Abstract Algebra (Course 311)

Course 311: Abstract Algebra Academic year 2007-08 D. R. Wilkins c David R. Wilkins 1997–2007 Copyright

Contents 1 Topics in Group Theory 1.1 Groups . . . . . . . . . . . . . . . . . . . . 1.2 Examples of Groups . . . . . . . . . . . . 1.3 Elementary Properties of Groups . . . . . 1.4 Subgroups . . . . . . . . . . . . . . . . . . 1.5 Cyclic Groups . . . . . . . . . . . . . . . . 1.6 Cosets and Lagrange’s Theorem . . . . . . 1.7 Normal Subgroups and Quotient Groups . 1.8 Homomorphisms . . . . . . . . . . . . . . 1.9 The Isomorphism Theorems . . . . . . . . 1.10 Group Actions, Orbits and Stabilizers . . . 1.11 Conjugacy . . . . . . . . . . . . . . . . . . 1.12 The Class Equation of a Finite Group . . . 1.13 Cauchy’s Theorem . . . . . . . . . . . . . 1.14 The Structure of p-Groups . . . . . . . . . 1.15 The Sylow Theorems . . . . . . . . . . . . 1.16 Some Applications of the Sylow Theorems 1.17 Simple Groups . . . . . . . . . . . . . . . 1.18 Solvable Groups . . . . . . . . . . . . . . .

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Course 311: Abstract Algebra Academic year 2007-08 Chapter 2: Rings and Polynomials D. R. Wilkins c David R. Wilkins 1997–2007 Copyright

Contents 2 Rings and Polynomials 2.1 Rings, Integral Domains and Fields . 2.2 Ideals . . . . . . . . . . . . . . . . . 2.3 Quotient Rings and Homomorphisms 2.4 The Characteristic of a Ring . . . . . 2.5 Polynomial Rings . . . . . . . . . . . 2.6 Gauss’s Lemma . . . . . . . . . . . . 2.7 Eisenstein’s Irreducibility Criterion .

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Course 311: Abstract Algebra Academic year 2007-08 D. R. Wilkins c David R. Wilkins 1997–2007 Copyright

Contents 3 Introduction to Galois Theory 3.1 Field Extensions and the Tower Law . . . . . . . . . 3.2 Algebraic Field Extensions . . . . . . . . . . . . . . . 3.3 Algebraically Closed Fields . . . . . . . . . . . . . . . 3.4 Ruler and Compass Constructions . . . . . . . . . . . 3.5 Splitting Fields . . . . . . . . . . . . . . . . . . . . . 3.6 Normal Extensions . . . . . . . . . . . . . . . . . . . 3.7 Separability . . . . . . . . . . . . . . . . . . . . . . . 3.8 Finite Fields . . . . . . . . . . . . . . . . . . . . . . . 3.9 The Primitive Element Theorem . . . . . . . . . . . . 3.10 The Galois Group of a Field Extension . . . . . . . . 3.11 The Galois correspondence . . . . . . . . . . . . . . . 3.12 Quadratic Polynomials . . . . . . . . . . . . . . . . . 3.13 Cubic Polynomials . . . . . . . . . . . . . . . . . . . 3.14 Quartic Polynomials . . . . . . . . . . . . . . . . . . 3.15 The Galois group of the polynomial x4 − 2 . . . . . . 3.16 The Galois group of a polynomial . . . . . . . . . . . 3.17 Solvable polynomials and their Galois groups . . . . . 3.18 A quintic polynomial that is not solvable by radicals

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Course 311: Abstract Algebra Academic year 2007-08 D. R. Wilkins c David R. Wilkins 1997–2007 Copyright

Contents 4 Commutative Algebra and Algebraic Geometry 4.1 Modules . . . . . . . . . . . . . . . . . . . . . . . 4.2 Noetherian Modules . . . . . . . . . . . . . . . . 4.3 Noetherian Rings and Hilbert’s Basis Theorem . . 4.4 Polynomial Rings in Several Variables . . . . . . . 4.5 Algebraic Sets and the Zariski Topology . . . . . 4.6 The Structure of Algebraic Sets . . . . . . . . . . 4.7 Maximal Ideals and Zorn’s Lemma . . . . . . . . 4.8 Prime Ideals . . . . . . . . . . . . . . . . . . . . . 4.9 Affine Varieties and Irreducibility . . . . . . . . . 4.10 Radical Ideals . . . . . . . . . . . . . . . . . . . . 4.11 Commutative Algebras of Finite Type . . . . . . 4.12 Zariski’s Theorem . . . . . . . . . . . . . . . . . . 4.13 Hilbert’s Nullstellensatz . . . . . . . . . . . . . .

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1

Topics in Group Theory

1.1

Groups

A binary operation ∗ on a set G associates to elements x and y of G a third element x ∗ y of G. For example, addition and multiplication are binary operations of the set of all integers. Definition A group G consists of a set G together with a binary operation ∗ for which the following properties are satisfied: • (x ∗ y) ∗ z = x ∗ (y ∗ z) for all elements x, y, and z of G (the Associative Law ); • there exists an element e of G (known as the identity element of G) such that e ∗ x = x = x ∗ e, for all elements x of G; • for each element x of G there exists an element x0 of G (known as the inverse of x) such that x ∗ x0 = e = x0 ∗ x (where e is the identity element of G). The order |G| of a finite group G is the number of elements of G. A group G is Abelian (or commutative) if x ∗ y = y ∗ x for all elements x and y of G. One usually adopts multiplicative notation for groups, where the product x ∗ y of two elements x and y of a group G is denoted by xy. The associative property then requires that (xy)z = x(yz) for all elements x, y and z of G. The identity element is often denoted by e (or by eG when it is necessary to specify explicitly the group to which it belongs), and the inverse of an element x of G is then denoted by x−1 . It is sometimes convenient or customary to use additive notation for certain groups. Here the group operation is denoted by +, the identity element of the group is denoted by 0, the inverse of an element x of the group is denoted by −x. By convention, additive notation is rarely used for non-Abelian groups. When expressed in additive notation the axioms for a Abelian group require that (x + y) + z = x + (y + z), x + y = y + x, x + 0 = 0 + x = x and x + (−x) = (−x) + x = 0 for all elements x, y and z of the group. We shall usually employ multiplicative notation when discussing general properties of groups. Additive notation will be employed for certain groups (such as the set of integers with the operation of addition) where this notation is the natural one to use.

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1.2

Examples of Groups

The sets of integers, rational numbers, real numbers and complex numbers are Abelian groups, where the group operation is the operation of addition. The sets of non-zero rational numbers, non-zero real numbers and nonzero complex numbers are also Abelian groups, where the group operation is the operation of multiplication. For each positive integer m the set Zm of congruence classes of integers modulo m is a group, where the group operation is addition of congruence classes. For each positive integer m the set Z∗m of congruence classes modulo m of integers coprime to m is a group, where the group operation is multiplication of congruence classes. In particular, for each prime number p the set Z∗p of congruence classes modulo p of integers not divisible by p is a group, where the group operation is multiplication of congruence classes. For each positive integer n the set of all nonsingular n × n matrices is a group, where the group operation is matrix multiplication. These groups are not Abelian when n ≥ 2. Let E n denote n-dimensional Euclidean space, so that E 2 denotes the Euclidean plane, and E 3 denotes three-dimensional Euclidean space. A geometrical figure may be represented as a subset S of E n . A symmetry of S is a transformation T : E n → E n of E n which sends straight lines to straight lines, preserves all lengths and angles, and has the property that T (S) = S. The collection of all symmetries of a geometrical figure is a group, the symmetry group of S, the group operation being that of composition of transformations. For any natural number n greater than 2, the the dihedral group D2n of order 2n is defined to be the symmetry group of a regular n-sided polygon in the Euclidean plane. It consists of rotations though an angle of 2πj/n about the centre of the polygon for j = 0, 1, 2, . . . , n − 1, together with the reflections in the n axes of symmetry of the polygon. The symmetries of a rectangle that is not a square constitute a group of order 4. This group consists of the identity transformation, reflection in the axis of symmetry joining the midpoints of the two shorter sides, reflection in the axis of symmetry joining the two longer sides, and rotation though an angle of π radians (180◦ ). If I denotes the identity transformation, A and B denote the reflections in the two axes of symmetry, and C denotes the rotation through π radians then A2 = B 2 = C 2 = I, AB = BA = C, AC = CA = B and BC = CB = A. This group is Abelian: it is often referred to as the Klein 4-group (or, in German, Kleinsche Viergruppe). The symmetries of a regular tetrahedron in 3-dimensional space constitute 2

a group. Any permutation of the vertices of the tetrahedron can be effected by an appropriate symmetry of the tetrahedron. Moreover each symmetry is completely determined by the permutation of the vertices which it induces. Therefore the group of symmetries of a regular tetrahedron is of order 24, since there are 24 permutations of a set with four elements. It turns out that this group is non-Abelian.

1.3

Elementary Properties of Groups

In what follows, we describe basic properties of a group G, using multiplicative notation and denoting the identity element of the group by the letter e. Lemma 1.1 A group G has exactly one identity element e satisfying ex = x = xe for all x ∈ G. Proof Suppose that f is an element of G with the property that f x = x for all elements x of G. Then in particular f = f e = e. Similarly one can show that e is the only element of G satisfying xe = x for all elements x of G. Lemma 1.2 An element x of a group G has exactly one inverse x−1 . Proof We know from the axioms that the group G contains at least one element x−1 which satisfies xx−1 = e and x−1 x = e. If z is any element of G which satisfies xz = e then z = ez = (x−1 x)z = x−1 (xz) = x−1 e = x−1 . Similarly if w is any element of G which satisfies wx = e then w = x−1 . In particular we conclude that the inverse x−1 of x is uniquely determined, as required. Lemma 1.3 Let x and y be elements of a group G. Then (xy)−1 = y −1 x−1 . Proof It follows from the group axioms that (xy)(y −1 x−1 ) = x(y(y −1 x−1 )) = x((yy −1 )x−1 ) = x(ex−1 ) = xx−1 = e. Similarly (y −1 x−1 )(xy) = e, and thus y −1 x−1 is the inverse of xy, as required. Note in particular that (x−1 )−1 = x for all elements x of a group G, since x has the properties that characterize the inverse of the inverse x−1 of x. Given an element x of a group G, we define xn for each positive integer n by the requirement that x1 = x and xn = xn−1 x for all n > 1. We also define x0 = e, where e is the identity element of the group, and we define x−n to be the inverse of xn for all positive integers n. 3

Theorem 1.4 Let x be an element of a group G. Then xm+n = xm xn and xmn = (xm )n for all integers m and n. Proof The identity xm+n = xm xn clearly holds when m = 0 and when n = 0. The identity xm+n = xm xn can be proved for all positive integers m and n by induction on n. The identity when m and n are both negative then follows from the identity x−m−n = x−n x−m on taking inverses. The result when m and n have opposite signs can easily be deduced from that where m and n both have the same sign. The identity xmn = (xm )n follows immediately from the definitions when n = 0, 1 or −1. The result when n is positive can be proved by induction on n. The result when n is negative can then be obtained on taking inverses. If additive notation is employed for an Abelian group then the notation ‘x ’ is replaced by ‘nx’ for all integers n and elements x of the group. The analogue of Theorem 1.4 then states that (m + n)x = mx + nx and (mn)x = m(n(x)) for all integers m and n. The associative law may be generalized to products of four or more elements of a group. n

Example Given four elements x1 , x2 , x3 and x4 of a group, the products ((x1 x2 )x3 )x4 ,

(x1 x2 )(x3 x4 ),

(x1 (x2 x3 ))x4 ,

x1 ((x2 x3 )x4 ),

x1 (x2 (x3 x4 ))

all have the same value. (Note that x1 x2 x3 x4 is by definition the value of the first of these expressions.) Two expressions, each specifying a finite product of elements of a group G, determine the same element of G if the same elements of G occur in both expressions, and in the same order. This result can be proved by induction on the number of elements of G making up such a product.

1.4

Subgroups

Definition Let G be a group, and let H be a subset of G. We say that H is a subgroup of G if the following conditions are satisfied: • the identity element of G is an element of H; • the product of any two elements of H is itself an element of H; • the inverse of any element of H is itself an element of H. A subgroup H of G is said to be proper if H 6= G. 4

Lemma 1.5 Let x be an element of a group G. Then the set of all elements of G that are of the form xn for some integer n is a subgroup of G. Proof Let H = {xn : n ∈ Z}. Then the identity element belongs to H, since it is equal to x0 . The product of two elements of H is itself an element of H, since xm xn = xm+n for all integers m and n (see Theorem 1.4). Also the inverse of an element of H is itself an element of H since (xn )−1 = x−n for all integers n. Thus H is a subgroup of G, as required. Definition Let x be an element of a group G. The order of x is the smallest positive integer n for which xn = e. The subgroup generated by x is the subgroup consisting of all elements of G that are of the form xn for some integer n. Lemma 1.6 Let H and K be subgroups of a group G. Then H ∩ K is also a subgroup of G. Proof The identity element of G belongs to H ∩ K since it belongs to the subgroups H and K. If x and y are elements of H ∩ K then xy is an element of H (since x and y are elements of H), and xy is an element of K, and therefore xy is an element of H ∩ K. Also the inverse x−1 of an element x of H ∩ K belongs to H and to K and thus belongs to H ∩ K, as required. More generally, the intersection of any collection of subgroups of a given group is itself a subgroup of that group.

1.5

Cyclic Groups

Definition A group G is said to be cyclic, with generator x, if every element of G is of the form xn for some integer n. Example The group Z of integers under addition is a cyclic group, generated by 1. Example Let n be a positive integer. The set Zn of congruence classes of integers modulo n is a cyclic group of order n with respect to the operation of addition. Example The group of all rotations of the plane about the origin through an integer multiple of 2π/n radians is a cyclic group of order n for all integers n. This group is generated by an anticlockwise rotation through an angle of 2π/n radians. 5

1.6

Cosets and Lagrange’s Theorem

Definition Let H be a subgroup of a group G. A left coset of H in G is a subset of G that is of the form xH, where x ∈ G and xH = {y ∈ G : y = xh for some h ∈ H}. Similarly a right coset of H in G is a subset of G that is of the form Hx, where x ∈ G and Hx = {y ∈ G : y = hx for some h ∈ H}. Note that a subgroup H of a group G is itself a left coset of H in G. Lemma 1.7 Let H be a subgroup of a group G. Then the left cosets of H in G have the following properties:— (i) x ∈ xH for all x ∈ G; (ii) if x and y are elements of G, and if y = xa for some a ∈ H, then xH = yH; (iii) if x and y are elements of G, and if xH ∩ yH is non-empty then xH = yH. Proof Let x ∈ G. Then x = xe, where e is the identity element of G. But e ∈ H. It follows that x ∈ xH. This proves (i). Let x and y be elements of G, where y = xa for some a ∈ H. Then yh = x(ah) and xh = y(a−1 h) for all h ∈ H. Moreover ah ∈ H and a−1 h ∈ H for all h ∈ H, since H is a subgroup of G. It follows that yH ⊂ xH and xH ⊂ yH, and hence xH = yH. This proves (ii). Finally suppose that xH ∩ yH is non-empty for some elements x and y of G. Let z be an element of xH ∩ yH. Then z = xa for some a ∈ H, and z = yb for some b ∈ H. It follows from (ii) that zH = xH and zH = yH. Therefore xH = yH. This proves (iii). Lemma 1.8 Let H be a finite subgroup of a group G. Then each left coset of H in G has the same number of elements as H. Proof Let H = {h1 , h2 , . . . , hm }, where h1 , h2 , . . . , hm are distinct, and let x be an element of G. Then the left coset xH consists of the elements xhj for j = 1, 2, . . . , m. Suppose that j and k are integers between 1 and m for which xhj = xhk . Then hj = x−1 (xhj ) = x−1 (xhk ) = hk , and thus j = k, since h1 , h2 , . . . , hm are distinct. It follows that the elements xh1 , xh2 , . . . , xhm are distinct. We conclude that the subgroup H and the left coset xH both have m elements, as required. 6

Theorem 1.9 (Lagrange’s Theorem) Let G be a finite group, and let H be a subgroup of G. Then the order of H divides the order of G. Proof Each element of G belongs to at least one left coset of H in G, and no element can belong to two distinct left cosets of H in G (see Lemma 1.7). Therefore every element of G belongs to exactly one left coset of H. Moreover each left coset of H contains |H| elements (Lemma 1.8). Therefore |G| = n|H|, where n is the number of left cosets of H in G. The result follows. Definition Let H be a subgroup of a group G. If the number of left cosets of H in G is finite then the number of such cosets is referred to as the index of H in G, denoted by [G: H]. The proof of Lagrange’s Theorem shows that the index [G: H] of a subgroup H of a finite group G is given by [G: H] = |G|/|H|. Corollary 1.10 Let x be an element of a finite group G. Then the order of x divides the order of G. Proof Let H be the set of all elements of G that are of the form xn for some integer n. Then H is a subgroup of G (see Lemma 1.5), and the order of H is the order of x. But the order of H divides G by Lagrange’s Theorem (Theorem 1.9). The result follows. Corollary 1.11 Any finite group of prime order is cyclic. Proof Let G be a group of prime order, and let x be some element of G that is not the identity element. Then the order of x is greater than one and divides the order of G. But then the order of x must be equal to the order of G, since the latter is a prime number. Thus G is a cyclic group generated by x, as required.

1.7

Normal Subgroups and Quotient Groups

Let A and B be subsets of a group G. The product AB of the sets A and B is defined by AB = {xy : x ∈ A and y ∈ B}. We denote {x}A and A{x} by xA and Ax, for all elements x of G and subsets A of G. The Associative Law for multiplication of elements of G ensures that (AB)C = A(BC) for all subsets A, B and C of G. We can therefore use the notation ABC to denote the products (AB)C and A(BC); 7

and we can use analogous notation to denote the product of four or more subsets of G. If A, B and C are subsets of a group G, and if A ⊂ B then clearly AC ⊂ BC and CA ⊂ CB. Note that if H is a subgroup of the group G and if x is an element of G then xH is the left coset of H in G that contains the element x. Similarly Hx is the right coset of H in G that contains the element x. If H is a subgroup of G then HH = H. Indeed HH ⊂ H, since the product of two elements of a subgroup H is itself an element of H. Also H ⊂ HH since h = eh for any element h of H, where e, the identity element of G, belongs to H. Definition A subgroup N of a group G is said to be a normal subgroup of G if xnx−1 ∈ N for all n ∈ N and x ∈ G. The notation ‘N / G’ signifies ‘N is a normal subgroup of G’. Definition A non-trivial group G is said to be simple if the only normal subgroups of G are the whole of G and the trivial subgroup {e} whose only element is the identity element e of G. Lemma 1.12 Every subgroup of an Abelian group is a normal subgroup. Proof Let N be a subgroup of an Abelian group G. Then xnx−1 = (xn)x−1 = (nx)x−1 = n(xx−1 ) = ne = n for all n ∈ N and x ∈ G, where e is the identity element of G. The result follows. Example Let S3 be the group of permutations of the set {1, 2, 3}, and let H be the subgroup of S3 consisting of the identity permutation and the transposition (1 2). Then H is not normal in G, since (2 3)−1 (1 2)(2 3) = (2 3)(1 2)(2 3) = (1 3) and (1 3) does not belong to the subgroup H. Proposition 1.13 A subgroup N of a group G is a normal subgroup of G if and only if xN x−1 = N for all elements x of G. Proof Suppose that N is a normal subgroup of G. Let x be an element of G. Then xN x−1 ⊂ N . (This follows directly from the definition of a normal subgroup.) On replacing x by x−1 we see also that x−1 N x ⊂ N , and thus N = x(x−1 N x)x−1 ⊂ xN x−1 . Thus each of the sets N and xN x−1 is contained in the other, and therefore xN x−1 = N . Conversely if N is a subgroup of G with the property that xN x−1 = N for all x ∈ G, then it follows immediately from the definition of a normal subgroup that N is a normal subgroup of G. 8

Corollary 1.14 A subgroup N of a group G is a normal subgroup of G if and only if xN = N x for all elements x of G. Proof Let N be a subgroup of G, and let x be an element of G. If xN x−1 = N then xN = (xN x−1 )x = N x. Conversely if xN = N x then xN x−1 = N xx−1 = N e = N , where e is the identity element of G. Thus xN = N x if and only if xN x−1 = N . It follows from Proposition 1.13 that a subgroup N of G is normal if and only if xN = N x for all elements x of G, as required. Let N be a normal subgroup of G. Corollary 1.14 shows that a subset of G is a left coset of N in G if and only if it is a right coset of N in G. We may therefore refer to the left and right cosets of a normal subgroup N as cosets of N in G (since it is not in this case necessary to distinguish between left and right cosets). Lemma 1.15 Let N be a normal subgroup of a group G and let x and y be elements of G. Then (xN )(yN ) = (xy)N . Proof If N is a normal subgroup of G then N y = yN , and therefore (xN )(yN ) = x(N y)N = x(yN )N = (xy)(N N ). But N N = N , since N is a subgroup of G. Therefore (xN )(yN ) = (xy)N , as required. Proposition 1.16 Let G be a group, and let N be a normal subgroup of G. Then the set of all cosets of N in G is a group under the operation of multiplication. The identity element of this group is N itself, and the inverse of a coset xN is the coset x−1 N for any element x of G. Proof Let x, y and z be any elements of G. Then the product of the cosets xN and yN is the coset (xy)N . The subgroup N is itself a coset of N in G, since N = eN . Moreover (xN )N = (xN )(eN ) = (xe)N = xN, N (xN ) = (eN )(xN ) = (ex)N = xN, (xN )(x−1 N ) = (xx−1 )N = N, (x−1 N )(xN ) = (x−1 x)N = N. for all elements x of G. Thus the group axioms are satisfied. Definition Let N be a normal subgroup of a group G. The quotient group G/N is defined to be the group of cosets of N in G under the operation of multiplication. 9

Example Consider the dihedral group D8 of order 8, which we represent as the group of symmetries of a square in the plane with corners at the points whose Cartesian co-ordinates are (1, 1), (−1, 1), (−1, −1) and (1, −1). Then D8 = {I, R, R2 , R3 , T1 , T2 , T3 , T4 }, where I denotes the identity transformation, R denotes an anticlockwise rotation about the origin through a right angle, and T1 , T2 , T3 and T4 denote the reflections in the lines y = 0, x = y, x = 0 and x = −y respectively. Let N = {I, R2 }. Then N is a subgroup of D8 . The left cosets of N in D8 are N , A, B and C, where A = {R, R3 },

B = {T1 , T3 },

C = {T2 , T4 }.

Moreover N , A, B and C are also the right cosets of N in D8 , and thus N is a normal subgroup of D8 . On multiplying the cosets A, B and C with one another we find that AB = BA = C, AC = CA = B and BC = CB = A. The quotient group D8 /N consists of the set {N, A, B, C}, with the group operation just described.

1.8

Homomorphisms

Definition A homomorphism θ: G → K from a group G to a group K is a function with the property that θ(g1 ∗ g2 ) = θ(g1 ) ∗ θ(g2 ) for all g1 , g2 ∈ G, where ∗ denotes the group operation on G and on K. Example Let q be an integer. The function from the group Z of integers to itself that sends each integer n to qn is a homomorphism. Example Let x be an element of a group G. The function that sends each integer n to the element xn is a homomorphism from the group Z of integers to G, since xm+n = xm xn for all integers m and n (Theorem 1.4). Lemma 1.17 Let θ: G → K be a homomorphism. Then θ(eG ) = eK , where eG and eK denote the identity elements of the groups G and K. Also θ(x−1 ) = θ(x)−1 for all elements x of G. Proof Let z = θ(eG ). Then z 2 = θ(eG )θ(eG ) = θ(eG eG ) = θ(eG ) = z. The result that θ(eG ) = eK now follows from the fact that an element z of K satisfies z 2 = z if and only if z is the identity element of K. Let x be an element of G. The element θ(x−1 ) satisfies θ(x)θ(x−1 ) = θ(xx−1 ) = θ(eG ) = eK , and similarly θ(x−1 )θ(x) = eK . The uniqueness of the inverse of θ(x) now ensures that θ(x−1 ) = θ(x)−1 . 10

An isomorphism θ: G → K between groups G and K is a homomorphism that is also a bijection mapping G onto K. Two groups G and K are isomorphic if there exists an isomorphism mapping G onto K. Example Let D6 be the group of symmetries of an equilateral triangle in the plane with vertices A, B and C, and let S3 be the group of permutations of the set {A, B, C}. The function which sends a symmetry of the triangle to the corresponding permutation of its vertices is an isomorphism between the dihedral group D6 of order 6 and the symmetric group S3 . Example Let R be the group of real numbers with the operation of addition, and let R+ be the group of strictly positive real numbers with the operation of multiplication. The function exp: R → R+ that sends each real number x to the positive real number ex is an isomorphism: it is both a homomorphism of groups and a bijection. The inverse of this isomorphism is the function log: R+ → R that sends each strictly positive real number to its natural logarithm. Here is some further terminology regarding homomorphisms: • A monomorphism is an injective homomorphism. • An epimorphism is a surjective homomorphism. • An endomorphism is a homomorphism mapping a group into itself. • An automorphism is an isomorphism mapping a group onto itself. Definition The kernel ker θ of the homomorphism θ: G → K is the set of all elements of G that are mapped by θ onto the identity element of K. Example Let the group operation on the set {+1, −1} be multiplication, and let θ: Z → {+1, −1} be the homomorphism that sends each integer n to (−1)n . Then the kernel of the homomorphism θ is the subgroup of Z consisting of all even numbers. Lemma 1.18 Let G and K be groups, and let θ: G → K be a homomorphism from G to K. Then the kernel ker θ of θ is a normal subgroup of G. Proof Let x and y be elements of ker θ. Then θ(x) = eK and θ(y) = eK , where eK denotes the identity element of K. But then θ(xy) = θ(x)θ(y) = eK eK = eK , and thus xy belongs to ker θ. Also θ(x−1 ) = θ(x)−1 = e−1 K = eK , and thus x−1 belongs to ker θ. We conclude that ker θ is a subgroup of K. Moreover ker θ is a normal subgroup of G, for if g ∈ G and x ∈ ker θ then θ(gxg −1 ) = θ(g)θ(x)θ(g)−1 = θ(g)θ(g −1 ) = eK . 11

If N is a normal subgroup of some group G then N is the kernel of the quotient homomorphism θ: G → G/N that sends g ∈ G to the coset gN . It follows therefore that a subset of a group G is a normal subgroup of G if and only if it is the kernel of some homomorphism. Proposition 1.19 Let G and K be groups, let θ: G → K be a homomorphism from G to K, and let N be a normal subgroup of G. Suppose that N ⊂ ker θ. Then the homomorphism θ: G → K induces a homomorphism ˆ G/N → K sending gN ∈ G/N to θ(g). Moreover θ: ˆ G/N → K is injective θ: if and only if N = ker θ. Proof Let x and y be elements of G. Now xN = yN if and only if x−1 y ∈ N . Also θ(x) = θ(y) if and only if x−1 y ∈ ker θ. Thus if N ⊂ ker θ then θ(x) = θ(y) whenever xN = yN , and thus θ: G → K induces a well-defined function ˆ G/N → K sending xN ∈ G/N to θ(x). This function is a homomorphism, θ: ˆ ˆ ˆ ˆ since θ((xN )(yN )) = θ(xyN ) = θ(xy) = θ(x)θ(y) = θ(xN )θ(yN ). Suppose now that N = ker θ. Then θ(x) = θ(y) if and only if xN = yN . ˆ G/N → K is injective. Conversely if θ: ˆ G/N → Thus the homomorphism θ: K is injective then N must be the kernel of θ, as required. Corollary 1.20 Let G and K be groups, and let θ: G → K be a homomorphism. Then θ(G) ∼ = G/ ker θ.

1.9

The Isomorphism Theorems

Lemma 1.21 Let G be a group, let H be a subgroup of G, and let N be a normal subgroup of G. Then the set HN is a subgroup of G, where HN = {hn : h ∈ H and n ∈ N }. Proof The set HN clearly contains the identity element of G. Let x and y be elements of HN . We must show that xy and x−1 belong to HN . Now x = hu and y = kv for some elements h and k of H and for some elements u and v of N . Then xy = (hk)(k −1 ukv). But k −1 uk ∈ N , since N is normal. It follows that k −1 ukv ∈ N , since N is a subgroup and k −1 ukv is the product of the elements k −1 uk and v of N . Also hk ∈ H. It follows that xy ∈ HN . We must also show that x−1 ∈ HN . Now x−1 = u−1 h−1 = h−1 (hu−1 h−1 ). Also h−1 ∈ H, since H is a subgroup of G, and hu−1 h−1 ∈ N , since N is a normal subgroup of G. It follows that x−1 ∈ HN , and thus HN is a subgroup of G, as required.

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Theorem 1.22 (First Isomorphism Theorem) Let G be a group, let H be a subgroup of G, and let N be a normal subgroup of G. Then HN ∼ H . = N N ∩H Proof Every element of HN/N is a coset of N that is of the form hN for some h ∈ H. Thus if ϕ(h) = hN for all h ∈ H then ϕ: H → HN/N is a surjective homomorphism, and ker ϕ = N ∩ H. But ϕ(H) ∼ = H/ ker ϕ ∼ (Corollary 1.20). Therefore HN/N = H/(N ∩ H) as required. Theorem 1.23 (Second Isomorphism Theorem) Let M and N be normal subgroups of a group G, where M ⊂ N . Then G ∼ G/M . = N N/M Proof There is a well-defined homomorphism θ: G/M → G/N that sends gM to gN for all g ∈ G. Moreover the homomorphism θ is surjective, and ker θ = N/M . But θ(G/M ) ∼ = (G/M )/ ker θ. Therefore G/N is isomorphic to (G/M ) / (N/M ), as required.

