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q. Then p cannot divide q  1, so q
;f.
1 mod p. By (i), G has a normal Sylow psubgroup, so G is not simple.
(3) By (i), G has a normal Sylow psubgroup P and a normal Sylow qsubgroup Q. Since P and Q are of prime order (p and q, respectively), they are cyclic. Ifx generates P and y generates Q, then xy = yx. [P and Q have trivial intersection because any member of the intersection has order dividing both p and q.] But then xy has order pq = 1G I. Thus G
= (xy).
126
Group Theory
We now look at the more complicated case I G in the next proof is very interesting.
I = p2q. The combinatorial argument
Proposition Suppose that the order of the finite group G is p2q, where p and q are distinct primes. Then G has either a normal Sylow psubgroup or a normal Sylow qsubgroup. Thus G is not simple.
Proof If the conclusion is false then np and nq are both greater than 1. By Sylow, nq divides p2, so nq = p or p2, and we will show that the second case leads to a contradiction .. A Sylow qsubgroup Q is of order q and is therefore cyclic. Furthermore, every element of Q except the identity is a generator of Q. Conversely, any element of order q generates a Sylow qsubgroup. Since the only divisors of q are 1 and q, any two distinct Sylow qsubgroups have trivial intersection. Thus the number of elements of G of order q is exactly n/q  I). If nq = p2, then the number of elements that are not of order q is p2q _ p2(q _ 1) = p2. Now let P be any Sylow psubgroup of G. Then I P I = p2, so no element of P can have order q (the orders must be 1, P or p2). Since there are only p2 elements of order unequal to q available, P takes care of all ofthem. Thus there cannot be another Sylow p subgroup, so np = 1, a contradiction. We conclude that nq must be p. Now by Sylow, nq == I mod q, hence p == 1 mod q, so p > q. But np divides q, a prime, so np = q. Since np == 1 mod p, we have q == 1 mod p, and consequently q > p. Our original assumption that both np and nq are greater than one has led inexorably to a contradiction. Problems
1.
Show that every group of order 15 is cyclic.
2.
If G/Z(G) is cyclic, show that G
3.
Show that for prime p, every group of order p2 is abelian.
4.
Let G be a group with
= Z(G), and therefore
I G I = pqr,
G is abelian.
where p, q and r are distinct primes and
(without loss of generality) p > q > r. Show that
IG I ~
1 + np(p  1) + n/q
 1) + n/r  I). 5.
Continuing Problem 4, if G is simple, show that np' nq and nr are all greater than I. Then show that np = qr' nq ~ p and nr ~ q.
6.
Show that a group whose order is the product of three distinct primes not simple.
7.
Let G be a simple group of order prm, where r ~ 1, m > 1, and the prime p does not divide m. Let n = np be the number of Sylow psubgroups of G. If H
IS
Group Theory
127
where P is a Sylow psubgroup of G, then [G : H] = n. Show that P cannot be normal in G (hence n > I), and conclude that I G I must divide n!.
= NG(P),
8.
If G is a group of order 250,000
= 2456, show that G is not simple.
COMPOSITION SERIES DEFINITION One Way tv Break Down a Group into Simpler Components is via a Subnormal Series
1 = Go :::! G 1 :::! ... :::! Gr
= G.
"Subnormal" means that each subgroup Gi is normal in its successor Gi+I. In a normal series, the Gi are required to be normal subgroups of the entire group G. For convenience, the trivial subgroup will be written as 1. Suppose that G i is not a maximal normal subgroup of G i + l , equivalently (by the correspondence theorem) Gi+/Gi is not simple. Then the original subnormal series can be refined by inserting a group Hsuch that Gi<JH<JGi+I. We can continue refining in the hope that the process will terminate (it always will ifG is finite). If all factors Gi+I/Gi are simple, we say that the group G has a composition series. [By convention, the trivial group has a composition series, namely itself.] The JordanH··older theorem asserts that if G has a composition series, the resulting composition length r and the composition factors Gi+/Gi are unique (up to isomorphism and rearrangement). Thus all refinements lead to essentially the same result. Simple groups therefore give important information about arbitrary groups; if G 1 and G2 have different composition factors, they cannot be isomorphic. Here is an example ofa composition series. Let 8 4 be the group of all permutations of {l, 2, 3, 4}, and A4 the subgroup of even permutations (normal in 8 4). Let V be the four group (normal in A 4 , in fact in 8 4 , by direct verification). Let Z 2 be any subgroup of Vof order 2. Then 1 <J Z 2 <J V <J A4 <J 84.
The proof of the JordanH··older theorem requires some technical machinery. LEMMA
1.
If K :::! H
~
G andfis a homomorphism on G, thenj(K) :::! j(ll).
2.
If K :::! H
~
G and N :::! G, then NK :::! NH.
3.
If A, B, C and D are subgroups of G with A :::! Band C :::! D, then A(B n C
:::! A(B n D), and by symmetry, C(D n A) :::! C(D n B).
128
Group Theory
4.
In (3), A(B n C) n B n D = C(D n A) n D n B. Equivalently, A(B n C) n D = C(D n A) n B.
Proof 1. For h E H, k E K, we havef{h)f(k)f{h)l
=f{hkh I )
E f{K).
2.
Letfbe the canonical map of G onto GIN. By (i) we have NKIN :s! NHIN. The result follows from the correspondence theorem.
3.
Apply (2) with G = B, N=A, K= B
4.
The two versions are equivalent because A(B n C) ::; Band C(D n A) ::; D. If x belongs to the set on the left, then x = ac for some a E A, c E B n C, and x also belongs to D. But x
n
C,H= B
n
D.
= c(c 1 ac) = ca* for some a*
E A
:s! B. Since xED
and c E C ::; D, we have a* E D, hence a* E DnA. Thus x = ca* E C(D n A), and since x = a c' with a E A ::; Band c E B n C ::; B, x E C(D n A) n B. Therefore the left side is a subset of the right side, and a symmetrical argument completes the proof.
The diagram below is helpful in visualizing the next result. B
D
A
C
To keep track of symmetry, take mirror images about the dotted line. Thus the group A will correspond to C, B to D, A(B n C) to C(D n A), and A(B n D) to C(D n B).
ZASSENHAUS LEMMA Let A, B, C and D be subgroups of G, with A :s! Band C :s! D. Then A(BnD) A(BnC)
rv
C(DnB) C(DnA)
Proof The quotient groups are welldefined. An element of the group on the left is of the form ayA(B n C), a E A, y E B n D. But ay = y(yl ay) = ya*, a* E A. Thus ayA(B n C) = ya* A(B n C) = yA(B n C). Similarly, an element of the right side is of the form zC(D n A) with zED n B = B n D. Thus if y, z E B n D, then yA(B n C) = zA(B n C) iffz1y E A(B n C) n B n D
this is equivalent to zly E C(D n A) n D n B iffyC(D n A) = zC(D n A).
Group Theory
129
Thus if h maps yA(B n C) to yC(D n A), th(.,{l h is a welldefined bijection from the left to the right side of Zassenhaus' equation. By definition of multiplication in a quotient group, h is an isomorphism. DEFINITION
If a subnormal series is refined by inserting H between G j and G j + I' let us allow H to coincide with G j or G j +I . If all such insertions are strictly between the "endgroups", we will speak of a proper refinement. Two series are equivalent if they have the same length and their factor groups are the same, up to isomorphism and rearrangement. SCHREIER REFINEMENT THEOREM
Let 1 = Ho ::S! HI ::S! ... ::S! Hr = G and 1 = Ko ::S! KI ::S! ... ::S! Ks = G be two subnormal series for the group G. Then the series have equivalent refinements. Proof Let Hi} = Hj(Hj+ I n
IS), Ki} = IS(IS+I
n HJ By Zassenhaus we have
Take A =Hj,B = H j+ I, C= IS ,D=IS+I. We can now construct equivalent refinements; the easiest way to see this is to look at a typical concrete example. The first refinement will have r blocks of length s, and the second will have s blocks of length r. Thus the length will be rs in both cases. With r = 2 and s = 3, we have 1 = Hoo ::S! HOI ::S! H02 ::S! H03 = HI = HIO ::S! HII ::S! H12 ::S! H I3 = H2 = G, 1 = Koo ::S! KIO ::S! K20 = KI = KOI ::S! Kil ::S! K2I = K2 = K02 ::S! K12 ::S! K22 = K3 = G. The corresponding factor groups are
HolHoo ~ Kl(/KOfY HoiHoi ~ KllKoI ' HolHo2 ~ KliKo2 HlllHIO ~ K201K IfY HIiHll ~ K2I IK ll , H13 IH12 ~ K2i K 12· (Notice the pattern; in each isomorphism, the first subscript in the numerator is increased by 1 and the second subscript is decreased by 1 in going from left to right. The subscripts in the denominator are unchanged.) The factor groups ofthe second series are a reordering of the factor groups of the first series. The hard work is now accomplished, and we have everything we need to prove the main result. JORDANHOLDER THEOREM
If G has a composition series S (in particular if G is finite), then any subnormal series R without repetition can be refined to a composition series. Furthermore, any two composition series for G are equivalent.
