondence between those K[G] modules whose underlying K-vector spaces have dimension n and the conjugacy classes of group homomorphisms ρ : G → Gln (K). (Such homomorphisms ρ are called n-dimensional representations of G over K.) Deduce that there is a one-to-one correspondence between the isomorphism classes of K[G]-modules of dimension 1 over K (which are a priori simple) and the set of group homomorphisms from G to K × . 5. Deduce from Problem 3 of Exercises 10.7.8 that if G is a finite abelian group and K is an algebraically closed field whose characteristic does not divide |G|, then the simple K[G] modules are precisely those with dimension 1 as vector spaces over K. 6. Write Q[D2n ], R[D2n ], and C[D2n ] as products of matrix rings over division rings. 7. Write R[Q8 ] as a product of matrix rings over division rings.

12.4

Homological Dimension

We can also characterize the infinitely generated modules over a semisimple ring. Proposition 12.4.1. Let A be a left semisimple ring. Then every left A-module is isomorphic to a direct sum of simple left ideals, and hence is projective, as well as being a semisimple module over A. Proof By Corollary 12.2.10, it suffices to show that every A-module is semisimple. Let M be an A-module and let N be a submodule. We define a partially ordered set whose elements are submodules N containing N , together with a retraction r : N → N for the inclusion of N in N . The partial ordering is given by setting (N , r ) ≤ (N , r ) if N ⊂ N and if r restricts to r on N . Zorn’s Lemma shows that there exists a maximal element (N , r ) of this set. We claim that N = M , and hence N is a direct summand of M . This will complete the proof, as N was an arbitrary submodule of M . Thus, suppose that m ∈ M does not lie in N and let fm : A → M be the A-module −1 homomorphism that takes 1 to m. Let a = fm (N ). Then A is an internal direct sum A = a+b. Since ker fm ⊂ a, fm exhibits Am as the internal direct sum (Am∩N )+fm (b). As a result, the sum N + fm (b) is direct. Now set r : N + fm (b) → N to be equal to r on N and to be 0 on fm (b). This contradicts the maximality of (N , r ), and the result follows. This now gives a characterization of semisimple rings. Corollary 12.4.2. Let A be a ring. Then the following conditions are equivalent.


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1. A is left semisimple. 2. Every left A-module is projective. 3. Every finitely generated left A-module is projective. Proof All that’s left to show is that the third condition implies the first. But if a is a left ideal, then A/a is finitely generated, and hence the sequence 0 → a → A → A/a → 0 splits. So a is a direct summand of A. In view of Corollary 12.4.2, semisimple rings may be characterized in terms of the following notion. Definition 12.4.3. We say that the ring A has global left homological dimension ≤ n if for each left A-module M there is an exact sequence 0 → Pn → · · · → P0 → M → 0 where Pi is projective for all i. We may now apply Corollary 12.4.2. Corollary 12.4.4. A ring is left semisimple if and only if it has global left homological dimension 0. Being hereditary is also equivalent to a homological condition. To show this, we shall generalize Proposition 12.1.5 to the case of infinitely generated modules. The following proposition is due to Irving Kaplansky. Proposition 12.4.5. Let F be a left free module over the left hereditary ring A. Then any submodule M ⊂ F is isomorphic to a direct sum of ideals, and hence is projective. Proof Let {ei | i ∈ I} be a basis for F . We shall put a well-ordering on I. This is a partial ordering with the property that any subset J ⊂ I has a smallest element, i.e., an element j ∈ J such that j ≤ k for all k ∈ J. We may do this by the Well-Ordering Principle, which is equivalent to the Axiom of Choice. For i ∈ I, let Fi be the submodule generated by {ej | j ≤ i} and let πi : Fi → A be induced by the projection onto the summand generated by ei (i.e., πi (ej ) = 0 for j = i and πi (ei ) = 1). Write Mi = M ∩ Fi , and let ai = πi (Mi ). Then we have a short exact sequence π

i 0 → Ki → Mi −→ ai → 0.

Here πi is the restriction of πi to Mi and Ki is the kernel of πi . Since A is hereditary, ai is projective, and the sequence splits. Write Ni ⊂ Mi for the image of ai under a section for πi . We claim that M is the internal direct sum of {Ni | i ∈ I}. To see this, suppose first that n1 + · · · + nk = 0, with nj ∈ Nij for j = 1, . . . , k and with i1 < · · · < ik . Then 0 = πik (n1 + · · · + nk ) = πik (nk ).


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∼ =

Since πik : Nik −→ aik , we must have nk = 0. Inductively, ni = 0 for all i. Thus, it suffices to show that the Ni generate M . We argue by contradiction. Let i ∈ I be the smallest element such that M is not contained in i j∈I Nj , and let m ∈ Mi be an element not in j∈I Nj . Let si : ai → Ni be the inverse of the isomorphism πi : Ni → ai . Write m = m − si (πi (m)). Then m is not in j∈I Nj . And πi (m ) = 0. Thus, m lies in the submodule of F generated by {ej | j < i}. Since m is a finite linear combination of these ej , it must lie in Fj ∩ M = Mj for some j < i, contradicting the minimality of i. Since P.I.D.s are hereditary and their ideals are all free, we obtain the following corollary. Corollary 12.4.6. Every projective module over a P.I.D. is free. Lemma 12.4.7. (Schanuel’s Lemma) Suppose given exact sequences i

f

0→K− →P − →M →0

and

i

f

0→L− →Q− →M →0

of A-modules, with P and Q projective. Then K ⊕ Q ∼ = L ⊕ P. Proof We shall make use of the pullback construction that’s discussed in the proof of Proposition 9.8.2: P ×M Q = {(x, y) ∈ P × Q | f (x) = g(y)}. We obtain a commutative diagram whose straight lines are exact: 0

0

0

/ N1 g / K

f N2 ı f ı / P ×M Q g f i / P

0

/ L i

/ Q g / M

/ 0

0 Here, f (x, y) = y and g(x, y) = x, while f and g are obtained by restricting f and g, respectively. We now make some claims the reader should check. We claim that in any such pullback diagram, the maps f and g are isomorphisms. Also, since f is surjective, so is f , and since g is surjective, so is g.


529

Since f is surjective and Q is projective, we have a split short exact sequence f

ı

0 → N1 − → P ×M Q − → Q → 0. Since g is an isomorphism, K ⊕ Q ∼ = P ×M Q. Since g is surjective and P is projective, we have a split short exact sequence j

g

0 → N2 − → P ×M A − → P → 0. Since f is an isomorphism, L ⊕ P ∼ = P ×M Q. We may now characterize hereditary rings in terms of homological dimension. Proposition 12.4.8. A ring A is left hereditary if and only if it has global left homological dimension ≤ 1. Proof If A is hereditary and M is a left A-module, let f : F → M be surjective, where F is free. Then we have a short exact sequence f

0→K→F − → M → 0. By Proposition 12.4.5, K is projective. Since M was arbitrary, we see that A has global left homological dimension ≤ 1. Conversely, suppose that A has global left homological dimension ≤ 1 and let a be a left ideal of A. Then our hypothesis provides an exact sequence 0 → P1 → P0 → A/a → 0, with P0 and P1 projective. Of course, 0 → a → A → Aa → 0 is exact, so Schanuel’s Lemma gives an isomorphism a ⊕ P0 ∼ = P1 ⊕ A. Thus, a is projective, and hence A is hereditary.

12.5

Hereditary Rings

Over a P.I.D., Proposition 12.1.5 shows that every finitely generated projective module is free. However, it is definitely not the case that every finitely generated projective module over an arbitrary hereditary domain is free. For instance, as we shall see in Section 12.7, the ring of integers (to be defined there) in a finite extension of Q is a hereditary Noetherian domain. However, these rings of integers turn out very rarely to be P.I.D.s. So the following proposition will give a recognition principle for non-free ideals in these rings, and hence a source of examples of finitely generated projective modules that are not free. Proposition 12.5.1. Let a be a nonzero ideal in the commutative ring A. Then a is free as an A-module if and only if it is free of rank 1, in which case it is a principal ideal (x) for some x ∈ A that is not a zero-divisor. In consequence, if A is a commutative ring in which every ideal is free, then A is a P.I.D.