1.10

Group Actions, Orbits and Stabilizers

Definition A left action of a group G on a set X associates to each g ∈ G and x ∈ X an element g.x of X in such a way that g.(h.x) = (gh).x and 1.x = x for all g, h ∈ G and x ∈ X, where 1 denotes the identity element of G. Given a left action of a group G on a set X, the orbit of an element x of X is the subset {g.x : g ∈ G} of X, and the stabilizer of x is the subgroup {g ∈ G : g.x = x} of G. Lemma 1.24 Let G be a finite group which acts on a set X on the left. Then the orbit of an element x of X contains [G: H] elements, where [G: H] is the index of the stabilizer H of x in G. Proof There is a well-defined function θ: G/H → X defined on the set G/H of left cosets of H in G which sends gH to g.x for all g ∈ G. Moreover this function is injective, and its image is the orbit of x. The result follows.

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1.11

Conjugacy

Definition Two elements h and k of a group G are said to be conjugate if k = ghg −1 for some g ∈ G. One can readily verify that the relation of conjugacy is reflexive, symmetric and transitive and is thus an equivalence relation on a group G. The equivalence classes determined by this relation are referred to as the conjugacy classes of G. A group G is the disjoint union of its conjugacy classes. Moreover the conjugacy class of the identity element of G contains no other element of G. A group G is Abelian if and only if all its conjugacy classes contain exactly one element of the group G. Definition Let G be a group. The centralizer C(h) of an element h of G is the subgroup of G defined by C(h) = {g ∈ G : gh = hg}. Lemma 1.25 Let G be a finite group, and let h ∈ G. Then the number of elements in the conjugacy class of h is equal to the index [G: C(h)] of the centralizer C(h) of h in G. Proof There is a well-defined function f : G/C(h) → G, defined on the set G/C(h) of left cosets of C(h) in G, which sends the coset gC(h) to ghg −1 for all g ∈ G. This function is injective, and its image is the conjugacy class of h. The result follows. Let H be a subgroup of a group G. One can easily verify that gHg −1 is also a subgroup of G for all g ∈ G, where gHg −1 = {ghg −1 : h ∈ H}. Definition Two subgroups H and K of a group G are said to be conjugate if K = gHg −1 for some g ∈ G. The relation of conjugacy is an equivalence relation on the collection of subgroups of a given group G.

1.12

The Class Equation of a Finite Group

Definition The centre Z(G) of a group G is the subgroup of G defined by Z(G) = {g ∈ G : gh = hg for all h ∈ G}.

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One can verify that the centre of a group G is a normal subgroup of G. Let G be a finite group, and let Z(G) be the centre of G. Then G \ Z(G) is a disjoint union of conjugacy classes. Let r be the number of conjugacy classes contained in G\Z(G), and let n1 , n2 , . . . , nr be the number of elements in these conjugacy classes. Then ni > 1 for all i, since the centre Z(G) of G is the subgroup of G consisting of those elements of G whose conjugacy class contains just one element. Now the group G is the disjoint union of its conjugacy classes, and therefore |G| = |Z(G)| + n1 + n2 + · · · + nr . This equation is referred to as the class equation of the group G. Definition Let g be an element of a group G. The centralizer C(g) of g is the subgroup of G defined by C(g) = {h ∈ G : hg = gh}. Proposition 1.26 Let G be a finite group, and let p be a prime number. Suppose that pk divides the order of G for some positive integer k. Then either pk divides the order of some proper subgroup of G, or else p divides the order of the centre of G. Proof Choose elements g1 , g2 , . . . , gr of G\Z(G), where Z(G) is the centre of G, such that each conjugacy class included in G \ Z(G) contains exactly one of these elements. Let ni be the number of elements in the conjugacy class of gi and let C(gi ) be the centralizer of gi for each i. Then C(gi ) is a proper subgroup of G, and |G| = ni |C(gi )|. Thus if pk divides |G| but does not divide the order of any proper subgroup of G then p must divide ni for i = 1, 2, . . . , r. Examination of the class equation |G| = |Z(G)| + n1 + n2 + · · · + nr now shows that p divides |Z(G)|, as required.

1.13

Cauchy’s Theorem

Theorem 1.27 (Cauchy) Let G be an finite group, and let p be a prime number that divides the order of G. Then G contains an element of order p. Proof We prove the result by induction on the order of G. Thus suppose that every finite group whose order is divisible by p and less than |G| contains an element of order p. If p divides the order of some proper subgroup of G then that subgroup contains the required element of order p. If p does not divide the order of any proper subgroup of G then Proposition 1.26 ensures that p divides the order of the centre Z(G) of G, and thus Z(G) cannot be a proper subgroup of G. But then G = Z(G) and the group G is Abelian. 15

Thus let G be an Abelian group whose order is divisible by p, and let H be a proper subgroup of G that is not contained in any larger proper subgroup. If |H| is divisible by p then the induction hypothesis ensures that H contains the required element of order p, since |H| < |G|. Suppose then that |H| is not divisible by p. Choose g ∈ G \ H, and let C be the cyclic subgroup of G generated by g. Then HC = G, since HC 6= H and HC is a subgroup of G containing H. It follows from the First Isomorphism Theorem (Theorem 1.22) that G/H ∼ = C/H ∩ C. Now p divides |G/H|, since |G/H| = |G|/|H| and p divides |G| but not |H|. Therefore p divides |C|. Thus if m = |C|/p then g m is the required element of order p. This completes the proof of Cauchy’s Theorem.

1.14

The Structure of p-Groups

Definition Let p be a prime number. A p-group is a finite group whose order is some power pk of p. Lemma 1.28 Let p be a prime number, and let G be a p-group. Then there exists a normal subgroup of G of order p that is contained in the centre of G. Proof Let |G| = pk . Then pk divides the order of G but does not divide the order of any proper subgroup of G. It follows from Proposition 1.26 that p divides the order of the centre of G. It then follows from Cauchy’s Theorem (Theorem 1.27) that the centre of G contains some element of order p. This element generates a cyclic subgroup of order p, and this subgroup is normal since its elements commute with every element of G. Proposition 1.29 Let G be a p-group, where p is some prime number, and let H be a proper subgroup of G. Then there exists some subgroup K of G such that H / K and K/H is a cyclic group of order p. Proof We prove the result by induction on the order of G. Thus suppose that the result holds for all p-groups whose order is less than that of G. Let Z be the centre of G. Then ZH is a well-defined subgroup of G, since Z is a normal subgroup of G. Suppose that ZH 6= H. Then H is a normal subgroup of ZH. The quotient group ZH/H is a p-group, and contains a subgroup K1 of order p (Lemma 1.28). Let K = {g ∈ ZH : gH ∈ K1 }. Then H / K and K/H ∼ = K1 , and therefore K is the required subgroup of G. Finally suppose that ZH = H. Then Z ⊂ H. Let H1 = {hZ : h ∈ H}. Then H1 is a subgroup of G/Z. But G/Z is a p-group, and |G/Z| < |G|, since |Z| ≥ p (Lemma 1.28). The induction hypothesis ensures the existence 16

of a subgroup K1 of G/Z such that H1 / K1 and K1 /H1 is cyclic of order p. Let K = {g ∈ G : gZ ∈ K1 }. Then H / K and K/H ∼ = K1 /H1 . Thus K is the required subgroup of G. Repeated applications of Proposition 1.29 yield the following result. Corollary 1.30 Let G be a finite group whose order is a power of some prime number p. Then there exist subgroups G0 , G1 , . . . , Gn of G, where G0 is the trivial subgroup and Gn = G, such that Gi−1 / Gi and Gi /Gi−1 is a cyclic group of order p for i = 1, 2, . . . , n.

1.15

The Sylow Theorems

Definition Let G be a finite group, and let p be a prime number dividing the order |G| of G. A p-subgroup of G is a subgroup whose order is some power of p. A Sylow p-subgroup of G is a subgroup whose order is pk , where k is the largest natural number for which pk divides |G|. Theorem 1.31 (First Sylow Theorem) Let G be a finite group, and let p be a prime number dividing the order of G. Then G contains a Sylow p-subgroup. Proof We prove the result by induction on the order of G. Thus suppose that all groups whose order is less than that of G contain the required Sylow p-subgroups. Let k be the largest positive integer for which pk divides |G|. If pk divides the order of some proper subgroup H of G then the induction hypothesis ensures that H contains the required Sylow p-subgroup of order pk . If pk does not divide the order of any proper subgroup of G then p divides the order of the centre Z(G) of G (Proposition 1.26). It follows from Cauchy’s Theorem (Theorem 1.27) that Z(G) contains an element of order p, and this element generates a normal subgroup N of G of order p. The induction hypothesis then ensures that G/N has a Sylow p-subgroup L of order pk−1 , since |G/N | = |G|/p. Let K = {g ∈ G : gN ∈ L}. Then |K| = p|L| = pk , and thus K is the required Sylow p-subgroup of G. Theorem 1.32 (Second Sylow Theorem) Let G be a finite group, and let p be a prime number dividing the order of G. Then all Sylow p-subgroups of G are conjugate, and any p-subgroup of G is contained in some Sylow psubgroup of G. Moreover the number of Sylow p-subgroups in G divides the order of G and is congruent to 1 modulo p.

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Proof Let K be a Sylow p-subgroup of G, and let X be the set of left cosets of K in G. Let H be a p-subgroup of G. Then H acts on X on the left, where h(gK) = hgK for all h ∈ H and g ∈ G. Moreover h(gK) = gK if and only if g −1 hg ∈ K. Thus an element gK of X is fixed by H if and only if g −1 Hg ⊂ K. Let |G| = pk m, where k and m are positive integers and m is coprime to p. Then |K| = pk . Now the number of left cosets of K in G is |G|/|K|. Thus the set X has m elements. Now the number of elements in any orbit for the action of H on X divides the order of H, since it is the index in H of the stabilizer of some element of that orbit (Lemma 1.24). But then the number of elements in each orbit must be some power of p, since H is a p-group. Thus if an element of X is not fixed by H then the number of elements in its orbit is divisible by p. But X is a disjoint union of orbits under the action of H on X. Thus if m0 denotes the number of elements of X that are fixed by H then m − m0 is divisible by p. Now m is not divisible by p. It follows that m0 6= 0, and m0 is not divisible by p. Thus there exists at least one element g of G such that g −1 Hg ⊂ K. But then H is contained in the Sylow p-subgroup gKg −1 . Thus every p-subgroup is contained in a Sylow p-subgroup of G, and this Sylow p-subgroup is a conjugate of the given Sylow p-subgroup K. In particular any two Sylow p-subgroups are conjugate. It only remains to show that the number of Sylow p-subgroups in G divides the order of |G| and is congruent to 1 modulo p. On applying the above results with H = K, we see that g −1 Kg = K for some g ∈ G if and only if gK is a fixed point for the action of K on X. But the number of elements g of G for which gK is a fixed point is m0 |K|, where m0 is the number of fixed points in X. It follows that the number of elements g of G for which g −1 Kg = K is pk m0 . But every Sylow p-subgroup of G is of the form g −1 Kg for some g ∈ G. It follows that the number n of Sylow p-subgroups in G is given by n = |G|/pk m0 = m/m0 . In particular n divides |G|. Now we have already shown that m − m0 is divisible by p. It follows that m0 is coprime to p, since m is coprime to p. Also m − m0 is divisible by m0 , since (m − m0 )/m0 = n − 1. Putting these results together, we see that m − m0 is divisible by m0 p, and therefore n − 1 is divisible by p. Thus n divides |G| and is congruent to 1 modulo p, as required.

1.16

Some Applications of the Sylow Theorems

Theorem 1.33 Let p and q be prime numbers, where p < q and q 6≡ 1 (mod p). Then any group of order pq is cyclic.

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Proof Let G be a group of order pq. It follows from the First Sylow Theorem that G contains Sylow subgroups Np and Nq of orders p and q respectively. Now the number np of Sylow p-subgroups divides pq and satisfies np ≡ 1 (mod p), by the Second Sylow Theorem. Clearly np cannot be divisible by p, and therefore either np = 1 or np = q. But q 6≡ 1 (mod p). It follows that np = 1. Thus the group G has just one subgroup of order p. Now, given any element g of G, the subgroups Np and gNp g −1 are of order p. It follows that gNp g −1 = Np for all elements g of G. Thus Np is a normal subgroup of G. A similar argument shows that Nq is also a normal subgroup of G, since p < q, and therefore p 6≡ 1 (mod q). Now Np ∩ Nq is a subgroup of both Np and Nq . It follows from Lagrange’s Theorem that the order of Np ∩ Nq divides both of the prime numbers p and q, and therefore |Np ∩ Nq | = 1 and Np ∩ Nq = {e}, where e is the identity element of G. Let x ∈ Np and y ∈ Nq . Then yx−1 y −1 ∈ Np and xyx−1 ∈ Nq , since Np and Nq are normal subgroups of G. But then xyx−1 y −1 ∈ Np ∩ Nq , since xyx−1 y −1 = x(yx−1 y −1 ) = (xyx−1 )y −1 , and therefore xyx−1 y −1 = e. Thus xy = yx for all x ∈ Np and y ∈ Nq . It follows easily from this that the function ϕ: Np × Nq → G which sends (x, y) ∈ Np × Nq to xy is a homomorphism. This homomorphism is injective, for if xy = e for some x ∈ Np and y ∈ Nq , then x = y −1 , and hence x ∈ Np ∩ Nq , from which it follows that x = e and y = e. But any injective homomorphism between two finite groups of the same order is necessarily an isomorphism. Therefore the function ϕ: Np × Nq → G is an isomorphism, and thus G ∼ = Np × Nq . Now any group whose order is prime number must be cyclic. Therefore the groups Np and Nq are cyclic. Let x be an element of Np that generates Np , and let y be an element of Nq that generates Nq . Then (x, y)n = (xn , y n ) for all integers n. It follows from this that the order of (x, y) cannot be equal to 1, p or q, and must therefore be equal to pq. Thus Np × Nq is a cyclic group generated by (x, y), and therefore G is a cyclic group, generated by xy, as required. Example Any finite group whose order is 15, 33, 35, 51, 65, 69, 85, 87, 91 or 95 is cyclic. Theorem 1.34 Let G be a group of order 2p where p is a prime number greater than 2. Then either the group G is cyclic, or else the group G is isomorphic to the dihedral group D2p of symmetries of a regular p-sided polygon in the plane.

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Proof It follows from the First Sylow Theorem, or from Cauchy’s Theorem, that the group G contains elements x and y whose orders are 2 and p respectively. The subgroup N generated by y is then a Sylow p-subgroup of G. Now it follows from the Second Sylow Theorem that the number of Sylow p-subgroups of G divides 2p and is congruent to 1 modulo p. There can therefore be only one such Sylow p-subgroup, since 2, p and 2p are not congruent to 1 modulo p. Now if g is any element of G then gN g −1 is a Sylow p-subgroup of G, and therefore gN g −1 = N . We deduce that N is a normal subgroup of G, of order p. Now consider the element xyx−1 of G. This must be an element of the normal subgroup N of G generated by y. Therefore xyx−1 = y k for some integer k. Moreover k is not divisible by p, since xyx−1 is not the identity element e of G. Then 2

y k = (y k )k = (xyx−1 )k = xy k x−1 = x(xyx−1 )x−1 = x2 yx−2 . But x2 = x−2 = e, since x is an element of G of order 2. It follows that 2 2 y k = y, and thus y k −1 = e. But then p divides k 2 − 1, since y is an element of order p. Moreover k 2 − 1 = (k − 1)(k + 1). It follows that either p divides k − 1, in which case xyx−1 = y, or else p divides k + 1, in which case xyx−1 = y −1 . In the case when xyx−1 = y we see that xy = yx, and one can then readily verify that the group G is a cyclic group of order 2p generated by xy. In the case when xyx−1 = y −1 the group G is isomorphic to the dihedral group D2p of order 2p. In this case the elements x and y generate G (since they generate a subgroup of G whose order divides 2p but is greater than p, and must therefore be equal to 2p). Under the isomorphism with the dihedral group D2p the element x corresponds to a reflection in one of the axes of symmetry of the regular p-sided polygon, and the element y corresponds to a rotation of that polygon about its centre through an angle of 2π/p radians. Theorem 1.35 Let p and q be prime numbers with p < q, and let d be the smallest positive integer for which pd ≡ 1 (mod q). If G is a group of order pk q, where 0 < k < d then G contains a normal subgroup of order q. If G is a group of order pd q then either G contains a normal subgroup of order q or else G contains a normal subgroup of order pd . Proof It follows from the First Sylow Theorem (or directly from Cauchy’s Theorem) that the group G contains at least one Sylow q-subgroup K, and this is of order q. If |G| = pk q then the number nq of such Sylow q-subgroups divides pk q and satisfies nq ≡ 1 (mod q), by the Second Sylow Theorem. 20

It follows that nq is coprime to q, and therefore nq = pj for some integer j satisfying 0 ≤ j ≤ k. If k < d then none of the integers p, p2 , . . . , pk are congruent to 1 modulo q, and therefore j = 0 and nq = 1. In this case there is just one Sylow q-subgroup K, and this is a normal subgroup. (Given any element g of G, the subgroup gKg −1 is a Sylow q-subgroup, and therefore gKg −1 = K.) If k = d then none of the integers pj with 0 < j < d are congruent to 1 modulo q, and therefore either nq = 1 or nq = pd . If nq = 1 then there is just one Sylow q-subgroup K, and this is a normal subgroup. If nq > 1 then nq = pd , and thus there are pd Sylow q-subgroups, and these are of order q. Now if Ki and Kj are two distinct subgroups of order q then Ki ∩ Kj is a proper subgroup of both Ki and Kj , and its order is a proper divisor of the order q of Ki and Kj , by Lagrange’s Theorem. But q is a prime number. It follows that Ki ∩ Kj = {e}, where e is the identity element of G. We deduce from this that no element of G of order q can belong to more than one subgroup of order q. But each subgroup of G of order q contains q − 1 elements of order q (namely all elements of that subgroup with the exception of the identity element). It follows that the group G contains pd (q − 1) elements of order q. Now |G| = pd q. It follows that G contains exactly pd elements that are not of order q. But it follows from the First Sylow Theorem that G contains at least one Sylow p-subgroup H, and this is of order pd . This subgroup must therefore contain all the elements of G that are not of order q. It follows that the group G cannot contain more than one such Sylow p-subgroup. This Sylow p-subgroup H is therefore a normal subgroup of G of order pd , as required.

1.17

Simple Groups

Definition A non-trivial group G is said to be simple if the only normal subgroups of G are the whole of G and the trivial subgroup {e} whose only element is the identity element e of G. Lemma 1.36 Any non-trivial Abelian simple group is a cyclic group whose order is a prime number. Proof Let G be a non-trivial Abelian simple group, and let x be an element of G that is not equal to the identity element e of G. All subgroups of an Abelian group are normal subgroups. Therefore the subgroup of G generated by x is a normal subgroup of G, and must therefore be the whole of G. Therefore G is a cyclic group, generated by the element x. Moreover all elements of G other than the identity element are generators of G, and are 21

therefore of order p, where p = |G|. Let d be a divisor of p. Then xd is an element of order p/d, since p/d is the smallest positive integer k for which xdk = e. It follows that either d = 1 or d = p (since the group G contains no element whose order is greater than 1 but less than p). It follows that the order p of G is a prime number, as required. Using the Sylow Theorems and related results, we can prove that any finite simple group whose order is less than 60 is a cyclic group of prime order. Now the prime numbers less than 60 are the following: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53 and 59. All groups of these orders are simple groups, and are cyclic groups. If p is a prime number greater than 2 then any group of order 2p is either a cyclic group or else is isomorphic to the dihedral group D2p of order 2p (Theorem 1.34). In either case such a group contains a normal subgroup of order p, and therefore not a simple group. In particular, there are no simple groups of orders 6, 10, 14, 22, 26, 34, 38, 46 or 58. If G is a group of order pk for some prime number p and for some integer k satisfying k ≥ 2, then G contains a normal subgroup of order p (Lemma 1.28). It follows that such a group is not simple. In particular, there are no simple groups of orders 4, 8, 16, 32, 9, 27, 25 and 49. Let G be a group of order pq, where p and q are prime numbers and p < q. Any Sylow q-subgroup of G is of order q, and the number of such Sylow q-subgroups must divide pq and be congruent to 1 modulo q. Now p cannot be congruent to 1 modulo q, since 1 < p < q. Therefore G has just one Sylow q-subgroup, and this is a normal subgroup of G of order q. It follows that such a group is not a simple group. In particular there are no simple groups of orders 15, 21, 33, 35, 39, 51, 55 or 57. (In particular it follows from Theorem 1.33 that any group whose order is 15, 33, 35, or 51 is a cyclic group.) It only remains to verify that there are no simple groups of orders 12, 18, 20, 24, 28, 30, 36, 40, 42, 44, 45, 48, 50, 52, 54 or 56. We can deal with many of these on applying Theorem 1.35. On applying this theorem with p = 2, q = 3 and d = 2, we see that there are no simple groups of orders 6 or 12. On applying the theorem with p = 2, q = 5 and d = 4, we see that there are no simple groups of orders 10, 20, 40 or 80. On applying the theorem with p = 2, q = 7 and d = 3, we see that there are no simple groups of orders 14, 28 or 56. On applying the theorem with p = 2, q = 11 we see that there are no simple groups of orders 22, 44 etc., on applying the theorem with p = 2, q = 13 we see that there are no simple groups of orders 26, 52 etc., and on applying the theorem with p = 3 and 22

q = 5, we see that there are no simple groups of orders 15, 45 etc. It now remains to verify that there are no simple groups of orders 18, 24, 30, 36, 42, 48, 50 or 54. Using the Second Sylow Theorem, we see that any group of order 18 has just one Sylow 3-subgroup. This Sylow 3-subgroup is then a normal group of order 9, and therefore no group of order 18 is simple. Similarly a group of order 50 has just one Sylow 5-subgroup, which is then a normal subgroup of order 25, and therefore no group of order 50 is simple. Also a group of order 54 has just one Sylow 3-subgroup, which is then a normal subgroup of order 27, and therefore no group of order 54 is simple. On applying the Second Sylow Theorem, we see the number of Sylow 7-subgroups of any group of order 42 must divide 42 and be congruent to 1 modulo 7. This number must then be coprime to 7 and therefore divide 6, since 42 = 7 × 6. But no divisor of 6 greater than 1 is coprime to 1 modulo 7. It follows that any group of order 42 has just one Sylow 7-subgroup, and this subgroup is therefore a normal subgroup of order 7. Thus no group of order 42 is simple. On applying the Second Sylow Theorem, we see that if a group of order 30 has more than one subgroup of order 3 then it must have 10 such subgroups, and must therefore have 20 elements of order 3 (since each subgroup of order 3 contains two elements of order 3, and the intersection of two distinct subgroups of order 3 must be the trivial subgroup). Similarly if a group of order 30 has more than one subgroup of order 5 then it must have 6 such subgroups, and must therefore have 24 elements of order 5. Obviously such a group cannot have both 20 elements of order 3 and 24 elements of order 5. Therefore it either has a single subgroup of order 3 or a single subgroup of order 5. This subgroup is normal. Therefore no group of order 30 is simple. In order to show that there are no simple groups of order less than 60, apart from the cyclic groups whose order is prime, it only remains to verify that there are no simple groups of orders 24, 36 and 48. In order to deal with these remaining cases, we need to make use of the following result. Lemma 1.37 Let H and K be subgroups of a finite group G. Then |H ∩ K| ≥

|H| |K| . |G|

Proof Let ϕ: H × K → G be the function with ϕ(h, k) = hk for all h ∈ H and k ∈ K. (This function is not in general a homomorphism.) Let (h1 , k1 ) and (h2 , k2 ) be elements of H × K. Then h1 k1 = h2 k2 if and only −1 −1 −1 if h−1 and 2 h1 = k2 k1 , in which case h2 h1 ∈ H ∩ K. But then h2 = h1 x k2 = xk1 for some element x of H ∩ K. Thus ϕ(h1 , k1 ) = ϕ(h2 , k2 ) if and 23

only if (h2 , k2 ) = (h1 x−1 , xk1 ) for some element x of H ∩ K. It follows that each element of the range ϕ(H × K) of the function ϕ is the image of exactly |H| |K| |H ∩ K| elements of H × K. It follows from this that ϕ(H × K) has |H ∩ K| elements. But ϕ(H × K) is a subset of G. Therefore |H| |K| ≤ |G|. |H ∩ K| The required inequality now follows directly. Let G be a finite group, and let H be a subgroup of index 2 in G (i.e., a subgroup with half as many elements as G). Then H is a normal subgroup of G. Indeed the subsets H and G \ H of G are the left cosets and are also the right cosets of H in G, and therefore the left cosets of H in G coincide with the right cosets. Example We now show that there are no simple groups of order 24. Let G be a group of order 24. Then G contains a Sylow 2-subgroup H of order 8. If this is the only Sylow 2-subgroup, then it is a normal subgroup, and therefore the group G is not simple. Otherwise the group G contains at least two distinct subgroups H and K of order 8. It then follows from Lemma 1.37 that |H ∩ K| ≥ 38 . But |H ∩ K| divides 8, by Lagrange’s Theorem, since H ∩ K is a subgroup of H and of K. Therefore |H ∩ K| = 4. It follows that H ∩ K is a subgroup of index 2 in H and K, and is therefore a normal subgroup of both H and K. Let J = {g ∈ G : g(H ∩ K)g −1 = H ∩ K}. Then J is a subgroup of G, and H ∩ K is a normal subgroup of J. Moreover H and K are subgroups of J, and therefore |J| is divisible by 8, by Lagrange’s Theorem. But J is a subgroup of G, and hence |J| divides 24. Also |J| > 8, since H (and K) are proper subgroups of J. It follows that |J| = 24, and therefore J = G. But then H ∩ K is a normal subgroup of G of order 4, and therefore G is not simple. An analogous argument shows that there are no simple groups of order 48: a group G of order 48 contains either a single Sylow 2-subgroup of order 16, which is then a normal subgroup of G, or else it contains a normal subgroup of order 8 which is the intersection of two distinct Sylow 2-subgroups of G. The following result will be needed in order to show that there are no simple groups of order 36. (It may be obtained as an immediate corollary of Proposition 1.29.) 24

Lemma 1.38 Let G be a group of order p2 where p is a prime number, and let H be a subgroup of G of order p. Then H is a normal subgroup of G. Proof Let J = {g ∈ G : gHg −1 = H}. Then J is a subgroup of G and H is a normal subgroup of J. We shall show that J = G. Now the centre Z(G) of G is contained in J. Moreover it follows from Lemma 1.28 that |Z(G)| is divisible by p. Were it the case that |J| = p then J = H = Z(G). But then J would consist of all elements of G for which gZ(G)g −1 = Z(G), and thus would be the whole of G, which is impossible. It follows that |J| = p2 (since |J| > p and |J| divides p2 ). But then J = G, and hence H is a normal subgroup of G, as required. Example We now show that there are no simple groups of order 36. Let G be a group of order 36. Then G contains a Sylow 3-subgroup H of order 9. If this is the only Sylow 3-subgroup, then it is a normal subgroup, and therefore the group G is not simple. Otherwise the group G contains at least two distinct subgroups H and K of order 9. It then follows from Lemma 1.37 that |H ∩ K| ≥ 94 . But |H ∩ K| divides 9, by Lagrange’s Theorem, since H ∩ K is a subgroup of H and of K. Therefore |H ∩ K| = 3. On applying Lemma 1.38 we see that H ∩ K is a normal subgroup of H and of K. Let J = {g ∈ G : g(H ∩ K)g −1 = H ∩ K}. Then J is a subgroup of G, and H ∩ K is a normal subgroup of J. Moreover H and K are subgroups of J, and therefore |J| is divisible by 9, by Lagrange’s Theorem. But J is a subgroup of G, and hence |J| divides 36. Also |J| > 9, since H (and K) are proper groups of J. It follows that either |J| = 18 or 36. If |J| = 36 then J = G and H ∩ K is a normal subgroup of G of order 3. If |J| = 18 then J is a subgroup of G of index 2, and is therefore a normal subgroup of order 18. We conclude that any group of order 36 contains at least one non-trivial normal subgroup. Therefore there are no simple groups of order 36. We have now shown that there are indeed no simple groups of order less than 60, other than the cyclic groups of prime order. There is a simple group of order 60 which is simple but is not cyclic. This group is the alternating group A5 , consisting of all even permutations of a set with five elements. Lemma 1.39 The alternating group A5 is simple.

25

Proof We regard A5 as the group even permutations of the set {1, 2, 3, 4, 5}. There are 60 such permutations: the identity permutation, twenty 3-cycles, twenty-four 5-cycles, and fifteen permutations that are products of two disjoint transpositions. (Such a product of disjoint transpositions is a permutation (a1 a2 )(a3 a4 ) that interchanges a1 with a2 and a3 with a4 for some distinct elements a1 , a2 , a3 and a4 of the set {1, 2, 3, 4, 5}.) Now each 3-cycle in A5 generates a Sylow 3-subgroup of order 3, and these subgroups are all conjugate to one another by the Second Sylow Theorem. It follows that any normal subgroup of A5 that contains at least one 3-cycle must contain all twenty 3-cycles, and thus its order must therefore be at least 21 (since it must also contain the identity element). Similarly each 5-cycle in A5 generates a Sylow 5-subgroup of order 5, and these subgroups are all conjugate to one another. Therefore any normal subgroup of A5 that contains at least one 5-cycle must contain all twenty four 5-cycles, and thus its order must be at least 25. Now if A5 were to contain a subgroup of order 30, this subgroup would be the kernel of a non-constant homomorphism ϕ: A5 → {1, −1} from A5 to the multiplicative group consisting of the numbers 1 and −1. But any 3-cycle or 5-cycle would have to belong to the kernel of this homomorphism, and therefore this kernel would contain at least 45 elements, which is impossible. We conclude that A5 cannot contain any subgroup of order 30. It follows from Lagrange’s Theorem that any normal subgroup of A5 that contains at least one 3-cycle or 5-cycle must be the whole of A5 . The group A5 contains 5 Sylow 2-subgroups, which are of order 4. One of these consists of the identity permutation, together with the three permutations (1 2)(3 4), (1 3)(2 4) and (1 4)(2 3). (Each of these permutations fixes the element 5.) There are four other such Sylow 2-subgroups, and all of the Sylow 2-subgroups are conjugate to one another. It follows that A5 does not contain any normal subgroup of order 4. Moreover A5 cannot contain any normal subgroup of order 2, since any element of order 2 belongs to one of the five Sylow 2-subgroups of order 4, and is therefore conjugate to elements of order 2 in the other Sylow 2-subgroups. Now any subgroup of A5 whose order is divisible by 3 must contain a 3-cycle by Cauchy’s Theorem. (Theorem 1.27.) Similarly any subgroup of A5 whose order is divisible by 5 must contain a 5-cycle. It follows that the order of any proper normal subgroup of A5 cannot be divisible by 3 or 5. But this order must divide 60. Therefore the order of any proper normal subgroup of A5 must be at most 4. But we have seen that A5 cannot contain any normal subgroup of order 4 or 2. Therefore any proper normal subgroup of A5 is trivial, and therefore A5 is simple.