130
Group Theory
Proof Rand S have equivalent refinements. Remove any repetitions from the refinements to produce equivalent refinements Ro and So without repetitions. But a composition series has no proper refinements, hence So = S, proving the first assertion. If R is also a composition series, then Ro = R as well, and R is equivalent to S. Problems
1.
Show that if G has a composition series, so does every normal subgroup of G.
2.
Give an example of a group that has no composition series.
3.
Give an example of two nonisomorphic groups with the same composition factors, up to rearrangement.
Problems 49 will prove that the alternating group An is simple for all n ~ 5. (A 1 and A2 are trivial and hence not simple; A3 is cyclic of order 3 and is therefore simple.) In these problems, N stands for a normal subgroup of An. ~
4.
Show that if n
5.
Show that if N contains a 3cycle, then it contains all 3cycles, so that N = An.
6.
From now on, assume that N is a proper normal subgroup of An' and n Show that no permutation in N contains a cycle of length 4 or more.
7.
Show that no permutation in N contains the product of two disjoint 3cycles. Thus in view of Problems 4, 5 and 6, every member of tv is the product of an even number of disjoint transpositions.
8.
In Problem 7, show that the number of transpositions in a nontrivial member of N must be at least 4.
9.
Finally, show that the assumption that N contains a product of 4 or more disjoint transpositions leads to a contradiction, proving that N = 1, so that An is simple. It follows that a composition series for Sn is 1 <J An <J Sn.
3, then An is generated by 3cycles. ~
5.
10. A chief series is a normal series without repetition that cannot be properly refined to another normal series. Show that if G has a chief series, then any normal series without repetition can be refined to a chief series. Furthermore, any two chief series of a given group are equivalent. 11. In a composition series, the factor groups Gi+/Gi are required to be simple. What is the analogous condition for a chief series?
SOLVABLE AND NILPOTENT GROUPS Solvable groups are so named because of their connection with solvability of polynomial equatio. To get started, we need a property of subgroups that is stronger than normality.
Group Theory
131
DEFINITION A subgroup H of the group G is characteristic (in G) iffor each automorphism/ofG, j{lI) = H. Thus/restricted to His an automorphism of H. Consequently, if His characteristic in G, then it is normal in G. If follows from the definition that if H is characteristic in K and K is characteristic in G, then H is characteristic in G. Another useful result is the following. 1.
If H is characteristic in K and K is normal in G, then H is normal in G.
To see this, observe that any inner automorphism of G maps K to itself, so restricts to an automorphism (not necessarily inner) of K. Further restriction to H results in an automorphism of H, and the result follows. The commutator subgroup G of a group G is the subgroup generated by all commutators [x, y] = xyx1yl. (Since [x, y]l = [y, x], G consists of all finite products of commutators.)
2.
G' is characteristic in G.
This follows because any automorphism/maps a commutator to a commutator:j[x, y] = [/(x), j(y)]. 3.
G is abelian if and only if G' is trivial.
This holds because [x, y] = 1 iff xy = yx. 4. GIG' is abelian. Thus forming the quotient of G by G', sometimes called modding out by G', in a sense "abelianizes" the group. For G 'xG y = G yG'x iff G 'xy = G yx iff xy(yx)l E G' iff xyx1yl E G', and this holds for all x and y by definition of G'. 5.
If N g G, then GIN is abelian if and only if G'
~
N.
The proof of (4) with G' replaced by N shows that GIN is abelian iff all commutators belong to N, that is, iff G' ~ N. The process of taking commutators can be iterated: G(O) = G, G(l) = G', G(2) = (G)', and in general, G(i+l)
= (G(i)),. i = 0, 1. 2, ....
Since G(i+l) is characteristic in G(i), an induction argument shows that each G(i) is characteristic, hence normal, in G. The group G is said to be solvable ifG(r) = 1 for some r. We then have a normal series 1 = G(r) 1 and N(f3) > 1, so
N(a) =N(f3)
= p. If a = x + iy, thenp =x2 + T.
140
Number Theory
Conversely, ifp isan odd prime andp=x2+y, thenp is congruent to 1 mod 4. (Ifx is even, then x2 == 0 mod 4, and if x is odd, thenx2 == 1 mod 4. We cannot have x and y both even or both odd, since p is odd.) It is natural to conjecture that we can identify those primes that can be represented as x2 +/
dIY, where d is a negative integer, by working in the ring Z
[Jd] .But the Gaussian integers (d
=  1) form a Euclidean domain, in particular a unique factorization domain. On the other hand, unique factorization fails for d :s;  3, so the above argument collapses. Difficulties of this sort led Kummer to invent "ideal numbers", which later became ideals at the hands of Dede kind. We will see that although a ring of algebraic integers need not be a UFO, unique factorization of ideals will always hold.
INTEGRAL EXTENSIONS If ElF is a field extension and a E E, then a is algebraic over F iff a is a root of a polynomial with coefficients in F. We can assume ifwe like that the polynomial is monic, and this turns out to be crucial in generalizing the idea to ring extensions.
DEFINITION In this chapter, unless otherwise specified, all rings are assumed commutative. LetA be a subring of the ring R, and letx E R. We say that x is integral over A if x is a root ofa monic polynomial/with coefficients in A. The equationJCX) = 0 is called an equation o/integral dependence for x over A. If x is a real or com plex number that is integral over Z, then x is called an algebraic integer. Thus for every integer d,
Jd is an algebraic integer, as is any nth
root of unity. (The monic polynomials are, respectively, X2 + d and xn  1.) In preparation for the next result on conditions equivalent to integrality, note that A [x], the set of polynomials in x with coefficients inA, is anAmodule. (The sum oftwo polynomials is a polynomial, and multiplying a polynomial by a member ofA produces another polynomial over A.) Proposition
Let A be a subring of R, with x
E
R. The following conditions are equivalent:
1.
x is integral over A;
2.
The Amodule A [x] is finitely generated;
3.
x belongs to a subring B of R such that A ~ Band B is a finitely generated Amodule.
Proof 1. implies (2). Ifx is a root of a monic polynomial over A of degree n, then xn and all higher powers of x can be expressed as linear combinations of lower powers of x. Thus 1, x, x 2, , ... ,
xn I generate A [x] over A.
141
Number Theory
2.
Implies (iii). Take B =A[x].
3.
Implies (i). If~I' ... , ~n generate B over A, thenx~j is a linear combination of the ~j'
say X~j=
~~=ICijBj' Thus if~ is a column vector whose components are the ~j,Iis an nby
n identity matrix, and C = [c ij]' then
(xI
C)~
= 0,
and if we premultiply by the adjoint matrix of xl  C (as in Cramer's rule), we get
[det(xI 
C)]I~
°
= 0,
hence [det(xI  C)]b = for every b E B. Since B is a ring we may set b = 1 and conclude that x is a root of the monic polynomial det(XI C) inA[X]. We are going to prove a transitivity property for integral extensions, and the following result will be helpful.
LEMMA LetA be a subringof R, with xI' ... , xn E R. If xI is integral over A,x2 is integral over A[xd, ... , andxn is integral over A[xl' ... , xnd, thenA[xl' ... , x n] is a finitely generated Amodule. Proof The n = 1. Going from n  1 to n amounts to proving that ifA, Band C are rings, with C a finitely generated Bmodule and B a finitely generated A module, then C is a finitely generated Amodule. This follows by a brief computation: r
C= ~BYj' j=!
s
B=
r
s
~AZk so C= ~~AYjZk' k=1
j=1 k=!
Transitivity of Integral Extensions Let A, Band C be subrings of R. If C is integral over B, that is, each element of C is integral over B, and B is integral over A, then C is integral over A.
Proof Let x E C, with xn + bn_lxn_1 + ... + blx + bo= 0, b j E B. Then x is integral over A[b o' .... bnd. Each b j is integral over A, hence over A [b o' ... , bjd. A [b o' ... , bn_l . x] is a finitely generated Amodule. x is integral over A.
Definition If A is a subring of R, the integral closure ofA in R is the set Ac of elements of R that are integral over A. Note that As Ac because each a E A is a root of Xa. We say that A is integrally closed in R if Ac =A. Ifwe simply say that A is integrally closed without reference to R, we assume that A is an integral domain with quotient field K, and A is integrally closed in K.
Number Theon)
142
Ifx andy are integral over A,A[x, y] is a finitely generated Amodule. Since x+ y, xyand xy belong to this module, they are integral over A. The important conclusion is that
Ac is a subring of R containingA. If we take the integral closure of the integral closure, we get nothing new.
Proposition The integral closure Ac of A in R is integrally closed in R. Proof By definition,A c is integral over A. Ifx is integral over A c' then,x is integral over A, so that x E Ac'
We can identify a large class of integrally closed rings.