530

Proof If a is free on more than one generator, then picking a pair of the basis elements induces an injective homomorphism from A2 into a. But a is a submodule of A, so we get an injection from A2 into A, contradicting Theorem 9.5.20. Thus, if an ideal a of A is free as an A-module, it must have rank 1, so there’s an ∼ = isomorphism f : A −→ a. And if x = f (1), we have a = (x). Since f is injective, x is not a zero-divisor. If every ideal is free, then every ideal is a principal, and it suffices to show that A is an integral domain. But if x ∈ A is a zero-divisor, then AnnA ((x)) = 0, so (x) cannot be free. Recall that a discrete valuation ring is a Euclidean domain which is local, and hence has at most one prime element, up to equivalence of primes. Corollary 12.5.2. A Noetherian hereditary local ring is a discrete valuation ring. Proof The hereditary property says that every ideal is projective, and the Noetherian property that every ideal is finitely generated. But finitely generated projectives over a local ring are free by Proposition 9.8.14, so Proposition 12.5.1 shows that our ring is a P.I.D. A P.I.D. which is not a field is local if and only if it has exactly one prime element (up to equivalence of primes). And Problem 9 of Exercises 8.1.37 shows that such a ring is Euclidean, and hence a discrete valuation ring. We now give another consequence of Proposition 12.5.1. Recall from Proposition 9.8.16 that every finitely generated projective module P over an integral domain A has constant rank, meaning that the rank of the free Ap -module Pp is independent of the choice of prime ideal p of A. Corollary 12.5.3. Let A be an integral domain and let a be a nonzero ideal of A that is finitely generated and projective as an A-module. Then a has constant rank 1. Proof Let p be a prime ideal of A. Because A is a domain, ap is a nonzero ideal of Ap . Since ap is finitely generated and free over Ap , the result follows from Proposition 12.5.1. Note the stipulation in the preceding that A be a domain is necessary. Corollary 12.5.4. Let A be a Noetherian, hereditary commutative ring. Then every localization of A is a discrete valuation ring. Proof Recall that every ideal of Ap has the form ap for an ideal a of A. Since A is hereditary and Noetherian, a is finitely generated and projective, so ap = Ap ⊗A a is a free Ap -module by Proposition 9.8.14. Thus, Ap is a P.I.D. by Proposition 12.5.1. Since Ap is local, Corollary 12.5.2 suffices. Exercises 12.5.5. 1. Show that a Noetherian hereditary commutative ring has Krull dimension ≤ 1.


12.6

531

Dedekind Domains

As we shall show in Theorem 12.6.8, a Dedekind domain is precisely a hereditary integral domain. However, the definition we shall give is more classical, and is interesting from the point of view of the development of ideal theory. The point is that a number field, meaning a finite extension of the rational numbers, Q, has a subring, called its ring of integers, which behaves somewhat like the integers do as a subring of Q. At one time, it was assumed that the ring of integers of a number field would be a P.I.D. Indeed, a false proof of Fermat’s Last Theorem was given on the basis of this false assumption. The truth of the matter is that the ring of integers of a number field is a Dedekind domain. We shall show this in Section 12.7. And in a Dedekind domain, the ideals behave in the same way that the principal ideals do in a U.F.D. Indeed, ideal theory began with this observation. Ideals were originally thought of as idealized numbers. Moving from elements to ideals, we shall make an analogy between prime elements and maximal ideals and an analogy between irreducible elements and prime ideals. Here, the operation on ideals that concerns us is the product of ideals: ab = {a1 b1 + · · · + ak bk | k ≥ 1, ai ∈ a, and bi ∈ b for i = 1, . . . , k} Recall from Lemma 7.6.9 that if p, a, and b are ideals in the commutative ring A, with p prime, then ab ⊂ p implies that either a ⊂ p or b ⊂ p. Since ab ⊂ a ∩ b, we obtain the following lemma, which displays the analogy between prime ideals and irreducible elements. Lemma 12.6.1. Let p be a prime ideal in the commutative ring A. Suppose given ideals a and b with p = ab. Then either p = a or p = b. Definition 12.6.2. A Dedekind domain is an integral domain in which every proper, nonzero ideal is a product of maximal ideals. One result comes immediately from Lemma 12.6.1. Recall that the Krull dimension of a commutative ring A is the number of inclusion maps in the longest possible chain of proper inclusions p0 ⊂ · · · ⊂ pn of prime ideals of A. Proposition 12.6.3. Every nonzero prime ideal in a Dedekind domain is maximal. Thus, a Dedekind domain that is not a field has Krull dimension 1. Proof Let p be a nonzero prime ideal in the Dedekind domain A. Then we may write p = m1 . . . mk for (not necessarily distinct) maximal ideals m1 , . . . , mk . But an induction using Lemma 12.6.1 now shows that p = mi for some i. The following observation is important. Lemma 12.6.4. Let A be an integral domain with fraction field K. Let a be a nonzero ideal of A and let f : a → A be an A-module homomorphism. Then there is an element α ∈ K such that f is the restriction to a of the A-module homomorphism from K to itself that’s induced by multiplication by α. In fact, for any nonzero element a ∈ a, we have f (a)/a = α.


532

Proof Let 0 = a ∈ a. Then for x ∈ a, we have af (x) = f (ax) = f (a)x, so f (x) = (f (a)/a) · x as an element of K. It is useful to establish as quickly as possible that the Dedekind domains are the hereditary integral domains. A key idea is the notion of invertible ideals. Definition 12.6.5. A nonzero ideal a in an integral domain is invertible if there is a nonzero ideal b such that ab is principal. The next proposition displays the relevance of invertibility to the hereditary property. Proposition 12.6.6. Let A be an integral domain. A nonzero ideal a of A is invertible if and only if it is projective as an A-module. If a is an invertible ideal, it is also finitely generated. Proof First suppose that a is invertible, with ab = (a). Then we can write a1 b1 + · · · + ak bk = a, with ai ∈ a and bi ∈ b for i = 1, . . . , n. Let f : Ak → a be the A-module homomorphism that takes ei to ai for 1 ≤ i ≤ k. We shall construct a section for f , showing that a is a finitely generated projective module over A. Let b ∈ b. Since ab = (a), we see that multiplication by b/a carries a into A. Thus, there is a well defined homomorphism s : a → Ak with bk x b1 x ,..., s(x) = a a for x ∈ a. Since a1 b1 + · · · + ak bk = a, we see that s is a section for f , as claimed. Conversely, suppose that a is projective. Let f : F → a be a surjective homomorphism, where F is the free A-module with basis {ei | i ∈ I}. Let s : a → F be a section of f . Since the direct sum embeds in the direct product, Lemma 12.6.4 shows there are elements αi ∈ K for each i ∈ I such that s(x) = i∈I (αi x)ei for each x ∈ a, where K is the field of fractions of A. But the sum must be finite, so all but finitely many of the αi must be 0. In particular, we see that f restricts to a surjection on a finitely generated free submodule of F , so we may assume that F is the finitely generated free module with basis e1 , . . . , en , and that s(x) = (α1 x)e1 + · · · + (αn x)en for all x ∈ a. Let f (ei ) = ai . Because s is a section of f , we have x = (α1 a1 )x + · · · + (αn an )x for all x ∈ a, and hence α1 a1 + · · · + αn an = 1. Let a be a common denominator for α1 , . . . , αn , and set bi = aαi ∈ A for 1 ≤ i ≤ n. Recall that multiplication by αi carries a into A. Thus, multiplication by bi carries a into (a). Note that a1 b1 + · · · + an bn = a. Thus, if we set b to be the ideal generated by b1 , . . . , bn , we have ab = (a), and hence a is invertible. Thus, projective ideals in an integral domain are finitely generated. Corollary 12.6.7. A hereditary integral domain is Noetherian.