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1.18

Solvable Groups

The concept of a solvable group was introduced into mathematics by Evariste Galois, in order to state and prove his fundamental general theorems concerning the solvability of polynomial equations. We now investigate the basic properties of such solvable groups. Definition A group G is said to be solvable (or soluble) if there exists a finite sequence G0 , G1 , . . . , Gn of subgroups of G, where G0 = {1} and Gn = G, such that Gi−1 is normal in Gi and Gi /Gi−1 is Abelian for i = 1, 2, . . . , n. Example The symmetric group Σ4 is solvable. Indeed let V4 be the Kleinsche Viergruppe consisting of the identity permutation ι and the permutations (12)(34), (13)(24) and (14)(23), and let A4 be the alternating group consisting of all even permutations of {1, 2, 3, 4}. Then {ι} / V4 / A4 / Σ4 , V4 is Abelian, A4 /V4 is cyclic of order 3, and Σ4 /A4 is cyclic of order 2. In order to prove certain basic results concerning solvable groups, we need to make use of the Isomorphism Theorems for groups, which may be described as follows. Lemma 1.40 Let G be a group, let H1 and H2 be subgroups of G, where H1 / H2 , and let J1 = H1 ∩ N , J2 = H2 ∩ N , K1 = H1 N/N and K2 = H2 N/N , where N is some normal subgroup of G. Then J1 / J2 and K1 / K2 . Moreover there exists a normal subgroup of H2 /H1 isomorphic to J2 /J1 , and the quotient of H2 /H1 by this normal subgroup is isomorphic to K2 /K1 . Proof It is a straightforward exercise to verify that J1 / J2 and K1 / K2 . Let θ: H2 → K2 be the surjective homomorphism sending h ∈ H2 to the coset hN . Now θ induces a well-defined surjective homomorphism ψ: H2 /H1 → K2 /K1 , since θ(H1 ) ⊂ K1 . Also θ−1 (K1 ) = H2 ∩ (H1 N ). But H2 ∩ (H1 N ) = H1 (H2 ∩ N ), for if a ∈ H1 , b ∈ N and ab ∈ H2 then b ∈ H2 ∩ N . Therefore ker ψ = θ−1 (K1 )/H1 = H1 (H2 ∩ N )/H1 ∼ = H2 ∩ N/H1 ∩ N = J2 /J1 by the First Isomorphism Theorem (Theorem 1.22). Moreover the quotient of H2 /H1 by the normal subgroup ker ψ is isomorphic to the image K2 /K1 of ψ. Thus ker ψ is the required normal subgroup of H2 /H1 . Proposition 1.41 Let G be a group, and let H be a subgroup of G. Then (i) if G is solvable then any subgroup H of G is solvable; (ii) if G is solvable then G/N is solvable for any normal subgroup N of G; 27

(iii) if N is a normal subgroup of G and if both N and G/N are solvable then G is solvable. Proof Suppose that G is solvable. Let G0 , G1 , . . . , Gm be a finite sequence of subgroups of G, where G0 = {1}, Gn = G, and Gi−1 / Gi and Gi /Gi−1 is Abelian for i = 1, 2, . . . , m. We first show that the subgroup H is solvable. Let Hi = H ∩ Gi for i = 0, 1, . . . , m. Then H0 = {1} and Hm = H. If u ∈ Hi and v ∈ Hi−1 then uvu−1 ∈ H, since H is a subgroup of G. Also uvu−1 ∈ Gi−1 , since u ∈ Gi−1 , v ∈ Gi and Gi−1 is normal in Gi . Therefore uvu−1 ∈ Hi−1 . Thus Hi−1 is a normal subgroup of Hi for i = 1, 2, . . . , m. Moreover Hi Gi ∩ H Gi−1 (Gi ∩ H) ∼ = = Hi−1 Gi−1 ∩ (Gi ∩ H) Gi−1 by the First Isomorphism Theorem (Theorem 1.22), and thus Hi /Hi−1 is isomorphic to a subgroup of the Abelian group Gi /Gi−1 . It follows that Hi /Hi−1 must itself be an Abelian group. We conclude therefore that the subgroup H of G is solvable. Now let N be a normal subgroup of G, and let Ki = Gi N/N for all i. Then K0 is the trivial subgroup of G/N and Km = G/N . It follows from Lemma 1.40 that Ki−1 / Ki and Ki /Ki−1 is isomorphic to the quotient of Gi /Gi−1 by some normal subgroup. But a quotient of any Abelian group must itself be Abelian. Thus each quotient group Ki /Ki−1 is Abelian, and thus G/N is solvable. Finally suppose that G is a group, N is a normal subgroup of G and both N and G/N are solvable. We must prove that G is solvable. Now the solvability of N ensures the existence of a finite sequence G0 , G1 , . . . , Gm of subgroups of N , where G0 = {1}, Gm = N , and Gi−1 / Gi and Gi /Gi−1 is Abelian for i = 1, 2, . . . , m. Also the solvability of G/N ensures the existence of a finite sequence K0 , K1 , . . . , Kn of subgroups of G/N , where K0 = N/N , Kn = G/N , and Ki−1 / Ki and Ki /Ki−1 is Abelian for i = 1, 2, . . . , n. Let Gm+i be the preimage of Ki under the the quotient homomorphism ν: G → G/N , for i = 1, 2, . . . , n. The Second Isomorphism Theorem (Theorem 1.23) ensures that Gm+i /Gm+i−1 ∼ = Ki /Ki−1 for all i > 0. Therefore G0 , G1 , . . . , Gm+n is a finite sequence of subgroups of G, where G0 = {1}, Gn = G, and Gi−1 / Gi and Gi /Gi−1 is Abelian for i = 1, 2, . . . , m + n. Thus the group G is solvable, as required. Example The alternating group A5 is simple. It follows that A5 is not solvable, since the definition of solvable groups ensures that that any simple solvable group is cyclic, and A5 is not cyclic. Now if n ≥ 5 the symmetric 28

group Σn of all permutations of a set of n elements contains a subgroup isomorphic to A5 . (Take as this subgroup the set of all even permutations of five of the elements permuted by the elements of Σn .) Moreover any subgroup of a solvable group is solvable (Proposition 1.41.) It follows therefore that the symmetric group Σn is not solvable when n ≥ 5.

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2

Rings and Polynomials

2.1

Rings, Integral Domains and Fields

Definition A ring consists of a set R on which are defined operations of addition and multiplication satisfying the following axioms: • x+y = y+x for all elements x and y of R (i.e., addition is commutative); • (x + y) + z = x + (y + z) for all elements x, y and z of R (i.e., addition is associative); • there exists an an element 0 of R (known as the zero element) with the property that x + 0 = x for all elements x of R; • given any element x of R, there exists an element −x of R with the property that x + (−x) = 0; • x(yz) = (xy)z for all elements x, y and z of R (i.e., multiplication is associative); • x(y + z) = xy + xz and (x + y)z = xz + yz for all elements x, y and z of R (the Distributive Law ). Lemma 2.1 Let R be a ring. Then x0 = 0 and 0x = 0 for all elements x of R. Proof The zero element 0 of R satisfies 0 + 0 = 0. Using the Distributive Law, we deduce that x0 + x0 = x(0 + 0) = x0 and 0x + 0x = (0 + 0)x = 0x. Thus if we add −(x0) to both sides of the identity x0 + x0 = x0 we see that x0 = 0. Similarly if we add −(0x) to both sides of the identity 0x + 0x = 0x we see that 0x = 0. Lemma 2.2 Let R be a ring. Then (−x)y = −(xy) and x(−y) = −(xy) for all elements x and y of R. Proof It follows from the Distributive Law that xy +(−x)y = (x+(−x))y = 0y = 0 and xy + x(−y) = x(y + (−y)) = x0 = 0. Therefore (−x)y = −(xy) and x(−y) = −(xy). A subset S of a ring R is said to be a subring of R if 0 ∈ S, a + b ∈ S, −a ∈ S and ab ∈ S for all a, b ∈ S. A ring R is said to be commutative if xy = yx for all x, y ∈ R. Not every ring is commutative: an example of a non-commutative ring is provided by the ring of n × n matrices with real or complex coefficients when n > 1. 30

A ring R is said to be unital if it possesses a (necessarily unique) non-zero multiplicative identity element 1 satisfying 1x = x = x1 for all x ∈ R. Definition A unital commutative ring R is said to be an integral domain if the product of any two non-zero elements of R is itself non-zero. Definition A field consists of a set K on which are defined operations of addition and multiplication satisfying the following axioms: • x+y = y+x for all elements x and y of K (i.e., addition is commutative); • (x + y) + z = x + (y + z) for all elements x, y and z of K (i.e., addition is associative); • there exists an an element 0 of K known as the zero element with the property that x + 0 = x for all elements x of K; • given any element x of K, there exists an element −x of K with the property that x + (−x) = 0; • xy = yx for all elements x and y of K (i.e., multiplication is commutative); • x(yz) = (xy)z for all elements x, y and z of K (i.e., multiplication is associative); • there exists a non-zero element 1 of K with the property that 1x = x for all elements x of K; • given any non-zero element x of K, there exists an element x−1 of K with the property that xx−1 = 1; • x(y + z) = xy + xz and (x + y)z = xz + yz for all elements x, y and z of K (the Distributive Law ). An examination of the relevant definitions shows that a unital commutative ring R is a field if and only if, given any non-zero element x of R, there exists an element x−1 of R such that xx−1 = 1. Moreover a ring R is a field if and only if the set of non-zero elements of R is an Abelian group with respect to the operation of multiplication. Lemma 2.3 A field is an integral domain.

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Proof A field is a unital commutative ring. Let x and y be non-zero elements of a field K. Then there exist elements x−1 and y −1 of K such that xx−1 = 1 and yy −1 = 1. Then xyy −1 x−1 = 1. It follows that xy 6= 0, since 0(y −1 x−1 ) = 0 and 1 6= 0. The set Z of integers is an integral domain with respect to the usual operations of addition and multiplication. The sets Q, R and C of rational, real and complex numbers are fields.

2.2

Ideals

Definition Let R be a ring. A subset I of R is said to be an ideal of R if 0 ∈ I, a + b ∈ I, −a ∈ I, ra ∈ I and ar ∈ I for all a, b ∈ I and r ∈ R. An ideal I of R is said to be a proper ideal of R if I 6= R. Note that an ideal I of a unital ring R is proper if and only if 1 6∈ I. Indeed if 1 ∈ I then r ∈ I for all r ∈ R, since r = r1. Lemma 2.4 A unital commutative ring R is a field if and only if the only ideals of R are {0} and R. Proof Suppose that R is a field. Let I be a non-zero ideal of R. Then there exists x ∈ I satisfying x 6= 0. Moreover there exists x−1 ∈ R satisfying xx−1 = 1 = x−1 x. Therefore 1 ∈ I, and hence I = R. Thus the only ideals of R are {0} and R. Conversely, suppose that R is a unital commutative ring with the property that the only ideals of R are {0} and R. Let x be a non-zero element of R, and let Rx denote the subset of R consisting of all elements of R that are of the form rx for some r ∈ R. It is easy to verify that Rx is an ideal of R. (In order to show that yr ∈ Rx for all y ∈ Rx and r ∈ R, one must use the fact that the ring R is commutative.) Moreover Rx 6= {0}, since x ∈ Rx. We deduce that Rx = R. Therefore 1 ∈ Rx, and hence there exists some element x−1 of R satisfying x−1 x = 1. This shows that R is a field, as required. The intersection of any collection of ideals of a ring R is itself an ideal of R. For if a and b are elements of R that belong to all the ideals in the collection, then the same is true of 0, a + b, −a, ra and ar for all r ∈ R. Let X be a subset of the ring R. The ideal of R generated by X is defined to be the intersection of all the ideals of R that contain the set X. Note that this ideal is well-defined and is the smallest ideal of R containing the set X (i.e., it is contained in every other ideal that contains the set X). 32

We denote by (f1 , f2 , . . . , fk ) the ideal of R generated by any finite subset {f1 , f2 , . . . , fk } of R. We say that an ideal I of the ring R is finitely generated if there exists a finite subset of I which generates the ideal I. Lemma 2.5 Let R be a unital commutative ring, and let X be a subset of R. Then the ideal generated by X coincides with the set of all elements of R that can be expressed as a finite sum of the form r1 x1 + r2 x2 + · · · + rk xk , where x1 , x2 , . . . , xk ∈ X and r1 , r2 , . . . , rk ∈ R. Proof Let I be the subset of R consisting of all these finite sums. If J is any ideal of R which contains the set X then J must contain each of these finite sums, and thus I ⊂ J. Let a and b be elements of I. It follows immediately from the definition of I that 0 ∈ I, a + b ∈ I, −a ∈ I, and ra ∈ I for all r ∈ R. Also ar = ra, since R is commutative, and thus ar ∈ I. Thus I is an ideal of R. Moreover X ⊂ I, since the ring R is unital and x = 1x for all x ∈ X. Thus I is the smallest ideal of R containing the set X, as required. Each integer n generates an ideal nZ of the ring Z of integers. This ideal consists of those integers that are divisible by n. Lemma 2.6 Every ideal of the ring Z of integers is generated by some nonnegative integer n. Proof The zero ideal is of the required form with n = 0. Let I be some non-zero ideal of Z. Then I contains at least one strictly positive integer (since −m ∈ I for all m ∈ I). Let n be the smallest strictly positive integer belonging to I. If j ∈ I then we can write j = qn + r for some integers q and r with 0 ≤ r < n. Now r ∈ I, since r = j − qn, j ∈ I and qn ∈ I. But 0 ≤ r < n, and n is by definition the smallest strictly positive integer belonging to I. We conclude therefore that r = 0, and thus j = qn. This shows that I = nZ, as required.

2.3

Quotient Rings and Homomorphisms

Let R be a ring and let I be an ideal of R. If we regard R as an Abelian group with respect to the operation of addition, then the ideal I is a (normal) subgroup of R, and we can therefore form a corresponding quotient group R/I whose elements are the cosets of I in R. Thus an element of R/I is of the form I + x for some x ∈ R, and I + x = I + x0 if and only if x − x0 ∈ I. If

33

x, x0 , y and y 0 are elements of R satisfying I + x = I + x0 and I + y = I + y 0 then (x + y) − (x0 + y 0 ) = (x − x0 ) + (y − y 0 ), xy − x0 y 0 = xy − xy 0 + xy 0 − x0 y 0 = x(y − y 0 ) + (x − x0 )y 0 . But x − x0 and y − y 0 belong to I, and also x(y − y 0 ) and (x − x0 )y 0 belong to I, since I is an ideal. It follows that (x + y) − (x0 + y 0 ) and xy − x0 y 0 both belong to I, and thus I + x + y = I + x0 + y 0 and I + xy = I + x0 y 0 . Therefore the quotient group R/I admits well-defined operations of addition and multiplication, given by (I + x) + (I + y) = I + x + y,

(I + x)(I + y) = I + xy

for all I +x ∈ R/I and I +y ∈ R/I. One can readily verify that R/I is a ring with respect to these operations. We refer to the ring R/I as the quotient of the ring R by the ideal I. Example Let n be an integer satisfying n > 1. The quotient Z/nZ of the ring Z of integers by the ideal nZ generated by n is the ring of congruence classes of integers modulo n. This ring has n elements, and is a field if and only if n is a prime number. Definition A function ϕ: R → S from a ring R to a ring S is said to be a homomorphism (or ring homomorphism) if and only if ϕ(x+y) = ϕ(x)+ϕ(y) and ϕ(xy) = ϕ(x)ϕ(y) for all x, y ∈ R. If in addition the rings R and S are unital then a homomorphism ϕ: R → S is said to be unital if ϕ(1) = 1 (i.e., ϕ maps the identity element of R onto that of S). Let R and S be rings, and let ϕ: R → S be a ring homomorphism. Then the kernel ker ϕ of the homomorphism ϕ is an ideal of R, where ker ϕ = {x ∈ R : ϕ(x) = 0}. The image ϕ(R) of the homomorphism is a subring of S; however it is not in general an ideal of S. An ideal I of a ring R is the kernel of the quotient homomorphism that sends x ∈ R to the coset I + x. Definition An isomorphism ϕ: R → S between rings R and S is a homomorphism that is also a bijection between R and S. The inverse of an isomorphism is itself an isomorphism. Two rings are said to be isomorphic if there is an isomorphism between them. 34

The verification of the following result is a straightforward exercise. Proposition 2.7 Let ϕ: R → S be a homomorphism from a ring R to a ring S, and let I be an ideal of R satisfying I ⊂ ker ϕ. Then there exists a unique homomorphism ϕ: R/I → S such that ϕ(I + x) = ϕ(x) for all x ∈ R. Moreover ϕ: R/I → S is injective if and only if I = ker ϕ. Corollary 2.8 Let ϕ: R → S be ring homomorphism. Then ϕ(R) is isomorphic to R/ ker ϕ.

2.4

The Characteristic of a Ring

Let R be a ring, and let r ∈ R. We may define n.r for all natural numbers n by recursion on n so that 1.r = r and n.r = (n − 1).r + r for all n > 0. We define also 0.r = 0 and (−n).r = −(n.r) for all natural numbers n. Then (m + n).r = m.r + n.r, (mn).r = m.(n.r),

n.(r + s) = n.r + n.s, (m.r)(n.s) = (mn).(rs)

for all integers m an n and for all elements r and s of R. In particular, suppose that R is a unital ring. Then the set of all integers n satisfying n.1 = 0 is an ideal of Z. Therefore there exists a unique nonnegative integer p such that pZ = {n ∈ Z : n.1 = 0} (see Lemma 2.6). This integer p is referred to as the characteristic of the ring R, and is denoted by charR. Lemma 2.9 Let R be an integral domain. Then either charR = 0 or else charR is a prime number. Proof Let p = charR. Clearly p 6= 1. Suppose that p > 1 and p = jk, where j and k are positive integers. Then (j.1)(k.1) = (jk).1 = p.1 = 0. But R is an integral domain. Therefore either j.1 = 0, or k.1 = 0. But if j.1 = 0 then p divides j and therefore j = p. Similarly if k.1 = 0 then k = p. It follows that p is a prime number, as required.

2.5

Polynomial Rings

Let R be a unital commutative ring. The set of all polynomials c0 + c1 x + c2 x2 + · · · + cn xn in an indeterminate x with coefficients c0 , . . . , cn in the ring R themselves constitute a ring, which we shall denote by R[x]. If the coefficient cn of 35

highest power of x is non-zero then the polynomial is said to be of degree n, and the coefficient cn is referred to as the leading coefficient of the polynomial. The polynomial is said to be monic if the leading coefficient cn is equal to the multiplicative identity element 1 of the ring R. Two polynomials with coefficients in the ring R are equal if and only if they are of the same degree and corresponding coefficients are equal. Polynomials may be added, subtracted and multiplied in the usual fashion. We now consider various properties of polynomials whose coefficients belong to a field K (such as the field of rational numbers, real numbers or complex numbers). Lemma 2.10 Let K be a field, and let f ∈ K[x] be a non-zero polynomial with coefficients in K. Then, given any polynomial h ∈ K[x], there exist unique polynomials q and r in K[x] such that h = f q + r and either r = 0 or else deg r < deg f . Proof If deg h < deg f then we may take q = 0 and r = h. In general we prove the existence of q and r by induction on the degree deg h of h. Thus suppose that deg h ≥ deg f and that any polynomial of degree less than deg h can be expressed in the required form. Now there is some element c of K for which the polynomials h(x) and cf (x) have the same leading coefficient. Let h1 (x) = h(x) − cxm f (x), where m = deg h − deg f . Then either h1 = 0 or deg h1 < deg h. The inductive hypothesis then ensures the existence of polynomials q1 and r such that h1 = f q1 + r and either r = 0 or else deg r < deg f . But then h = f q + r, where q(x) = cxm + q1 (x). We now verify the uniqueness of q and r. Suppose that f q + r = f q + r, where q, r ∈ K[x] and either r = 0 or deg r < deg f . Then (q − q)f = r − r. But deg((q − q)f ) ≥ deg f whenever q 6= q, and deg(r − r) < deg f whenever r 6= r. Therefore the equality (q − q)f = r − r cannot hold unless q = q and r = r. This proves the uniqueness of q and r. Any polynomial f with coefficients in a field K generates an ideal (f ) of the polynomial ring K[x] consisting of all polynomials in K[x] that are divisible by f . Lemma 2.11 Let K be a field, and let I be an ideal of the polynomial ring K[x]. Then there exists f ∈ K[x] such that I = (f ), where (f ) denotes the ideal of K[x] generated by f . Proof If I = {0} then we can take f = 0. Otherwise choose f ∈ I such that f 6= 0 and the degree of f does not exceed the degree of any non-zero polynomial in I. Then, for each h ∈ I, there exist polynomials q and r in K[x] 36

such that h = f q + r and either r = 0 or else deg r < deg f . (Lemma 2.10). But r ∈ I, since r = h − f q and h and f both belong to I. The choice of f then ensures that r = 0 and h = qf . Thus I = (f ). Definition Polynomials f1 , f2 , . . . , fk with coefficients in some field K. are said to be coprime if there is no non-constant polynomial that divides all of them. Theorem 2.12 Let f1 , f2 , . . . , fk be coprime polynomials with coefficients in some field K. Then there exist polynomials g1 , g2 , . . . , gk with coefficients in K such that f1 (x)g1 (x) + f2 (x)g2 (x) + · · · + fk (x)gk (x) = 1. Proof Let I be the ideal in K[x] generated by f1 , f2 , . . . , fk . It follows from Lemma 2.11 that the ideal I is generated by some polynomial d. Then d divides all of f1 , f2 , . . . , fk and is therefore a constant polynomial, since these polynomials are coprime. It follows that I = K[x]. But the ideal I of K[x] generated by f1 , f2 , . . . , fk coincides with the subset of K[x] consisting of all polynomials that may be represented as finite sums of the form f1 (x)g1 (x) + f2 (x)g2 (x) + · · · + fk (x)gk (x) for some polynomials g1 , g2 , . . . , gk . It follows that the constant polynomial with value 1 may be expressed as a sum of this form, as required. Definition A non-constant polynomial f with coefficients in a field K is said to be irreducible over K if it is not divisible by any non-constant polynomial of lower degree with coefficients in K. Any polynomial with coefficients in a field K may be factored as a product of irreducible polynomials. This is easily proved by induction on the degree of the polynomial, for if a non-constant polynomial is not itself irreducible, then it can be factored as a product of polynomials of lower degree. Lemma 2.13 Let K be a field. Then the ring K[x] of polynomials with coefficients in K contains infinitely many irreducible polynomials. Proof Let f1 , f2 , . . . , fk ∈ K[x] be irreducible polynomials, and let g = f1 f2 · · · fk + 1. Then g is not divisible by f1 , f2 , . . . , fk , and therefore no irreducible factor of g is divisible by any of f1 , f2 , . . . , fk . It follows that K[x] must contain irreducible polynomials distinct from f1 , f2 , . . . , fk . Thus the number of irreducible polynomials in K[x] cannot be finite. 37

The proof of Lemma 2.13 is a direct analogue of Euclid’s proof of the existence of infinitely many prime numbers. Proposition 2.14 Let f , g and h be polynomials with coefficients in some field K. Suppose that f is irreducible over K and that f divides the product gh. Then either f divides g or else f divides h. Proof Suppose that f does not divide g. We must show that f divides h. Now the only polynomials that divide f are constant polynomials and multiples of f . No multiple of f divides g. Therefore the only polynomials that divide both f and g are constant polynomials. Thus f and g are coprime. It follows from Proposition 2.12 that there exist polynomials u and v with coefficients in K such that 1 = ug + vf . Then h = ugh + vf h. But f divides ugh + vf h, since f divides gh. It follows that f divides h, as required. Proposition 2.15 Let K be a field, and let (f ) be the ideal of K[x] generated by an irreducible polynomial f with coefficients in K. Then K[x]/(f ) is a field. Proof Let I = (f ). Then the quotient ring K[x]/I is commutative and has a multiplicative identity element I +1. Let g ∈ K[x]. Suppose that I +g 6= I. Now the only factors of f are constant polynomials and constant multiples of f , since f is irreducible. But no constant multiple of f can divide g, since g 6∈ I. It follows that the only common factors of f and g are constant polynomials. Thus f and g are coprime. It follows from Proposition 2.12 that there exist polynomials h, k ∈ K[x] such that f h + gk = 1. But then (I +k)(I +g) = I +1 in K[x]/I, since f h ∈ I. Thus I +k is the multiplicative inverse of I + g in K[x]/I. We deduce that every non-zero element of K[x]/I is invertible, and thus K[x]/I is a field, as required.

2.6

Gauss’s Lemma

We shall show that a polynomial with integer coefficients is irreducible over Q if and only if it cannot be expressed as a product of polynomials of lower degree with integer coefficients. Definition A polynomial with integer coefficients is said to be primitive if there is no prime number that divides all the coefficients of the polynomial Lemma 2.16 (Gauss’s Lemma) Let g and h be polynomials with integer coefficients. If g and h are both primitive then so is gh. 38

Proof Let g(x) = b0 + b1 x + b2 x2 + · · · + br xr and h(x) = c0 + c1 x + c2 x2 + · · · + cs xs , and let g(x)h(x) = a0 + a1 x + a2 x2 + · · · + ar+s xr+s . Let p be a prime number. Then the polynomials g and h must both have at least one coefficient that is not divisible by p. Let j and k be the smallest values of i for which p does not divide bi and ci respectively. Then aj+k −bj ck is divisible j−1 k−1 P P by p, since aj+k − bj ck = bi cj+k−i + bj+k−i ci , where p divides bi for all i=0

i=0

i < j and p divides ci for all i < k. But p does not divide bj ck since p does not divide either bj or ck . Therefore p does not divide the coefficient aj+k of gh. This shows that the polynomial gh is primitive, as required. Proposition 2.17 A polynomial with integer coefficients is irreducible over the field Q of rational numbers if and only if it cannot be factored as a product of polynomials of lower degree with integer coefficients. Proof Let f be a polynomial with integer coefficients. If f is irreducible over Q then f clearly cannot be factored as a product of polynomials of lower degree with integer coefficients. Conversely suppose that f cannot be factored in this way. Let f (x) = g(x)h(x), where g and h are polynomials with rational coefficients. Then there exist positive integers r and s such that the polynomials rg(x) and sh(x) have integer coefficients. Let the positive integers u and v be the highest common factors of the coefficients of the polynomials rg(x) and sh(x) respectively. Then rg(x) = ug∗ (x) and sh(x) = vh∗ (x), where g∗ and h∗ are primitive polynomials with integer coefficients. Then (rs)f (x) = (uv)g∗ (x)h∗ (x). We now show that f (x) = mg∗ (x)h∗ (x) for some integer m. Let l be the smallest divisor of rs such that lf (x) = mg∗ (x)h∗ (x) for some integer m. We show that l = 1. Suppose that it were the case that l > 1. Then there would exist a prime factor p of l. Now p could not divide m, since otherwise (l/p)f (x) = (m/p)g∗ (x)h∗ (x), which contradicts the definition of l. Theorefore p would have to divide each coefficient of g∗ (x)h∗ (x), which is impossible, since it follows from Gauss’s Lemma (Lemma 2.16) that the product g∗ h∗ of the primitive polynomials g∗ and h∗ is itself a primitive polynomial. Therefore l = 1 and f (x) = mg∗ (x)h∗ (x). Now f does not factor as a product of polynomials of lower degree with integer coefficients. Therefore either deg f = deg g∗ = deg g, or else deg f = deg h∗ = deg h, Thus f is irreducible over Q, as required.

2.7

Eisenstein’s Irreducibility Criterion

Proposition 2.18 (Eisenstein’s Irreducibility Criterion) Let f (x) = a0 + a1 x + a2 x2 + · · · + an xn 39

be a polynomial of degree n with integer coefficients, and let p be a prime number. Suppose that • p does not divide an , • p divides a0 , a1 , . . . , an−1 , • p2 does not divide a0 . Then the polynomial f is irreducible over the field Q of rational numbers. Proof Suppose that f (x) = g(x)h(x), where g and h are polynomials with integer coefficients. Let g(x) = b0 + b1 x + b2 x2 + · · · + br xr and h(x) = c0 +c1 x+c2 x2 +· · ·+cs xs . Then a0 = b0 c0 . Now a0 is divisible by p but is not divisible by p2 . Therefore exactly one of the coefficients b0 and c0 is divisible by p. Suppose that p divides b0 but does not divide c0 . Now p does not divide all the coefficients of g(x), since it does not divide all the coefficients of f (x). Let j be the smallest value of i for which p does not divide bi . Then p divides j−1 P aj − bj c0 , since aj − bj c0 = bi cj−i and bi is divisible by p when i < j. But i=0

bj c0 is not divisible by p, since p is prime and neither bj nor c0 is divisible by p. Therefore aj is not divisible by p, and hence j = n and deg g ≥ n = deg f . Thus deg g = deg f and deg h = 0. Thus the polynomial f does not factor as a product of polynomials of lower degree with integer coefficients, and therefore f is irreducible over Q (Proposition 2.17).