Proposition If A is a UFD, then A is integrally closed. Proof If x belongs to the quotient fieldK, then we can write x = alb where a, b
E
A, with
a and b relatively prime. Ifx is integral over A, then there is an equation of the form
°
(a/b)n + an_l(a/b)n1 + ... + al(a/b) + a o = A. Multiplying by bn, we have an + bc = 0, with C EA. Thus b divides all, which
with a i E cannot happen for relatively prime a and b unless b has no prime factors at all, in other words, b is a unit. Butthen x = ab I EA. We can now discuss one of the standard setups for doing algebraic number theory.
DEFINITION A numberfield is a subfield L of the complex numbers C such that L is a finite extension of the rationals Q. Thus the elements of L are algebraic numbers. The integral closure of Z in L is called the ring of algebraic integers (or simply integers) of L.
Problems
I.
Assume that B is a field, and let a be ,ft nonzero element of A. Then since aI there is an equation of the form (aI)n + cn_l(aI)n1 + ... + CiaI + Co
2.
=
E
B,
°
with c i E A. Show that aI E A, proving that A is a field. Now assume that A is a field, and let b be a nonzero element of B, A [b] is a finitedimensional vector space over A. Letfbe the Alinear transformation on this vector space given by multiplication by b, in other words,j{z) = bz, z E A[b]. Show thatfis injective.
143
Number Theon; 3.
Show thatJis surjective as well, and conclude that B is a field. In Problems 46, letA be a subring of B, with B integral over A. Let Q be a prime ideal of B and let P = Q (\ A.
4.
Show that P is a prime ideal of A, and that AlP can be regarded as a subring of BIQ.
5.
Show that BIQ is integral over AlP.
6.
Show that P is a maximal ideal of A if and only if Q is a maximal ideal of B.
Quadratic Extensions of the Rationals We will determine the algebraic integers of L = Q (Jd), where d is a squarefree integer (a product of distinct primes). The restriction on d involves no loss of generality; for example, Q(~)= Q( J). The minimal polynomial of Jd over Q isX2d, which has roots ±Jd.
The extension UQ is Galois, and the Galois group consists ofthe identity and the automorphism
cr(a + b Jd ) = a  b Jd, a, b
E
Q.
A remark on notation: To make sure that there is no confusion between algebraic integers and ordinary integers, we will use the term rational integer for a member of z. Lemma If a and b are rational numbers, then a + b Jd is an algebraic integer if and only if2a and 2 a  db 2 belong to z. In this case, 2b is also in z.
Proof Let x
= a + b Jd ' so that cr(x) = a  b Jd . Then x + cr(x) = 2a E Q and xcr(x) =
a2 + db 2 E Q. Now if x is an algebraic integer, then x is a root of a monic polynomialJ E Z [X]. Butj{cr(x» = cr(f(x» since cr is an automorphism, so cr(x) is also a root ofJand hence an algebraic integer. 2a and a2  db 2 are also algebraic integers, as well as rational numbers. Z is integrally closed, so 2a and a 2  db 2 belong to Z . The converse holds because a + b.Jd is a root of (X  a)2 = db 2, i.e., X2  2aX + a2  db 2 = o. Now if2a and a2  db 2 are rational integers, then (2a)2 +d(2b)2 = 4(a 2  db 2) E Z, so
d(2b)2 + IE. If2b 1. IE, thel1 its denominator would include a prime factor p, which would appear as p2 in the denominator of (2b)2. Multiplication of (2b)2 by d cannot cancel the p2 because d is squarefree, and the result follows. Corollary The set B of algebraic integers of Q ( Jd ), d squarefree, can be described as follows.
1.
If d
;d 1 mod 4, then B consists of all a + b Jd ' a, b E
IE;
144
Number Theory 2.
If d == 1 mod 4, then B consists of all
u
v r.
2 + i" d ,U, v E Z,
where
U
and v have the
same parity (both even or both odd). [Note that since d is squarefree, it is not divisible by 4, so the condition in (i) is d == 2 or 3 mod 4.]
Proof The algebraic integers are ofthe form
v r. 2U + 2" d where u, v E
2
Z and
4u  4dv
2
E Z,
i.e., u 2  dv 2 == 0 mod 4. It follows that u and v have the same parity. (The square of an even number is congruent to 0 and the square of an odd number to 1 mod 4.) Moreover, the "both .. u v odd" case can only occur when d == 1 mod 4. The "both even" case IS eqUIvalent to 2' 2 E Z, and the result follows. We can express these results in a more convenient form. We will show in that the set B of algebraic integers in any number field L is a free Z module of rank n = [L : Q]. A basis for this module is called an integral basis or Z basis for B.
THEOREM Let B be the algebraic integers of Q (Jd)' d squarefree.
;£
1.
If d
2.
If d == 1 mod 4, then 1 and
1 mod 4, then 1 and
Proof (i) 1 and irrational. (ii) 1 and v Jd
Jd
Jd
form an integral basis of B;
1
2 (1 + Jd) form an integral basis.
span B over Z, and they are linearly indegendent because
1
Jd
is
1
2 (l + Jd) are algebraic int~gers. To show that they span B, consider 2 (u +
), where u and v have the same parity. Then
uv
with 2 and v in Z . Finally, to show linear independence, assume that a, b
a+b[~(l +Jd)j= o. Then 2a + b + b Jd = 0, which forces a = b = O.
E
Z and
Number Theory
145
Problems
1.
Let L = Q (a.), where a. is a root of the irreducible quadratic X2 + bX + c, with b, c E
Q. Show that L
= Q ( Jd ) for some squarefree integer d. Thus the analysis of
this section covers all possible quadratic extensions of Q . 2.
Show that the quadratic extensions Q ( Jd ), d squarefree, are all distinct.
3.
Continuing Problem 2, show that in fact no two distinct quadratic extensions of Q are Q isomorphic.
Cyclotomic fields do not exhibit the same behavior. Let ron = ei2:rrln , a primitive nth root of unity. By a direct computation, we have w~n
=ron' and
= Q (roo).
4.
Show that if n is odd, then Q (ron)
5.
If X is an algebraic integer, show that the minimal polynomial of x over Q has coefficients in Z. (This will be a consequence of the general theory to be developed in this chapter, but it is accessible now without heavy machinery.) Consequently, an algebraic integer that belongs to Q in fact belongs to Z. (The minimal polynomial
ofr
E
Q over Q isXr.)
6.
Give an example of a quadratic extension of Q that is also a cyclotomic extension.
7.
Show that an integral basis for the ring of algebraic integers of a number field Lis, in particular, a basis for Lover Q.
NORMSAND TRACES DEFINITION If ElF is a field extension of finite degree n, then in particular, E is an ndimensional vector space over F, and the machinery of basic linear algebra becomes available. Ifx is any element of E, we can study the Flineartransfonnation m(x) given by multiplication by x, that is, m(x)y= xy. We define the norm and the trace of x, relative to the extension ElF, as
N[EIF](x) = det m(x) and llEIF](x) = trace m(x). We will write N(x) and T(x) if ElF is understood. If the matrixA{x) = [ai/x)] represents m(x) with respect to some basis for E over F, then the nonn ofx is the detenninant ofA (x) and the trace ofx is the trace ofA(x), that is, the sum of the main diagonal entries. The characteristic polynomial of x is defined as the characteristic polynomial ofthe matrix A (x), that is,
146
Number Theory
char [E/F](x) = det[XI A (x)] where I is an n by n identity matrix. If ElF is understood, we will refer to the characteristic i)D!ynomial o/x, written char(x).
Example Let E = C and F= 1R. A basis for cover 1R is {I, i} and, with x = a + bi' we have (a + bi)(l) = a(l) + b(i) and (a + bi)(i) =  b(l) + a(i).
Thus a A(a + bi) = [h
hj a
The norm, trace and characteristic polynomial of a + bi are N(a + hi) = a 2 + b2, T(a + bi) = 2a, char (a + bi) = X2  2aX + a 2 + b 2.
The computation is exactly the same if E = Q (i) and F= Q. Notice that the coefficient of
Xis minus the trace and the constant term is the norm. In general, it follows from the definition of characteristic polynomial that char (x)
= xn 
T(x)xn I + ... + (+ l)n N(x).
[The only terms muItiplying.xn 1 in the expansion of the determinant are Gjj(x), i = 1, ... , n. Set X = 0 to show that the constant term of char(x) is (1 )n det A (x).]
LEMMA If E is an extension of F and x to F. If a E F, then N(a)
= an,
T(a)
E
= na,
E, then N(x), T(x) and the coefficients of char(x) belong
and char (a)
= (X + a)n.
Proof The first assertion follows because the entries of the matrix A(x) are in F. The second statement holds because if a E F, the matrix representing multiplication by a is al.
It is natural to look for a connection between the characteristic polynomial of x and the minimal polynomial ofx over F.