533

In the study of Dedekind domains, it is often convenient to work with a generalized notion of ideals. If A is a domain with fraction field K, we may consider the A-submodules of K. If a and b are A-submodules of K, we define their product, ab, as if they were ideals: ab = {α1 β1 + · · · + αk βk | k ≥ 1, αi ∈ a, and βi ∈ b for i = 1, . . . , k}. This product is easily seen to be associative and commutative and to satisfy the distributive law with respect to the usual sum operation on submodules. Certain of the A-submodules of K are generally known as fractional ideals. We shall confine our formal treatment of fractional ideals to the case where A is a Dedekind domain, which we treat in Exercises 12.6.23. We now give a characterization of Dedekind domains. Theorem 12.6.8. Let A be an integral domain. Then the following conditions are equivalent. 1. For any inclusion a ⊂ b of ideals in A, there is an ideal c such that a = bc. 2. A is hereditary. 3. A is a Dedekind domain. Proof The first condition implies the second, as if a is a nonzero ideal of A and if 0 = a ∈ a, then (a) ⊂ a, so there is an ideal b with ab = (a). Thus, every nonzero ideal of A is invertible, and hence A is hereditary. The second condition implies the first as follows. Let a ⊂ b with a = 0. Since A is hereditary, b is invertible. Say bc = (x). Then, as A-submodules of the fraction field of A, we have (1/x)c · b = A. Since a ⊂ b, (1/x)c · a is an A-submodule of A, hence an ideal. Write (1/x)c · a = d. But then bd = Aa = a, and hence the first condition holds. Thus, the first two conditions are equivalent. Now assume the third, so that A is a Dedekind domain. The result is obvious if A is a field, so we assume it is not. We first claim that every maximal ideal of a is invertible. To see this, let m be a maximal ideal, and let 0 = a ∈ m. Then (a) is a product of maximal ideals, say (a) = m1 . . . mk . But then m1 . . . mk ⊂ m. By Lemma 7.6.9, we must have mi ⊂ m for some i. Since mi and m are maximal, this gives m = mi . But mi is a factor if the principal ideal (a), so m is invertible. Clearly, a product of invertible ideals is invertible. Since every proper, nonzero ideal in a Dedekind domain is a product of maximal ideals, every such ideal is invertible. Thus, Dedekind domains are hereditary. It suffices to show that the first two conditions imply the third. Thus, let a be a proper, nonzero ideal in the hereditary domain A. If a is maximal, there’s nothing to show. Otherwise, a ⊂ m1 , with m1 maximal. By condition 1, we have a = m1 a1 for some ideal a1 . By Problem 4 of Exercises 10.4.11 (which is an immediate consequence of Corollary 10.4.10), the inclusion a ⊂ a1 is proper. We continue inductively, writing a = m1 . . . mn an for n ≥ 1, until we arrive at the point where an is maximal. Since a ⊂ a1 ⊂ · · · ⊂ an are proper inclusions, an must be maximal for some finite n, since hereditary domains are Noetherian.


534

The first thing that we should establish now is the uniqueness of the factorization of ideals in a Dedekind domain. The hereditary property makes this easier. The next lemma is obvious if the ideal a is principal. But the result extends to invertible ideals, as they are factors of principal ideals. Lemma 12.6.9. Let a be an invertible ideal in a domain A, and let b and c be ideals such that ab = ac. Then b = c. Proposition 12.6.10. Let A be a Dedekind domain. Then any proper, nonzero ideal has a unique factorization of the form a = mr11 . . . mrkk where m1 , . . . , mk are distinct maximal ideals and the exponents ri are positive for i = 1, . . . , k. Here, uniqueness is up to a reordering of the factors mr11 , . . . , mrkk . Proof Suppose given two such decompositions mr11 . . . mrkk = ns11 . . . nsl l . Then the left-hand side shows that the ideal in question is contained in m1 . So Lemma 7.6.9 shows that ni ⊂ m1 for some i. Renumbering the n’s we may assume that n1 ⊂ m1 , but since both are maximal, this gives n1 = m1 . By Lemma 12.6.9, this implies that m1r1 −1 . . . mrkk = ns11 −1 . . . nsl l , and the result follows by induction. The first condition of Theorem 12.6.8 shows that if a and b nonzero ideals of a Dedekind domain, then a ⊂ b if and only if a = bc for some ideal c of A. Unique factorization now gives a corollary. Corollary 12.6.11. Let A be a Dedekind domain and let a = mr11 . . . mrkk , where m1 , . . . , mk are distinct maximal ideals of A, and each ri is positive. Then the ideals of A that contain a are precisely those of the form ms11 . . . mskk , where 0 ≤ si ≤ ri for all i. We wish now to learn more about the structure of modules over a Dedekind domain. Proposition 12.6.12. Let M be a finitely generated module over a Dedekind domain. Then M is projective if and only if it is torsion-free. Proof By Proposition 12.1.5, every finitely generated projective module over a Dedekind domain is a direct sum of ideals, and hence is torsion-free. Conversely, let M be a finitely generated torsion-free A-module, with A a Dedekind domain. Then the canonical map η : M → M(0) is injective, where M(0) is the localization of M at (0). Thus, M is a finitely generated A-submodule of a finitely generated vector space over the fraction field, K, of A. Let e1 , . . . , en be a K-basis for M(0) . Since M is finitely generated, we can clear the denominators of the coefficients with respect to this basis of the generators of M , producing a nonzero a ∈ A such that aM lies in the free A-module generated by e1 , . . . , en . In particular, M is isomorphic to a submodule of An , and hence is projective by Proposition 12.1.5.


535

Note, then, that any finitely generated submodule of a torsion-free A-module is projective, and hence flat. Thus, we obtain the next corollary from the proof given in Corollary 9.5.12. Corollary 12.6.13. Let A be a Dedekind domain. Then an A-module is flat if and only if it is torsion-free. In particular, every finitely generated flat module is projective. More information about the ideals will be useful if we are to get a better understanding of the classification of the finitely generated projective modules over a Dedekind domain. The decomposition as a product of maximal ideals has a great deal of redundancy as far as describing the isomorphism classes of the ideals as A-modules. After all, if A is a P.I.D., then, independent of this decomposition, any two ideals are isomorphic as modules. Lemma 12.6.14. Let a, b, c, and d be ideals of the domain A, and suppose that a ∼ =c and b ∼ = d as A-modules. Then ab ∼ = cd. Proof Let K be the fraction field of A. Lemma 12.6.4 gives elements α, β ∈ K with αa = c and βb = d. So αβab = cd. Definition 12.6.15. Let A be a Dedekind domain. The class group, Cl(A), is the abelian group whose elements are the isomorphism classes, [a], of nonzero ideals of A, and whose multiplication is given by [a][b] = [ab]. Of course, [A] is the identity element of Cl(A), and the inverse of [a] is given by [b], where ab is principal. Class groups are quite difficult to compute. However, one thing is clear. Lemma 12.6.16. Let A be a Dedekind domain. Then Cl(A) is the trivial group if and only if A is a P.I.D. Next, we shall show that any ideal of a Dedekind domain is generated by at most two elements. Recall that a principal ideal ring is a commutative ring in which every ideal is principal. Proposition 12.6.17. Let a be a proper, nonzero ideal in the Dedekind domain A. Then the quotient ring A/a is a principal ideal ring. Proof Write a = mr11 . . . mrkk , with m1 , . . . , mk distinct maximal ideals of A, and ri > 0 for all i. Then Corollary 12.6.11 shows that the ideals of A/a are those of the form b/a, where b = ms11 . . . mskk with 0 ≤ si ≤ ri for all i. For such a b, suppose we find b ∈ A such that b ∈ msi i but b ∈ msi i +1 for all i. Then a calculation of exponents shows that (b) + a ⊂ b, but (b) + a cannot lie in any smaller ideal containing a. Thus, (b) + a = b, and b/a = (b). To find such a b, choose, for each i, a bi ∈ msi i such that bi ∈ misi +1 . Since mi is s +1 the only maximal ideal containing misi +1 , we have misi +1 + mj j = A for i = j. By the Chinese Remainder Theorem, there is a b ∈ A such that b ≡ bi mod msi i +1 for all i. This is our desired b. Corollary 12.6.18. Let a be a non-principal ideal in the Dedekind domain A, and let a be any nonzero element in a. Then there is an element b ∈ a such that a is the ideal generated by a and b.