40

3

Introduction to Galois Theory

3.1

Field Extensions and the Tower Law

Let K be a field. An extension L: K of K is an embedding of K in some larger field L. Definition Let L: K and M : K be field extensions. A K-homomorphism θ: L → M is a homomorphism of fields which satisfies θ(a) = a for all a ∈ K. A K-monomorphism is an injective K-homomorphism. A K-isomorphism is a bijective K-homomorphism. A K-automorphism of L is a K-isomorphism mapping L onto itself. Two extensions L1 : K and L2 : K of a field K are said to be K-isomorphic (or isomorphic) if there exists a K-isomorphism ϕ: L1 → L2 between L1 and L2 . If L: K is a field extension then we can regard L as a vector space over the field K. If L is a finite-dimensional vector space over K then we say that the extension L: K is finite. The degree [L: K] of a finite field extension L: K is defined to be the dimension of L considered as a vector space over K. Proposition 3.1 (The Tower Law) Let M : L and L: K be field extensions. Then the extension M : K is finite if and only if M : L and L: K are both finite, in which case [M : K] = [M : L][L: K]. Proof Suppose that M : K is a finite field extension. Then L, regarded as a vector space over K, is a subspace of the finite-dimensional vector space M , and therefore L is itself a finite-dimensional vector space over K. Thus L: K is finite. Also there exists a finite subset of M which spans M as a vector space over K, since M : K is finite, and this finite subset must also span M over L, and thus M : L must be finite. Conversely suppose that M : L and L: K are both finite extensions. Let x1 , x2 , . . . , xm be a basis for L, considered as a vector space over the field K, and let y1 , y2 , . . . , yn be a basis for M , considered as a vector space over the field L. Note that m = [L: K] and n = [M : L]. We claim that the set of all products xi yj with i = 1, 2, . . . , m and j = 1, 2, . . . , n is a basis for M , considered as a vector space over K. First we show that the elements xi yj are linearly independent over K. m P n P Suppose that λij xi yj = 0, where λij ∈ K for all i and j. Then i=1 j=1

m P

λij xi ∈ L for all j, and y1 , y2 , . . . , yn are linearly independent over L,

i=1

41

and therefore

m P

λij xi = 0 for j = 1, 2, . . . , n. But x1 , x2 , . . . , xm are linearly

i=1

independent over K. It follows that λij = 0 for all i and j. This shows that the elements xi yj are linearly independent over K. Now y1 , y2 , . . . , yn span M as a vector space over L, and therefore any n P element z of M can be written in the form z = µj yj , where µj ∈ L for j=1

all j. But each µj can be written in the form µj = for all i and j. But then z =

m P n P

m P

λij xi , where λij ∈ K

i=1

λij xi yj . This shows that the products

i=1 j=1

xi yj span M as a vector space over K, and thus {xi yj : 1 ≤ i ≤ m and 1 ≤ j ≤ n} is a basis of M , considered as a vector space over K. We conclude that the extension M : K is finite, and [M : K] = mn = [M : L][L: K], as required. Let L: K be a field extension. If A is any subset of L, then the set K ∪ A generates a subfield K(A) of L which is the intersection of all subfields of L that contain K ∪ A. (Note that any intersection of subfields of L is itself a subfield of L.) We say that K(A) is the field obtained from K by adjoining the set A. We denote K({α1 , α2 , . . . , αk }) by K(α1 , α2 , . . . , αk ) for any finite subset {α1 , α2 , . . . , αk } of L. In particular K(α) denotes the field obtained by adjoining some element α of L to K. A field extension L: K is said to be simple if there exists some element α of L such that L = K(α).

3.2

Algebraic Field Extensions

Definition Let L: K be a field extension, and let α be an element of L. If there exists some non-zero polynomial f ∈ K[x] with coefficients in K such that f (α) = 0, then α is said to be algebraic over K; otherwise α is said to be transcendental over K. A field extension L: K is said to be algebraic if every element of L is algebraic over K. Lemma 3.2 A finite field extension is algebraic.

42

Proof Let L: K be a finite field extension, and let n = [L: K]. Let α ∈ L. Then either the elements 1, α, α2 , . . . , αn are not all distinct, or else these elements are linearly dependent over the field K (since a linearly independent subset of L can have at most n elements.) Therefore there exist c0 , c1 , c2 , . . . , cn ∈ K, not all zero, such that c0 + c1 α + c2 α2 + · · · + cn αn = 0. Thus α is algebraic over K. This shows that the field extension L: K is algebraic, as required. Definition A polynomial f with coefficients in some field or unital ring is said to be monic if its leading coefficient (i.e., the coefficient of the highest power of x occurring in f (x) with a non-zero coefficient) is equal to 1. Lemma 3.3 Let K be a field and let α be an element of some extension field L of K. Suppose that α is algebraic over K. Then there exists a unique irreducible monic polynomial m ∈ K[x], with coefficients in K, characterized by the following property: f ∈ K[x] satisfies f (α) = 0 if and only if m divides f in K[x]. Proof Let I = {f ∈ K[x] : f (α) = 0}. Then I is a non-zero ideal of K[x]. Now there exists some polynomial m with coefficients in K which generates the ideal I (Lemma 2.11). Moreover, by dividing m by its leading coefficient, if necessary, we can ensure that m is a monic polynomial. Then f ∈ K[x] satisfies f (α) = 0 if and only if m divides f . Suppose that m = gh where g, h ∈ K[x]. Then 0 = m(α) = g(α)h(α). But then either g(α) = 0, in which case m divides g, or else h(α) = 0, in which case m divides h. The polynomial m is thus irreducible over K. The polynomial m is uniquely determined since if some monic polynomial m also satisfies the required conditions then m and m divide one another and therefore m = m. Definition Let K be a field and let L be an extension field of K. Let α be an element of L that is algebraic over K. The minimum polynomial m of α over K is the unique irreducible monic polynomial m ∈ K[x] with coefficients in K characterized by the following property: f ∈ K[x] satisfies f (α) = 0 if and only if m divides f in K[x]. Note that if f ∈ K[x] is an irreducible monic polynomial, and if α is a root of f in some extension field L of K, then f is the minimum polynomial of α over K. 43

Theorem 3.4 A simple field extension K(α): K is finite if and only if α is algebraic over K, in which case [K(α): K] is the degree of the minimum polynomial of α over K. Proof Suppose that the field extension K(α): K is finite. It then follows from Lemma 3.2 that α is algebraic over K. Conversely suppose that α is algebraic over K. Let R = {f (α) : f ∈ K[x]}. Now f (α) = 0 if and only if the minimum polynomial m of α over K divides f . It follows that f (α) = 0 if and only if f ∈ (m), where (m) is the ideal of K[x] generated by m. The ring homomorphism from K[x] to R that sends f ∈ K[x] to f (α) therefore induces an isomorphism between the quotient ring K[x]/(m) and the ring R. But K[x]/(m) is a field, since m is irreducible (Proposition 2.15). Therefore R is a subfield of K(α) containing K ∪ {α}, and hence R = K(α). Let z ∈ K(α). Then z = g(α) for some g ∈ K[x]. But then there exist polynomials l and f belonging to K[x] such that g = lm + f and either f = 0 or deg f < deg m (Lemma 2.10). But then z = f (α) since m(α) = 0. Suppose that z = h(α) for some polynomial h ∈ K[x], where either h = 0 or deg h < deg m. Then m divides h−f , since α is a zero of h−f . But if h−f were non-zero then its degree would be less than that of m, and thus h − f would not be divisible by m. We therefore conclude that h = f . Thus any element z of K(α) can be expressed in the form z = f (α) for some uniquely determined polynomial f ∈ K[x] satisfying either f = 0 or deg f < deg m. Thus if n = deg m then 1, α, α2 . . . , αn−1 is a basis of K(α) over K. It follows that the extension K(α): K is finite and [K(α): K] = deg m, as required. Corollary 3.5 A field extension L: K is finite if and only if there exists a finite subset {α1 , α2 , . . . , αk } of L such that αi is algebraic over K for i = 1, 2, . . . , k and L = K(α1 , α2 , . . . , αk ). Proof Suppose that the field extension L: K is a finite. Then it is algebraic (Lemma 3.2). Thus if {α1 , α2 , . . . , αk } is a basis for L, considered as a vector space over K, then each αi is algebraic and L = K(α1 , α2 , . . . , αk ). Conversely suppose that L = K(α1 , α2 , . . . , αk ), where αi is algebraic over K for i = 1, 2, . . . , k. Let Ki = K(α1 , α2 , . . . , αi ) for i = 1, 2, . . . , k. Clearly Ki−1 (αi ) ⊂ Ki for all i > 1, since Ki−1 ⊂ Ki and αi ∈ Ki . Also Ki ⊂ Ki−1 (αi ), since Ki−1 (αi ) is a subfield of L containing K ∪ {α1 , α2 , . . . , αi } We deduce that Ki = Ki−1 (αi ) for i = 2, 3, . . . , k. Moreover αi is clearly algebraic over Ki−1 since it is algebraic over K, and K ⊂ Ki−1 . It follows from Theorem 3.4 that the field extension Ki : Ki−1 is finite for each i. Using the Tower Law (Proposition 3.1), we deduce that L: K is a finite extension, as required. 44

Corollary 3.6 Let M : L and L: K be algebraic field extensions. Then M : K is an algebraic field extension. Proof Let α be an element of M . We must show that α is algebraic over K. Now there exists some non-zero polynomial f ∈ L[x] with coefficients in L such that f (α) = 0, since M : L is algebraic. Let β1 , β2 , . . . , βk be the coefficients of f (x), and let L0 = K(β1 , β2 , . . . , βk ). Now each βi is algebraic over K (since L: K is algebraic). Thus L0 : K is finite. Moreover α is algebraic over L0 , since the coefficients of the polynomial f belong to L0 , and thus L0 (α): L0 is finite (Theorem 3.4). It follows from the Tower Law (Proposition 3.1) that L0 (α): K is finite. But then K(α): K is finite, and hence α is algebraic over K, as required.

3.3

Algebraically Closed Fields

Definition A field K is said to be algebraically closed if, given any nonconstant polynomial f ∈ K[x] with coefficients in K, there exists some α ∈ K satisfying f (α) = 0. The field C of complex numbers is algebraically closed. This result is the Fundamental Theorem of Algebra. Lemma 3.7 Let K be an algebraically closed field, and let L: K be an algebraic extension of K. Then L = K. Proof Let α ∈ L, and let mα ∈ K[x] be the minimal polynomial of α over K. Then the polynomial mα (x) has a root a in K, and is therefore divisible by the polynomial x − a. It follows that mα (x) = x − a, since mα (x) is an irreducible monic polynomial. But then α = a, and therefore α ∈ K. This shows that every element of L belongs to K, and thus L = K, as required.

3.4

Ruler and Compass Constructions

One can make use of the Tower Law in order to prove the impossibility of performing a number of geometric constructions in a finite number of steps using straightedge and and compasses alone. These impossible constructions include the following: • the trisection of an arbitrary angle; • the construction of the edge of a cube having twice the volume of some given cube; 45

• the construction of a square having the same area as a given circle. Definition Let P0 and P1 be the points of the Euclidean plane given by P0 = (0, 0) and P1 = (1, 0). We say that a point P of the plane is constructible using straightedge and compasses alone if P = Pn for some finite sequence P0 , P1 , . . . , Pn of points of the plane, where P0 = (0, 0), P1 = (1, 0) and, for each j > 1, the point Pj is one of the following:— • the intersection of two distinct straight lines, each passing through at least two points belonging to the set {P0 , P1 , . . . , Pj−1 }; • the point at which a straight line joining two points belonging to the set {P0 , P1 , . . . , Pj−1 } intersects a circle which is centred on a point of this set and passes through another point of the set; • the point of intersection of two distinct circles, where each circle is centred on a point of the set {P0 , P1 , . . . , Pj−1 } and passes through another point of the set. Constructible points of the plane are those that can be constructed from the given points P0 and P1 using straightedge (i.e., unmarked ruler) and compasses alone. Theorem 3.8 Let (x, y) be a constructible point of the Euclidean plane. Then [Q(x, y): Q] = 2r for some non-negative integer r. Proof Let P = (x, y) and let P0 , P1 , . . . , Pn be a finite sequence of points of the plane with the properties listed above. Let K0 = K1 = Q and Kj = Kj−1 (xj , yj ) for j = 2, 3, . . . , n, where Pj = (xj , yj ). Straightforward coordinate geometry shows that, for each j, the real numbers xj and yj are both roots of linear or quadratic polynomials with coefficients in Kj−1 . It follows that [Kj−1 (xj ): Kj−1 ] = 1 or 2 and [Kj−1 (xj , yj ): Kj−1 (xj )] = 1 or 2 for each j. It follows from the Tower Law (Proposition 3.1) that [Kn : Q] = 2s for some non-negative integer s. But [Kn : Q] = [Kn : Q(x, y)][Q(x, y): Q]. We deduce that [Q(x, y): Q] divides 2s , and therefore [Q(x, y): Q] = 2r for some non-negative integer r. One can apply this criterion to show that there is no geometrical construction that enables one to trisect an arbitrary angle using straightedge and compasses alone. The same method can be used to show the impossibility of ‘duplicating a cube’ or ‘squaring a circle’ using straightedge and compasses alone. 46

Example We show that there is no geometrical construction for the trisection of an angle of π3 radians (i.e., 60◦ ) using straightedge and compasses alone. Let a = cos π9 and b = sin π9 . Now the point (cos π3 , sin π3 ) (i.e, the √ point ( 12 , 12 3)) is constructible. Thus if an angle of π3 radians could be trisected using straightedge and compasses alone, then the point (a, b) would be constructible. Now cos 3θ = cos θ cos 2θ − sin θ sin 2θ = cos θ(cos2 θ − sin2 θ) − 2 sin2 θ cos θ = 4 cos3 θ − 3 cos θ for any angle θ. On setting θ = π9 we deduce that 4a3 − 3a = 12 and thus 8a3 − 6a − 1 = 0. Now 8a3 − 6a − 1 = f (2a − 1), where f (x) = x3 + 3x2 − 3. An immediate application of Eisenstein’s criterion for irreducibility shows that the polynomial f is irreducible over the field Q of rational numbers, and thus [Q(a): Q] = [Q(2a − 1): Q] = 3. It now follows from Theorem 3.8 that the point (cos π9 , sin π9 ) is not constructible using straightedge and compasses alone. Therefore it is not possible to trisect an angle of π3 radians using straightedge and compasses alone. It follows that there is no geometrical construction for the trisection of an arbitrary angle using straightedge and compasses alone. Example It is not difficult to see that if it were possible to construct two √ √ 3 3 points in the plane a distance 2 apart, then the point ( 2, 0) would be constructible. But it follows from Theorem 3.8 that this is impossible, √ 3 since 2√is a root of the irreducible monic polynomial x3 − 2, and therefore [Q( 3 2), Q] = 3. We conclude that there is no geometric construction using straightedge and compasses alone that will construct from a line segment in the plane a second line segment such that a cube with the second line segment as an edge will have twice the volume of a cube with the first line segment as an edge. Example It can be shown that π is not algebraic over the field Q of rational √ numbers. Therefore π is not algebraic over Q. It then follows from Theorem 3.8 it is not possible to give a geometrical construction for obtaining a square with the same area as a given circle, using straightedge and compasses alone. (Thus it is not possible to ‘square the circle’ using straightedge and compasses alone.) Lemma 3.9 If the endpoints of any line segment in the plane are constructible, then so is the midpoint.

47

Proof Let P and Q be constructible points in the plane. Let S and T be the points where the circle centred on P and passing through Q intersects the circle centred on Q and passing through P . Then S and T are constructible points in the plane, and the point R at which the line ST intersects the line P Q is the midpoint of the line segment P Q. Thus this midpoint is a constructible point. Lemma 3.10 If any three vertices of a parallelogram in the plane are constructible, then so is the fourth vertex. Proof Let the vertices of the parallelogram listed in anticlockwise (or in clockwise) order be A, B, C and D, where A, B and D are constructible points. We must show that C is also constructible. Now the midpoint E of the line segment BD is a constructible point, and the circle centred on E and passing though A will intersect the line AE in the point C. Thus C is a constructible point, as required. Theorem 3.11 Let K denote the set of all real numbers x for which the point (x, 0) is constructible using straightedge and compasses alone. Then K is a subfield of the field of real numbers, and a point (x, y) of the plane is constructible using straightedge and compass √ alone if and only if x ∈ K and y ∈ K. Moreover if x ∈ K and x > 0 then x ∈ K. Proof Clearly 0 ∈ K and 1 ∈ K. Let x and y be real numbers belonging to K. Then (x, 0) and (y, 0) are constructible points of the plane. Let M be the midpoint of the line segment whose endpoints are (x, 0) and (y, 0). Then M is constructible (Lemma 3.9), and M = ( 21 (x + y), 0). The circle centred on M and passing through the origin intersects the x-axis at the origin and at the point (x + y, 0). Therefore (x + y, 0) is a constructible point, and thus x + y ∈ K. Also the circle centred on the origin and passing through (x, 0) intersects the x-axis at (−x, 0). Thus (−x, 0) is a constructible point, and thus −x ∈ K. We claim that if x ∈ K then the point (0, x) is constructible. Now if x ∈ K and x 6= 0 then (x, 0) and (−x, 0) are constructible points, and the circle centred on (x, 0) and passing through (−x, 0) intersects the circle centred on (−x, 0) and passing through (x, 0) in two √ √ points that lie on the y-axis. These two points (namely (0, 3x) and (0, − 3x)) are constructible, and therefore the circle centred on the origin and passing though (x, 0) intersects the y-axis in two constructible points which are (0, x) and (0, −x). Thus if x ∈ K then the point (0, x) is constructible. Let x and y be real numbers belonging to K. Then the points (x, 0), (0, y) and (0, 1) are constructible. The point (x, y − 1) is then constructible, 48

since it is the fourth vertex of a parallelogram which has three vertices at the constructible points (x, 0), (0, y) and (0, 1) (Lemma 3.10). But the line which passes through the two constructible points (0, y) and (x, y − 1) intersects the x-axis at the point (xy, 0). Therefore the point (xy, 0) is constructible, and thus xy ∈ K. Now suppose that x ∈ K, y ∈ K and y 6= 0. The point (x, 1 − y) is constructible, since it is the fourth vertex of a parallelogram with vertices at the constructible points (x, 0), (0, y) and (0, 1). The line segment joining the constructible points (0, 1) and (x, 1 − y) intersects the x-axis at the point (xy −1 , 0). Thus xy −1 ∈ K. The above results show that K is a subfield of the field of real numbers. Moreover if x ∈ K and y ∈ K then the point (x, y) is constructible, since it is the fourth vertex of a rectangle with vertices at the constructible points (0, 0), (x, 0) and (0, y). Conversely, suppose that the point (x, y) is constructible. We claim that the point (x, 0) is constructible and thus x ∈ K. This result is obviously true if y = 0. If y 6= 0 then the circles centred on the points (0, 0) and (1, 0) and passing through (x, y) intersect in the two points (x, y) and (x, −y). The point (x, 0) is thus the point at which the line passing through the constructible points (x, y) and (x, −y) intersects the x-axis, and is thus itself constructible. The point (0, y) is then the fourth vertex of a rectangle with vertices at the constructible points (0, 0), (x, 0) and (x, y), and thus is itself constructible. The circle centred on the origin and passing though (0, y) intersects the x-axis at (y, 0). Thus (y, 0) is constructible, and thus y ∈ K. We have thus shown that a point (x, y) is constructible using straightedge and compasses alone if and only if x ∈ K and y ∈ K. Suppose that x ∈ K and that x > 0. Then 21 (1 − x) ∈ K. Thus if C = (0, 12 (1 − x)) then C is a constructible point. Let (u, 0) be the point at which the circle centred on C and passing through the constructible point (0, 1) intersects the x-axis. (The circle does intersect the x-axis since it passes through (0, 1) and (0, −x), and x > 0.) The radius of this circle is 12 (1 + x)), and therefore 14 (1 − x)2 + u2 = 14 (1 + x)2 (Pythagoras’ Theorem.) But then 2 u √ = x. But (u, 0) is a constructible point. Thus if x ∈ K and x > 0 then x ∈ K, as required. The above theorems can be applied to the problem of determining whether or not it is possible to construct a regular n-sided polygon with a straightedge and compass, given its centre and one of its vertices. The impossibility of trisecting an angle of 60◦ shows that a regular 18-sided polygon is not constructible using straightedge and compass. Now if one can construct a regular n-sided polygon then one can easily construct a regular 2n-sided polygon by bisecting the angles of the n-sided polygon. Thus the problem 49

reduces to that of determining which regular polygons with an odd number of sides are constructible. Moreover it is not difficult to reduce down to the case where n is a power of some odd prime number. Gauss discovered that a regular 17-sided polygon was constructible in 1796, when he was 19 years old. Techniques of Galois Theory show that the regular n-sided polygon is constructible using straightedge and compass if and only if n = 2s p1 p2 · · · pt , where p1 , p2 , . . . , pt are distinct Fermat primes: a Fermat prime is a prime number that is of the form 2k +1 for some integer k. If k = uv, where u and v are positive integers and v is odd, then 2k + 1 = wv + 1 = (w + 1)(wv−1 − wv−2 + · · · − w + 1), where w = 2u , and hence m 2k + 1 is not prime. Thus any Fermat prime is of the form 22 + 1 for some non-negative integer m. Fermat observed in 1640 that Fm is prime when m ≤ 4. These Fermat primes have the values F0 = 3, F1 = 5, F2 = 17, F3 = 257 and F4 = 65537. Fermat conjectured that all the numbers Fm were prime. However it has been shown that Fm is not prime for any integer m between 5 and 16. Moreover F16 = 265536 + 1 ≈ 1020000 . Note that the five Fermat primes 3, 5, 17, 257 and 65537 provide only 32 constructible regular polygons with an odd number of sides. It is not difficult to see that the geometric problem of constructing a regular n-sided polygon using straightedge and compasses is equivalent to the algebraic problem of finding a formula to express the nth roots of unity in the complex plane in terms of integers or rational numbers by means of algebraic formulae which involve finite addition, subtraction, multiplication, division and the successive extraction of square roots. Thus the problem is closely related to that of expressing the roots of a given polynomial in terms of its coefficients by means of algebraic formulae which involve only finite addition, subtraction, multiplication, division and the successive extraction of pth roots for appropriate prime numbers p.

3.5

Splitting Fields

Definition Let L: K be a field extension, and let f ∈ K[x] be a polynomial with coefficients in K. The polynomial f is said to split over L if f is a constant polynomial or if there exist elements α1 , α2 , . . . , αn of L such that f (x) = c(x − α1 )(x − α2 ) · · · (x − αn ), where c ∈ K is the leading coefficient of f . We see therefore that a polynomial f ∈ K[x] splits over an extension field L of K if and only if f factors in L[x] as a product of constant or linear factors. 50

Definition Let L: K be a field extension, and let f ∈ K[x] be a polynomial with coefficients in K. The field L is said to be a splitting field for f over K if the following conditions are satisfied:— • the polynomial f splits over L; • the polynomial f does not split over any proper subfield of L that contains the field K. Lemma 3.12 Let M : K be a field extension, and let f ∈ K[x] be a polynomial with coefficients in K. Suppose that the polynomial f splits over M . Then there exists a unique subfield L of M which is a splitting field for f over K. Proof Let L be the intersection of all subfields M 0 of M containing K with the property that the polynomial f splits over M 0 . One can readily verify that L is the unique splitting field for f over K contained in M . The Fundamental Theorem of Algebra ensures that a polynomial f ∈ Q[x] with rational coefficients always splits over the field C of complex numbers. Thus some unique subfield L of C is a splitting field for f over Q. Note that if the polynomial f ∈ K[x] splits over an extension field M of K, and if α1 , α2 , . . . , αn are the roots of the polynomial f in M , then the unique splitting field of f over K contained in M is the field K(α1 , α2 , . . . , αn ) obtaining on adjoining the roots of f to K. √ Example The field Q( 2) is a splitting field for the polynomial x2 − 2 over Q. We shall prove below that splitting fields always exist and that any two splitting field extensions for a given polynomial over a field K are isomorphic. Given any homomorphism σ: K → M of fields, we define σ∗ (a0 + a1 x + · · · + an xn ) = σ(a0 ) + σ(a1 )x + · · · + σ(an )xn for all polynomials a0 + a1 x + · · · + an xn with coefficients in K. Note that σ∗ (f + g) = σ∗ (f ) + σ∗ (g) and σ∗ (f g) = σ∗ (f )σ∗ (g) for all f, g ∈ K[x]. Theorem 3.13 (Kronecker) Let K be a field, and let f ∈ K[x] be a nonconstant polynomial with coefficients in K. Then there exists an extension field L of K and an element α of L for which f (α) = 0.

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Proof Let g be an irreducible factor of f , and let L = K[x]/(g), where (g) is the ideal of K[x] generated by g. For each a ∈ K let i(a) = a + (g). Then i: K → L is a monomorphism. We embed K in L on identifying a ∈ K with i(a). Now L is a field, since g is irreducible (Proposition 2.15). Let α = x+(g). Then g(α) is the image of the polynomial g under the quotient homomorphism from K[x] to L, and therefore g(α) = 0. But g is a factor of the polynomial f . Therefore f (α) = 0, as required. Corollary 3.14 Let K be a field and let f ∈ K[x]. Then there exists a splitting field for f over K. Proof We use induction on the degree deg f of f . The result is trivially true when deg f = 1 (since f then splits over K itself). Suppose that the result holds for all fields and for all polynomials of degree less than deg f . Now it follows from Theorem 3.13 that there exists a field extension K1 : K of K and an element α of K1 satisfying f (α) = 0. Moreover f (x) = (x − α)g(x) for some polynomial g with coefficients in K(α). Now deg g < deg f . It follows from the induction hypothesis that there exists a splitting field L for g over K(α). Then f splits over L. Suppose that f splits over some field M , where K ⊂ M ⊂ L. Then α ∈ M and hence K(α) ⊂ M . But M must also contain the roots of g, since these are roots of f . It follows from the definition of splitting fields that M = L. Thus L is the required splitting field for the polynomial f over K. Any two splitting fields for a given polynomial with coefficients in a field K are K-isomorphic. This result is a special case of the following theorem. Theorem 3.15 Let K1 and K2 be fields, and let σ: K1 → K2 be an isomorphism between K1 and K2 . Let f ∈ K1 [x] be a polynomial with coefficients in K1 , and let L1 and L2 be splitting fields for f and σ∗ (f ) over K1 and K2 respectively. Then there exists an isomorphism τ : L1 → L2 which extends σ: K1 → K2 . Proof We prove the result by induction on [L1 : K1 ]. The result is trivially true when [L1 : K1 ] = 1. Suppose that [L1 : K1 ] > 1 and the result holds for splitting field extensions of lower degree. Choose a root α of f in L1 \K1 , and let m be the minimum polynomial of α over K1 . Then m divides f and σ∗ (m) divides σ∗ (f ), and therefore σ∗ (m) splits over L2 . Moreover the polynomial σ∗ (m) is irreducible over K2 , since σ: K1 → K2 induces an isomorphism between the polynomial rings K1 [x] and K2 [x]. Choose a root β of σ∗ (m). 52

Let g and h be polynomials with coefficients in K1 . Now g(α) = h(α) if and only if m divides g − h. Similarly σ∗ (g)(β) = σ∗ (h)(β) if and only if σ∗ (m) divides σ∗ (g) − σ∗ (h). Therefore σ∗ (g)(β) = σ∗ (h)(β) if and only if g(α) = h(α), and thus there is a well-defined isomorphism ϕ: K1 (α) → K2 (β) which sends g(α) to σ∗ (g)(β) for any polynomial g with coefficients in K. Now L1 and L2 are splitting fields for the polynomials f and σ∗ (f ) over the fields K1 (α) and K2 (β) respectively, and [L1 : K1 (α)] < [L1 : K1 ]. The induction hypothesis therefore ensures the existence of an isomorphism τ : L1 → L2 extending ϕ: K1 (α) → K2 (β). Then τ : L1 → L2 is the required extension of σ: K1 → K2 . Corollary 3.16 Let L: K be a splitting field extension, and let α and β be elements of L. Then there exists a K-automorphism of L sending α to β if and only if α and β have the same minimum polynomial over K. Proof Suppose that there exists a K-automorphism σ of L which sends α to β. Then h(β) = σ(h(α)) for all polynomials h ∈ K[x] with coefficients in K. Therefore h(α) = 0 if and only if h(β) = 0. It follows that α and β must have the same minimum polynomial over K. Conversely suppose that α and β are elements of L that have the same minimum polynomial m over K. Let h1 and h2 be polynomials with coefficients in K. Now h1 (α) = h2 (α) if and only if h1 − h2 is divisible by the minimum polynomial m. It follows that h1 (α) = h2 (α) if and only if h1 (β) = h2 (β). Therefore there is a well-defined K-isomorphism ϕ: K(α) → K(β) that sends h(α) to h(β) for all polynomials h with coefficients in K. Then ϕ(α) = β. Now L is the splitting field over K for some polynomial f with coefficients in K. The field L is then a splitting field for f over both K(α) and K(β). It follows from Theorem 3.15 that the K-isomorphism ϕ: K(α) → K(β) extends to a K-automorphism τ of L that sends α to β, as required.