Proposition char [E/F](x) = [min (x,
F)Y
where r = [E : F(x)]. Proof First assume that r = 1, so that E = F(x). By the CayleyHamilton theorem, the linear transformation m(x) satisfies char(x), and since m(x) is mUltiplication by x, x itselfis a root
147
Number Theon)
of char(x). Thus min(x, F) divides char(x). But both polynomials have degree n, and the result follows. In the general case, letYI' ... , Y s be a basis for F(x) over F, and letz l, ... , zr be a basis for E over F(x). Then the Y;Zj form a basis for E over F. LetA = A (x) be the matrix representing mUltiplication by x in the extensionF(xYF, sothatXYj= Lk ak;yk, andx(y;z) = Lk ak; (ykz). Order the basis for ElF aSYlzl' Y2 z l' ... , YsZl;Yl z 2' Y2 z 2' ... , Y~2; ···;Ylz" Y2z" ... , Y~r. Then m(x) is represented in ElF as
A 0
o
A
0 0
o
0
A
Thus char [EIF](x) = [det(XIAW, which by the r= 1 case coincides with [min(x, F)Y. Corollary
Let [E: F] = n, and [F(x) : F] = d. Letx l , ... , xd be the roots of min (x, F) in a splitting field (counting mUltiplicity). Then ]nld
d
N(x)= [Ux/
d
,T(x) = ;~x/
and d
char (x) =
nld
IU
(X x;)
Proof The formula for the characteristic polynomial. The norm is ( l)n times the constant term of char (x), hence is
(I )"(_1)"
[u xl
d
Finally, if min (x, F) =.xrJ + adI.xzI1 + ... + alX + ao' then the coefficient ofxnI in [min(x, F)]nld is
; ad_I =  ;
L:I x; . Since the trace is the negative of this coefficient, the result
follows. If E is a separable extension ofF, there are very useful alternative expressions for the trace and norm.
Number Theon)
148
Proposition
Let E/F be a separable extension of degree n, and let 0"1' ... , O"n be the distinct monomorphisms of E into an algebraic closure of E, or equally well into a normal extension L of F containing E. Then T[E I F](x) =
n
n
i=1
i=l
L: (Ji(X) and N[E/FJ(x) = IT (Ji(x).
Consequently, T(ax + by) = aT(x) + bT (y) and N(xy) = N(x)N(y) for x, y E E, a, b E F. Proof Each of the d distinctFembeddings 't j of F(x) into L takes x into a unique conjugatex j , n and extends to exactly d = [E: F(x)] Fembeddings of E into L, all of which also take x to xi" Thus
and n
d
IT (Jj(x) IT T;(X) i=l
n/d
=N(x)
i=1
The linearity of Tand the multiplicativity ofNhold without any assumption of separability, we have m(ax + by) = am (x) + bm(y) and m(xy) = m(x) ° m(y). Corollary
If F
~
K
~
E, where E/F is finite and separable, then
T[E/F] = T[KlF]OT[EIK] and N[EIF] = N[K/F] ON [ElK]. Proof Let 0"1' ... , O"n be the distinct Fembeddings of K into L, and let 't 1, ... , 't m be the distinct Kembeddings of E into L, where L is the normal closure of E over F. UF is Galois, each mapping O"j and 't. extends to an automorphism of L. Therefore it makes sense to allow the . mappings to be co'mposed.
Now each O"{). is an Fembedding of E into L, and the number of mappings is m!, = [E: K][K
: FJ = [E : FJ. Furthermore, the O"j'tJ are distinct when restricted to E. For if O"j'tj = O"k't/ on E,
149
Number Theory
hence on K, then CJi = 'tk on K (because CJi=: 't[= the identity on K). Thus i = k, so that 't. = 't[ on E. But then} = I. lIKlF](lIElK])(x) = lltJF](x). The norm is handled the same way, with sums replaced by products.
Corollary If ElF is a finite separable extension, then llEIF](x) cannot be 0 for all x
Proof If:n:x) = 0 for all x, .L:;=lo/(x)
E
E.
= 0 for all x. This contradicts Dedekind's lemma.
If ElF is finite and separable, then the "trace form" [the bilinear form (x, y) ~ lIElF](xy)] is nondegenerate, i.e., if T(xy) = 0 for all y, then x = O. For if x =1= 0, :n:xo) =1= 0, and T(xy) = 0 for all y, choose y so that xy = Xo to reach a contradiction.
SETUP FOR ALGEBRAIC NUMBER THEORY Let A be an integral domain with quotient field K, and let L be a finite separable extension of K. Let B be the set of elements of L that are integral over A, that is, B is the integral closure of A in L. The diagram below summarizes all the information.
LB
I
I
KA In the most important special case, A = Z, K = Q, L is a number field, and B is the ring of algebraic integers of L.
Proposition If x E B, then the coefficients of char [L/K](x) and min (x, K) are integral over A. In particular, llVK] (x) and N[L/K] (x) are integral over A. IfA is integrally closed, the coefficients belongtoA.
Proof The coefficients ofmin(x, K) are sums of products of the roots Xi' it suffices to show that the Xi are integral over A. Each Xi is a conjugate ofx over K, there is a Kisomorphism 't : K(x) ~ K(x) such that 'tlx) = Xi' Ifwe apply 't to an equation of integral dependence for x i i over A, we get an equation ofintegral dependence for Xi over A. Problems
1.
If E = Q ( .Jd ) and x = a + b.Jd
2.
If E = Q (9) where 9 is a root of the irreducible cubic X3  3X + 1, find the norm
E
E, find the norm and trace of x.
and trace of 92 . 3.
Find the trace of"the primitive 6th root of unity E in the cyclotomic extension Q6'
150
Number Theon) We will now prove Hilbert's Theorem 90: If ElF is a cyclic extension with [E : F] = nand
Galois group G = {l, cr, ... , crn I } generated by cr, and x
= 1 if and only ifthere exists Y E
1.
N(x)
2.
T(x) = 0 if and only ifthere exists Z
3.
E
E
E, then
E such that x =y/cr(y);
E such that x = Z  cr(z).
Prove the "if' parts of (i) and (ii) by direct computation.
By Dedekind's lemma, 1, cr, cr2, ... , ~I are linearly independent over E, so 1 + xcr + xcr(x)cr2 + ... + xcr(x) ... ~2(x)~1
is not identically 0 on E. 4.
If T(x) = 0, show that co  cr(co) = xT(u).
5.
If z = w/T(u), show that z  cr(z)
6.
In Hilbert's Theorem 90, are the elements y and z unique?
7.
Let e be a root of.x4  2 over Q. Find the trace over Q ofe,
8.
Continuing Problem 9, show that
= x, proving the "only if' part of (ii).
J3
e2, e 3 and J3 e,.
cannot belong to Q [e].
DISCRIMINANT We have met the discriminant of a polynomial in connection with Galois theory. There is also a discriminant in algebraic number theory. The two concepts are unrelated at first glance, but there is a connection between them. We assume the basic AKLB setup, with n = [L : K].
DEFINITION The discriminant ofthe ntuple x
= (xI' ... , x n ) of elements of Lis
D(x) = det(l1VK](xf). Thus we form a matrix whose if element is the trace of Xfj , and take the determinant ofthe matrix. D(x) belongs to K. If the Xi are in B, D(x) is integral over A, hence belongs to A ifA is integrally closed. The discriminant behaves quite reasonably under linear transformation:
LEMMA· IfY = Cx, where C is an n by n matrix over K and x and yare ntuples written as column vectors, then D(y) = (det C) 2D(x).
Number Theory
151
Proof The trace ofy'ys is
1 L CnT(x,xj )CSj
T[L CnCSjXiXj = ',j
',j
hence
(T(YrYs» = C(T(xf)C' where C' is the transpose of C. The result follows upon taking determinants, Here is an alternative expression for the discriminant.
LEMMA Let 0'1' ,'" O'n be the Kembeddings of L into an algebraic closure of L. Then D(x) = [det( O'i(~)]2. Thus we form the matrix whose if element is O'/{x), take the determinant and square the result.
Proof T(x .) = I>'k(XiXj ) = L:O"k (x;)O"k (xj ) fJ
k
k
so ifC is the matrix whose
if entry is O'/(x), then
(T(xf) = C'C and again the result follows upon taking determinants. The discriminant "discriminates" between bases and nonbases, as follows. Proposition
If x = (x I'
Proof If
' ",
x n ), then the xi form a basis for Lover K if and only if D(x) :;:. O.
L: j CjX
j
= 0, with the cj
E
K and not all 0, then}
L: j CjO"i(X) = 0 for all i, so
the columns of the matrix B = (O'i(x) are linearly dependent. Thus linear dependence of the xi implies that D = O. Conversely, assume that the xi are linearly independent (and therefore a basis since n = [L : K]). If D = 0, then the rows of B are linearly dependent, so for some c i E K, not all O,wehave L,ciO";(Xj ) =Oforall}.Sincethex}formabasis,wehave L,c,O";(u) =0 for all U E
L, so the monomorphisms O'i are linearly dependent. This contradicts Dedekind's lemma.