536

Proof A/(a) is a principal ideal ring. Just set b equal to any element of a such that a/(a) = (b). We now wish to classify the projective modules over a Dedekind domain. The next lemma is basic. Lemma 12.6.19. Let A be any commutative ring and let a and b be ideals of A which are relatively prime in the sense that a + b = A. Then ab = a ∩ b. Proof Clearly, ab ⊂ a ∩ b. But if 1 = a + b with a ∈ a and b ∈ b, then for any c ∈ a ∩ b, we have c = ac + cb ∈ ab. Proposition 12.6.20. Let a and b be ideals in the Dedekind domain A. Then a⊕b∼ = A ⊕ ab. Proof First assume that a and b are relatively prime. Then, as the reader may check, there’s an exact sequence ψ

ϕ

0→a∩b− →a⊕b− → A → 0, where ψ(x) = (x, −x), and ϕ(a, b) = a + b. Since A is free, the sequence splits, giving an isomorphism a ⊕ b ∼ = A ⊕ (a ∩ b), and the result follows from Lemma 12.6.19. Thus, for an arbitrary pair of ideals a and b, it suffices, by Lemma 12.6.14, to find an ideal a that is isomorphic to a as an A-module, such that a and b are relatively prime. Let c be an ideal such that ac is principal. Say ac = (a). Applying Proposition 12.6.17 to the ideal c/bc of A/bc, we see there’s an element c ∈ c such that c = bc + (c). Multiplying both sides by a, we get (a) = ac

= abc + ca = ab + ca

We can then divide both sides by a. We see that a = (c/a)a is an ideal of A isomorphic to a as a module, and b + a = A. We are now able to completely classify the finitely generated projective modules over a Dedekind domain modulo an understanding of the isomorphism classes of ideals. Thus, we reduce the classification of the projective modules to the calculation of the class group Cl(A). The next theorem is due to Steinitz. Theorem 12.6.21. Let A be a Dedekind domain. Suppose given nonzero ideals ai , 1 ≤ i ≤ n and bi , 1 ≤ i ≤ k, of A. Then there is an A-module isomorphism a1 ⊕ · · · ⊕ an ∼ = b1 ⊕ · · · ⊕ b k if and only if n = k and the product ideals a1 . . . an and b1 . . . bk are isomorphic as modules. Proof By Proposition 12.6.20, a1 ⊕ · · · ⊕ an ∼ = An−1 ⊕ (a1 . . . an ), and a similar isomorphism holds for the b’s. So if n = k and a1 . . . an ∼ = b1 . . . bk , then the direct sum of the a’s is isomorphic to the direct sum of the b’s as claimed.


537

Conversely, suppose given an isomorphism ∼ =

f : a1 ⊕ · · · ⊕ an −→ b1 ⊕ · · · ⊕ bk . We first claim that n = k. To see this, note that if c is any nonzero ideal of A and if K is the field of fractions of A, then the localization, K ⊗A c, of c at (0) is a onedimensional vector space over K by Corollary 12.5.3. Thus, f ⊗ 1K is an isomorphism from an n-dimensional vector space over K to an k-dimensional one, so n = k. Next, note that by Lemma 12.6.4, each composite ιj

f

π

i aj −→ a1 ⊕ · · · ⊕ an − → b1 ⊕ · · · ⊕ bn −→ bi

is induced by multiplication by an element αij ∈ K. Here, ιj is the canonical inclusion in the direct sum and πi is the projection onto the i-th factor. Thus, if M = (αij ) ∈ Mn (K), then, if we write the n-tuples in a1 ⊕ · · · ⊕ an vertically, we see that ⎛ ⎞ ⎛ ⎞ a1 a1 ⎜ ⎟ ⎜ ⎟ f ⎝ ... ⎠ = M · ⎝ ... ⎠ . an

an

Note that f −1 is also represented by a matrix in Mn (K), which, by the naturality of the passage between mappings and matrices, together with the fact that each of the above direct sums contains a K-basis for K n , must be M −1 . We claim that multiplication by det M carries a1 . . . an into b1 . . . bn , and that multiplication by det(M −1 ) carries b1 . . . bn into a1 . . . an . Since det M and det(M −1 ) are inverse elements of K (Corollary 10.3.2), this implies that multiplication by det M gives an isomorphism 1 . . . bn , and the proof will be complete. ⎛ from a1 . . . an to b⎞ a1 0 . . . 0 ⎜ 0 a2 . . . 0 ⎟ ⎟ ⎜ Let M = ⎜ ⎟ be the diagonal matrix whose entries are a1 , . . . , an , .. ⎠ ⎝ . 0 0 . . . an where ai ∈ ai for i = 1, . . . , n. Then the ij-th entry of M M lies in bi for all i, j. Thus, by the formula of Corollary 10.2.6, we see that det(M M ) ∈ b1 . . . bn . But det(M M ) = (det M ) · (det M ) = (det M ) · a1 . . . an , by Propositions 10.3.1 and 10.3.7. Since elements of the form a1 . . . an generate a1 . . . an , det M · a1 . . . an ⊂ b1 . . . bn , as claimed. The same proof shows that multiplication by det(M −1 ) carries b1 . . . bn into a1 . . . an , so the result follows. And this gives the following, weaker result. ) 0 (A) is Corollary 12.6.22. Let A be a Dedekind domain. Then the reduced K-group K isomorphic to the class group Cl(A). Thus, we get an isomorphism K0 (A) ∼ = Z ⊕ Cl(A).


538

) 0 (A) with the kernel of Proof Let K be the fraction field of A. Then we can identify K ) 0 (A) are the formal the natural map K0 (A) → K0 (K). In particular, the elements of K differences [P ] − [Q] where the K-ranks of P and Q are equal. Note that if Q ⊕ Q ∼ = An is free, then [P ] − [Q] = [P ⊕ Q ] − [An ], ) 0 (A) may be written as [P ] − [An ] for some P such that K ⊗A P so every element of K is an n-dimensional vector space over K. Now Proposition 12.6.20 shows that if P is isomorphic to the direct sum a1 ⊕ · · · ⊕ an of ideals, then [P ] − [An ] = [(a1 . . . an ) ⊕ An−1 ] − [An ] = [a1 . . . an ] − [A]. ) 0 (A) It is now easy to see that there is a surjective group homomorphism ϕ : Cl(A) → K given by ϕ([a]) = [a] − [A]. By Lemma 9.9.8, [a] ∈ ker ϕ if and only if there is an isomorphism a ⊕ An ∼ = A ⊕ An for n sufficiently large. But Theorem 12.6.21 says this is impossible unless a ∼ = A, in which case [a] is the trivial element of Cl(A). Exercises 12.6.23. 1. Show that a Dedekind domain is a U.F.D. if and only if it is a P.I.D. 2. Let A be a Dedekind domain with fraction field K. A fractional ideal of A is defined to be a finitely generated A-submodule of K. (a) Show that an A-submodule i of A is a fractional ideal if and only if ai ⊂ A for some a ∈ A. Deduce that every fractional ideal is isomorphic as an A-module to an ideal of A. (b) Show that the nonzero fractional ideals of A form an abelian group I(A) under the product operation for A-submodules of K. (c) Show that there is an exact sequence p

→ Cl(A) → 0. 0 → A× → K × → I(A) − Here K × maps onto the principal fractional ideals in I(A), and for any fractional ideal i, p(i) = [a] for any ideal a of A that is isomorphic to i as an A-module. 3. Let a and b be ideals of the Dedekind domain A. Show that the natural map a ⊗A b → ab localizes to be an isomorphism at each prime ideal of A, and hence is an isomorphism of A-modules. ) 0 (A) to be the kernel of K0 (A) to K0 (K), and (a) Deduce that if we consider K ) 0 (A) in hence an ideal of K0 (A), then the product of any two elements of K the ring structure of K0 (A) is 0.