3.6

Normal Extensions

Definition A field extension L: K is said to be normal if every irreducible polynomial in K[x] with at least one root in L splits over L. Note that a field extension L: K is normal if and only if, given any element α of L, the minimum polynomial of α over K splits over L. Theorem 3.17 Let K be a field, and let L be an extension field of K. Then L is a splitting field over K for some polynomial with coefficients in K if and only if the field extension L: K is both finite and normal. 53

Proof Suppose that L: K is both finite and normal. Then there exist algebraic elements α1 , α2 , . . . , αn of L such that L = K(α1 , α2 , . . . , αn ) (Corollary 3.5). Let f (x) = m1 (x)m2 (x) · · · mn (x), where mj ∈ K[x] is the minimum polynomial of αj over K for j = 1, 2, . . . , n. Then mj splits over L since mj is irreducible and L: K is normal. Thus f splits over L. It follows that L is a splitting field for f over K, since L is obtained from K by adjoining roots of f . Conversely suppose that L is a splitting field over K for some polynomial f ∈ K[x]. Then L is obtained from K by adjoining the roots of f , and therefore the extension L: K is finite. (Corollary 3.5). Let g ∈ K[x] be irreducible, and let M be a splitting field for the polynomial f g over L. Then L ⊂ M and the polynomials f and g both split over M . Let β and γ be roots of g in M . Now the polynomial f splits over the fields L(β) and L(γ). Moreover if f splits over any subfield of M containing K(β) then that subfield must contain L (since L is a splitting field for f over K) and thus must contain L(β). We deduce that L(β) is a splitting field for f over K(β). Similarly L(γ) is a splitting field for f over K(γ). Now there is a well-defined K-isomorphism σ: K(β) → K(γ) which sends h(β) to h(γ) for all polynomials h with coefficients in K, since two such polynomials h1 and h2 take the same value at a root of the irreducible polynomial g if and only if their difference h1 −h2 is divisible by g. This isomorphism σ: K(β) → K(γ) extends to an K-isomorphism τ : L(β) → L(γ) between L(β) and L(γ), since L(β) and L(β) are splitting fields for f over the field K(β) and K(γ) respectively (Theorem 3.15). Thus the extensions L(β): K and L(γ): K are isomorphic, and [L(β): K] = [L(γ): K]. But [L(β): K] = [L(β): L][L: K] and [L(γ): K] = [L(γ): L][L: K] by the Tower Law (Theorem 3.1). It follows that [L(β): L] = [L(γ): L]. In particular β ∈ L if and only if γ ∈ L. This shows that that any irreducible polynomial with a root in L must split over L, and thus L: K is normal, as required.

3.7

Separability

Let K be a field. We recall that nk is defined inductively for all integers n and for all elements k of K so that 0k = 0 and (n + 1)k = nk + k for all n ∈ Z and k ∈ K. Thus 1k = k, 2k = k + k, 3k = k + k + k etc., and (−n)k = −(nk) for all n ∈ Z. Definition Let K be a field, and let f ∈ K[x] be a polynomial with coeffin P cients c0 , c1 , . . . , cn in K, where f (x) = cj xj . The formal derivative Df j=0

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of f is defined by the formula (Df )(x) =

n P

jcj xj−1 .

j=1

(The definition of formal derivative given above is a purely algebraic definition, applying to polynomials with coefficients in any field whatsoever, which corresponds to the formula for the derivative of a polynomial with real coefficients obtained by elementary calculus.) Let K be a field. One can readily verify by straightforward calculation that D(f + g) = Df + Dg and D(f g) = (Df )g + f (Dg) for all f ∈ K[x]. If f is a constant polynomial then Df = 0. Let K be a field, and let f ∈ K[x]. An element α of an extension field L of K is said to be a repeated zero if (x − α)2 divides f (x). Proposition 3.18 Let K be a field, and let f ∈ K[x]. The polynomial f has a repeated zero in a splitting field for f over K if and only if there exists a non-constant polynomial with coefficients in K that divides both f and its formal derivative Df in K[x]. Proof Suppose that f ∈ K[x] has a repeated root α in a splitting field L. Then f (x) = (x − α)2 h(x) for some polynomial h ∈ L[x]. But then (Df )(x) = 2(x − α)h(x) + (x − α)2 (Dh)(x) and hence (Df )(α) = 0. It follows that the minimum polynomial of α over K is a non-constant polynomial with coefficients in K which divides both f and Df . Conversely let f ∈ K[x] be a polynomial with the property that f and Df are both divisible by some non-constant polynomial g ∈ K[x]. Let L be a splitting field for f over K. Then g splits over L (since g is a factor of f ). Let α ∈ L be a root of g. Then f (α) = 0, and hence f (x) = (x − α)e(x) for some polynomial e ∈ L[x]. On differentiating, we find that (Df )(x) = e(x) + (x − α)De(x). But (Df )(α) = 0, since g(α) = 0 and g divides Df in K[x]. It follows that e(α) = (Df )(α) = 0, and thus e(x) = (x − α)h(x) for some polynomial h ∈ L[x]. But then f (x) = (x − α)2 h(x), and thus the polynomial f has a repeated root in the splitting field L, as required. Definition Let K be a field. An irreducible polynomial in K[x] is said to be separable over K if it does not have repeated roots in a splitting field. A polynomial in K[x] is said to separable over K if all its irreducible factors are separable over K. A polynomial is said to be inseparable if it is not separable.

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Corollary 3.19 Let K be a field. An irreducible polynomial f is inseparable if and only if Df = 0. Proof Let f ∈ K[x] be an irreducible polynomial. Suppose that f is inseparable. Then f has a repeated root in a splitting field, and it follows from Proposition 3.18 that there exists a non-constant polynomial g in K[x] dividing both f and its formal derivative Df . But then g = cf for some non-zero element c of K, since f is irreducible, and thus f divides Df . But if Df were non-zero then deg Df < deg f , and thus f would not divide Df . Thus Df = 0. Conversely if Df = 0 then f divides both f and Df . It follows from Proposition 3.18 that f has a repeated root in a splitting field, and is thus inseparable. Definition An algebraic field extension L: K is said to be separable over K if the minimum polynomial of each element of L is separable over K. Suppose that K is a field of characteristic zero. Then n.k 6= 0 for all n ∈ Z and k ∈ K satisfying n 6= 0 and k 6= 0. It follows from the definition of the formal derivative that Df = 0 if and only if f ∈ K[x] is a constant polynomial. The following result therefore follows immediately from Corollary 3.19. Corollary 3.20 Suppose that K is a field of characteristic zero. Then every polynomial with coefficients in K is separable over K, and thus every field extension L: K of K is separable.

3.8

Finite Fields

Lemma 3.21 Let K be a field of characteristic p, where p > 0. Then (x + y)p = xp + y p and (xy)p = xp y p for all x, y ∈ K. Thus the function x 7→ xp is a monomorphism mapping the field K into itself. p

Proof The Binomial Theorem tells us that (x + y) =

p   X p

j

xj y p−j , where

j=0     p p p(p − 1) · · · (p − j + 1) for j = 1, 2, . . . , p. The de= 1 and = j! 0 j nominator of each binomial coefficient must divide the numerator, since this coefficient is an integer. Now the characteristic p of K is a prime number. Moreover if 0 < j < p then p is a factor of the numerator but is not a factor of the denominator. It follows from the Fundamental Theorem of Arithmetic

56

  p that p divides for all j satisfying 0 < j < p. But px = 0 for all x ∈ K, j since charK = p. Therefore (x + y)p = xp + y p for all x, y ∈ K. The identity (xy)p = xp y p is immediate from the commutativity of K. Let K be a field of characteristic p, where p > 0. The monomorphism x 7→ xp is referred to as the Frobenius monomorphism of K. If K is finite then this monomorphism is an automorphism of K, since any injection mapping a finite set into itself must be a bijection. Theorem 3.22 A field K has pn elements if and only if it is a splitting field n for the polynomial xp − x over its prime subfield Fp , where Fp ∼ = Z/pZ. Proof Suppose that K has q elements, where q = pn . If α ∈ K \ {0} then αq−1 = 1, since the set of non-zero elements of K is a group of order q − 1 with respect to multiplication. It follows that αq = α for all α ∈ K. Thus all elements of K are roots of the polynomial xq − x. This polynomial must therefore split over K, since its degree is q and K has q elements. Moreover the polynomial cannot split over any proper subfield of K. Thus K is a splitting field for this polynomial. Conversely suppose that K is a splitting field for the polynomial f over Fp , where f (x) = xq − x and q = pn . Let σ(α) = αq for all α ∈ K. Then σ: K → K is a monomorphism, being the composition of n successive applications of the Frobenius monomorphism of K. Moreover an element α of K is a root of f if and only if σ(α) = α. It follows from this that the roots of f constitute a subfield of K. This subfield is the whole of K, since K is a splitting field. Thus K consists of the roots of f . Now Df (x) = qxq−1 − 1 = −1, since q is divisible by the characteristic p of Fp . It follows from Proposition 3.18 that the roots of f are distinct. Therefore f has q roots, and thus K has q elements, as required. Let K be a finite field of characteristic p. Then K has pn elements, where n = [K: Fp ], since any vector space of dimension n over a field of order p must have exactly pn elements. The following result is now a consequence of the existence of splitting fields (Corollary 3.14) and the uniqueness of splitting fields up to isomorphism (Theorem 3.15) Corollary 3.23 There exists a finite field GF(pn ) of order pn for each prime number p and positive integer n. Two finite fields are isomorphic if and only if they have the same number of elements.

57

The field GF(pn ) is referred to as the Galois field of order pn . The non-zero elements of a field constitute a group under multiplication. We shall prove that all finite subgroups of the group of non-zero elements of a field are cyclic. It follows immediately from this that the group of non-zero elements of a finite field is cyclic. For each positive integer n, we denote by ϕ(n) the number of integers x X satisfying 0 ≤ x < n that are coprime to n. We show that the sum ϕ(d) d|n

of ϕ(d) taken over all divisors of a positive integer n is equal to n. Lemma 3.24 Let n be a positive integer. Then

X

ϕ(d) = n.

d|n

Proof If x is an integer satisfying X0 ≤ x < n then (x, n) = n/d for some divisor d of n. It follows that n = nd , where nd is the number of integers x d|n

satisfying 0 ≤ x < n for which (x, n) = n/d. Thus it suffices to show that nd = ϕ(d) for each divisor d of n. Let d be a divisor of n, and let a = n/d. Given any integer x satisfying 0 ≤ x < n that is divisible by a, there exists an integer y satisfying 0 ≤ y < d such that x = ay. Then (x, n) = (ay, ad) = a(y, d). It follows that the integers x satisfying 0 ≤ x < n for which (x, n) = a are those of the form ay, where y is an integer, 0 ≤ y < d and (y, d) = 1. It follows that there are exactly ϕ(d) integersX x satisfying 0 ≤ x < n for which (x, n) = n/d, and thus nd = ϕ(d) and n = ϕ(d), as required. d|n

The set of all non-zero elements of a field is a group with respect to the operation of multiplication. Theorem 3.25 Let G be a finite subgroup of the group of non-zero elements of a field. Then the group G is cyclic. Proof Let n be the order of the group G. It follows from Lagrange’s Theorem that the order of every element of G divides n. For each divisor dX of n, let ψ(d) denote the number of elements of G that are of order d. Clearly ψ(d) = n. d|n

Let g be an element of G of order d, where d is a divisor of n. The elements 1, g, g 2 , . . . , g d−1 are distinct elements of G and are roots of the polynomial xd − 1. But a polynomial of degree d with coefficients in a field has at most d roots in that field. Therefore every element x of G satisfying xd = 1 is g k 58

for some uniquely determined integer k satisfying 0 ≤ k < d. If k is coprime to d then g k has order d, for if (g k )n = 1 then d divides kn and hence d divides n. Conversely if g k has order d then d and k are coprime, for if e is a common divisor of k and d then (g k )d/e = g d(k/e) = 1, and hence e = 1. Thus if there exists at least one element g of G that is of order d then the elements of G that are of order d are the elements g k for those integers k satisfying 0 ≤ k < d that are coprime to d. It follows that if ψ(d) > 0 then ψ(d) = ϕ(d), where ϕ(d) is the number of integers k satisfying 0 ≤ k < d that are coprime to d. X Now 0 ≤ ψ(d) ≤ ϕ(d) for each divisor d of n. But ψ(d) = n and d|n

X

ϕ(d) = n. It follows that ψ(d) = φ(d) for each divisor d of n. In

d|n

particular ψ(n) = ϕ(n) ≥ 1. Thus there exists an element of G whose order is the order n of G. This element generates G, and thus G is cyclic, as required. Corollary 3.26 The group of non-zero elements of a finite field is cyclic.

3.9

The Primitive Element Theorem

Theorem 3.27 (Primitive Element Theorem) Every finite separable field extension is simple. Proof Let L: K be a finite separable field extension. Suppose that K is a finite field. Then L is also a finite field, since it is a finite-dimensional vector space over K. The group of non-zero elements of L is therefore generated by a single non-zero element θ of L (Corollary 3.26). But then L = K(θ) and thus L: K is simple. This proves the Primitive Element Theorem in the case where the field K is finite. Next suppose that L = K(β, γ), where K is infinite, β and γ are algebraic over K and L: K is separable. Let N be a splitting field for the polynomial f g, where f and g are the minimum polynomials of β and γ respectively over K. Then f and g both split over N . Let β1 , β2 , . . . , βq be the roots of f in N , and let γ1 , γ2 , . . . , γr be the roots of g in N , where β1 = β and γ1 = γ. The separability of L: K ensures that γk 6= γj when k 6= j. Now K is infinite. We can therefore choose c ∈ K so that c 6= (βi − β)/(γ − γj ) for any i and j with j 6= 1. Let h(x) = f (θ − cx), where θ = β + cγ. Then h is a polynomial in the indeterminate x with coefficients in K(θ) which satisfies h(γ) = f (β) = 0. Moreover h(γj ) 6= 0 whenever j 6= 1, since θ − cγj 6= βi for all i and j with j 6= 1. Thus γ is the only 59

common root of g and h. It follows that x − γ is a highest common factor of g and h in the polynomial ring K(θ)[x], and therefore γ ∈ K(θ). But then β ∈ K(θ), since β = θ − cγ and c ∈ K. It follows that L = K(θ). It now follows by induction on m that if L = K(α1 , α2 , . . . , αm ), where K is infinite, α1 , α2 , . . . , αm are algebraic over K, and L: K is separable, then the extension L: K is simple. Thus all finite separable field extensions are simple, as required.

3.10

The Galois Group of a Field Extension

Definition The Galois group Γ(L: K) of a field extension L: K is the group of all automorphisms of the field L that fix all elements of the subfield K. Lemma 3.28 If L: K is a finite separable field extension then |Γ(L: K)| ≤ [L: K]. Proof It follows from the Primitive Element Theorem (Theorem 3.27) that there exists some element α of L such that L = K(α). Let λ be an element of L. Then λ = g(α) for some polynomial g with coefficients in K. But then σ(λ) = g(σ(α)) for all σ ∈ Γ(L: K), since the coefficients of g are fixed by σ. It follows that each automorphism σ in Γ(L: K) is uniquely determined once σ(α) is known. Let f be the minimum polynomial of α over K. Then f (σ(α)) = σ(f (α)) = 0 for all σ ∈ Γ(L: K) since the coefficients of f are in K and are therefore fixed by σ. Thus σ(α) is a root of f . It follows that the order |Γ(L: K)| of the Galois group is bounded above by the number of roots of f that belong to L, and is thus bounded above by the degree deg f of f . But deg f = [L: K] (Theorem 3.4). Thus |Γ(L: K)| ≤ [L: K], as required. Definition Let L be a field, and let G be a group of automorphisms of L. The fixed field of G is the subfield K of L defined by K = {a ∈ L : σ(a) = a for all σ ∈ G}. Proposition 3.29 Let L be a field, let G be a finite group of automorphisms of L, and let K be the fixed field of G. Then each element α of L is algebraic over K, and the minimum polynomial of α over K is the polynomial (x − α1 )(x − α2 ) · · · (x − αk ), where α1 , α2 , . . . , αk are distinct and are the elements of the orbit of α under the action of G on L. 60

Proof Let f (x) = (x − α1 )(x − α2 ) · · · (x − αk ). Then the polynomial f is invariant under the action of G, since each automorphism in the group G permutes the elements α1 , α2 , . . . , αk and therefore permutes the factors of f amongst themselves. It follows that the coefficients of the polynomial f belong to the fixed field K of G. Thus α is algebraic over K, as it is a root of the polynomial f . Now, given any root αi of f , there exists some σ ∈ G such that αi = σ(α). Thus if g ∈ K[x] is a polynomial with coefficients in K which satisfies g(α) = 0 then g(αi ) = σ(g(α)) = 0, since the coefficients of g are fixed by σ. But then f divides g. Thus f is the minimum polynomial of α over K, as required. Definition A field extension is said to be a Galois extension if it is finite, normal and separable. Theorem 3.30 Let L be a field, let G be a finite subgroup of the group of automorphisms of L, and let K be the fixed field of G. Then the field extension L: K is a Galois extension. Moreover G is the Galois group Γ(L: K) of L: K and |G| = [L: K]. Proof It follows from Proposition 3.29 that, for each α ∈ L, the minimum polynomial of α over K splits over L and has no multiple roots. Thus the extension L: K is both normal and separable. Let M be any field satisfying K ⊂ M ⊂ L for which the extension M : K is finite. The extension M : K is separable, since L: K is separable. It follows from the Primitive Element Theorem (Theorem 3.27) that the extension M : K is simple. Thus M = K(α) for some α ∈ L. But then [M : K] is equal to the degree of the minimum polynomial of α over K (Theorem 3.4). It follows from Proposition 3.29 that [M : K] is equal to the number of elements in the orbit of α under the action of G on L. Therefore [M : K] divides |G| for any intermediate field M for which the extension M : K is finite. Now let the intermediate field M be chosen so as to maximize [M : K]. If λ ∈ L then λ is algebraic over K, and therefore [M (λ): M ] is finite. It follows from the Tower Law (Theorem 3.1) that [M (λ): K] is finite, and [M (λ): K] = [M (λ): M ][M : K]. But M has been chosen so as to maximize [M : K]. Therefore [M (λ): K] = [M : K], and [M (λ): M ] = 1. Thus λ ∈ M . We conclude that M = L. Thus L: K is finite and [L: K] divides |G|. The field extension L: K is a Galois extension, since it has been shown to be finite, normal and separable. Now G ⊂ Γ(L: K) and |Γ(L: K)| ≤ [L: K] (Lemma 3.28). Therefore |Γ(L: K)| ≤ [L: K] ≤ |G| ≤ |Γ(L: K)|, and thus G = Γ(L: K) and |G| = [L: K], as required. 61

Theorem 3.31 Let Γ(L: K) be the Galois group of a finite field extension L: K. Then |Γ(L: K)| divides [L: K]. Moreover |Γ(L: K)| = [L: K] if and only if L: K is a Galois extension, in which case K is the fixed field of Γ(L: K). Proof Let M be the fixed field of Γ(L: K). It follows from Theorem 3.30 that L: M is a Galois extension and |Γ(L: K)| = [L: M ]. Now [L: K] = [L: M ][M : K] by the Tower Law (Theorem 3.1). Thus |Γ(L: K)| divides [L: K]. If |Γ(L: K)| = [L: K] then M = K. But then L: K is a Galois extension and K is the fixed field of Γ(L: K). Conversely suppose that L: K is a Galois extension. We must show that |Γ(L: K)| = [L: K]. Now the extension L: K is both finite and separable. It follows from the Primitive Element Theorem (Theorem 3.27) that there exists some element θ of L such that L = K(θ). Let f be the minimum polynomial of θ over K. Then f splits over L, since f is irreducible and the extension L: K is normal. Let θ1 , θ2 , . . . , θn be the roots of f in L, where θ1 = θ and n = deg f . If σ is a K-automorphism of L then f (σ(θ)) = σ(f (θ)) = 0, since the coefficients of the polynomial f belong to K and are therefore fixed by σ. Thus σ(θ) = θj for some j. We claim that, for each root θj of f , there is exactly one K-automorphism σj of L satisfying σj (θ) = θj . Let g(x) and h(x) be polynomials with coefficients in K. Suppose that g(θ) = h(θ). Then g − h is divisible by the minimum polynomial f of θ. It follows that g(θj ) = h(θj ) for any root θj of f . Now every element of L is of the form g(θ) for some g ∈ K[x], since L = K(θ). We deduce therefore that there is a well-defined function σj : L → L with the property that σj (g(θ)) = g(θj ) for all g ∈ K[x]. The definition of this function ensures that it is the unique automorphism of the field L that fixes each element of K and sends θ to θj . Now the roots of the polynomial f in L are distinct, since f is irreducible and L: K is separable. Moreover the order of the Galois group Γ(L: K) is equal to the number of roots of f , since each root determines a unique element of the Galois group. Therefore |Γ(L: K)| = deg f . But deg f = [L: K] since L = K(θ) and f is the minimum polynomial of θ over K (Theorem 3.4). Thus |Γ(L: K)| = [L: K], as required.

3.11

The Galois correspondence

Proposition 3.32 Let K, L and M be fields satisfying K ⊂ M ⊂ L. Suppose that L: K is a Galois extension. Then so is L: M . If in addition M : K is normal, then M : K is a Galois extension. Proof Let α ∈ L and let fK ∈ K[x] and fM ∈ M [x] be the minimum polyomials of α over K and M respectively. Then fK splits over L, since fK 62

is irreducible over K and L: K is a normal extension. Also the roots of fK in L are distinct, since L: K is a separable extension. But fM divides fK , since fK (α) = 0 and the coefficients of fK belong to M . It follows that fM also splits over L, and its roots are distinct. We deduce that the finite extension L: M is both normal and separable, and is therefore a Galois extension. The finite extension M : K is clearly separable, since L: K is separable. Thus if M : K is a normal extension then it is a Galois extension. Proposition 3.33 Let L: K be a Galois extension, and let M be a field satisfying K ⊂ M ⊂ L. Then the extension M : K is normal if and only if σ(M ) = M for all σ ∈ Γ(L: K). Proof Let α be an element of M , and let f ∈ K[x] be the minimum polynomial of α over K. Now K is the fixed field of the Galois group Γ(L: K), since the field extension L: K is a Galois extension (Theorem 3.31). It follows that the polynomial f splits over L, and the roots of f are the elements of the orbit of α under the action of Γ(L: K) on L (Proposition 3.29). Therefore f splits over M if and only if σ(α) ∈ M for all σ ∈ Γ(L: K). Now the extension M : K is normal if and only if the minimum polynomial of any element of M over K splits over M . It follows that the extension M : K is normal if and only if σ(M ) ⊂ M for all σ ∈ Γ(L: K). But if σ(M ) ⊂ M for all σ ∈ Γ(L: K) then σ −1 (M ) ⊂ M and M = σ(σ −1 (M )) ⊂ σ(M ) and thus σ(M ) = M for all σ ∈ Γ(L: K). Therefore the extension M : K is normal if and only if σ(M ) = M for all σ ∈ Γ(L: K). Corollary 3.34 Let L: K be a Galois extension, and let M be a field satisfying K ⊂ M ⊂ L. Suppose that the extension M : K is normal. Then the restriction σ|M to M of any K-automorphism σ of L is a K-automorphism of M . Proof Let σ ∈ Γ(L: K) be a K-automorphism of L. We see from Proposition 3.33 that σ(M ) = M . Similarly σ −1 (M ) = M . It follows that the restrictions σ|M : M → M and σ −1 |M : M → M of σ and σ −1 to M are Khomomorphisms mapping M into itself. Moreover σ −1 |M : M → M is the inverse of σ|M : M → M . Thus σ|M : M → M is an isomorphism, and is thus a K-automorphism of M , as required. Theorem 3.35 (The Galois Correspondence) Let L: K be a Galois extension of a field K. Then there is a natural bijective correspondence between fields M satisfying K ⊂ M ⊂ L and subgroups of the Galois group Γ(L: K) of the extension L: K. If M is a field satisfying K ⊂ M ⊂ L then the subgroup of Γ(L: K) corresponding to M is the Galois group Γ(L: M ) of the extension 63

L: M . If G is a subgroup of Γ(L: K) then the subfield of L corresponding to G is the fixed field of G. Moreover the extension M : K is normal if and only if Γ(L: M ) is a normal subgroup of the Galois group Γ(L: K), in which case Γ(M : K) ∼ = Γ(L: K)/Γ(L: M ). Proof Let M be a subfield of L containing K. Then L: M is a Galois extension (Proposition 3.32). The existence of the required bijective correspondence between fields M satisfying K ⊂ M ⊂ L and subgroups of the Galois group Γ(L: K) follows immediately from Theorem 3.30 and Theorem 3.31. Let M be a field satisfying K ⊂ M ⊂ L. Now the extension M : K is normal if and only if σ(M ) = M for all σ ∈ Γ(L: K). (Proposition 3.33). Let H = Γ(L: M ). Then M = σ(M ) if and only if H = σHσ −1 , since M and σ(M ) are the fixed fields of H and σHσ −1 respectively, and there is a bijective correspondence between subgroups of the Galois group Γ(L: K) and their fixed fields. Thus the extension M : K is normal if and only if Γ(L: M ) is a normal subgroup of Γ(L: K). Finally suppose that M : K is a normal extension. For each σ ∈ Γ(L: K), let ρ(σ) be the restriction σ|M of σ to M . Then ρ: Γ(L: K) → Γ(M : K) is a group homomorphism whose kernel is Γ(L: M ). We can apply Theorem 3.30 to the extension M : K to deduce that ρ(Γ(L: K)) = Γ(M : K), since the fixed field of ρ(Γ(L: K)) is K. Therefore the homomorphism ρ: Γ(L: K) → Γ(M : K) induces the required isomorphism between Γ(L: K)/Γ(L: M ) and Γ(M : K).

3.12

Quadratic Polynomials

We consider the problem of expressing the roots of a polynomial of low degree in terms of its coefficients. Then the well-known procedure for locating the roots of a quadratic polynomial with real or complex coefficients generalizes to quadratic polynomials with coefficients in a field K whose characteristic does not equal 2. Given a quadratic polynomial ax2 + bx + c with coefficients a and b belonging to some such field K, let us adjoin to K an element δ satisfying δ 2 = b2 − 4ac. Then the polynomial splits over K(δ), and its roots are (−b ± δ)/(2a). We shall describe below analogous procedures for expressing the roots of cubic and quartic polynomials in terms of their coefficients.

3.13

Cubic Polynomials

Consider a cubic polynomial x3 + ax2 + bx + c, where the coefficients a, b and c belong to some field K of characteristic zero. If f (x) = x3 + ax2 + bx + c 2 3 then f (x − 31 a) = x3 − px − q, where p = 13 a2 − b and q = 13 ba − 27 a − c. It 64

therefore suffices to restrict our attention to cubic polynomials of the form x3 − px − q, where p and q belong to K. Let f (x) = x3 − px − q, and let u and v be elements of some splitting field for f over K. Then f (u + v) = u3 + v 3 + (3uv − p)(u + v) − q. Suppose that 3uv = p. Then f (u + v) = u3 + p3 /(27u3 ) − q. Thus f (u + p/(3u)) = 0 if and only if u3 is a root of the quadratic polynomial x2 − xq + p3 /27. Now the roots of this quadratic polynomial are r q q2 p3 ± − , 4 27 2 and the product of these roots is p3 /27. Thus if one of these roots is equal to u3 then the other is equal to v 3 , where v = p/(3u). It follows that the roots of the cubic polynomial f are s s r r 2 3 3 q 3 q q p q2 p3 + − + − − 2 4 27 2 4 27 where the two cube roots must be chosen so as to ensure that their product is equal to 13 p. It follows that the cubic polynomial x3 − px − q splits over the 1 3 field K(, ξ, ω), where 2 = 14 q 2 − 27 p and ξ 3 = 12 q +  and where ω satisfies 3 ω = 1 and ω 6= 1. The roots of the polynomial in this extension field are α, β and γ, where α=ξ+

p , 3ξ

β = ωξ + ω 2

p , 3ξ

γ = ω2ξ + ω3

p . 3ξ

Now let us consider the possibilities for the Galois group Γ(L: K), where L is a splitting field for f over K. Now L = K(α, β, γ), where α, β and γ are the roots of f . Also a K-automorphism of L must permute the roots of f amongst themselves, and it is determined by its action on these roots. Therefore Γ(L: K) is isomorphic to a subgroup of the symmetric group Σ3 (i.e., the group of permutations of a set of 3 objects), and thus the possibilities for the order of Γ(L: K) are 1, 2, 3 and 6. It follows from Corollary 3.16 that f is irreducible over K if and only if the roots of f are distinct and the Galois group acts transitively on the roots of f . By considering all possible subgroups of Σ3 it is not difficult to see that f is irreducible over K if and only if |Γ(L: K)| = 3 or 6. If f splits over K then |Γ(L: K)| = 1. If f factors in K[x] as the product of a linear factor and an irreducible quadratic factor then |Γ(L: K)| = 2. 65

Let δ = (α−β)(α−γ)(β −γ). Then δ 2 is invariant under any permutation of α β and γ, and therefore δ 2 is fixed by all automorphisms in the Galois group Γ(L: K). Therefore δ 2 ∈ K. The element δ 2 of K is referred to as the discriminant of the polynomial f . A straightforward calculation shows that if f (x) = x3 − px − q then δ 2 = 4p3 − 27q 2 . Now δ changes sign under any permutation of the roots α, β and γ that transposes two of the roots whilst leaving the third root fixed. But δ ∈ K if and only if δ is fixed by all elements of the Galois group Γ(L: K), in which case the Galois group must induce only cyclic permutations of the roots α, β and γ. Therefore Γ(L: K) is isomorphic to the cyclic group of order 3 if and only if f is irreducible and the discriminant 4p3 − 27q 2 of f has a square root in the field K. If f is irreducible but the discriminant does not have a square root in K then Γ(L: K) is isomorphic to the symmetric group Σ3 , and |Γ(L: K)| = 6.

3.14

Quartic Polynomials

We now consider how to locate the roots of a quartic polynomial with coefficients in a field K of characteristic zero. A substitution of the form x 7→ x−c, where c ∈ K, will reduce the problem to that of locating the roots α, β, γ and δ of a quartic polynomial f of the form f (x) = x4 − px2 − qx − r in some splitting field L. Now the roots α, β, γ and δ of the quartic polynomial x4 − px2 − qx − r, must satisfy the equation (x − α)(x − β)(x − γ)(x − δ) = x4 − px2 − qx − r. Equating coefficients of x, we find that α + β + γ + δ = 0, and p = −(αβ + αγ + αδ + βγ + βδ + γδ), q = βγδ + αγδ + αβδ + αβγ, r = −αβγδ. Let λ = (α + β)(γ + δ) = −(α + β)2 = −(γ + δ)2 , µ = (α + γ)(β + δ) = −(α + γ)2 = −(β + δ)2 , ν = (α + δ)(β + γ) = −(α + δ)2 = −(β + γ)2 . 66

We shall show that λ + µ + ν, µν + λν + λµ and λµν can all be expressed in terms of p, q and r. To do this we eliminate α from the above expressions using the identity α + β + γ + δ = 0. We find p = = q = = r =

(β + γ + δ)(β + γ + δ) − γδ − βδ − βγ β 2 + γ 2 + δ 2 + γδ + βδ + βγ, βγδ − (β + γ + δ)(γδ + βδ + βγ) −(β 2 γ + β 2 δ + γ 2 β + γ 2 δ + δ 2 β + δ 2 γ) − 2βγδ, β 2 γδ + γ 2 βδ + δ 2 βγ.