152
Number Theory
We now make the connection between the discriminant defined above and the discriminant of a polynomial defined previously. Proposition
Assume that L = K(x), and let/be the minimal polynomial ofx over K. Let D be the discriminant of the basis 1, x, x2, ... , x n I for Lover K. Then D is the discriminant of the polynomialf
Proof Let xI' ... ,
XII
be the roots of/in a splitting field, with xI = x. Let (Jj be the K
embedding that takes X to xi' i = 1, ... , n. Then (Jj(x) = x/ ,D is the square of the determinant of the matrix nl
xl
Xf
xl
X2
xi
X2
Xn
x2 n
nI Xn
nl
and the result follows from the formula for a Vandermonde determinant. Corollary
Under the hypothesis,
(l)(~) N[LI K](f'(x))
D=
wherej' is the derivative off
Proof Let a
= (_l/~L
D=
II (Xi _Xj)2 = all (Xi Xj )= allll (Xi Xj). I<j
I""j
i
i""j
ButftX) = (X Xl) ... (X xn ), so j'(x.)
=
L:ll (X Xj) k
1
j""k
withXreplaced by xi" WhenXis replaced by Xj' only the k= iterm is nonzero, hence j'(X.) 1
= D(xiXj )"'"
Consequently, n
D
= all/'(xi )· i=1
).
153
Number Theory
But
so, D
= aN[L/K](f'(x)).
The discriminant of an integral basis for a number field has special properties. We will get at these results by considering the general AKLB setup, adding some additional conditions as we go along.
LEMMA There is a basis for L/K consisting entirely of elements of B.
Proof Let xI' ... , xn be a basis for Lover K. Each Xj is algebraic over K, and therefore satisfies a polynomial equation of the fonn
with am i= 0 and the a j EA. (Initially, we only have a j
E
K, but then a j is the ratio of two
elements in A, and we can form a common denominator.) Multiply the equation by a:;:I to obtain an equation of integral dependence for Yj = arrfi over A. The Yi form the desired basis.
THEOREM Suppose we have a nondegenerate symmetric bilinear form on an ndimensional vector space V, written for convenience using inner product notation (x, y).lfx l , ... , xn is any basis for V, then there is a basisYI' ... , Yn for V, called the dual basis referred to V, such that
!
1,
(Xi'
YJ') = fly = 0,
i = j; .
l:;t!:
. J.
This is a standard (and quite instructive) result in linear algebra, and it will be developed in the exercises.
THEOREM If A is a principal ideal domain, thenB is a free Amodule ofrankn.
Proof The trace is a nondegenerate symmetric bilinear fonn on the ndimensional vector space Lover K. A is integrally closed, the trace of any element of L belongs to A. Now let x I' .... xn be any basis for Lover K consisting of algebraic integers, and let Y I' ... , Yn be the dual basis referred to L.lfz E B, then we can writez= 2:=~=lajYJ with aj trace ofx;Z belongs toA, and we also have
E
K. We know that the
154
Number Theory
Thus each a i belongs toA, so that B is anAsubmodule of the free Amodule EBJ=1 AYj. B is a free Amodule of rank at most n. B contains a basis for Lover K, and if we wish, we can assume that this basis isx l , ... , x n . Then B contains the free Amodule EBJ=1 Ax), so the rank of
B as an Amodule is at least n, and hence exactly n.
Corollary The set B of algebraic integers in any number field L is a free Z module of rank n = [L :
Q]. Therefore B has an integral basis. The discriminant is the same for every integral basis; it is known as the field discriminant. Proof Take A = Z to show that B has an integral basis. The transformation matrix C between two integral bases is invertible, and both C and C I have rational integer coefficients. Take detenninants in the equation CC I = Ito conclude that detC is a unit in Z . Therefore det C =±], all integral bases have the same discriminant. Problems
Let x I' ... , xn be a basis for the vector space V, and let (x, y) be a nondegenerate symmetric bilinearfonn on V.
l.
For any y E V, the mapping x ~ (x. y) is a linear fonn 1(1), i.e., a linear map from V to the field of scalars. Show that the linear transformation y ~ l(y) from V to V*, the dual space of V (i.e., the space of all linear fonns on V), is injective.
2.
Show that any linear fonn on V is l(y) for some Y
3.
Let.li, ... , in be the dual basis corresponding to x I' ... , x n' Thus each/; belongs to
4.
V* (not V) and~(xi) = basis referred to V. Show that xi =
oij'
E
V.
If.0 = l(y), show that YI' ... , Y n is the
re~uired
dual
L~=I(Xi,X)Yj' Thus in order to compute the dual basis referred
to Vin tenns of the original basis, we must invert the matrix ((xi' x).
5.
A matrix C with coefficients in Z is said to be unimodular if its detenninant is ±] . Show that C is unimodular if and only if C is invertible and its inverse has coefficients in z.
6.
Show that the field discriminant of the quadratic extension Q ( Jd), d squarefree, is
Number Theory
155 D=
14d d
ifd;E'lmod4; ifd= Imod4.
7.
Let x" ... , xn be arbitrary algebraic integers in a number field, and consider the detenninant of the matrix (aj(x)). The direct expansion of the detenninant has n! terms. Let P be the sum of those terms in the expansion that have plus signs in front of them, and N the sum of those terms prefixed by minus signs. Thus the discriminant D of (x" ... , xn ) is (P  N)2. Show that P + Nand PN are fixed by each a j , and deduce that P + Nand PN are rational numbers.
8.
Continuing Problem 7, show that P + Nand PN are rational integers.
9.
Continuing Problem 8, prove Stickelberger's theorem: D == 0 or 1 mod 4.
10.
Let L be a number field of degree n over Q, and let Y" ... , Yn be a basis for Lover
Q consisting of algebraic integers. Let x" ... , xn be an integral basis. Show that if the discriminant D(y), ... , Y n ) is squarefree, then each xi can be expressed as a linear combination of the Yj with integer coefficients. ] 1.
Continuing Problem ]0, show that if D(y" ... , Y n) is squarefree, theny, .... , Y n is an integral basis.
] 2.
Is the converse of the result of Problem 11 true?
13.
In the standard AKLB setup, show that L is the quotient field of B.
NOETHERIAN AND ARTINIAN MODULES AND RINGS DEFINITION In this section, rings are not assumed commutative. LetMbe an Rmodule, and suppose that we have an increasing sequence of submodules M, :::; M2 :::; M3 :::; ... , or a decreasing sequence M, 2: M2 2: M3 2: .... We say that the sequence stabilizes iffor some t, Mt = Mt+' =Mt+2 = .... The question of stabilization of sequences of sub modules appears in a fundamental way in many areas of abstract algebra and its applications. The module M is said to satisfy the ascending chain condition (ace) if every increasing sequence of submodules stabilizes; M satisfies the descending chain condition (dec) if every decreasing sequence of submodules stabilizes.
Proposition The following conditions on an Rmodule Mare equivalent, and define a Noetherian module: ].
M satisfies the acc;
2.
Every nonempty collection of submodules of Mhas a maximal element (with respect to inclusion).
156
Number Theon)
The following conditions on Mare equivalent, and define an Artinian module:
1.
M satisfies the dcc;
2.
Every nonempty collection of submodules of Mhas a minimal element.
Proof Assume (1), and let S be a nonempty collection of submodules. Choose MI
E
S.
If MI is maximal, we are finished; otherwise we have MI < M2 for some M2 E S. Ifwe continue inductively, the process must terminate at a maximal element; otherwise the acc would be violated. Conversely, assume (2), and let MI ~ M2 ~ .... The sequence must stabilize; otherwise {MI' M 2, ... } would be a nonempty collection of submodules with no maximal element. The proofis exactly the same in the Artinian case, with all inequalities reversed. There is another equivalent condition in the Noetherian case.
Proposition M is Noetherian iff every submodule of M is finitely generated.
Proof If the sequence MI
~
M2
~
... does not stabilize, let N =
U~l Mr'
Then N is a
submodule of M, and it cannot be finitely generated. For if xI' ... , Xs generate N, then for sufficiently large t, all the xi belong to Mr But then N ~~~ MI+ I ~ ... ~N, so~=Mt+1 = ... Conversely, assume that the acc holds, and letN ~ M. IfN:t:O, choose xI EN. If Rxl =N, then N is finitely generated. Otherwise, there exists x 2 ~ Rx l • If xl and x 2 generate N, we are
Rx I + Rx2 . The acc forces the process to terminate at
finished. Otherwise, there exists x3
~
some stage t, in which case xI'
generate N.
... , XI
The analogous equivalent condition in the Artinian case is that every quotient module MIN isfinitely cogenerated, that is, if the intersection of a collection of submodules of MIN is 0, then there is a finite subcollection whose intersection is 0.
DEFINITION A ring R is Noetherian [resp. Artin ian] if it is Noetherian [resp. Artinian] as a module over itself. Ifwe need to distinguish between R as a left, as opposed to right, Rmodule, we will refer to a left Noetherian and a right Noetherian ring, and similarly for Artinian rings. Examples
1.
If F is a field, then the polynomial ring F[XJ is Noetherian (another special case of Example 1) but not Artinian. A descending chain of ideals that does not stabilize is (X) :::> (X2) :::> (X3) ::;) ....
Number Theon;
2.