539

(b) Show that the elements of K0 (A) that map to the multiplicative identity of K0 (K) form a subgroup of K0 (A)× isomorphic to Cl(A). (This group of units in K0 (A) is known as the Picard group, Pic(A), of A.) 4. Let a1 , . . . , ak be nonzero ideals of the Dedekind domain A. Show that ΛkA (a1 ⊕ · · · ⊕ ak ) ∼ = a1 . . . ak . Use this to give a shorter proof of Theorem 12.6.21.

12.7

Integral Dependence

Definitions 12.7.1. Let A ⊂ B be an inclusion of commutative rings. We say that b ∈ B is integral over A if it is a root of a monic polynomial with coefficients in A. We say that B itself is integral over A if every element of B is integral over A. We also say, in this situation, that B is integrally dependent on A. We can characterize integral elements as follows. Recall that an A-module M is faithful if AnnA (M ) = 0. Proposition 12.7.2. Let A ⊂ B be an inclusion of commutative rings and let b ∈ B. Then the following conditions are equivalent. 1. b is integral over A. 2. A[b] is finitely generated as an A-module. 3. There is a subring C ⊂ B, containing both A and b, such that C is finitely generated as an A-module. 4. There is a faithful A[b]-module M that is finitely generated as an A-module. Proof Suppose that b is a root of the monic polynomial f (X) ∈ A[X]. Then A[b] is a quotient of A[X]/(f (X)), which is a free module of rank deg f over A by Proposition 7.7.35. Thus, the first condition implies the second. Clearly, the second condition implies the third, and the third implies the fourth by taking M = C. Thus, assume given an A[b] module M as in the fourth statement. Let n x1 , . . . , xn generate M over A. Then we may write bxj = i=1 aij xi for each j, where aij ∈ A for all i, j. Let M = (aij ) ∈ Mn (A), and let fb ∈ EndA (M ) be induced by multiplication by b. Then the Generalized Cayley–Hamilton Theorem shows that chM (fb ) = 0 in EndA (M ). But M is a faithful A[b]-module, and hence the natural map A[b] → EndA (M ) (i.e., the A-algebra homomorphism that takes b to fb ) is injective. Thus, chM (b) = 0 in B, and hence b is a root of the monic polynomial chM (X) ∈ A[X]. Clearly, if A ⊂ B ⊂ C and if c ∈ C is integral over A, then it is integral over B. Thus, an induction on the second condition above gives the next corollary. Corollary 12.7.3. Let A ⊂ B be an inclusion of commutative rings and let b1 , . . . , bn ∈ B be integral over A. Then A[b1 , . . . , bn ] is finitely generated as an A-module. Definitions 12.7.4. Let A ⊂ B be an inclusion of commutative rings. Then the integral closure of A in B is the set of elements in B that are integral over A. Note that by Corollary 12.7.3 and Proposition 12.7.2, the integral closure of A in B is a subring of B containing A. We say that A is integrally closed in B if the integral closure of A in B is just A itself. An integral domain A is said to be an integrally closed domain if it is integrally closed in its field of fractions.


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We’ve already seen a number of examples of integrally closed domains. Lemma 12.7.5. A unique factorization domain is integrally closed. Proof Let A be a U.F.D. with fraction field K, and suppose that α ∈ K is integral over A. Write α = a/b, with a and b relatively prime, and let f (X) = X k +ak−1 X k−1 +· · ·+a0 be a monic polynomial over A with f (α) = 0. Then clearing denominators, we get ak + ak−1 ak−1 b + · · · + a0 bk = 0. The result is that b must divide ak . Since a and b are relatively prime, this forces b to be a unit, and hence α ∈ A. Corollary 12.7.6. Suppose given inclusions A ⊂ B ⊂ C of commutative rings, such that B is integral over A. Then any element of C that is integral over B is also integral over A. Proof Let f (X) = X k + bk−1 X k−1 + · · · + b0 be a monic polynomial over B which has c is a root. Then A[b1 , . . . , bn ] is finitely generated as an A-module. Since c is integral over A[b1 , . . . , bn ], A[b1 , . . . , bn , c] is finitely generated as an A-module as well. We now give an important special case of integral closures. Proposition 12.7.7. Let A be a domain with fraction field K, and let L be an algebraic extension of K. Let B be the integral closure of A in L. Then the field of fractions of B is equal to L. In fact, for each α ∈ L, there is an a ∈ A such that aα ∈ B. Also, B is integrally closed in L, and hence B is an integrally closed domain. Proof Let α ∈ L. Since L is algebraic over K, α is a root of a polynomial with coefficients in K. In fact, by clearing denominators, we see that α is a root of a polynomial f (X) = an X n + · · · + a0 with coefficients in A. n−1 ai X i . Then g(an α) = an−1 f (α) = 0, so an α is integral Set g(X) = X n + i=0 an−i−1 n n over A. It suffices to show that B is integrally closed. But if α ∈ L is integral over B, then it is integral over A by Corollary 12.7.6, and hence lies in B. Lemma 12.7.8. Let A be an integrally closed domain with fraction field K. Let L be a finite extension of K, and let B be the integral closure of A in L. Then the norm and trace behave as follows: NL/K (B) ⊂ A

and

trL/K (B) ⊂ A.

Proof Let L be any extension of K and let σ : L → L be an embedding over K. Then if b ∈ L is a root of the monic polynomial f (X) ∈ A[X], σ(b) is also a root of f . By Proposition 11.14.3, both NL/K (b) and trL/K (b) are integral over A for each b ∈ B. Since these elements lie in K, the result follows from the fact that A is integrally closed.

Lemma 12.7.9. Let A be an integrally closed domain with fraction field K. Let L be a finite extension of K, and let B be the integral closure of A in L. Let b ∈ B. Then the (monic) minimal polynomial, f (X), of b over K lies in A[X].


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Proof Let L be an algebraic closure of L. Then we have a factorization f (X) = (X − b1 )r1 . . . (X − bk )rk in L[X]. For each i, there is an embedding of L in L that carries b to bi , so each bi must be integral over A. The coefficients of f (X) are polynomials in the roots bi , and hence are integral over A. Since A is integrally closed and the coefficients lie in K, the result follows. Corollary 12.7.10. Let A be an integrally closed domain with fraction field K. Let L be an extension field of K, and suppose that α ∈ L is integral over A. Then A[α] is a free A-module of rank [K(α) : K]. Recall that a number field is a finite extension of the rational numbers, Q. Definitions 12.7.11. Let K be a number field, or, more generally, an algebraic extension of Q. Then the integral closure of Z in K is known as the ring of integers of K, and is denoted by O(K). The elements of O(K) are known as the algebraic integers in K. As an example, we calculate the ring of integers of a quadratic extension of Q. Recall from √ Problem 8 of Exercises 8.2.24 that every quadratic extension of Q has the form Q( n), where n is a square-free integer. Proposition 12.7.12. Let n = 1 be a square-free integer. Then 2 √ Z[6 n] 7 if n ≡ 1 mod 4 √ √ O(Q( n)) = Z 1+2 n if n ≡ 1 mod 4. √ √ Proof Write K = Q( n)√and let √ α = a + b n with a, b ∈ Q. Since the nontrivial element of Gal(K/Q) takes n to − n, Proposition 11.14.3 gives √ √ and trK/Q (α)=(a + b √n) + (a − b√ n)=2a NK/Q (α)= (a + b n) · (a − b n) =a2 − nb2 . Note that if b = 0, then the minimal polynomial of α over Q is X 2 − trK/Q (α)X + NK/Q (α). This, together with Lemma 12.7.8, shows that α ∈ O(K) if and only if 2a ∈ Z and a2 − nb2 ∈ Z. Suppose α ∈ O(K). Then a = c/2 with c ∈ Z. Recall that n is square-free. Thus, 2 2 the norm formula now gives b = d/2 with d ∈ Z, and further √ gives c − nd ≡ 0 mod 4. If c is even, then so is d, so that a, b ∈ Z, giving α ∈ Z[ n]. If c is odd, then so is d, and hence c√2 ≡ d2 ≡ 1 mod 4. But this forces n ≡ 1 mod 4. Thus, if n ≡ 1 mod 4, then O(K) = Z[ n], as claimed. √ If n ≡ 1 mod 4 and c and d are odd, then α = (c + d n)/2 is easily seen√ to have integral√norm and trace, and hence to lie√in O(K). Note that since √ 1 and n lie in Z[(1 + n)/2], it is easy to see that (c + d n)/2 also lies in Z[(1 + n)/2] if c and d are odd. The result follows. Note that negative values of n (including n = −1) are included in the above. Rings of integers of number fields are the central object of study in the field of algebraic number theory. There are many unanswered questions about them. Of particular interest to many researchers is the study of the rings of integers of the cyclotomic extensions of Q. It is a fact that O(Q(ζn )) = Z[ζn ] for all integers n. We shall prove this presently if n is prime.