Then   2 2 2 λ + µ + ν = − (γ + δ) + (β + δ) + (β + γ)   = −2 β 2 + γ 2 + δ 2 + γδ + βδ + βγ 2

2

λ +µ +ν

2

p2

= −2p, = (γ + δ)4 + (β + δ)4 + (β + γ)4 = γ 4 + 4γ 3 δ + 6γ 2 δ 2 + 4γδ 3 + δ 4 + β 4 + 4β 3 δ + 6β 2 δ 2 + 4βδ 3 + δ 4 + β 4 + 4β 3 γ + 6β 2 γ 2 + 4βγ 3 + γ 4 = 2(β 4 + γ 4 + δ 4 ) + 4(β 3 γ + β 3 δ + γ 3 β + γ 3 δ + δ 3 β + δ 3 γ) + 6(γ 2 δ 2 + β 2 δ 2 + β 2 γ 2 ), = β 4 + γ 4 + δ 4 + 3(γ 2 δ 2 + β 2 δ 2 + β 2 γ 2 ) + 4(β 2 γδ + γ 2 βδ + δ 2 βγ) + 2(β 3 γ + β 3 δ + γ 3 β + γ 3 δ + δ 3 β + δ 3 γ).

Therefore λ2 + µ2 + ν 2 = 2p2 − 8(β 2 γδ + γ 2 βδ + δ 2 βγ) = 2p2 − 8r. But 4p2 = (λ + µ + ν)2 = λ2 + µ2 + ν 2 + 2(µν + λν + λµ) Therefore µν + λν + λµ = 2p2 − 21 (λ2 + µ2 + ν 2 ) = p2 + 4r. 67

Finally, we note that  2 λµν = − (γ + δ)(β + δ)(β + γ) . Now (γ + δ)(β + δ)(β + γ) = β 2 γ + β 2 δ + γ 2 β + γ 2 δ + δ 2 β + δ 2 γ + 2βγδ = −q. (α + β)(α + γ)(α + δ) = −(γ + δ)(β + δ)(β + γ) = q. Therefore λµν = −(−q)2 = −q 2 . Thus λ, µ and ν are the roots of the resolvent cubic x3 + 2px2 + (p2 + 4r)x + q 2 . √ √ One can then verify that the roots of f take the form 12 ( −λ + −µ + √ √ √ √ −ν), where these square roots are chosen to ensure that −λ −µ −ν = q. (It should be noted that there are four possible ways in which the square roots can be chosen to satisfy this condition; these yield all four roots of the polynomial f .) We can therefore determine the roots of f in an appropriate splitting field once we have expressed the quantities λ, µ and ν in terms of the coefficients of the polynomial. Remark Any permutation of the roots of the quartic x4 − px2 − qx − r, will permute the roots λ, µ and ν of the resolvent cubic g(x) = (x − λ)(x − µ)(x − ν) amongst themselves, and will therefore permute the factors of g. Therefore the coefficients of g are fixed by all elements of the Galois group Γ(L: K) and therefore must belong to the ground field K. As we have seen from the calculations above, these coefficients can be expressed in terms of p, q, r.

3.15

The Galois group of the polynomial x4 − 2

We shall apply the Galois correspondence to investigate the structure of the splitting field for the polynomial x4 − 2 over the field Q of rational numbers. 68

A straightforward application of Eisenstein’s Irreducibility Criterion (Proposition 2.18) shows that the polynomial x4 − 2 is irreducible over Q. Let ξ be 4 the unique positive real number satisfying ξ 4 = 2. Then the roots of x√ −2 in the field C of complex numbers are ξ, iξ, −ξ and −iξ, where i = −1. Thus if L = Q(ξ, i) then L is a splitting field for the polynomial x4 − 2 over Q. Now the polynomial x4 − 2 is the minimum polynomial of ξ over Q, since this polynomial is irreducible. We can therefore apply Theorem 3.4 to deduce that [Q(ξ): Q] = 4. Now i does not belong to Q(ξ), since Q(ξ) ⊂ R. Therefore the polynomial x2 + 1 is the minimum polynomial of i over Q(ξ). Another application of Theorem 3.4 now shows that [L: Q(ξ)] = [Q(ξ, i): Q(ξ)] = 2. It follows from the Tower Law (Theorem 3.1) that [L: Q] = [L: Q(ξ)][Q(ξ): Q] = 8. Moreover the extension L: Q is a Galois extension, and therefore its Galois group Γ(L: Q) is a group of order 8 (Theorem 3.31). Another application of the Tower Law now shows that [L: Q(i)] = 4, since [L: Q] = [L: Q(i)][Q(i): Q] and [Q(i): Q] = 2. Therefore the minimum polynomial of ξ over Q(i) is a polynomial of degree 4 (Theorem 3.4). But ξ is a root of x4 −2. Therefore x4 −2 is irreducible over Q(i), and is the minimum polynomial of ξ over Q(i). Corollary 3.16 then ensures the existence of an automorphism σ of L that sends ξ ∈ L to iξ and fixes each element of Q(i). Similarly there exists an automorphism τ of L that sends i to −i and fixes each element of Q(ξ). (The automorphism τ is in fact the restriction to L of the automorphism of C that sends each complex number to its complex conjugate.) Now the automorphisms σ, σ 2 , σ 3 and σ 4 fix i and therefore send ξ to iξ, −ξ, −iξ and ξ respectively. Therefore σ 4 = ι, where ι is the identity automorphism of L. Similarly τ 2 = ι. Straightforward calculations show that τ σ = σ 3 τ , and (στ )2 = (σ 2 τ )2 = (σ 3 τ )2 = ι. It follows easily from this that Γ(L: Q) = {ι, σ, σ 2 , σ 3 , τ, στ, σ 2 τ, σ 3 τ }, and Γ(L: Q) is isomorphic to the dihedral group of order 8 (i.e., the group of symmetries of a square in the plane). The Galois correspondence is a bijective correspondence between the subgroups of Γ(L: Q) and subfields of L that contain Q. The subfield of L corresponding to a given subgroup of Γ(L: Q) is set of all elements of L that are fixed by all the automorphisms in the subgroup. One can verify that the correspondence between subgroups of Γ(L: Q) and their fixed fields is as

69

follows:—

3.16

Subgroup of Γ(L: Q)

Fixed field

Γ(L: K) {ι, σ, σ 2 , σ 3 } {ι, σ 2 , τ, σ 2 τ } {ι, σ 2 , στ, σ 3 τ } {ι, σ 2 } {ι, τ } {ι, σ 2 τ } {ι, στ } {ι, σ 3 τ } {ι}

Q Q(i) √ Q( √2) Q(i √ 2) Q( 2, i) Q(ξ) Q(iξ) Q((1 − i)/ξ) Q((1 + i)/ξ) Q(ξ, i)

The Galois group of a polynomial

Definition Let f be a polynomial with coefficients in some field K. The Galois group ΓK (f ) of f over K is defined to be the Galois group Γ(L: K) of the extension L: K, where L is some splitting field for the polynomial f over K. We recall that all splitting fields for a given polynomial over a field K are K-isomorphic (see Theorem 3.15), and thus the Galois groups of these splitting field extensions are isomorphic. The Galois group of the given polynomial over K is therefore well-defined (up to isomorphism of groups) and does not depend on the choice of splitting field. Lemma 3.36 Let f be a polynomial with coefficients in some field K and let M be an extension field of K. Then ΓM (f ) is isomorphic to a subgroup of ΓK (f ). Proof Let N be a splitting field for f over M . Then N contains a splitting field L for f over K. An element σ of Γ(N : M ) is an automorphism of N that fixes every element of M and therefore fixes every element of K. Its restriction σ|L to L is then a K-automorphism of L (Corollary 3.34). Moreover (σ ◦ τ )|L = (σ|L ) ◦ (τ |L ) for all σ, τ ∈ Γ(N : M ). Therefore there is a group homomorphism from Γ(N : M ) to Γ(L: K) which sends an automorphism σ ∈ Γ(N : M ) to its restriction σ|L to L. Now if σ ∈ Γ(N : M ) is in the kernel of this group homomorphism from Γ(N : M ) to Γ(L: K) then σ|L must be the identity automorphism of L. But f splits over L, and therefore all the roots of f are elements of L. It follows that σ(α) = α for each root α of f . The fixed field of σ must therefore be the whole of N , since M is contained in the fixed field of σ, and N is 70

a splitting field for f over M . Thus σ must be the identity automorphism of N . We conclude therefore that the group homomorphism from Γ(N : M ) to Γ(L: K) sending σ ∈ Γ(N : M ) to σ|L is injective, and therefore maps Γ(N : M ) isomorphically onto a subgroup of Γ(L: K). The result therefore follows from the definition of the Galois group of a polynomial. Let f be a polynomial with coefficients in some field K and let the roots of f is some splitting field L be α1 , α2 , . . . , αn . An element σ of Γ(L: K) is a K-automorphism of L, and therefore σ permutes the roots of f . Moreover two automorphism σ and τ in the Galois group Γ(L: K) are equal if and only if σ(αj ) = τ (αj ) for j = 1, 2, . . . , n, since L = K(α1 , α2 , . . . , αn ). Thus the Galois group of a polynomial can be represented as a subgroup of the group of permutations of its roots. We deduce immediately the following result. Lemma 3.37 Let f be a polynomial with coefficients in some field K. Then the Galois group of f over K is isomorphic to a subgroup of the symmetric group Σn , where n is the degree of f .

3.17

Solvable polynomials and their Galois groups

Definition We say that a polynomial with coefficients in a given field is solvable by radicals if the roots of the polynomial in a splitting field can be constructed from its coefficients in a finite number of steps involving only the operations of addition, subtraction, multiplication, division and extraction of nth roots for appropriate natural numbers n. It follows from the definition above that a polynomial with coefficients in a field K is solvable by radicals if and only if there exist fields K0 , K1 , . . . , Km such that K0 = K, the polynomial f splits over Km , and, for each integer i between 1 and m, the field Ki is obtained on adjoining to Ki−1 an element αi with the property that αipi ∈ Ki−1 for some positive integer pi . Moreover we can assume, without loss of generality that p1 , p2 , . . . , pm are prime numbers, since an nth root α of an element of a given field can be adjoined that field by successively adjoining powers αn1 , αn2 , . . . , αnk of α chosen such that n/n1 is prime, ni /ni−1 is prime for i = 2, 3, . . . , k, and nk = 1. We shall prove that a polynomial with coefficients in a field K of characteristic zero is solvable by radicals if and only if its Galois group ΓK (f ) over K is a solvable group. Let L be a field, and let p be a prime number that is not equal to the characteristic of L. Suppose that the polynomial xp − 1 splits over L. Then the polynomial xp − 1 has distinct roots, since its formal derivative pxp−1 is 71

non-zero at each root of xp − 1. An element ω of L is said to be a primitive pth root of unity if ω p = 1 and ω 6= 1. The primitive pth roots of unity are the roots of the polynomial xp−1 +xp−2 +· · ·+1, since xp −1 = (x −1)(xp−1 + xp−2 + · · · + 1). Also the group of pth roots of unity in L is a cyclic group over order p which is generated by any primitive pth root of unity. Lemma 3.38 Let K be a field, and let p be a prime number that is not equal to the characteristic of K. If ω is a primitive pth root of unity in some extension field of K then the Galois group of the extension K(ω): K is Abelian. Proof Let L = K(ω). Then L is a splitting field for the polynomial xp − 1. Let σ and τ be K-automorphisms of L. Then σ(ω) and τ (ω) are roots of xp −1 (since the automorphisms σ and τ permute the roots of this polynomial) and therefore there exist non-negative integers q and r such that σ(ω) = ω q and τ (ω) = ω r . Then σ(τ (ω)) = ω qr = τ (σ(ω)). But there is at most one K-automorphism of L sending ω to ω qr . It follows that σ ◦ τ = τ ◦ σ. Thus the Galois group Γ(L: K) is Abelian, as required. Lemma 3.39 Let K be a field of characteristic zero and let M be a splitting field for the polynomial xp − c over K, where p is some prime number and c ∈ K. Then the Galois group Γ(M : K) of the extension M : K is solvable. Proof The result is trivial when c = 0, since M = K in this case. Suppose c 6= 0. The roots of the polynomial xp − c are distinct, and each pth root of unity is the ratio of two roots of xp − c. Therefore M = K(α, ω), where αp = c and ω is some primitive pth root of unity. Now K(ω): K is a normal extension, since K(ω) is a splitting field for the polynomial xp − 1 over K (Theorem 3.17). On applying the Galois correspondence (Theorem 3.35), we see that Γ(M : K(ω)) is a normal subgroup of Γ(M : K), and Γ(M : K)/Γ(M : K(ω)) is isomorphic to Γ(K(ω): K). But Γ(K(ω): K) is Abelian (Lemma 3.38). It therefore suffices to show that Γ(M : K(ω)) is also Abelian. Now the field M is obtained from K(ω) by adjoining an element α satisfying αp = c. Therefore each automorphism σ in Γ(M : K(ω)) is uniquely determined by the value of σ(α). Moreover σ(α) is also a root of xp − c, and therefore σ(α) = αω j for some integer j. Thus if σ and τ are automorphisms of M belonging to Γ(M : K(ω)), and if σ(α) = αω j and τ (α) = αω k , then σ(τ (α)) = τ (σ(α)) = αω j+k , since σ(ω) = τ (ω) = ω. Therefore σ ◦ τ = τ ◦ σ. We deduce that Γ(M : K(ω)) is Abelian, and thus Γ(M : K) is solvable, as required. 72

Lemma 3.40 Let f be a polynomial with coefficients in a field K of characteristic zero, and let K 0 = K(α), where α ∈ K 0 satisfies αp ∈ K for some prime number p. Then ΓK (f ) is solvable if and only if ΓK 0 (f ) is solvable. Proof Let N be a splitting field for the polynomial f (x)(xp − c) over K, where c = αp . Then N contains a splitting field L for f over K and a splitting field M for xp − c over K. Then N : K, L: K and M : K are Galois extensions. The Galois correspondence (Theorem 3.35) ensures that Γ(N : L) and Γ(N : M ) are normal subgroups of Γ(N : K). Moreover Γ(L: K) is isomorphic to Γ(N : K)/Γ(N : L), and Γ(M : K) is isomorphic to Γ(N : K)/Γ(N : M ). Now M and N are splitting fields for the polynomial xp − c over the fields K and L respectively. It follows from Lemma 3.39 that Γ(M : K) and Γ(N : L) are solvable. But if H is a normal subgroup of a finite group G then G is solvable if and only both H and G/H are solvable (Proposition 1.41). Therefore Γ(N : K) is solvable if and only if Γ(N : M ) is solvable. Also Γ(N : K) is solvable if and only if Γ(L: K) is solvable. It follows that Γ(N : M ) is solvable if and only if Γ(L: K) is solvable. But Γ(N : M ) ∼ = ΓM (f ) and Γ(L: K) ∼ = ΓK (f ), since L and N are splitting fields for f over K and M respectively. Thus ΓM (f ) is solvable if and only if ΓK (f ) is solvable. Now M is also a splitting field for the polynomial xp − c over K 0 , since 0 K = K(α), where α is a root of the polynomial xp − c. The above argument therefore shows that ΓM (f ) is solvable if and only if ΓK 0 (f ) is solvable. Therefore ΓK (f ) is solvable if and only if ΓK 0 (f ) is solvable, as required. Theorem 3.41 Let f be a polynomial with coefficients in a field K of characteristic zero. Suppose that f is solvable by radicals. Then the Galois group ΓK (f ) of f is a solvable group. Proof The polynomial f is solvable by radicals. Therefore there exist fields K0 , K1 , . . . , Km such that K0 = K, the polynomial f splits over Km , and, for each integer i between 1 and m, the field Ki is obtained on adjoining to Ki−1 an element αi with the property that αipi ∈ Ki−1 for some prime number pi . Now ΓKm (f ) is solvable, since it is the trivial group consisting of the identity automorphism of Km only. Also Lemma 3.40 ensures that, for each i > 0, ΓKi (f ) is solvable if and only if ΓKi−1 (f ) is solvable. It follows that ΓK (f ) is solvable, as required. Lemma 3.42 Let p be a prime number, let K be a field whose characteristic is not equal to p, and let L: K be a Galois extension of K of degree p. Suppose that the polynomial xp − 1 splits over K. Then there exists α ∈ L such that L = K(α) and αp ∈ K. 73

Proof The Galois group Γ(L: K) is a cyclic group of order p, since its order is equal to the degree p of the extension L: K. Let σ be a generator of Γ(L: K), let β be an element of L \ K, and let αj = β0 + ω j β1 + ω 2j β2 + · · · + ω (p−1)j βp−1 for j = 0, 1, . . . , p − 1, where β0 = β, βi = σ(βi−1 ) for i = 1, 2, . . . , p − 1, and ω is a primitive pth root of unity contained in K. Now σ(αj ) = ω −j αj for j = 0, 1, . . . , p − 1, since σ(ω) = ω, σ(βp−1 ) = β0 and ω p = 1. Therefore σ(αjp ) = αjp and hence αjp ∈ K for j = 0, 1, 2, . . . , p − 1. But α0 + α1 + α2 + · · · + αp−1 = pβ, since ω j is a root of the polynomial 1 + x + x2 + · · · + xp−1 for all integers j that are not divisible by p. Moreover pβ ∈ L \ K, since β ∈ L \ K and p 6= 0 in K. Therefore at least one of the elements α0 , α1 , . . . , αp−1 belongs to L \ K. Let α = αj , where αj ∈ L \ K. It follows from the Tower Law (Theorem 3.1) that [K(α), K] divides [L: K]. But [L: K] = p and p is prime. It follows that L = K(α). Moreover αp ∈ K, as required. Theorem 3.43 Let f be a polynomial with coefficients in a field K of characteristic zero. Suppose that the Galois group ΓK (f ) of f over K is solvable. Then f is solvable by radicals. Proof Let ω be a primitive pth root of unity. Then ΓK(ω) (f ) is isomorphic to a subgroup of ΓK (f ) (Lemma 3.36) and is therefore solvable (Proposition 1.41). Moreover f is solvable by radicals over K if and only if f is solvable by radicals over K(ω), since K(ω) is obtained from K by adjoining an element ω whose pth power belongs to K. We may therefore assume, without loss of generality, that K contains a primitive pth root of unity for each prime p that divides |ΓK (f )|. The result is trivial when |ΓK (f )| = 1, since in that case the polynomial f splits over K. We prove the result by induction on the degree |ΓK (f )| of the Galois group. Thus suppose that the result holds when the order of the Galois group is less than |ΓK (f )|. Let L be a splitting field for f over K. Then L: K is a Galois extension and Γ(L: K) ∼ = ΓK (f ). Now the solvable group Γ(L: K) contains a normal subgroup H for which the corresponding quotient group Γ(L: K)/H is a cyclic group of order p for some prime number p dividing |Γ(L: K)|. Let M be the fixed field of H. Then Γ(L: M ) = H and Γ(M : K) ∼ = Γ(L: K)/H. (Theorem 3.35), and therefore [M : K] = |Γ(L: K)/H| = p. It follows from Lemma 3.42 that M = K(α) for some element α ∈ M satisfying αp ∈ K. Moreover ΓM (f ) ∼ = H, and H is solvable, since any subgroup of 74

a solvable group is solvable (Proposition 1.41). The induction hypothesis ensures that f is solvable by radicals when considered as a polynomial with coefficients in M , and therefore the roots of f lie in some extension field of M obtained by successively adjoining radicals. But M is obtained from K by adjoining the radical α. Therefore f is solvable by radicals, when considered as a polynomial with coefficients in K, as required. On combining Theorem 3.41 and Theorem 3.43, we see that a polynomial with coefficients in a field K of characteristic zero is solvable by radicals if and only if its Galois group ΓK (f ) over K is a solvable group.

3.18

A quintic polynomial that is not solvable by radicals

Lemma 3.44 Let p be a prime number and let f be a polynomial of order p with rational coefficients. Suppose that f has exactly p − 2 real roots and is irreducible over the field Q of rational numbers. Then the Galois group of f over Q is isomorphic to the symmetric group Σp . Proof If α is a root of f then [Q(α): Q] = p since f is irreducible and deg f = p (Theorem 3.4). Thus if L is a splitting field extension for f over Q then [L: Q] = [L: Q(α)][Q(α): Q] by the Tower Law (Proposition 3.1) and therefore [L: Q] is divisible by p. But [L: Q] is the order of the Galois group G of f , and therefore |G| is divisible by p. It follows from a basic theorem of Cauchy that G must contain at least one element of order p. Moreover an element of G is determined by its action on the roots of f . Thus an element of G is of order p if and only if it cyclically permutes the roots of f . The irreducibility of f ensures that f has distinct roots (Corollary 3.20). Let α1 and α2 be the two roots of f that are not real. Then α1 and α2 are complex conjugates of one another, since f has real coefficients. We have already seen that G contains an element of order p which cyclically permutes the roots of f . On taking an appropriate power of this element, we obtain an element σ of G that cyclically permutes the roots of f and sends α1 to α2 . We label the real roots α3 , α4 , . . . , αp of f so that αj = σ(αj−1 ) for j = 2, 3, 4, . . . , p. Then σ(αp ) = α1 . Now complex conjugation restricts to a Q-automorphism τ of L that interchanges α1 and α2 but fixes αj for j > 2. But if 2 ≤ j ≤ p then σ 1−j τ σ j−1 transposes the roots αj−1 and αj and fixes the remaining roots. But transpositions of this form generate the whole of the group of permutations of the roots. Therefore every permutation of the roots of f is realised by some element of the Galois group G of f , and thus G∼ = Σp , as required. 75

Example Consider the quintic polynomial f where f (x) = x5 − 6x + 3. Eisenstein’s Irreducibility Criterion (Proposition 2.18) can be used to show that f is irreducible over Q. Now f (−2) = −17, f (−1) = 8, f (1) = −2 and f (2) = 23. The Intermediate Value Theorem ensures that f has at least 3 distinct real roots. If f had at least 4 distinct real roots then Rolle’s Theorem would ensure that the number of distinct real roots of f 0 and f 00 would be at least 3 and 2 respectively. But zero is the only root of f 00 since f 00 (x) = 20x3 . Therefore f must have exactly 3 distinct real roots. It follows from Lemma 3.44 that the Galois group of f is isomorphic to the symmetric group Σ5 . This group is not solvable. Theorem 3.41 then ensures that the polynomial f is not solvable by radicals over the field of rational numbers. The above example demonstrates that there cannot exist any general formula for obtaining the roots of a quintic polynomial from its coefficients in a finite number of steps involving only addition, subtraction, multiplication, division and the extraction of nth roots. For if such a general formula were to exist then every quintic polynomial with rational coefficients would be solvable by radicals.

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4

Commutative Algebra and Algebraic Geometry

4.1

Modules

Definition Let R be a unital commutative ring. A set M is said to be a module over R (or R-module) if (i) given any x, y ∈ M and r ∈ R, there are well-defined elements x + y and rx of M , (ii) M is an Abelian group with respect to the operation + of addition, (iii) the identities r(x + y) = rx + ry, (rs)x = r(sx),

(r + s)x = rx + sx, 1x = x

are satisfied for all x, y ∈ M and r, s ∈ R. Example If K is a field, then a K-module is by definition a vector space over K. Example Let (M, +) be an Abelian group, and let x ∈ M . If n is a positive integer then we define nx to be the sum x + x + · · · + x of n copies of x. If n is a negative integer then we define nx = −(|n|x), and we define 0x = 0. This enables us to regard any Abelian group as a module over the ring Z of integers. Conversely, any module over Z is also an Abelian group. Example Any unital commutative ring can be regarded as a module over itself in the obvious fashion. Let R be a unital commutative ring, and let M be an R-module. A subset L of M is said to be a submodule of M if x + y ∈ L and rx ∈ L for all x, y ∈ L and r ∈ R. If M is an R-module and L is a submodule of M then the quotient group M/L can itself be regarded as an R-module, where r(L + x) ≡ L + rx for all L + x ∈ M/L and r ∈ R. The R-module M/L is referred to as the quotient of the module M by the submodule L. Note that a subset I of a unital commutative ring R is a submodule of R if and only if I is an ideal of R. Let M and N be modules over some unital commutative group R. A function ϕ: M → N is said to be a homomorphism of R-modules if ϕ(x+y) = 77

ϕ(x)+ϕ(y) and ϕ(rx) = rϕ(x) for all x, y ∈ M and r ∈ R. A homomorphism of R-modules is said to be an isomorphism if it is invertible. The kernel ker ϕ and image ϕ(M ) of any homomorphism ϕ: M → N are themselves Rmodules. Moreover if ϕ: M → N is a homomorphism of R-modules, and if L is a submodule of M satisfying L ⊂ ker ϕ, then ϕ induces a homomorphism ϕ: M/L → N . This induced homomorphism is an isomorphism if and only if L = ker ϕ and N = ϕ(M ). Definition Let M1 , M2 , . . . , Mk be modules over a unital commutative ring R. The direct sum M1 ⊕ M2 ⊕ · · · ⊕ Mk is defined to be the set of ordered k-tuples (x1 , x2 , . . . , xk ), where xi ∈ Mi for i = 1, 2, . . . , k. This direct sum is itself an R-module: (x1 , x2 , . . . , xk ) + (y1 , y2 , . . . , yk ) = (x1 + y1 , x2 + y2 , . . . , xk + yk ), r(x1 , x2 , . . . , xk ) = (rx1 , rx2 , . . . , rxk ) for all xi , yi ∈ Mi and r ∈ R. If K is any field, then K n is the direct sum of n copies of K. Definition Let M be a module over some unital commutative ring R. Given any subset X of M , the submodule of M generated by the set X is defined to be the intersection of all submodules of M that contain the set X. It is therefore the smallest submodule of M that contains the set X. An Rmodule M is said to be finitely-generated if it is generated by some finite subset of itself. Lemma 4.1 Let M be a module over some unital commutative ring R, and let {x1 , x2 , . . . , xk } be a finite subset of M . Then the submodule of M generated by this set consists of all elements of M that are of the form r1 x 1 + r2 x 2 + · · · + rk x k for some r1 , r2 , . . . , rk ∈ R. Proof The subset of M consisting of all elements of M of this form is clearly a submodule of M . Moreover it is contained in every submodule of M that contains the set {x1 , x2 , . . . , xk }. The result follows.

78

4.2

Noetherian Modules

Definition Let R be a unital commutative ring. An R-module M is said to be Noetherian if every submodule of M is finitely-generated. Proposition 4.2 Let R be a unital commutative ring, and let M be a module over R. Then the following are equivalent:— (i) (Ascending Chain Condition) if L1 ⊂ L2 ⊂ L3 ⊂ · · · is an ascending chain of submodules of M then there exists an integer N such that Ln = LN for all n ≥ N ; (ii) (Maximal Condition) every non-empty collection of submodules of M has a maximal element (i.e., an submodule which is not contained in any other submodule belonging to the collection); (iii) (Finite Basis Condition) M is a Noetherian R-module (i.e., every submodule of M is finitely-generated). Proof Suppose that M satisfies the Ascending Chain Condition. Let C be a non-empty collection of submodules of M . Choose L1 ∈ C. If C were to contain no maximal element then we could choose, by induction on n, an ascending chain L1 ⊂ L2 ⊂ L3 ⊂ · · · of submodules belonging to C such that Ln 6= Ln+1 for all n, which would contradict the Ascending Chain Condition. Thus M must satisfy the Maximal Condition. Next suppose that M satisfies the Maximal Condition. Let L be an submodule of M , and let C be the collection of all finitely-generated submodules of M that are contained in L. Now the zero submodule {0} belongs to C, hence C contains a maximal element J, and J is generated by some finite subset {a1 , a2 , . . . , ak } of M . Let x ∈ L, and let K be the submodule generated by {x, a1 , a2 , . . . , ak }. Then K ∈ C, and J ⊂ K. It follows from the maximality of J that J = K, and thus x ∈ J. Therefore J = L, and thus L is finitely-generated. Thus M must satisfy the Finite Basis Condition. Finally suppose that M satisfies the Finite Basis Condition. Let L1 ⊂ L2 ⊂ L3 ⊂ · · · be an ascending chain of submodules of M , and let L be the +∞ S union Ln of the submodules Ln . Then L is itself an submodule of M . n=1

Indeed if a and b are elements of L then a and b both belong to Ln for some sufficiently large n, and hence a + b, −a and ra belong to Ln , and thus to L, for all r ∈ M . But the submodule L is finitely-generated. Let {a1 , a2 , . . . , ak } be a generating set of L. Choose N large enough to ensure that ai ∈ LN for i = 1, 2, . . . , k. Then L ⊂ LN , and hence LN = Ln = L for all n ≥ N . Thus M must satisfy the Ascending Chain Condition, as required. 79

Proposition 4.3 Let R be a unital commutative ring, let M be an R-module, and let L be a submodule of M . Then M is Noetherian if and only if L and M/L are Noetherian. Proof Suppose that the R-module M is Noetherian. Then the submodule L is also Noetherian, since any submodule of L is also a submodule of M and is therefore finitely-generated. Also any submodule K of M/L is of the form {L + x : x ∈ J} for some submodule J of M satisfying L ⊂ J. But J is finitely-generated (since M is Noetherian). Let x1 , x2 , . . . , xk be a finite generating set for J. Then L + x1 , L + x2 , . . . , L + xk is a finite generating set for K. Thus M/L is Noetherian. Conversely, suppose that L and M/L are Noetherian. We must show that M is Noetherian. Let J be any submodule of M , and let ν(J) be the image of J under the quotient homomorphism ν: M → M/L, where ν(x) = L + x for all x ∈ M . Then ν(J) is a submodule of the Noetherian module M/L and is therefore finitely-generated. It follows that there exist elements x1 , x2 , . . . , xk of J such that ν(J) is generated by L + x1 , L + x2 , . . . , L + xk . Also J ∩ L is a submodule of the Noetherian module L, and therefore there exists a finite generating set y1 , y2 , . . . , ym for J ∩ L. We claim that {x1 , x2 , . . . , xk , y1 , y2 , . . . , ym } is a generating set for J. Let z ∈ J. Then there exist r1 , r2 , . . . , rk ∈ R such that ν(z) = r1 (L + x1 ) + r2 (L + x2 ) + · · · + rk (L + xk ) = L + r1 x1 + r2 x2 + · · · + rk xk . But then z −(r1 x1 +r2 x2 +· · ·+rk xk ) ∈ J ∩L (since L = ker ν), and therefore there exist s1 , s2 , . . . , sm such that z − (r1 x1 + r2 x2 + · · · + rk xk ) = s1 y1 + s2 y2 + · · · + sm ym , and thus z=

k X

ri x i +

i=1

m X

si y i .

j=1

This shows that the submodule J of M is finitely-generated. We deduce that M is Noetherian, as required. 80

Corollary 4.4 The direct sum M1 ⊕ M2 ⊕ · · · ⊕ Mk of Noetherian modules M1 , M2 , . . . Nk over some unital commutative ring R is itself a Noetherian module over R. Proof The result follows easily by induction on k once it has been proved in the case k = 2. Let M1 and M2 be Noetherian R-modules. Then M1 ⊕{0} is a Noetherian submodule of M1 ⊕ M2 isomorphic to M1 , and the quotient of M1 ⊕ M2 by this submodule is a Noetherian R-module isomorphic to M2 . It follows from Proposition 4.3 that M1 ⊕ M2 is Noetherian, as required. One can define also the concept of a module over a non-commutative ring. Let R be a unital ring (not necessarily commutative), and let M be an Abelian group. We say that M is a left R-module if each r ∈ R and m ∈ M determine an element rm of M , and the identities r(x + y) = rx + ry,

(r + s)x = rx + sx,

(rs)x = r(sx),

1x = x

are satisfied for all x, y ∈ M and r, s ∈ R. Similarly we say that M is a right R-module if each r ∈ R and m ∈ M determine an element mr of M , and the identities (x + y)r = xr + yr,

x(r + s) = xr + xs,

x(rs) = (xr)s,

x1 = x

are satisfied for all x, y ∈ M and r, s ∈ R. (If R is commutative then the distinction between left R-modules and right R-modules is simply a question of notation; this is not the case if R is non-commutative.)