157
The ring FIXl, X2, ... ] of polynomials over F in infinitely many variables is neither Artinian nor Noetherian. A descending chain of ideals that does not stabilize is constructed as in Example 3, and an ascending chain of ideals that does not stabilize is
Remark
The following observations will be useful in deriving properties of Noetherian and Artinian modules. IfN ::; M, then a submodule L of M that contains N can always be written in the form K + N for some submodule K. (K = L is one possibility.) By the correspondence theorem,
(K l +N)IN= (K2+N)/Nimplies Kl +N=K2 +Nand (K l + N)IN ::; (K2 + N)IN implies Kl + N ::; K2 + N Proposition
If N is a submodule ofM, then Mis Noetherian [resp. Artinian] if and only ifNand MIN are Noetherian [resp. Artinian]. Proof Assume M is Noetherian. Then N is Noetherian since a submodule of N must also be a submodule of M. An ascending chain of submodules of J..1IN looks like (Ml + N)IN ::; (M2 + N)IN ::; .... But then the M j + N form an ascending sequence of submodules of M, which must stabilize. Consequently, the sequence (M; + N)IN, i = 1, 2, ... , must stabilize.
Conversely, assume that N and MIN are Noetherian, and let Ml ::; M2 ::; ... be an increasing sequence of submodules of M. Take i large enough so that both sequences {M; ( l N} and {Mj + N} have stabilized. If?, E M i+ t , then x + N E Mj + 1 + N = M j + N, so x =y + z where Y E M j and ZEN. Thus x  Y E M j+ t ( l N = M j ( l N, and since Y E M j we have x E M j as well. Consequently, Mi = Mi+I and the sequence of M;' s has stabilized. The Artinian case is handled by reversing inequalities (and interchanging indices i and i + 1 in the second half of the proof). Corollary
If M t , ... , Mn are Noetherian [resp. Artinian] Rmodules, then so is M t EB M2 EB ... EB Mn' Proof It suffices to consider n = 2 (induction will take care of higher values of n). The submodule N = M t of M = M t EB M2 is Noetherian by hypothesis, and MIN ~ M2 is also Noetherian (apply the first isomorphism theorem to the ~tural projection of M onto M2). Mis Noetherian. The Artinian case is done the same way. Corollary
If M is a finitely generated module over the Noetherian [resp. Artinian] ring R, then Mis N .)etherian [resp. Artinian]
158
Number Theory
Proof Mis a quotient of a free module L offinite rank. Since L is the direct sum of a finite number of copies of R.
Ascending and descending chains of submoduies are reminiscent ofnormal and subnormal series in group theory, and in fact we can make a precise connection. DEFINITION
A series of length n for a module M is a sequence of the form
M = Mo
~ lv/l ~ ... ~
Mn = O.
The series is called a composition series if each factor module MIM;+ I is simple. Thus we are requiring the series to have no proper refinement. Two series are equivalent if they have the same length and the same factor modules, up to isomorphism and rearrangement. By convention, the zero module has a composition series, namely {O} itself. JORDANHOLDER THEOREM FOR MODULES
If M has a composition series, then any two composition series for M are equivalent. Furthermore, any strictly decreasing sequence of submodules can be refined to a composition series. Proof The development of the lordanH"older theorem for groups can be taken over verbatim if we change multiplicative to additive notation. In particular, we can reproduce the preliminary lemma, the Zassenhaus lemma, the Schreier refinement theorem, and the lordanH'older Theorem. We need not worry about normality of subgroups because in an abelian group, all subgroups are normal. As an example of the change in notation, the Zassenhaus lemma becomes A+(BnD) '" C+(DnB) A+(BnC) = C+(DnA)'
This type ofproofcan be irritating, because it forces readers to look at the earlier development and make sure that everything does carry over. A possible question is "Why can't a composition series S of length n coexist with an infinite ascending or descending chain?" But if such a situation occurs, we can form a series T for M of length n +1. By Schreier, Sand T have equivalent refinements. 'Since S has no proper refinements, and equivalent refinement have the same length, we have n ~ n + 1, a contradiction. We can now relate the ascending and descending chain conditions to composition series. THEOREM
The Rmodule Mhas a composition series if and only if Mis both Noetherian and Artinian.
159
Number Theon)
Proof Assume thatMis Noetherian and Artinian. Assuming (without loss of generality) that M E 0, Mo = Mhas a maximal proper submodule MI' Now MI is Noetherian, so if M J "* 0, then M J has a maximal proper submodule M 2 . Continuing inductively, we must reach 0 at some point because Mis Artinian. By construction, each MIM;+J is simple, and we have a composition series for M. Here is a connection with algebraic number theory. Proposition
In the basic AKLB setup, assume that A is integrally closed. IfA is a Noetherian ring, then so is B. In particular, the ring of algebraic integers in a number field is Noetherian.
Proof B is a submodule ofa free Amodule M offinite rank. (The assumption that A is a PID is used to show that A is integrally closed, and we have this by hypothesis. The PID assumption is also used to show that B is a free Amodule, but we do not need this in the present argument.) Mis Noetherian, B is a Noetherian Amodule. An ideal of B is, in particular, anAsubmodule of B, hence is finitely generated over A and therefore over B. Thus B is a Noetherian
ring. Problems
1.
Let p be a fixed prime, and let A be the abelian group of all rational numbers alp"', n = 0, 1, ... , a E Z, where all calculations are modulo 1, in other words, A is a subgroup of Q / Z . Let An be the subgroup {O, that A is not a Noetherian Z module.
liP"',
2/p"', ... , (p" + 1YP"'}. Show
2.
Continuing Problem 1, if B is a proper subgroup of A, show that B must be one of the An' Thus A is an Artinian Z module. [This situation cannot arise for rings, where Artinian implies Noetherian.]
3.
If V is a vector space, show that V is finitedimensional iff V is Noetherian iff V is Artinian iff V has a composition series.
4.
Define the length of a module M [notation 1(M)] as the length of a compos;tion series for M. (If M has no composition series, take 1(M) = 00.) Suppose that we have a short exact sequence
O+N1M~M/N+O Show that 1(M) is finite if and only if leN) and [(MIN) are both finite. 5.
Show that I is additive, that is, 1(M) = leN) + 1(M/N).
6.
Let S be a subring of the ring R, and assume that S is a Noetherian ring. If R is finitely generated as a module over S, show that R is also a Noetherian ring.
160
Number Theory
7. 8.
Let R be a ring, and assume that the polynomial ring R[X] is Noetherian. Does it follow that R is Noetherian? Show that a module Mis Artinian if and only if every quotient module MIN is finitely cogenerated. FRACTIONAL IDEALS
Our goal is to establish unique factorization of ideals in a Dedekind domain, and to do this we will need to generalize the notion ofideal. First, some preliminaries. DEFINITION
2:;
If II' ... , In are ideals, the product II '" In is the set of all finite sums ali a2; ...ani , where aki E Ik , k = 1, ... , n. It follows from the definition that the product is an ideal contained in each ~. LEMMA
If P is a prime ideal that contains a product II ... In of ideals, then P contains ~ for some j. Proof Ifnot, let aj E ~ \ P,j = 1, ... , n. Then a l ... an belongs to II'" In ~ P, and since P is prime, some aj belongs to P, a contradiction.
Proposition If I is a nonzero ideal of the Noetherian integral domain R, then I contains a product of nonzero prime ideals. Proof Assume the contrary. If S is the collection of all nonzero ideals that do not contain a product of nonzero prime ideals, then since R is Noetherian, S has a maximal element J, and J cannot be prime because it belongs to S. Thus there are ele.nents a, b E R with a ¢ J, b ¢ J, and ab E J. By maximalityofJ, the idealsJ+ Ra andJ+ Rb each contain a product of nonzero prime ideals, hence so does (J + Ra)(J + Rb) ~ J + Rab =J. This is a contradiction. [Notice that we must use the fact that a product of nonzero ideals is nonzero, and this is where the hypothesis that R is an integral domain comes in.]
Corollary If I is an ideal ofthe Noetherian ring R (not necessarily an integral domain), then I contains a product of prime ideals. Proof Ideals in the ring of integers are ofthe form n Z , the set of multiples of n. A set of the form
2"3 z is not an ideal because it is not a subset of Z , yet it behaves in a similar manner.
The set is closed under addition and multiplication by an integer, and it becomes an ideal ofZ if we simply multiply all the elements by 2. It will be profitable to study sets of this type.
Number Theory
161
DEFINITION
Let R be an integral domain, with Kits quotient field, and let Ibe an Rsubmodule of K. We say that I is afractional ideal of R if rI ~ R for some nonzero r E R. We will call r a denominator of 1. An ordinary ideal of R is a fractional ideal (take r = 1), and will often be referred to an as integral ideal.
LEMMA 1.
If I is a finitely generated Rsubmodule of K, then I is a fractional ideal.
2.
If R is Noetherian and I is a fractional ideal of R, then I is a finitely generated Rsubmodule of K.
3.