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Lemma 12.7.13. Let a be the principal ideal of O(Q(ζp )) generated by 1 − ζp , for a prime number, p. Then a ∩ Z = (p) and trQ(ζp )/Q (a) ⊂ (p). Proof Since Φp (X) = 1 + X + · · · + X p−1 , we see that p = Φp (1)

= (1 − ζp ) . . . (1 − ζpp−1 ) = NQ(ζp )/Q (1 − ζp ).

Since NQ(ζp )/Q : O(Q(ζp )) → Z is a homomorphism of multiplicative monoids, we see that 1 − ζp cannot be a unit in O(Q(ζp )). Thus, the principal ideal a = (1 − ζp ) is proper, and hence a ∩ Z is a proper ideal of Z containing p. Since (p) is maximal, a ∩ Z = (p). For k > 1, we have 1 − ζpk = (1 − ζp ) · (1 + ζp + · · · + ζpk−1 ) in O(Q(ζp )), so that 1 − ζpk ∈ a. Thus, if σ ∈ Gal(Q(ζp )/Q), then σ(1 − ζp ) ∈ a. So if α ∈ O(Q(ζp )), Proposition 11.14.3 gives trQ(ζp )/Q (α(1 − ζp )) = σ(α)σ(1 − ζp ) σ∈Gal(Q(ζp )/Q)

∈

a ∩ Z = (p).

Proposition 12.7.14. For any prime p, O(Q(ζp )) = Z[ζp ]. Proof For p = 2, the result is trivial, so assume that p is odd. Let α = a0 + a1 ζp + · · · + ap−2 ζpp−2 be an element of O(Q(ζp )), with ai ∈ Q for all i. We shall show that ai ∈ Z for all i. Note first that α(1 − ζp ) = a0 (1 − ζp ) + a1 (ζp − ζp2 ) + · · · + ap−2 (ζpp−2 − ζpp−1 ). By Proposition 11.14.3, if (k, p) = 1, we have trQ(ζp )/Q (ζpk )

= ζp + ζp2 + · · · + ζpp−1 = −1,

since ζp is a root of 1 + X + · · · + X p−1 . Thus, if 1 ≤ k ≤ p − 2, ζpk − ζpk+1 lies in the kernel of trQ(ζp )/Q . And since trQ(ζp )/Q (1) = [Q(ζp ) : Q] = p − 1, we see that trQ(ζp )/Q (1 − ζp ) = p. Thus, trQ(ζp )/Q (α(1 − ζp ))

= trQ(ζp )/Q (a0 (1 − ζp )) = a0 · p,

By Lemma 12.7.13, trQ(ζp )/Q (α(1 − ζp )) ∈ (p) ⊂ Z, so a0 ∈ Z. Now, ζp−1 = ζpp−1 ∈ O(Q(ζp )), so ζp−1 (α − a0 ) is in O(Q(ζp )). By the argument given for a0 , we see that a1 ∈ Z. By induction, ai ∈ Z for all i. We shall see presently that the ring of integers in a number field is a Dedekind domain. It behooves us to see how the concepts of integral dependence and of Dedekind domains interrelate. First, we shall show that Dedekind domains are integrally closed.


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Lemma 12.7.15. Let A ⊂ B be an inclusion of commutative rings and let C be the integral closure of A in B. Let S be a multiplicative subset of A. Then S −1 C is the integral closure of S −1 A in S −1 B. Proof The integral closure of S −1 A in S −1 B is a subring of S −1 B containing both S −1 A and the image of C under the canonical map η : B → S −1 B. Thus, it must contain S −1 C. Now suppose that b/s is integral over S −1 A. Since s lies in A, we see that η(b) must also lie in the integral closure of S −1 A. Recall from Problem 9 of Exercises 7.11.27 that we may assume 1 ∈ S, so that η(b) = b/1. It suffices to show that there’s an element of the form bt in C with t ∈ S, as b/s = (bt)/(st). Suppose, then, that b/1 is a root of f (X) = X n +

an−1 n−1 a0 X + ··· + , sn−1 s0

where ai ∈ A and si ∈ S for all i. Now set t = s0 . . . sn−1 and set ti = bt/1 is a root of

!

j =i sj .

Then

g(X) = X n + an−1 tn−1 X n−1 + · · · + ai ti tn−i−1 X i + · · · + a0 t0 tn−1 , which we may think of as a polynomial with coefficients in A. So we are done if η is injective. In the general case, g(bt) gives an element of A that lies in the kernel of η, so n−1 that u · g(bt) = 0 for some u ∈ S. But if we write g(X) = X n + i=0 ai X i , we see that n−1 btu is a root of X n + i=0 ai un−i X i , and the result follows. Proposition 12.7.16. Let A be an integral domain. Then the following conditions are equivalent. 1. A is integrally closed. 2. Ap is integrally closed for all primes p of A. 3. Am is integrally closed for all maximal ideals m of A. Proof If K is the fraction field, then K = Kp for each prime p. Thus, the fact that the first condition implies the second is immediate from Lemma 12.7.15, while the second condition immediately implies the third. Suppose the third condition holds, and let C be the integral closure of A in the fraction field K. By Lemma 12.7.15, the third condition implies that the inclusion Am ⊂ Cm is surjective for each maximal ideal m of A. By Corollary 7.11.26, this implies that the inclusion A ⊂ C is also surjective, and hence A is integrally closed. Corollary 12.5.4 shows that every localization of a Dedekind domain is a discrete valuation ring. So Lemma 12.7.5 shows that every localization of a Dedekind domain is integrally closed. Thus, Proposition 12.7.16 gives the following corollary. Corollary 12.7.17. Dedekind domains are integrally closed. We next wish to examine the relationship between integral dependence and Krull dimension. Lemma 12.7.18. Let A ⊂ B be an inclusion of domains such that B is integral over A. Let b be any nonzero ideal of B. Then b ∩ A = 0.