4.3

Noetherian Rings and Hilbert’s Basis Theorem

Let R be a unital commutative ring. We can regard the ring R as an Rmodule, where the ring R acts on itself by left multiplication (so that r . r0 is the product rr0 of r and r0 for all elements r and r0 of R). We then find that a subset of R is an ideal of R if and only if it is a submodule of R. The following result therefore follows directly from Proposition 4.2. Proposition 4.5 Let R be a unital commutative ring. Then the following are equivalent:— (i) (Ascending Chain Condition) if I1 ⊂ I2 ⊂ I3 ⊂ · · · is an ascending chain of ideals of R then there exists an integer N such that In = IN for all n ≥ N ; 81

(ii) (Maximal Condition) every non-empty collection of ideals of R has a maximal element (i.e., an ideal which is not contained in any other ideal belonging to the collection); (iii) (Finite Basis Condition) every ideal of R is finitely-generated. Definition A unital commutative ring is said to be a Noetherian ring if every ideal of the ring is finitely-generated. A Noetherian domain is a Noetherian ring that is also an integral domain. Note that a unital commutative ring R is Noetherian if it satisfies any one of the conditions of Proposition 4.5. Corollary 4.6 Let M be a finitely-generated module over a Noetherian ring R. Then M is a Noetherian R-module. Proof Let {x1 , x2 , . . . , xk } be a finite generating set for M . Let Rk be the direct sum of k copies of R, and let ϕ: Rk → M be the homomorphism of R-modules sending (r1 , r2 , . . . , rk ) ∈ Rk to r1 x 1 + r2 x 2 + · · · + rk x k . It follows from Corollary 4.4 that Rk is a Noetherian R-module (since the Noetherian ring R is itself a Noetherian R-module). Moreover M is isomorphic to Rk / ker ϕ, since ϕ: Rk → M is surjective. It follows from Proposition 4.3 that M is Noetherian, as required. If I is a proper ideal of a Noetherian ring R then the collection of all proper ideals of R that contain the ideal I is clearly non-empty (since I itself belongs to the collection). It follows immediately from the Maximal Condition that I is contained in some maximal ideal of R. Lemma 4.7 Let R be a Noetherian ring, and let I be an ideal of R. Then the quotient ring R/I is Noetherian. Proof Let L be an ideal of R/I, and let J = {x ∈ R : I + x ∈ L}. Then J is an ideal of R, and therefore there exists a finite subset {a1 , a2 , . . . , ak } of J which generates J. But then L is generated by I + ai for i = 1, 2, . . . , k. Indeed every element of L is of the form I + x for some x ∈ J, and if x = r1 a1 + r2 a2 + · · · + rk ak , where r1 , r2 , . . . , rk ∈ R, then I + x = r1 (I + a1 ) + r2 (I + a2 ) + · · · + rk (I + ak ), as required. 82

Hilbert showed that if R is a field or is the ring Z of integers, then every ideal of R[x1 , x2 , . . . , xn ] is finitely-generated. The method that Hilbert used to prove this result can be generalized to yield the following theorem. Theorem 4.8 (Hilbert’s Basis Theorem) If R is a Noetherian ring, then so is the polynomial ring R[x]. Proof Let I be an ideal of R[x], and, for each non-negative integer n, let In denote the subset of R consisting of those elements of R that occur as leading coefficients of polynomials of degree n belonging to I, together with the zero element of R. Then In is an ideal of R. Moreover In ⊂ In+1 , for if p(x) is a polynomial of degree n belonging to I then xp(x) is a polynomial of degree n+1 belonging to I which has the same leading coefficient. Thus I0 ⊂ I1 ⊂ I2 ⊂ · · · is an ascending chain of ideals of R. But the Noetherian ring R satisfies the Ascending Chain Condition (see Proposition 4.5). Therefore there exists some natural number m such that In = Im for all n ≥ m. Now each ideal In is finitely-generated, hence, for each n ≤ m, we can choose a finite set {an,1 , an,2 , . . . , an,kn } which generates In . Moreover each generator an,i is the leading coefficient of some polynomial qn,i of degree n belonging to I. Let J be the ideal of R[x] generated by the polynomials qn,i for all 0 ≤ n ≤ m and 1 ≤ i ≤ kn . Then J is finitely-generated. We shall show by induction on deg p that every polynomial p belonging to I must belong to J, and thus I = J. Now if p ∈ I and deg p = 0 then p is a constant polynomial whose value belongs to I0 (by definition of I0 ), and thus p is a linear combination of the constant polynomials q0,i (since the values a0,i of the constant polynomials q0,i generate I0 ), showing that p ∈ J. Thus the result holds for all p ∈ I of degree 0. Now suppose that p ∈ I is a polynomial of degree n and that the result is true for all polynomials p in I of degree less than n. Consider first the case when n ≤ m. Let b be the leading coefficient of p. Then there exist c1 , c2 , . . . , ckn ∈ R such that b = c1 an,1 + c2 an,2 + · · · + ckn an,kn , since an,1 , an,2 , . . . , an,kn generate the ideal In of R. Then p(x) = c1 qn,1 (x) + c2 qn,2 (x) + · · · + ck qn,k (x) + r(x), where r ∈ I and deg r < deg p. It follows from the induction hypothesis that r ∈ J. But then p ∈ J. This proves the result for all polynomials p in I satisfying deg p ≤ m. Finally suppose that p ∈ I is a polynomial of degree n where n > m, and that the result has been verified for all polynomials of degree less than n. 83

Then the leading coefficient b of p belongs to In . But In = Im , since n ≥ m. As before, we see that there exist c1 , c2 , . . . , ckm ∈ R such that b = c1 am,1 + c2 am,2 + · · · + ckn am,km , since am,1 , am,2 , . . . , am,km generate the ideal In of R. Then p(x) = c1 xn−m qm,1 (x) + c2 xn−m qm,2 (x) + · · · + ck xn−m qm,k (x) + r(x), where r ∈ I and deg r < deg p. It follows from the induction hypothesis that r ∈ J. But then p ∈ J. This proves the result for all polynomials p in I satisfying deg p > m. Therefore I = J, and thus I is finitely-generated, as required. Theorem 4.9 Let R be a Noetherian ring. Then the ring R[x1 , x2 , . . . , xn ] of polynomials in the indeterminates x1 , x2 , . . . , xn with coefficients in R is a Noetherian ring. Proof It is easy to see to see that R[x1 , x2 , . . . , xn ] is naturally isomorphic to R[x1 , x2 , . . . , xn−1 ][xn ] when n > 1. (Any polynomial in the indeterminates x1 , x2 , . . . , xn with coefficients in the ring R may be viewed as a polynomial in the indeterminate xn with coefficients in the polynomial ring R[x1 , x2 , . . . , xn−1 ].) The required results therefore follows from Hilbert’s Basis Theorem (Theorem 4.8) by induction on n. Corollary 4.10 Let K be a field. Then every ideal of the polynomial ring K[x1 , x2 , . . . , xn ] is finitely-generated.

4.4

Polynomial Rings in Several Variables

A monomial in the independent indeterminates X1 , X2 , . . . , Xn is by definition an expression of the form X1i1 X2i2 · · · Xnin , where i1 , i2 , . . . , in are nonnegative integers. Such monomials are multiplied according to the rule

X1i1 X2i2 · · · Xnin X1j1 X2j2 · · · Xnjn = X1i1 +j1 X2i2 +j2 · · · Xnin +jn . A polynomial p in the independent indeterminates with coefficients in some ring R is by definition a formal linear combination of the form r1 m 1 + r2 m 2 + · · · + rk m k where r1 , r2 , . . . , rk ∈ R and m1 , m2 , . . . , mk are monomials in X1 , X2 , . . . , Xn . The coefficients r1 , r2 , . . . , rk of this polynomial are uniquely determined, 84

provided that the monomials m1 , m2 , . . . , mk are distinct. Such polynomials are added and multiplied together in the obvious fashion. In particular ! ! k l k X l X X X ri m i sj m0j = (ri sj )(mi m0j ), i=1

j=1

i=1 j=1

where the product mi m0j of the monomials mi and m0j is defined as described above. The set of all polynomials in the independent indeterminates X1 , X2 , . . . , Xn with coefficients in the ring R is itself a ring, which we denote by R[X1 , X2 , . . . , Xn ]. Example The polynomial 2X1 X23 − 6X1 X2 X32 is the product of the polynomials 2X1 X2 and X22 − 3X32 in the ring Z[X1 , X2 , X3 ] of polynomials in X1 , X2 , X3 with integer coefficients. Lemma 4.11 Let R be an integral domain. Then the ring R[x] of polynomials in the indeterminate x with coefficients in R is itself an integral domain, and deg(pq) = deg p + deg q for all non-zero polynomials p, q ∈ R[x]. Proof The integral domain R is commutative, hence so is R[x]. Moreover R[x] is unital, and the multiplicative identity element of R[x] is the constant polynomial whose coefficient is the multiplicative identity element 1 of the unital ring R. Let p and q be polynomials in R[x], and let ak and bl be the leading coefficients of p and q respectively, where k = deg p and l = deg q. Now p(x)q(x) = ak bl xk+l + terms of lower degree. Moreover ak bl 6= 0, since ak 6= 0, bl 6= 0, and the ring R of coefficients is an integral domain. Thus if p 6= 0 and q 6= 0 then pq 6= 0, showing that R[x] is an integral domain, and deg(pq) = k + l = deg p + deg q, as required. Let p be a polynomial in the indeterminates X1 , X2 , . . . , Xn with coefficients in the ring R, where n > 1. By collecting together terms involving Xnj for each non-negative integer j, we can write the polynomial p in the form p(X1 , X2 , . . . , Xn ) =

k X

pj (X1 , X2 , . . . , Xn−1 )Xnj

j=0

where pj ∈ R[X1 , X2 , . . . , Xn−1 ] for j = 0, 1, . . . , k. Now the right hand side of the above identity can be viewed as a polynomial in the indeterminate Xn with coefficients p1 , p2 , . . . , pk in the ring R[X1 , . . . , Xn−1 ]. Moreover the 85

polynomial p uniquely determines and is uniquely determined by the polynomials p1 , p2 , . . . , pk . It follows from this that the rings R[X1 , X2 , . . . , Xn ] and R[X1 , X2 , . . . , Xn−1 ][Xn ] are naturally isomorphic and can be identified with one another. We can use the identification in order to prove results concerning the structure of the polynomial ring R[X1 , X2 , . . . , Xn ] by induction on the number n of independent indeterminates X1 , X2 , . . . , Xn . For example, the following result follows directly by induction on n, using Lemma 4.11. Lemma 4.12 Let R be an integral domain. Then the ring R[X1 , X2 , . . . , Xn ] is also an integral domain. A monomial X1i1 X2i2 · · · Xnin is said to be of degree d, where d is some non-negative integer, if i1 + i2 + · · · + in = d. Definition Let R be a ring. A polynomial p ∈ R[X1 , X2 , . . . , Xn ] is said to be homogeneous of degree d if it can be expressed as a linear combination of monomials of degree d with coefficients in the ring R. Any polynomial p ∈ R[X1 , X2 , . . . , Xn ] can be decomposed as a sum of the form p(0) + p(1) + · · · + p(k) , where k is some sufficiently large non-negative integer and each polynomial p(i) is a homogeneous polynomial of degree i. The homogeneous polynomial p(i) is referred to as the homogeneous component of p of degree i; it is uniquely determined by p. A non-zero polynomial p is said to be of degree d if p(d) 6= 0 and p(i) = 0 for all i > d. The degree of a non-zero polynomial p is denoted by deg p. Lemma 4.13 Let R be a ring, and let p and q be non-zero polynomials belonging to R[X1 , X2 , . . . , Xn ]. Then deg(p + q) ≤ max(deg p, deg q), provided that p + q 6= 0, deg(pq) ≤ deg p + deg q, provided that pq 6= 0. Moreover if R is an integral domain then pq 6= 0 and deg(pq) = deg p + deg q. Proof The inequality (p + q) ≤ max(deg p, deg q) is obvious. Also p(i) q (j) is homogeneous of degree i + j for all i and j, since the product of a monomial of degree i and a monomial of degree j is a monomial of degree i + j. The inequality deg(pq) ≤ deg p + deg q follows immediately. Now suppose that R is an integral domain. Let k = deg p and l = deg q. Then the homogeneous component (pq)(k+l) of pq of degree k + l is given by (pq)(k+l) = p(k) q (l) . But R[X1 , X2 , . . . , Xn ] is an integral domain (see Lemma 4.12), and p(k) and q (l) are both non-zero. It follows that (pq)(k+l) 6= 0, and thus deg(pq) = deg p + deg q, as required. 86

4.5

Algebraic Sets and the Zariski Topology

Throughout this section, let K be a field. Definition We define affine n-space An over the field K to be the set K n of all n-tuples (x1 , x2 , . . . , xn ) with x1 , x2 , . . . , xn ∈ K. Where it is necessary to specify explicitly the field K involved, we shall denote affine n-space over the field K by An (K). Thus An (R) = Rn , and An (C) = Cn . Definition A subset of n-dimensional affine space An is said to be an algebraic set if it is of the form {(x1 , x2 , . . . , xn ) ∈ An : f (x1 , x2 , . . . , xn ) = 0 for all f ∈ S} for some subset S of the polynomial ring K[X1 , X2 , . . . , Xn ]. Example Any point of An is an algebraic set. Indeed, given any point (a1 , a2 , . . . , an ) of An , let fi (X1 , X2 , . . . , Xn ) = Xi − ai for i = 1, 2, . . . , n. Then the given point is equal to the set {(x1 , x2 , . . . , xn ) ∈ An : fi (x1 , x2 , . . . , xn ) = 0 for i = 1, 2, . . . , n}. Example The circle {(x, y) ∈ A2 (R) : x2 + y 2 = 1} is an algebraic set in the plane A2 (R). Let λ: K n → K be a linear functional on the vector space K n (i.e., a linear transformation from K n to K). It follows from elementary linear algebra that there exist b1 , b2 , . . . , bn ∈ K such that λ(x1 , x2 , . . . , xn ) = b1 x1 + b2 x2 + · · · + bn xn for all (x1 , x2 , . . . , xn ) ∈ K n . Thus if λ1 , λ2 , . . . , λk are linear functionals on K n , and if c1 , c2 , . . . , ck are suitable constants belonging to the field K then {(x1 , x2 . . . , xn ) ∈ An : λi (x1 , x2 , . . . , xn ) = ci for i = 1, 2, . . . , k} is an algebraic set in An . A set of this type is referred to as an affine subspace of An . It is said to be of dimension n − k, provided that the linear functionals λ1 , λ2 , . . . , λk are linearly independent. It follows directly from elementary linear algebra that, if we we identify affine n-space An with the vector space K n , then a subset of An is an m-dimensional affine subspace if and only if it is a translate of some m-dimensional vector subspace of K n (i.e., it is of the form v + W where v is a point of An and W is some m-dimensional vector subspace of K n ). 87

Lemma 4.14 Let V be an algebraic set in An , and let L be a one-dimensional affine subspace of An . Then either L ⊂ V or else L ∩ V is a finite set. Proof The affine subspace L is a translate of a one-dimensional subspace of K n , and therefore there exist vectors v and w in K n such that L = {v + wt : t ∈ K} (on identifying n-dimensional affine space An with the vector space K n ). Now we can write V = {(x1 , x2 , . . . , xn ) ∈ An : f (x1 , x2 , . . . , xn ) = 0 for all f ∈ S}, where S is some subset of the polynomial ring K[X1 , X2 , . . . , Xn ]. Now either each polynomial belonging to S is zero throughout L, in which case L ⊂ V , or else there is some f ∈ S which is non-zero at some point of L. Define g ∈ K[t] by the formula g(t) = f (v1 + w1 t, v2 + w2 t, . . . , vn + wn t) (where vi and wi denote the ith components of the vectors v and w for i = 1, 2, . . . , n). Then g is a non-zero polynomial in the indeterminate t, and therefore g has at most finitely many zeros. But g(t) = 0 whenever the point v + wt of L lies in V . Therefore L ∩ V is finite, as required. Example The sets {(x, y) ∈ A2 (R) : y = sin x} and {(x, y) ∈ A2 (R) : x ≥ 0} are not algebraic sets in A2 (R), since the line y = 0 is not contained in either of these sets, yet the line intersects these sets at infinitely many points of the set. Given any subset S of K[X1 , X2 , . . . , Xn ], we denote by V (S) the algebraic set in An defined by V (S) = {x ∈ An : f (x) = 0 for all f ∈ S}. Also, given any f ∈ K[X1 , X2 , . . . , Xn ], we define V (f ) = V ({f }). Given any subset Z of An , we define I(Z) = {f ∈ K[X1 , X2 , . . . , Xn ] : f (x) = 0 for all x ∈ Z}. Clearly S ⊂ I(V (S)) for all subsets S of K[X1 , X2 , . . . , Xn ], and Z ⊂ V (I(Z)) for all subsets Z of An . If S1 and S2 are subsets of K[X1 , X2 , . . . , Xn ] satisfying S1 ⊂ S2 then V (S2 ) ⊂ V (S1 ). Similarly, if Z1 and Z2 are subsets of An satisfying Z1 ⊂ Z2 then I(Z2 ) ⊂ I(Z1 ). 88

Lemma 4.15 V (I(V (S))) = V (S) for all subsets S of K[X1 , X2 , . . . , Xn ], and similarly I(V (I(Z))) = I(Z) for all subsets Z of An . Proof It follows from the observations above that V (S) ⊂ V (I(V (S))), since Z ⊂ V (I(Z)) for all subsets Z of An . But also S ⊂ I(V (S)), and hence V (I(V (S))) ⊂ V (S). Therefore V (I(V (S))) = V (S). An analogous argument can be used to show that I(V (I(Z))) = I(Z) for all subsets Z of An . Let I and J be ideals of a unital commutative ring R. We denote by IJ the ideal of R consisting of those elements of R that can be expressed as finite sums of the form i1 j1 + i2 j2 + · · · + ir jr with i1 , i2 , . . . , ir ∈ I and j1 , j2 , . . . , jr ∈ J. (One can readily verify that IJ is indeed an ideal of R.) Proposition 4.16 Let R = K[X1 , X2 , . . . , Xn ] for some field K. Then (i) V ({0}) = An and V (R) = ∅;
T P (ii) λ∈Λ V (Iλ ) = V λ∈Λ Iλ for every collection {Iλ : λ ∈ Λ} of ideals of R; (iii) V (I) ∪ V (J) = V (I ∩ J) = V (IJ) for all ideals I and J of R. Thus there is a well-defined topology on An (known as the Zariski topology) whose closed sets are the algebraic sets in An . Proof (i) is immediate.P
P I , and therefore V I If µ ∈ Λ then I ⊂ λ λ µ λ∈Λ λ∈Λ
T T⊂ V (Iµ ). Thus P V λ∈Λ Iλ ⊂ λ∈Λ V (Iλ ). Conversely if x is a point of λ∈Λ V (Iλ ) then fP(x) = 0 for allTλ ∈ Λ and f ∈ P Iλ , and
therefore f (x) = T 0 for all f ∈ λ∈Λ λ∈Λ V (Iλ ) ⊂ V λ∈Λ Iλ . It follows that λ∈Λ V (Iλ ) =
P Iλ . Thus V λ∈Λ Iλ . This proves (ii). Let I and J be ideals of R. Then I ∩J ⊂ I, I ∩J ⊂ J and IJ ⊂ I ∩J, and thus V (I) ⊂ V (I ∩ J), V (J) ⊂ V (I ∩ J) and V (I ∩ J) ⊂ V (IJ). Therefore V (I) ∪ V (J) ⊂ V (I ∩ J) ⊂ V (IJ). If x is a point of An which does not belong to V (I) ∪ V (J) then there exist polynomials f ∈ I and g ∈ J such that f (x) 6= 0 and g(x) 6= 0. But then f g ∈ IJ and f (x)g(x) 6= 0, and therefore x 6∈ V (IJ). Therefore V (IJ) ⊂ V (I) ∪ V (J). We conclude that V (I) ∪ V (J) = V (I ∩ J) = V (IJ). 89

This proves (iii). Let us define a topology on An whose open sets in An are the complements of algebraic sets. We see from (i) that ∅ and An are open. Moreover it follows from (ii) that any union of open sets is open, and it follows from (iii), using induction on the number of sets, that any finite intersection of open sets is open. Thus the topology is well-defined. Definition The Zariski topology on an algebraic set V in An is the topology whose open sets are of the form V \V (I) for some ideal I of K[X1 , X2 , . . . , Xn ]. It follows from Proposition 4.16 that the Zariski topology on an algebraic set V is well-defined and is the subspace topology on V induced by the topology on An whose closed sets are the algebraic sets in An . Moreover a subset V1 of V is closed if and only if V1 is itself an algebraic set. (This follows directly from the fact that the intersection of two algebraic sets is itself an algebraic set.) Example Any finite subset of An is an algebraic set. This follows from the fact that any point in An is an algebraic set, and any finite union of algebraic sets is an algebraic set. In general, the Zariski topology on an algebraic set V is not Hausdorff. It can in fact be shown that an algebraic set in An is Hausdorff (with respect to the Zariski topology) if and only if it consists of a finite set of points in An .

4.6

The Structure of Algebraic Sets

Let K be a field. We shall apply Hilbert’s Basis Theorem in order to study the structure of algebraic sets in n-dimensional affine space An over the field K. We shall continue to use the notation for algebraic sets in An and corresponding ideals of the polynomial ring that was established earlier. The following result is a direct consequence of the Hilbert Basis Theorem. Proposition 4.17 Let V be an algebraic set in An . Then there exists a finite collection f1 , f2 , f3 , . . . of polynomials in n independent indeterminates such that V = {x ∈ An : fi (x) = 0 for i = 1, 2, . . . , k}. Proof The set V is an algebraic set, and therefore V = V (I) for some ideal I of K[X1 , X2 , . . . , Xn ]. Moreover it follows from Corollary 4.10 that I is generated by some finite set {f1 , f2 , . . . , fk } of polynomials. But then V = V ({f1 , f2 , . . . , fk }), and thus V is of the required form. 90

A algebraic hypersurface in An is a algebraic set of An of the form V (f ) for some non-constant polynomial f ∈ K[X1 , X2 , . . . , Xn ], where V (f ) = {x ∈ An : f (x) = 0}. Corollary 4.18 Every proper algebraic set in An is the intersection of a finite number of algebraic hypersurfaces. Proof The empty set in An can be represented as an intersection of two hyperplanes (e.g., x1 = 0 and x1 = 1). Suppose therefore that the proper algebraic set V is non-empty. It follows from Proposition 4.17 that there exists a finite set {f1 , f2 , . . . , fk } polynomials belonging to K[X1 , X2 , . . . , Xn ] such that V = V ({f1 , f2 , . . . , fk }). Moreover the polynomials f1 , f2 , . . . , fk cannot all be zero, since V 6= An ; we can therefore assume (by removing the zero polynomials from the list) that the polynomials f1 , f2 , . . . , fk are non-zero. They must then all be non-constant, since V is non-empty. But then V = V (f1 ) ∩ V (f2 ) ∩ · · · ∩ V (fk ), as required. Proposition 4.19 Let C be a collection of subsets of An that are open with respect to the Zariski topology on An . Then there exists a finite collection D1 , D2 , . . .S , Dk of open sets belonging to C such that D1 ∪ D2 ∪ · · · ∪ Dk is the union D∈C D of all the open sets D belonging to C. Proof It follows from the definition of the Zariski topology that, for each open set D belonging to C, therePexists an ideal ID of K[X1 , X2 , . . . , Xn ] such that D = An \ V (ID ). Let I = D∈C ID . Then \ [ [ D = (An \ V (ID )) = An \ V (ID ) D∈C D∈C D∈C X  ID = An \ V (I) = An \ V D∈C

(see Proposition 4.16). Now the ideal I is finitely-generated (Corollary 4.10). Moreover there exists a finite generating set {f1 , f2 , . . . , fk } for I with the property that each generator fi belongs to one of the ideals ID , since if we are given any finite generating set for I, then each of the generators can be expressed as a finite sum of elements taken from the ideals ID , and the collection of all these elements constitutes a finite generating set for I which is of the required form. Choose D1 , D2 , . . . , Dk ∈ C such that fi ∈ IDi for i = 1, 2, . . . , k. Then I = ID1 + ID2 + · · · + IDk , 91

and thus [

n

D∈C

n

D = A − V (I) = A − V

k X

! IDi

=

i=1

k [

Di ,

i=1

as required. We recall that a topological space is compact if and only if every open cover of that space has a finite subcover. The following result therefore follows directly from Proposition 4.19. Corollary 4.20 Every subset of An is compact with respect to the Zariski topology.

4.7

Maximal Ideals and Zorn’s Lemma

Definition Let R be a ring. A proper ideal I of R is said to be maximal if the only ideals J of R satisfying I ⊂ J ⊂ R are J = I and J = R. Lemma 4.21 A proper ideal I of a unital commutative ring R is maximal if and only if the quotient ring R/I is a field. Proof Let I be a proper ideal of the unital commutative ring R. Then the quotient ring R/I is unital and commutative. Moreover there is a one-toone correspondence between ideals L of R/I and ideals J of R satisfying I ⊂ J ⊂ R: if J is any ideal of R satisfying I ⊂ J ⊂ R, and if L is the corresponding ideal of R/I then I + x ∈ L if and only if x ∈ J. We deduce that I is a maximal ideal of R if and only if the only ideals of R/I are the zero ideal {I} and R/I itself. It follows from Lemma 2.4 that I is a maximal ideal of R if and only if R/I is a field. We claim that every proper ideal of a ring R is contained in at least one maximal ideal. In order to prove this result we shall make use of Zorn’s Lemma concerning the existence of maximal elements of partially ordered sets. Definition Let S be a set. A partial order ≤ on S is a relation on S satisfying the following conditions:— (i) x ≤ x for all x ∈ S (i.e., the relation ≤ is reflexive), (ii) if x, y, z ∈ S satisfy x ≤ y and y ≤ z then x ≤ z (i.e., the relation ≤ is transitive), 92

(iii) if x, y ∈ S satisfy x ≤ y and y ≤ x then x = y (i.e., the relation ≤ is antisymmetric). Neither of the conditions x ≤ y or y ≤ x need necessarily be satisfied by arbitrary elements x and y of a partially ordered set S. A subset C of S is said to be totally ordered if one or other of the conditions x ≤ y and y ≤ x holds for each pair {x, y} of elements of C. Example Let S be a collection of subsets of some given set. Then S is partially ordered with respect to the relation ⊂ (where A, B ∈ S satisfy A ⊂ B if and only if A is a subset of B). Example The set N of natural numbers is partially ordered with respect to the relation |, where n|m if and only if n divides m. Let ≤ be the ordering relation on a partially ordered set S. An element u of S is said to be an upper bound for a subset B of S if x ≤ u for all x ∈ B. An element m of S is said to be maximal if the only element x of S satisfying m ≤ x is m itself. The following result is an important theorem in set theory. Zorn’s Lemma. Let S be a non-empty partially ordered set. Suppose that there exists an upper bound for each totally ordered subset of S. Then S contains a maximal element. We use Zorn’s lemma in order to prove the following existence theorem for maximal ideals. Theorem 4.22 Let R be a unital ring, and let I be a proper ideal of R. Then there exists a maximal ideal M of R satisfying I ⊂ M ⊂ R. Proof Let S be the set of all proper ideals J of R satisfying I ⊂ J. The set S is non-empty, since I ∈ S, and is partially ordered by the inclusion relation ⊂. We claim that there exists an upper bound for any totally ordered subset C of S. Let L be the union of all the ideals belonging to some totally ordered subset C of S. We claim that L is itself a proper ideal of R. Let a and b be elements of L. Then there exist proper ideals J1 and J2 belonging to C such that a ∈ J1 and b ∈ J2 . Moreover either J1 ⊂ J2 or else J2 ⊂ J1 , since the subset C of S is totally ordered. It follows that a + b belongs either to J1 or else to J2 , and thus a + b ∈ L. Similarly −a ∈ L, ra ∈ L and ar ∈ L for all r ∈ R. We conclude that L is an ideal of R. Moreover 1 6∈ L, since the 93

elements of C are proper ideals of R, and therefore 1 6∈ J for every J ∈ C. It follows that L is a proper ideal of R satisfying I ⊂ L. Thus L ∈ S, and L is an upper bound for C. The conditions of Zorn’s Lemma are satisfied by the partially ordered set S. Therefore S contains a maximal element M . This maximal element is the required maximal ideal of R containing the ideal I. Corollary 4.23 Every unital ring has at least one maximal ideal. Proof Apply Theorem 4.22 with I = {0}.