If I and J are fractional ideals with denominators rand s respectively, then In J, I
+ J and IJ are fractional ideals with respective denominators r (or s), rs and rs. Proof 1. If xl = a/b I, ... , xn = a,/bn generate I and b = b I ... bn, then bl ~ R. 2. IfrI ~ R, then I ~ rIR. As an Rmodule, rIR is isomorphic to R and is therefore Noetherian. Consequently, I is finitely generated.
3.
It follows from the definition that the intersection, sum and product of fractional ideals are fractional ideals. The assertions about denominators are proved by noting that reI n J) ~ rI ~ R, rs(l + J) ~ rI + sJ ~ R, and rsIJ = (rl)(sJ) ~ R.
The product oftwo nonzero fractional ideals is a nonzero fractional ideal, and the multiplication is associative (since multiplication in R is associative). There is an identity element, namely R, since RI ~ I = 11 ~ RI. We will show that if R is a Dedekind domain, then every nonzero fractional ideal has a multiplicative inverse, so the nonzero fractional ideals fonn a group. DEFINITION
A Dedekind domain is an integral domain R such that 1.
R is Noetherian,
2.
R is integrally closed, and
3.
Every nonzero prime ideal of R is maximal.
Every PID is a Dedekind domain. We will prove that the algebraic integers of a number field fonn a Dedekind domain. But as we know, the ring of algebraic integers need not be a PID, or even a UFD.
LEMMA Let I be a nonzero prime ideal of the Dedekind domain R, and let J = {x ThenRcJ.
E
K: xl ~ R}.
162
Number Theory
Proof Since R1 ~ R, it follows that R is a subset of J. Pick a nonzero element a E I, so that I contains the principal ideal Ra. Let n be the smallest positive integer such that Ra contains a product Pl'" P n of n nonzero prime ideals. Since R is Noetherian, there is such an n, and I contains one of the Pi' say Pl' But in a Dedekind domain, every nonzero prime ideal is maximal, so 1= Pl' Assuming n ~ 2, set II = P 2'" P n' so that Ra
II with b
rt
i
II by minimality of n. Choose b
E
Ra. Now III ~ Ra, in particular, Ib ~ Ra, hence Iba 1 ~ R. (Note that a has an
inverse in K, but not necessarily in R.) Thus ba1 E J, but ba1
rt R, for ifso, b ERa, contradicting
the choice of b. The case n = 1 must be handled separately. In this case, PI = I ~ Ra ~ PI' so 1= Ra. Thus Ra is a proper ideal, and we can choose b E R with b Ra. Then ba1 R, but ba1I = baIRa = bR ~ R, so ba 1 E J.
rt
rt
Proposition Let/be a nonzero prime ideal ofthe Dedekind domain R, and letJ = {x E K: xl ~ R}. Then J is a fractional ideal and IJ = R.
Proof By definition, J is an Rsubmodule of K. If r is a nonzero element of I and x E J, then rx E R, so rJ ~ Rand J is a fractional ideal. Now IJ ~ R by definition of J, so IJ is an integral ideal. Since 1= lR ~ lJ ~ R, maximality of I implies that either lJ = lor lJ = R. In the latter case, we are finished, so assume lJ = 1. If x E J, then xl ~ lJ = I, and by induction, xnI ~ I for all n = 1, 2, .... Let r be any nonzero element of 1. Then rxn E xnl ~ I ~ R, so R[x] is a fractional ideal. Since R is Noetherian, R[x] is a finitely generated Rsubmodule of K. x is integral over R. But R, a Dedekind domain, is integrally closed, so x E R. Therefore J ~ R.
Problems 1.
Show that a proper ideal P is prime if and only if for all ideals A and B, P ~ AB implies that P ~ A or P ~ B.
We are going to show that ifan ideal lis contained in the union of the prime ideals PI' ... ,
P n, then lis contained in some Pi" Equivalently, if for all i = 1, ... , n, we have I
i
Pi' then I
i
U:I ~ .There is no problem when n = 1, so assume the result holds for n  1 prime ideals. By the induction hypothesis, for each i there exists Xi
E
I with Xi
rf Pj ,j:t i.
2.
Show that we can assume without loss of generality that Xi
3.
Continuing Problem 2, let x
E Pi
for all i.
= L:~=IXI ...XIIXi+I ...Xn' Show that x
U:l ~ ,completing the proof.
E
I but x
rt
163
Number Theory
4.
If I and J are relatively prime ideals (l + J
= R),
show that 1J = I
E
J. More
generally, if II' ... , In are relatively prime in pairs, show that II .. , In = n~/=1 1/.. . 5.
Show that if a Dedekind domain R is a UFD, then R is a PID.
6.
Let R be an integral domain with quotient field K. If K is a fractional ideal of R, show that R = K.
7.
Let PI and P2 be relatively prime ideals in the ring R. Show that P{ and relatively prime for arbitrary positive integers rand s.
N
are
UNIQUE FACTORIZATION OF IDEALSIN A DEDEKIND DOMAIN In the previous section, we inverted nonzero prime ideals in a Dedekind domain. We must now extend this result to nonzero fractional ideals.
THEOREM If I is a nonzero fractional ideal of the Dedekind domain R, then I can be factored uniquely as
P,.'" ~n2 ...p"nr where the Pi are prime ideals and the ni are integers. Consequently, the nonzero
fractional ideals form a group under multiplication. Proof First consider the existence of such a factorization. Without loss of generality, we can restrict to integral ideals. [Note that if r 7: 0 and rI ~ R, then If (Rrtl(r.l).] By convention, we regard R as the product of the empty collection of prime ideals, so let S be the set of all nonzero proper ideals of R that cannot be factored in the given form, with all ni positive integers. [This trick will yield the useful result that the factorization of integral ideals only involves positive exponents.] Since R is Noetherian, S, ifnonempty, has a maximal element 10, which is contained in a maximal ideal 1. Ihas an inverse fractional ideal J.
10 = loR ~ Ie! ~ IJ= R. Therefore Ie! is an integral ideal, and we claim that 10 c Ie!. For if10 = Ie!, By maximality of 10 , Ie! is a product of prime ideals, say Ie!= PI'" Pr(with repetition allowed). Multiply both sides by the prime ideal I to conclude that 10 is a product of prime ideals, contradicting 10 E S. Thus S must be empty, and the existence ofthe desired factorization is established. To prove uniqueness, suppose that we have two prime factorizations _ nil Qts RI'" " 'pn, r !ttl···· s
where again we may assume without loss of generality that all exponents are positive. [If
pn appears, mUltiply both sides by pn.] Now PI contains the product ofthe ~ni ,P I contains Qj for some j. By maximality of Qp PI = Qp and we may renumber so that PI = QI' Multiply by the
164
Number Theory
inverse of P 1 (a fractional ideal, but there is no problem) to cancel PI and inductively to complete the proof.
Ql' and continue
COROLLARY A nonzero fractional ideal I is an integral ideal if and only if all exponents in the prime factorization of I are nonnegative.
Corollary Denote by np (1) the exponent ofthe prime ideal P in the factorization of 1. (If P does not appear, take np (1) = 0.) If II and 12 are nonzero fractional ideals, then II ~ 12 ifand only iffor every prime ideal P of R, np (11) ~ n p (12).
Proof We have 12 ~ II iff P,
np
(12) 
np
(11) 2::
/2/1 1 ~
~ R, this happens iff for every
o.
DEFINITION Let II and 12 be nonzero integral ideals. We say that II divides 12 if 12
= JII for some
integral ideal J. Just as with integers, an equivalent statement is that each prime factor of II is a factor of 12.
Corollary If II and 12 are nonzero integral ideals, then II divides 12 if and only if II ~ 12. In other words, for these ideals,
THEOREM In the basic AKLB setup, if A is a Dedekind domain, then so is B. In particular, the ring of algebraic integers in a number field is a Dedekind domain. In addition, B is a finitely generated Amodule and the quotient field of B is L.
Proof B is integrally closed in L. With Xi replaced by an arbitrary element of L, shows that L is the quotient field of B. Therefore B is integrally closed. B is a Noetherian ring, and the proof shows that B is a Noetherian, hence finitely generated, Amodule. It remains to prove that every nonzero prime ideal Qof B is maximal. Choose any nonzero element x of Q. Since x
E
B, x satisfies a polynomial equation
165
Number Theon)
x" + an_1x"1 + .... + a1x + ao = 0 with the ai E A. Ifwe take the positive integer n as small as possible, then ao * 0 by minimality of n. Solving for ao' we see that aoE Bx u A s Q 11 A, so P = Q 11 A =1= O. But Pis the preimage ofthe prime ideal Q under the inclusion map ofA into B. Therefore P is a nonzero prime, hence maximal, ideal of the Dedekind domainA. Consequently, AlP is a field. Now AlP can be identified with a subring of the integral domain BIQ via y + P ~ Y + Q. Moreover, BIQ is integral over AlP. [B is integral over A, and we can simply use the same equation of integral dependence.] BIQ is a field, so Q is a maximal ideal. Problems
The ring B of algebraic integers in Q ( H) is
z [H]. We will show that Z [ H] is
not a unique factorization domain. Consider the factorization (1 + H)(1
.J=5) = (2)(3).