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n−1 Proof Let b be a nonzero element of b and let f (X) = X n + i=0 ai X i be a monic polynomial over A with f (b) = 0. Note that if a0 = 0, then f is divisible by X. Since B is a domain, the result of this division will still have b as a root. By induction, we can find a monic polynomial over A with a nonzero constant term with b as a root. But the constant term is then easily seen to be an element of b ∩ A. Recall that for any ring homomorphism f : A → B between commutative rings and any prime ideal p of B, f −1 (p) is a prime ideal of A. Proposition 12.7.19. Let A ⊂ B be an inclusion of commutative rings such that B is integral over A. Let q 0 ⊂ q1 ⊂ · · · ⊂ qn be a sequence of proper inclusions of prime ideals of B. Then the inclusions (q0 ∩ A) ⊂ (q1 ∩ A) ⊂ · · · ⊂ (qn ∩ A) are also proper. Thus, the Krull dimension of B is less than or equal to the Krull dimension of A. Proof It suffices to assume that n = 1 above. Thus, suppose we have a proper inclusion q0 ⊂ q1 of prime ideals of B. Suppose, then, that q0 ∩ A = q1 ∩ A = p. Then we have an inclusion of domains A/p ⊂ B/q0 . And the prime ideal q1 = q1 /q0 of the latter has the property that q1 ∩ A/p = 0. But B/q0 is integral over A/p, so this contradicts Lemma 12.7.18. We wish to show the opposite implication with regard to the Krull dimensions of A and B. Here is a start. Lemma 12.7.20. Let A ⊂ B be an inclusion of domains such that B is integral over A. Then B is a field if and only if A is a field. Proof One direction is immediate from Proposition 12.7.19, as a domain is a field if and only if it has Krull dimension 0. Thus, if A is a field, so is B. Conversely, suppose that B is a field, and let 0 = a ∈ A. Then a−1 lies in B, and hence is integral over A. We obtain an equation a−n = −an−1 a−(n−1) − · · · − a0 with coefficients in A. Multiplying both sides of this equation by an−1 , we get a−1 = −an−1 − an−2 a − · · · − a0 an−1 , so a−1 lies in A. Since a was arbitrary, A is a field. Corollary 12.7.21. Let A ⊂ B be an inclusion of commutative rings such that B is integral over A. Let m be a maximal ideal of B. Then m ∩ A is a maximal ideal of A. Proof Since m ∩ A is prime in A, we have an inclusion A/(m ∩ A) ⊂ B/m of domains, where B/m is integral over A/(m ∩ A). Since B/m is a field, A/(m ∩ A) must be also. Proposition 12.7.22. Let A ⊂ B be an inclusion of commutative rings such that B is integral over A. Let p be a prime ideal of A. Then there is a prime ideal q of B with q ∩ A = p.


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Proof We have a commutative diagram A η Ap

⊂

/ B η

⊂ / Bp

.

By Lemma 12.7.15, Bp is integral over Ap . Thus, if m is a maximal ideal of Bp , Corollary 12.7.21 shows that m ∩ Ap must be the unique maximal ideal, pp , of Ap . Since p = η −1 (pp ), the result follows from the commutativity of the square. We can now prove what’s known as the Going Up Theorem of Cohen and Seidenberg. Theorem 12.7.23. (Going Up Theorem) Let A ⊂ B be an inclusion of commutative rings such that B is integral over A. Suppose given a sequence of inclusions p0 ⊂ · · · ⊂ pn of prime ideals of A, together with a “lift” of the beginning part of the sequence to a sequence of prime ideals of B: Thus, we are given a sequence q 0 ⊂ · · · ⊂ qk of prime ideals of B, with k < n, such that qi ∩ A = pi . Then we may extend this to a lift of the entire sequence, producing prime ideals qi of B for i = k + 1, . . . , n such that qi ∩ A = pi for all i, and so that q0 ⊂ · · · ⊂ q n . Proof An easy induction shows that we may assume that n = 1 and k = 0. Now pass to the inclusion A/p0 ⊂ B/q0 of domains, and apply Proposition 12.7.22 to the prime ideal p1 /p0 of A/p0 . Corollary 12.7.24. Let A ⊂ B be an inclusion of commutative rings such that B is integral over A. Then the Krull dimensions of A and B are equal in the sense that if either is finite, so is the other, and the two dimensions then agree. Proof By Proposition 12.7.19, it suffices to show that if A has a chain of proper inclusions of prime ideals of length n, then so does B. But this is immediate from Proposition 12.7.22 and the Going Up Theorem. We shall now give a characterization of Dedekind domains to the effect that they are the Noetherian, integrally closed domains of Krull dimension ≤ 1. First, we shall reexamine the notion of invertibility. Definition 12.7.25. Let A be an integral domain with fraction field K and let a be a nonzero ideal of A. We write a−1 for the A-submodule of K given by a−1 = {α ∈ K | αa ⊂ A}. Note that for a ∈ a, we have aa−1 ⊂ A. Thus, a−1 is isomorphic as an A-module to an ideal of A. This property constitutes one definition for a fractional ideal over a general domain. Recall that there is a product operation on the A-submodules of K, generalizing the product of ideals. The inverse A-submodule of an ideal a satisfies the following important property.


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Lemma 12.7.26. Let a be a nonzero ideal of the domain A. Then a is invertible if and only if aa−1 = A. Proof Let 0 = a ∈ a. Then aa−1 = (a)a−1 is an ideal of A. If aa−1 = A, we have a · (aa−1 ) = (a), so a is invertible. Conversely, if a is invertible, there is an ideal b such that ab = (a) for some a ∈ A. But then a · ((1/a)b) = A, so (1/a)b ⊂ a−1 , and A = a · ((1/a)b) ⊂ aa−1 ⊂ A, so aa−1 = A. Recall from Proposition 7.6.16 that Noetherian rings possess a stronger maximality principle than is found in general rings: If S is any nonempty family of ideals in a Noetherian ring, then there are maximal elements in S. This permits a sort of downward induction on ideals in a Noetherian ring. To show that a given property holds, one may argue by contradiction, choosing a maximal element in the set of ideals in which the property doesn’t hold, and deriving a contradiction. Here is an illustration of this technique. Lemma 12.7.27. Let A be a Noetherian commutative ring. Then every ideal of A contains a product of prime ideals. If A is a Noetherian domain but not a field, then every nonzero ideal of A contains a product of nonzero prime ideals. Proof We shall treat the statement in the first paragraph. The proof of the other is similar. Let a be maximal in the collection of ideals that do not contain a product of prime ideals. Then a is neither A nor a prime ideal of A. In particular, we can find a, b ∈ A such that neither a nor b lies in a, but ab lies in a. Thus, a + (a) and a + (b) strictly contain a. By the maximality of a, this says that both a + (a) and a + (b) contain products of prime ideals. But (a + (a))(a + (b)) ⊂ a, so a must also contain a product of prime ideals. Thus, the set of ideals that do not contain a product of primes must be empty. Another example of Noetherian induction will be useful in our characterization of Dedekind domains. Proposition 12.7.28. Let A be a Noetherian domain such that every maximal ideal of A is invertible. Then A is a Dedekind domain. Proof We claim that every proper, nonzero ideal is a product of maximal ideals. Supposing this false, let a be a maximal element in the set of proper, nonzero ideals that are not products of maximal ideals. Then a cannot be maximal, and hence must be contained in a maximal ideal m. Our hypothesis says that m is invertible, and hence m−1 m = A by Lemma 12.7.26. Since a ⊂ m, m−1 a ⊂ m−1 m, and hence is an ideal of A. Since A ⊂ m−1 , a ⊂ m−1 a. Note now that m−1 a properly contains a, as otherwise ma = a, and hence Problem 4 of Exercises 10.4.11 shows that a = 0. Also, m−1 a = A, as otherwise, a = m. So the maximality of a shows that m−1 a is a product of maximal ideals. But then a = m · m−1 a is also a product of maximal ideals, and we obtain our desired contradiction. Recall that Dedekind domains are Noetherian and integrally closed. Thus, the next lemma will allow us to close in on our new characterization of Dedekind domains.