4.8

Prime Ideals

Definition Let R be a unital ring. A proper ideal I is said to be prime if, given any ideals J and K satisfying JK ⊂ I, either J ⊂ I or K ⊂ I. The following result provides an alternative description of prime ideals of a ring that is both unital and commutative. Lemma 4.24 Let R be a unital commutative ring. An proper ideal I of R is prime if and only if, given any elements x and y of R satisfying xy ∈ I, either x ∈ I or y ∈ I. Proof Let I be a proper ideal of R. Suppose that I has the property that, given any elements x and y of R satisfying xy ∈ I, either x ∈ I or y ∈ I. Let J and K be ideals of R neither of which is a subset of the ideal I. Then there exist elements x ∈ J and y ∈ K which do not belong to I. But then xy belongs to JK but does not belong to I. Thus the ideal JK is not a subset of I. This shows that the ideal I is prime. Conversely, suppose that I is a prime ideal of R. Let x and y be elements of R satisfying xy ∈ I, and let J and K be the ideals generated by x and y respectively. Then J = {rx : r ∈ R},

K = {ry : r ∈ R},

since R is unital and commutative (see Lemma 2.5). It follows easily that JK = {rxy : r ∈ R}. Now xy ∈ I. It follows that JK ⊂ I. But I is prime. Therefore either J ⊂ I or K ⊂ I, and thus either x ∈ I or y ∈ I. Example Let n be a natural number. Then the ideal nZ of the ring Z of integers is a prime ideal if and only if n is a prime number. For an integer j belongs to the ideal nZ if and only if n divides j. Thus the ideal nZ is prime 94

if and only if, given any integers j and k such that n divides jk, either n divides j or n divides k. But it follows easily from the Fundamental Theorem of Arithmetic that a natural number n has this property if and only if n is a prime number. (The Fundamental Theorem of Arithmetic states that any natural number can be factorized uniquely as a product of prime numbers.) Lemma 4.25 An ideal I of a unital commutative ring R is prime if and only if the quotient ring R/I is an integral domain. Proof If I is a proper ideal of the unital commutative ring R then the quotient ring R/I is both unital and commutative. Moreover the zero element of R/I is I itself (regarded as a coset of I in R). Thus R/I is an integral domain if and only if, given elements x and y of R such that (I +x)(I +y) = I, either I + x = I or I + y = I. But (I + x)(I + y) = I + xy for all x, y ∈ R, and I + x = I if and only if x ∈ I. We conclude that R/I is an integral domain if and only if I is prime, as required. Lemma 4.26 Every maximal ideal of a unital commutative ring R is a prime ideal. Proof Let M be a maximal ideal of R. Then the quotient ring R/M is a field (see Lemma 4.21). In particular R/M is an integral domain, and hence M is a prime ideal.

4.9

Affine Varieties and Irreducibility

Definition A topological space Z is said to be reducible if it can be decomposed as a union F1 ∪ F2 of two proper closed subsets F1 and F2 . (A subset of Z is proper if it is not the whole of Z.) A topological space Z is said to be irreducible if it cannot be decomposed as a union of two proper closed subsets. Lemma 4.27 Let Z be a topological space. The following are equivalent:— (i) Z is irreducible, (ii) the intersection of any two non-empty open sets in Z is non-empty, (iii) every non-empty open subset of Z is dense. Moreover a subset A of a topological space Z is irreducible (with respect to the subspace topology) if and only if its closure A is irreducible.

95

Proof The topological space Z is irreducible if and only if the union of any two proper closed subsets of Z is a proper subset of Z. Now the complement of any proper closed set is a non-empty open set, and vica versa. Thus on taking complements we see that Z is irreducible if and only if the intersection of any two non-empty open subsets of Z is a non-empty subset of Z. This shows the equivalence of (i) and (ii). The equivalence of (ii) and (iii) follows from the fact that a subset of Z is dense if and only if it has non-empty intersection with every non-empty open set in Z. Let A be a subset of Z. It follows directly from the definition of the subspace topology on A that A is irreducible if and only if, given any closed sets F1 and F2 such that A ⊂ F1 ∪ F2 then either A ⊂ F1 or A ⊂ F2 . Now if F is any closed subset of Z then A ⊂ F if and only if A ⊂ F . It follows that A is irreducible if and only if A is irreducible. It follows immediately from Lemma 4.27 that a non-empty irreducible topological space is Hausdorff if and only if it consists of a single point. Lemma 4.28 Any irreducible topological space is connected. Proof A topological space Z is connected if and only if the only subsets of Z that are both open and closed are the empty set ∅ and the whole set Z. Thus suppose that the topological space Z were not connected. Then there would exist a non-empty proper subset U of Z that was both open and closed. Let V = Z \ U . Then U and V would be disjoint non-empty open sets. It would then follow from Lemma 4.27 that Z could not be irreducible. Lemma 4.29 Let V be an algebraic set, and let V1 be a proper algebraic subset of V . Then there exists f ∈ K[X1 , X2 , . . . , Xn ] such that f (x) = 0 for all x ∈ V1 but f 6∈ I(V ). Proof The inclusion V1 ⊂ V implies that I(V ) ⊂ I(V1 ). Now V = V (I(V )) and V1 = V (I(V1 )). Thus if V1 is a proper subset of V then I(V ) 6= I(V1 ), and hence there exists f ∈ I(V1 ) such that f 6∈ I(V ). Then f is the required polynomial. Proposition 4.30 A non-empty algebraic set V in An is irreducible (with respect to the Zariski topology) if and only if the ideal I(V ) is a prime ideal of K[X1 , X2 , . . . , Xn ].

96

Proof Suppose that the algebraic set V is irreducible. Let f and g be polynomials in K[X1 , X2 , . . . , Xn ] with the property that f g ∈ I(V ). Then V ⊂ V (f )∪V (g), since, given any point of V , one or other of the polynomials f and g must be zero at that point. Let V1 = V ∩ V (f ) and V2 = V ∩ V (g). Then V1 and V2 are algebraic subsets of V , and V = V1 ∪ V2 . Therefore either V = V1 or V = V2 , since the irreducible algebraic set V cannot be expressed as a union of two proper algebraic subsets. It follows that either f ∈ I(V ) or else g ∈ I(V ). Thus I(V ) is prime, by Lemma 4.24. Conversely, suppose that V is reducible. Then there exist proper algebraic subsets V1 and V2 of V such that V = V1 ∪V2 . It then follows from Lemma 4.29 that there exist polynomials f and g in K[X1 , X2 , . . . , Xn ] such that f (x) = 0 for all x ∈ V1 , g(x) = 0 for all x ∈ V2 , and neither f nor g belongs to I(V ). But then f (x)g(x) = 0 for all x ∈ V , since V = V1 ∪V2 , and hence f g ∈ I(V ). Thus the ideal I(V ) is not prime. Definition An affine algebraic variety is an irreducible algebraic set in An . Theorem 4.31 Every algebraic set in An can be expressed as a finite union of affine algebraic varieties. Proof Let C be the collection of all ideals I of K[X1 , X2 , . . . , Xn ] with the property that the corresponding algebraic set V (I) cannot be expressed as a finite union of affine varieties. We claim that C cannot contain any maximal element. Let I be an ideal of K[X1 , X2 , . . . , Xn ] belonging to C. Then the algebraic set V (I) cannot itself be an affine variety, and therefore there must exist proper algebraic subsets V1 and V2 of V such that V (I) = V1 ∪ V2 . Let I1 = I(V1 ) and I2 = I(V2 ). Then I(V (I)) ⊂ I1 and I(V (I)) ⊂ I2 , since V1 ⊂ V (I) and V2 ⊂ V (I). Also I ⊂ I(V (I)). It follows that I ⊂ I1 and I ⊂ I2 . Moreover V (I1 ) = V1 and V (I2 ) = V2 , since V1 and V2 are algebraic sets (see Lemma 4.15), and thus V (I1 ) 6= V (I) and V (I2 ) 6= V (I). It follows that I 6= I1 and I 6= I2 . Thus I is a proper subset of both I1 and I2 . Now V1 and V2 cannot both be finite unions of affine varieties, since V (I) is not a finite union of affine varieties. Thus one or other of the ideals I1 and I2 must belong to the collection C. It follows that no ideal I belonging to C can be maximal in C. But every non-empty collection of ideals of the Noetherian ring K[X1 , X2 , . . . , Xn ] must have a maximal element (see Proposition 4.5). Therefore C must be empty, and thus every algebraic set in An is a finite union of affine varieties, as required. We shall show that every algebraic set in An has an essentially unique representation as a finite union of affine varieties. 97

Lemma 4.32 Let V1 , V2 , . . . , Vk be algebraic sets in An , and let W be an affine variety satisfying W ⊂ V1 ∪ V2 ∪ · · · ∪ Vk . Then W ⊂ Vi for some i. Proof The affine variety W is the union of the algebraic sets W ∩ Vi for i = 1, 2, . . . , k. It follows from the irreducibility of W that the algebraic sets W ∩ Vi cannot all be proper subsets of W . Hence W = W ∩ Vi for some i, and hence W ⊂ Vi , as required. Proposition 4.33 Let V be an algebraic set in An , and let V = V1 ∪ V2 ∪ · · · Vk , where V1 , V2 , . . . , Vk are affine varieties, and Vi 6⊂ Vj for any j 6= i. Then V1 , V2 , . . . , Vk are uniquely determined by V . Proof Suppose that V = W1 ∪W2 ∪· · · Wm , where W1 , W2 , . . . , Wm are affine varieties, and Wi 6⊂ Wj for any j 6= i. Now it follows from Lemma 4.32 that, for each integer i between 1 and k, there exists some integer σ(i) between 1 and m such that Vi ⊂ Wσ(i) . Similarly, for each integer j between 1 and m, there exists some integer τ (j) between 1 and k such that Wj ⊂ Vτ (j) . Now Vi ⊂ Wσ(i) ⊂ Vτ (σ(i)) , But Vi 6⊂ Vi0 for any i0 6= i. It follows that i = τ (σ(i)) and Vi = Wσ(i) . Similarly Wj ⊂ Vτ (j) ⊂ Wσ(τ (j)) , and thus j = σ(τ (j)) and Wj = Vτ (j) . We deduce that σ: {1, 2, . . . , k} → {1, 2, . . . , m} is a bijection with inverse τ , and thus k = m. Moreover Vi = Wσ(i) , and thus the varieties V1 , V2 , . . . , Vk are uniquely determined by V , as required. Let V be an algebraic set, and let V = V1 ∪V2 ∪· · · Vk , where V1 , V2 , . . . , Vk are affine varieties, and Vi 6⊂ Vj for any j 6= i. The varieties V1 , V2 , . . . , Vk are referred to as the irreducible components of V .

4.10

Radical Ideals

Definition Let R be a unital commutative ring. An ideal I of R is said to be a radical ideal if every element x of R with the property that xm ∈ I for some natural number m belongs to I. Lemma 4.34 Every prime ideal of a unital commutative ring R is a radical ideal. Proof Let I be a prime ideal. Suppose that x ∈ R satisfies xm ∈ I. If m = 1 then we are done. If not, then either x ∈ I or xm−1 ∈ I, since I is prime. Thus it follows by induction on m that x ∈ I. Thus I is a radical ideal. 98

√ Lemma 4.35 Let I be an ideal of a unital commutative ring R, and let I m denote the set of all elements √ x of R with the property that x ∈ I√for some natural number m. Then I is a radical ideal of R. Moreover I = I if and only if I is a radical ideal of R. √ Proof Let x and y be elements of I. Then there exist natural numbers m and n such that xm ∈ I and y n ∈ I. Now m+n

(x + y)

=

m+n X i=0

m+n i



xi y m+n−i ,

(where x0 = 1 = y 0 ), and moreover, given any value of i between 0 and m + n, either i ≥ m or m + n − i ≥ n, so that either xi ∈ √ I or y m+n−i ∈√I. √ Therefore (x + y)m+n√∈ I, and thus x + y ∈ I. Also √ −x ∈ I and rx ∈ I for all r ∈ R. Thus I is an ideal of R. Clearly I is a radical ideal, and √ I = I if and only if I is a radical ideal. √ The ideal I is referred to as the radical of the ideal I. Lemma 4.36 Let Z be a subset of An . Then I(Z) is a radical ideal of the polynomial ring K[X1 , X2 , . . . , Xn ]. Moreover Z = V (I(Z)) if and only if Z is an algebraic set in An . Proof Note that if g and h are polynomials belonging to K[X1 , X2 , . . . , Xn ] which are zero throughout the set Z then the same is true of the polynomials g + h, −g and f g for all f ∈ K[X1 , X2 , . . . , Xn ]. Therefore I is an ideal of K[X1 , X2 , . . . , Xn ]. Moreover g m is identically zero on Z if and only if the same is true of g. Therefore the ideal I(Z) is a radical ideal. If Z = V (I(Z)) then Z is clearly an algebraic set. Conversely, if Z is an algebraic set then Z = V (S) for some subset S of K[X1 , X2 , . . . , Xn ], and therefore V (I(Z)) = V (I(V (S))) = V (S) = Z, by Lemma 4.15, as required. Lemma 4.37 Let S be a subset of the polynomial ring K[X1 , X√2 , . . . , Xn ], and let I be the ideal generated by S. Then V (S) = V (I) = V ( I), where √ I is the radical of the ideal I. Thus every algebraic set in An is of the form V (I) for some radical ideal I of K[X1 , X2 , . . . , Xn ].

99

Proof The ideal I(V (S)) of K[X1 , X2 , . . . , Xn ] contains the set S. Therefore √ I ⊂ I(V (S)), where I is the ideal generated by S. Moreover if f ∈ I then f m ∈ I for some natural number m, and thus f m ∈ I(V (S)). But I(V (S)) is a radical ideal (see Lemma 4.36). Therefore f ∈ I(V (S)). Thus √ S ⊂ I ⊂ I ⊂ I(V (S)). It follows that √ V (I(V (S))) ⊂ V ( I) ⊂ V (I) ⊂ V (S). But√ V (I(V (S))) = V (S) (see Lemma 4.15). Therefore V (S) = V (I) = V ( I), as required.

4.11

Commutative Algebras of Finite Type

Definition Let K be a field. A unital ring R is said to be a K-algebra if K ⊂ R, the multiplicative identity elements of K and R coincide, and ab = ba for all a ∈ K and b ∈ R. It follows from this definition that a unital commutative ring R is a Kalgebra if K ⊂ R and K and R have the same multiplicative identity element. Note that if L: K is a field extension, then the field L is a unital K-algebra. Definition Let K be a field, and let R1 and R2 be K-algebras. A ring homomorphism ϕ: R1 → R2 is said to be a K-homomorphism if ϕ(k) = k for all k ∈ K. Given any subset A of a unital commutative K-algebra R, we denote by K[A] the subring of R generated by K ∪ A (i.e., the smallest subring of R containing K ∪ A). In particular, if a1 , a2 , . . . , ak are elements of R then we denote by K[a1 , a2 , . . . , ak ] the subring of R generated by K ∪{a1 , a2 , . . . , ak }. If R = K[A] then we say that the set A generates the K-algebra R. Note that any element of K[a1 , a2 , . . . , ak ] is of the form f (a1 , a2 , . . . , ak ) for some polynomial f in k independent indeterminates with coefficients in K. Indeed the set of elements of R that are of this form is a subring of R, and is clearly the smallest subring of R containing K ∪ {a1 , a2 , . . . , ak }. Definition Let K be a field. A unital commutative ring R is said to be a Kalgebra of finite type if K ⊂ R, the identity elements of K and R coincide, and there exists a finite subset a1 , a2 , . . . , ak of R such that R = K[a1 , a2 , . . . , ak ].

100

Lemma 4.38 Let K be a field. Then every K-algebra of finite type is a Noetherian ring. Proof Let R be a K-algebra of finite type. Then there exist a1 , a2 , . . . , ak ∈ R such that R = K[a1 , a2 , . . . , ak ]. Now it follows from the Hilbert Basis Theorem that the ring K[X1 , X2 , . . . , Xk ] of polynomials in the independent indeterminates X1 , X2 , . . . , Xk with coefficients in K is a Noetherian ring (see Corollary 4.10). Moreover R ∼ = K[X1 , X2 , . . . , Xk ]/a, where a is the kernel of the homomorphism ε: K[X1 , X2 , . . . , Xk ] → R that sends f ∈ K[X1 , X2 , . . . , Xk ] to f (a1 , a2 , . . . , ak ). (Note that the homomorphism ε is surjective; indeed the image of this homomorphism is a subring of R containing K and ai for i = 1, 2, . . . , k, and is therefore the whole of R.) Thus R is isomorphic to the quotient of a Noetherian ring, and is therefore itself Noetherian (see Lemma 4.7). If K(α): K is a simple algebraic extension then K(α) is a K-algebra of finite type. Indeed K(α) is a finite-dimensional vector space over K (see Theorem 3.4). If a1 , a2 , . . . , ak span K(α) as a vector space over K then clearly K(α) = K[a1 , a2 , . . . , ak ].

4.12

Zariski’s Theorem

Proposition 4.39 Let K and L be fields, with K ⊂ L. Suppose that L: K is a simple field extension and that L is a K-algebra of finite type. Then the extension L: K is finite. Proof The field L is a K-algebra of finite type, and therefore there exist elements β1 , β2 , . . . , βm of L such that L = K[β1 , β2 , . . . , βm ]. Also the field extension L: K is simple, and therefore L = K(α) for some element α of K. Now, given any element β of L there exist polynomials f and g in K(x) such that g(α) 6= 0 and β = f (α)g(α)−1 . Indeed one may readily verify that the set of elements of L that may be expressed in the form f (α)g(α)−1 for some polynomials f, g ∈ K[X] with g(α) 6= 0 is a subfield of L which contains K ∪ {α}. It is therefore the whole of L, since L = K(α). It follows that there exist polynomials fi and gi in K[X] such that gi (α) 6= 0 and βi = fi (α)gi (α)−1 for i = 1, 2, . . . , m. Let e(x) = g1 (x)g2 (x) . . . , gm (x). We shall show that if the element α of L were not algebraic over K then every irreducible polynomial with coefficients in K would divide e(x), 101

Let p ∈ K[X] be an irreducible polynomial with coefficients in K, where p(α) 6= 0. Now L = K[β1 , β2 , . . . , βm ], and therefore every element of L is expressible as a polynomial in β1 , β2 , . . . , βm with coefficients in K. Thus there exists some polynomial Hp in m indeterminates, with coefficents in K, such that p(α)−1 = Hp (β1 , β2 , . . . , βm ). Let d be the total degree of H. One can readily verify that e(α)d Hp (β1 , β2 , . . . , βm ) = q(α), for some polynomial q(x) with coefficients in K. But then p(α)q(α) = e(α)d , and therefore α is a zero of the polynomial pq − ed . If it were the case that α were not algebraic over K then this polynomial pq − ed would be the zero polynomial, and thus p(x)q(x) = e(x)d . But it follows from Proposition 2.14 that an irreducible polynomial divides a product of polynomials if and only if it divides at least one of the factors. Therefore the irreducible polynomial p would be an irreducible factor of the polynomial e, and so would be an irreducible factor of one of the polynomials g1 , g2 , . . . , gm . We see therefore that if α were not algebraic over K then the polynomial e would be divisible by every irreducible polynomial in K[X]. But this is impossible, because a given polynomial in K[X] can have only finitely many irreducible factors, whereas K[X] contains infinitely many irreducible polynomials (Lemma 2.13). We conclude therefore that α must be algebraic over K. But any simple algebraic field extension is finite (Theorem 3.4). Therefore L: K is finite, as required. Lemma 4.40 Suppose that K ⊂ A ⊂ B, where A and B are unital commutative rings, and B is both a K-algebra of finite type and a finitely generated A-module. Then A is also a K-algebra of finite type. Proof There exist α1 , α2 , . . . , αm ∈ B such that B = K[α1 , α2 , . . . , αm ], since B is a K-algebra of finite type. Also there exist β1 , β2 , . . . , βn ∈ B such that B = Aβ1 + Aβ2 + · · · + Aβn , since B is a finitely generated A-module. Moreover we can P choose β1 = 1. But then there exist elements λqi of A such that αq = ni=1 λP qi βi for q = 1, 2, . . . , n. Also there exist elements µijk of A such that βi βj = nk=1 µijk βk for i, j = 1, 2 . . . , n. Let S = {λqi : 1 ≤ q ≤ m, 1 ≤ i ≤ n} ∪ {µijk : 1 ≤ i, j, k ≤ n},

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let A0 = K[S], and let B0 = A0 β1 + A0 β2 + · · · + A0 βn . Now each product βi βj is a linear combination of β1 , β2 , . . . , βn with coefficients µijk in A0 , and therefore βi βj ∈ B0 for all i and j. It follows from this that the product of any two elements of B0 must itself belong to B0 . Therefore B0 is a subring of B. Now K ⊂ B0 , since K ⊂ A0 and β1 = 1. Also αq ∈ B0 for q = 1, 2, . . . , m. But B = K(α1 , α2 , · · · αm ). It follows that B0 = B, and therefore B is a finitely-generated A0 -module. Now any K-algebra of finite type is a Noetherian ring (Lemma 4.38). It follows that A0 is a Noetherian ring, and therefore any finitely-generated module over A0 is Noetherian (see Corollary 4.6). In particular B is a Noetherian A0 -module, and therefore every submodule of B is a finitelygenerated A0 -module. In particular, A is a finitely-generated A0 -module. Let γ1 , γ2 , . . . , γp be a finite collection of elements of A that generate A as an A0 -module. Then any element a of A can be written in the form a = a1 γ1 + a2 γ2 + · · · + ap γp , where al ∈ A0 for l = 1, 2, . . . , p. But each element of A0 can be expressed as a polynomial in the elements λqi and µijk with coefficients in K. It follows that each element of A can be expressed as a polynomial in the elements λqi , µijk and γl (with coefficients in K), and thus A = K[T ], where T = S ∪ {γl : 1 ≤ l ≤ p}. Thus A is a K-algebra of finite type, as required. Theorem 4.41 (Zariski) Let L: K be a field extension. Suppose that the field L is a K-algebra of finite type. Then L: K is a finite extension of K. Proof We prove the result by induction on the number of elements required to generate L as a K-algebra. Thus suppose that L = K[α1 , α2 , . . . , αn ], and that the result is true for all field extensions L1 : K1 with the property that L1 is generated as a K1 -algebra by fewer than n elements (i.e., there exist elements β1 , β2 , . . . , βm of L1 , where m < n, such that L1 = K1 [β1 , β2 , . . . , βm ]). Let K1 = K(α1 ). Then L = K1 [α2 , α3 , · · · , αn ]. It follows from the induction hypothesis that L: K1 is a finite field extension (and thus L is a finitely-generated K1 -module). It then follows from Lemma 4.40 that K1 is a K-algebra of finite type. But the extension K1 : K is a simple extension. It therefore follows from Proposition 4.39 that the extension K1 : K is finite. Thus both L: K1 and K1 : K are finite extensions. It follows from the Tower Law (Proposition 3.1) that L: K is a finite extension, as required. 103

4.13

Hilbert’s Nullstellensatz

Proposition 4.42 Let K be an algebraically closed field, let R be a commutative K-algebra of finite type, and let m be a maximal ideal of R. Then there exists a surjective K-homomorphism ξ: R → K from R to K such that m = ker ξ. Proof Let L = R/m, and let ϕ: R → L denote the quotient homomorphism. Then L is a field (Lemma 4.21). Now m = ker ϕ and 1 6∈ m, and therefore ϕ|K 6= 0. It follows that m ∩ K is a proper ideal of the field K. But the only proper ideal of a field is the zero ideal (Lemma 2.4). Therefore m ∩ K = {0}. It follows that the restriction of ϕ to K is injective and maps K isomorphically onto a subfield of L. Let K1 = ϕ(K), and let ι: K → K1 be the isomorphism obtained on restricting ϕ: R → L to K. Then L: K1 is a field extension, and L is a K1 -algebra of finite type. It follows from Zariski’s Theorem (Theorem 4.41) that L: K1 is a finite field extension. But then L = K1 , since the field K1 is algebraically closed (Lemma 3.7). Let ξ = ι−1 ◦ ϕ. Then ξ: R → K is the required K-homomorphism from R to K. Theorem 4.43 Let K be an algebraically closed field, and let R be a commutative K-algebra of finite type. Let a be a proper ideal of R. Then there exists a K-homomorphism ξ: R → K from R to K such that a ⊂ ker ξ. Proof Every proper ideal of R is contained in some maximal ideal (Theorem 4.22). Let m be a maximal ideal of R with a ⊂ m. It follows from Proposition 4.42 that m = ker ξ for some K-homomorphism ξ: R → K. Then a ⊂ ker ξ, as required. Theorem 4.44 (Weak Nullstellensatz) Let K be an algebraically closed field, and let a be a proper ideal of the polynomial ring K[X1 , X2 , . . . , Xn ], where X1 , X2 , . . . , Xn are independent indeterminates. Then there exists some point (a1 , a2 , . . . , an ) of An (K) such that f (a1 , a2 , . . . , an ) = 0 for all f ∈ a. Proof Let R = K[X1 , X2 , . . . , Xn ]. Then R is a K-algebra of finite type. It follows from Theorem 4.43 that there exists a K-homomorphism ξ: R → K such that a ⊂ ker ξ. Let ai = ξ(Xi ) for i = 1, 2, . . . , n. Then ξ(f ) = f (a1 , a2 , . . . , an ) for all f ∈ R. It follows that f (a1 , a2 , . . . , an ) = 0 for all f ∈ a, as required. Theorem 4.45 (Strong Nullstellensatz) Let K be an algebraically closed field, let a be an ideal of the polynomial ring K[X1 , X2 , . . . , Xn ], and let f ∈

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K[X1 , X2 , . . . , Xn ] be a polynomial with the property that f (x1 , x2 , . . . , xn ) = 0 for all (x1 , x2 , . . . , xn ) ∈ V (a), where V (a) = {(x1 , x2 , . . . , xn ) ∈ An (K) : g(x1 , x2 , . . . , xn ) = 0 for all g ∈ a}. Then f r ∈ a for some natural number r. Proof Let R = K[X1 , X2 , . . . , Xn ], and let S denote the ring R[Y ] of polynomials in a single indeterminate Y with coefficients in the ring R. Then S can be viewed as the ring K[X1 , X2 , . . . , Xn , Y ] of polynomials in the n + 1 indeterminate indeterminates X1 , X2 , . . . , Xn , Y with coefficients in the field K. The ideal a of R determines a corresponding ideal b of S consisting of those elements of S that are of the form g0 + g1 Y + g2 Y 2 + · · · + gr Y r with g0 , g1 , . . . , gr ∈ a. (Thus the ideal b consists of those elements of the ring S that can be considered as polynomials in the indeterminate Y with coefficients in the ideal a of R.) Let f ∈ R be a polynomial in the indeterminates X1 , X2 , . . . , Xn with the property that f (x1 , x2 , . . . , xn ) = 0 for all (x1 , x2 , . . . , xn ) ∈ V (a), and let c be the ideal of S defined by c = b + (1 − f Y ). (Here (1 − f Y ) denotes the ideal of the polynomial ring S generated by the polynomial 1 − f (X1 , X2 , . . . , Xn )Y .) Let V (c) be the subset of (n + 1)dimensional affine space An+1 (K) consisting of all points (x1 , x2 , . . . , xn , y) ∈ An+1 (K) with the property that h(x1 , x2 , . . . , xn , y) = 0 for all h ∈ c. We claim that V (c) = ∅. Let (x1 , x2 , . . . , xn , y) be a point of V (b). Then g(x1 , x2 , . . . , xn ) = 0 for all g ∈ a, and therefore (x1 , x2 , . . . , xn ) ∈ V (a). But the polynomial f has the value zero at each point of V (a). It follows that the polynomial 1 − f Y has the value 1 at each point of V (b), and therefore V (c) = V (b) ∩ V (1 − f Y ) = ∅. It now follows immediately from the Weak Nullstellensatz (Theorem 4.44) that c cannot be a proper ideal of S, and therefore 1 ∈ c. Thus there exists a polynomial h belonging to the ideal b of S such that h − 1 ∈ (1 − f Y ). Moreover this polynomial h is of the form h(X1 , X2 , . . . , Xn , Y ) =

r X j=0

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gj (X1 , X2 , . . . , Xn )Y j ,

where g1 , g2 , . . . , gn ∈ a. Let g ∈ a be defined by g =

r P

gj f r−j . Now g − f r = g − f r h + f r (h − 1).

j=0

Also r

g−f h=

r X

gj f r−j (1 − f j Y j ) ∈ (1 − f Y ),

j=0

since the polynomial 1 − f j Y j is divisible by the polynomial 1 − f Y for all positive integers j. It follows that g−f r ∈ (1−f Y ). But the polynomial g−f r is a polynomial in the indeterminates X1 , X2 , . . . , Xn , and, if non-zero, would be of degree zero when considered as a polynomial in the indeterminate Y with coefficients in the ring R. Also any non-zero element of the ideal (1 − f Y ) of S is divisible by the polynomial 1 − f Y , and is therefore of strictly positive degree when considered as a polynomial in the indeterminate Y with coefficients in R. We conclude, therefore that g − f r = 0. But g ∈ a. Therefore f r ∈ a, as required.

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