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Lemma 12.7.29. Let A be a Noetherian, integrally closed domain that is not a field. Let m be a maximal ideal of A that is not invertible. Then the inverse fractional ideal satisfies m−1 = A. Proof From the definition of m−1 we see that A ⊂ m−1 and m−1 m ⊂ A. Thus, we have inclusions m ⊂ m−1 m ⊂ A of ideals. Since m is maximal, one of these inclusions is the identity, and since m is not invertible, we must have m = m−1 m. Let α ∈ m−1 . Since m−1 m = m, we must have αm ⊂ m, and hence m is an A[α]module. Since A is Noetherian, m is finitely generated as an A-module. Since every nonzero A[α]-submodule of the fraction field K is faithful, Proposition 12.7.2 shows that α is integral over A. Since A is integrally closed, this says α ∈ A, and the result follows. We can now give our new characterization of Dedekind domains. Theorem 12.7.30. A domain is a Dedekind domain if and only if it is Noetherian, integrally closed, and has Krull dimension ≤ 1. Proof That Dedekind domains have these three properties is given in Corollary 12.6.7 (via Theorem 12.6.8), Corollary 12.7.17, and Proposition 12.6.3. For the converse, Lemma 12.7.29 and Proposition 12.7.28 show that if A is a Noetherian, integrally closed domain of Krull dimension 1, then A is Dedekind provided we can show that for any maximal ideal m of A, there is an element α of the inverse fractional ideal m−1 that does not lie in A. (To show the existence of such an α, all we shall need is that A is a Noetherian domain of Krull dimension 1.) Let a ∈ m with (a) = m. Since A is Noetherian, Lemma 12.7.27 shows that (a) contains a product of nonzero prime ideals, which are maximal, since A has Krull dimension 1. Let r be the smallest integer such that (a) contains a product m1 . . . mr of r maximal ideals. Since (a) = m, r > 1. Since m1 . . . mr ⊂ (a) ⊂ m, one of the mi must lie in m because m is prime. Since both mi and m are maximal, we obtain that m = mi . For simplicity, assume i = 1. By the minimality of r, we can find b ∈ m2 . . . mr such that b ∈ (a). Now bm ⊂ m1 . . . mr ⊂ (a), so (b/a)m ⊂ A. Since b ∈ (a), a doesn’t divide b, and hence α = (b/a) is not in A. But α ∈ m−1 , and the result follows. We can now address the question of rings of integers. Theorem 12.7.31. Let A be a Dedekind domain with fraction field K. Let L be a finite separable extension field of K, and let B be the integral closure of A in L. Then B is a Dedekind domain with fraction field L. In addition, B is finitely generated and projective as an A-module, with Ap ⊗A B being a free Ap -module of rank [L : K] for each prime ideal p of A. In particular, if A is a P.I.D., then B is a free A-module of rank [L : K]. If m is a maximal ideal of B, then m = m ∩ A is a maximal ideal of A, and B/m is a finite extension of A/m, with [B/m : A/m] ≤ [L : K].


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Proof B is integrally closed by Proposition 12.7.7, and has Krull dimension ≤ 1 by Proposition 12.7.19. Once we show that B is a finitely generated A-module, we will know that B is Noetherian, at which point Theorem 12.7.30 will show that B is Dedekind, as claimed. Thus, it suffices to verify the assertions of the second and third paragraphs of the statement. Let L be an algebraic closure of L and let σ1 , . . . , σn be all the distinct embeddings (as a field) of L in L that restrict to the identity on K. Since L is separable over K, n = [L : K] by Proposition 11.4.12. Recall from Proposition 11.14.3 that the trace function trL/K : L → K of L over K is given by trL/K (α) = σ1 (α) + · · · + σn (α), because L is separable over K. Since the σi are ring homomorphisms, trL/K is easily seen to be a homomorphism of vector spaces over K. By Lemma 12.7.8, it restricts to an A-module homomorphism, trL/K : B → A. Now define : L → HomK (L, K) tr is an isomorphism of K-vector spaces. To see by tr(α)(β) = trL/K (αβ). We claim that tr is injective. this, since both vector spaces have dimension n, it suffices to show that tr Thus, we must show that for each nonzero α in L, there exists a β ∈ L such that trL/K (αβ) = 0. Thus, since L is a field, it suffices to show that trL/K : L → K is not the trivial homomorphism. Since trL/K is a sum of distinct embeddings, this follows immediately from Corollary 11.9.3. Recall from Proposition 12.7.7 that for any α ∈ L there is an a ∈ A such that aα ∈ B. In particular, beginning with any K-basis of L, we can construct a new basis whose elements all lie in B. Let b1 , . . . , bn be such a basis, and define f : L → K n by f (α) = (trL/K (b1 α), . . . , trL/K (bn α)). Since b1 , . . . , bn is a basis for L over K, any element in the kernel of f must also lie in Thus, f is an isomorphism of K-vector spaces. the kernel of tr. But trL/K maps B into A, so f is easily seen to restrict to an embedding of Amodules, f : B → An . Thus, since A is hereditary and Noetherian, B is a finitely generated projective A-module by Proposition 12.1.5. In fact, B is isomorphic as an A-module to a direct sum a1 ⊕ · · · ⊕ ak of ideals of A. If B is the direct sum of k nonzero ideals, then Ap ⊗A B is easily seen to have rank k over Ap for each prime ideal p of A. But Proposition 12.7.7 shows that every element of L has the form b/a for b in B and a ∈ A, so that L = K ⊗A B. Thus, k = n. If m is a maximal ideal of B, then m = m ∩ A is maximal in A by Corollary 12.7.21. Write S ⊂ A for the complement of m in A. We’ve just seen that S −1 B is a free S −1 Amodule of rank [L : K], so it suffices to show that the natural maps A/m → S −1 A/S −1 m and B/m → S −1 B/S −1 m are isomorphisms. The former is an isomorphism by Proposition 7.11.24, and the latter is an isomorphism by a similar argument. Without the assumption of separability, we shall show the following. Theorem 12.7.32. Let A be a Dedekind domain with fraction field K. Let L be a finite extension of K and let B be the integral closure of A in L. Then B is a Dedekind domain.


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Proof Let L0 be the separable closure of K in L. (See Problem 2 of Exercises 11.4.20.) Then L0 is separable over K, and L is purely inseparable over L0 . Let C be the integral closure of A in L0 . Then Theorem 12.7.31 shows C to be a Dedekind domain. Since B is the integral closure of C in L, we may assume that K = L0 and that L is a purely inseparable extension of K. By Problem 2 of Exercises 11.4.20, the minimal polynomial of any element β ∈ L e over K has the form X p − a for some a ∈ K. By Lemma 12.7.9, we have a ∈ A if and only if β ∈ B. Since the degrees pe must all divide [L : K], there is an integer r such r r that β p ∈ K for all β ∈ L, and β ∈ B if and only if β p ∈ A. Let L be an algebraic closure of L and let ϕ : L → L be the Frobenius homomorphism. = (ϕr )−1 (K) and let A = (ϕr )−1 (A). Then the above shows that L ⊂ K, and Let K Since ϕ is a homomorphism of fields, A is isomorphic to A, and hence is a B = L ∩ A. Dedekind domain. By Theorem 12.6.8 and Proposition 12.6.6, it suffices to show that every nonzero ideal is Dedekind, we know that Ab is invertible as an ideal of A. b of B is invertible. Since A By Lemma 12.7.26, this means that if we set | γ · Ab ⊂ A}, M = {γ ∈ K = A. This is easily seen to imply that there are elements γi ∈ M and bi ∈ b then M · Ab k such that i=1 γi bi = 1. r r k Applying ϕr , we obtain that i=1 δi bi = 1, where δi = γip bpi −1 . By Lemma 12.7.26, it suffices to show that each δi ∈ L and that δi b ⊂ B. Since ϕr (γi ) ∈ K and bi ∈ b, δi ∈ L, as desired. The result now follows since if b ∈ b, r ∩ L = B. δi b = (γi bi )p −1 (γi b) lies in A Exercises 12.7.33. 1. Show that the assertion of Lemma 12.7.18 becomes false if we remove the assumption that A and B are domains. √ 2. Let n be a square-free integer. What is the integral closure of Z[1/2] in Q( n)? 3. Let A be a Dedekind domain and let B be the integral closure of A in a finite extension of its field of fractions. Let p be a nonzero prime ideal of A. Show that Bp is a proper ideal of B. Suppose Bp = qr11 . . . qrkk , where q1 , . . . , qk are distinct prime ideals of B and each ri > 0. (We say that p ramifies in B if ri > 1 for some i.) Find all ideals b of B with the property that b ∩ A = p. 4. Let A be an integrally closed domain and let B be the integral closure of A in a finite extension, L, of its field of fractions, K. Show that b ∈ B × if and only if NL/K (b) ∈ A× . 5. Let p be a prime and let r > 0. Let a be the principal ideal of Z[ζpr ] generated by r−1 1 − ζpr . Show that a(p−1)p = (p), the principal ideal of Z[ζpr ] generated by p. Deduce (unless p = 2 and r = 1) that p ramifies in O(Q(ζpr )). 6. Let n > 1 be an integer with more than one prime divisor. Show that 1 − ζn is a unit in Z[ζn ].

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