1.
By computing norms, verify that all four of the above factors are irreducible.
2.
Show that the only units of Bare ± 1.
3.
Show that no factor on one side of the above equation is an associate of a factor on the other side, so unique factorization fails.
4.
We can use the prime factorization of ideals in a Dedekind domain to compute the greatest common divisor and the least common multiple of two nonzero ideals I and J, exactly as with integers. Show that the greatest common divisor of I and J is 1+ J and the least common multiple is III J.
5.
A Dedekind domain R comes close to being a PID in the following sense. (All ideals are assumed nonzero.) If I is an integral ideal, in fact if I is a fractional ideal, show that there is an integral ideal J such that IJ is a principal ideal of R.
6.
Show that the ring of algebraic integers in Q ( .J17 ) is not a unique factorization domain.
7.
In Problem 6, the only algebraic integers of norm 1 are ±l. Show that this property does not hold for the algebraic integers in Q ( H).
ARITHMETIC IN DEDEKIND DOMAINS Unique factorization of ideals in a Dedekind domain permits calculations that are analogous to familiar manipulations involving ordinary integers. In this section, we illustrate some ofthe ideas.
166
Number Theory
Let PI' ... , Pn be distinct nonzero prime ideals of the Oedekind domain R, and letJ = PI'" Pn' Let Qj be the product ofthe Pj with P j omitted, that is, .
Q; = PI'" PjIPj+1 ••• Pn· (If n = 1, we take Qj = R.) If I is any nonzero ideal of R, then by unique factorization, IQj:J IJ. For each i= 1, ... , n,choose an e1ementaibelongingtoIQ jbut nottoIJ, and leta= I:;=laj.
LEMMA a
E
Ibut for each i, a ¢ IPj' (In particular, a:t: 0.)
Proof Since each ai belongs to IQj ~ I, we have a E 1. Now ai cannot belong to IP j, for if so, a j E IP j~ IQj' which is the least common mUltiple of IP j and IQj. But by definition of Ql~ the least common multiple is simply IJ, and this contradicts the choice of a j • We break up the sum defining a as follows:
a = (a l + ... + aj_l) + a j + (a j+1 + ... + an)' Ifj:t: i, then aj E IQj~ IP j, so the first and third terms belong to IP j. Since a j ¢ IP j, as found above, we have a ¢ IP j • Proposition
Let I be a nonzero ideal ofthe Oedekind domain R. Then there is a nonzero ideal I' such that II' is a principal ideal (a). Moreover, ifJis an arbitrary nonzero ideal of R, I' can be chosen to be relatively prime to J.
Proof Let PI' ... , Pn be the distinct prime divisors of J. Then a E I, so (a) ~ 1. Since divides means contains, I divides (a), so (a) = II' for some nonzero ideal 1. If I is divisible by Pl~ then 1= P10 for some nonzero ideal 10, and (a) = IP10' Consequentl}{, a E IP j, contradicting. Corollary
A Dedekind domain with only finitely many prime ideals is a PIO.
Proof LetJbe the product of all the nonzero prime ideals. If lis any nonzero ideal, there is a nonzero ideal I' such that II' is a principal ideal (a), with I' relatively prime to J. But then the set of prime factors of I is empty, so that I' = R. Thus (a) = II' = IR = 1. The next result shows that a Oedekind domain is not too far away from a principal ideal domain. Corollary
Let Ibe a nonzero ideal of the Dedekind domain R, and let a be any nonzero element of1. Then I can be generated by two elements, one of which is a.
167
Number Theory
Proof Since a E I, we have (a) ~ I, so I divides (a), say (a) = IJ. There is a nonzero ideal I' such that II' is a principal ideal (b) and I' is relatively prime toJ. If gcd stands for greatest common divisor, then the ideal generated by a and b is gcd((a), (b)) = gcd(l1, II) = I since gcd (1, I) = (1). Problems
1.
Let I(R) be the group of nonzero fractional ideals of a Dedekind domain R. If peR) is the subset of I(R) consisting of all nonzero principal fractional ideals Rx, x E K, show that peR) is a subgroup of I(R). The quotient group C(R) = I(R)/P(R) is called the ideal class group of R. Since R is commutative, C(R) is abelian, and it can be shown that in the number field case, C(R) is finite.
2.
Continuing Problem 1, show that C(R) is trivial iff R is a PID.
We will now go through the factorization of an ideal in a number field. The necessary background is developed in a course in algebraic number theory, but some ofthe manipulations are accessible to us now. By (7.2.3), the ring B of algebraic integers of the number field Q ( H)
z [H]. (Note that  5 == 3 mod 4.) Ifwe wish to factor the ideal (2) = 2B in B, the idea is to factor Xl + 5 mod 2, and the result is x 2 +5 == (x + 1)2 mod 2. Identifying x with H ,we form is
the ideal P2 = (2, 1 + H), which turns out to be prime. The desired factorization is (2) = P22 . This technique works if B = Z [a], where the number field Lis Q (a).
pl.
3.
Show that 1 
H
4.
Show that 2
pl
5.
Expand
6.
Following the technique suggested in the above problems, factor x 2 + 5 mod 3, and
E
E P 2'
and conclude that 6
hence (2) ~
pl = (2, 1 + H)(2,
E
pl 1 + 'J=5), and conclude that
pl
~ (2).
conjecture that the prime factorization of (3) in the ring of integers of Q ( H) is (3) = P 3P'3 for appropriate P3 and P'3 . 7.
With P3 and P3' as found in Problem 6, verify that (3) = P3P3'.
PADIC NUMBERS We will give a very informal introduction to this basic area of number theory. Throughout the discussion, p is a fixed prime.
168
Number Theory
DEFINITION A padic integer can be described in several ways. One representation is via a series
x = ao + alP + a2p 2 + .... , a j
E
Z.
(Let's ignore the problem of convergence for now.) The partial sums are xn = ao +a IP + + a,p", so that xn  xnl = a,p". A padic integer can also be defined as a sequence ofintegers x = {xo' xl' ... ,} satisfying
xn ==xn_l modpn, n = 1, 2, .... Given a sequence satisfying (2), we can recover the coefficients of the series by
The sequences X and yare regarded as defining the same padic integer iff xn == Yn mod pn+ I , n = 0, 1, .... Replacing each xn by the smallest nonnegative integer congruent to it mod ~+l is equivalent to restricting the a j in (1) to {O, 1, ... , P  I}. [We call this the standard representation.] The limiting case in some sense of an expansion in base p. Sums and products ofpadic integers can be defined by polynomial multiplication is used. With the representation, we take
x + Y = {xn + y n}, xy = {xnYn}' With addition and multiplication defined in this way, we get the ring ofpadic integers, denoted by Op' (A more common notation is Z p' with the ring of integers modulo p written as Z /p Z . Since the integers mod p occur much more frequently in this text than the padic integers, and Z p is a bit simpler than Z /p Z , I elected to use Z p for the integers mod p.) The rational integers Z form a subring ofOp viax = {x, x, x, ... }. We now identify the units ofOp '
Proposition The padic integer x = {xn} is a unit ofOp (also called apadic unit) ifand only ifxo mod p. In particular, a rational integer a is a padic unit if and only if a
;£ 0
;£ 0 mod p.
Proof If(a o+ alP + ... )(bo + blP+ ... ) = 1, then aob o = 1, so ao ;£ 0 modp, proving the "only if' part. Thus assume thatx o ;£ 0 modp. xn == x n l == ... xomodp, soxn ;£ 0 modp. Thereforexn and~l are relatively prime, so there existsYn such thatxJln== 1 modpn+l' hence modpn' Now,xn==x n_l modpn' so xnYn1 ==xnIYn1 == 1 modpn' ThusxnYn ==xnYn1 modpn' so Yn == Ynl mod Pn" The sequence Y = {yn} is therefore a padic integer, and by construction, Xy = 1.
169
Number Theory
Corollary Every nonzero padic integer has the form x = JIIu where n 2: 0 and u is a padic unit. Consequently, Sp is an integral domain. Furthermore, Sp has only one prime elementp, and every x E Sp is a power ofp, up to multiplication by a umt. Proof The series representation for x has a nonzero term aJ/" oflowest degree n, where an can be taken between 1 and p  1. Factor out JII to obtain x =JIIu, where u is a unit.
DEFINITION The quotient field QpofSp is called the field ofpadic numbers. Each a E Qphas the form fJ"'u, where m is an integer (possibly negative) and u is a unit in 9p . Thus a has a "Laurent expansion"
Another representation is a = xlp, where x is a padic integer and r 2:
o. This version is
convenient for doing addition and multiplication in Q p. The rationals Q are a subfield of Q p. To see this, let alb be a rational number in lowest terms (a and b relatively prime). Ifp does not divide b, then b is a unit of9p. Since a E IE ~ 9p' we have alb ~ 9p. If b = plb 'where p does not divide b " we can factor out pI and reduce to the previous case. Thus alb always belongs to QP' and alb E Sp iffP does not divide b. Rational numbers belonging to Sp are sometimes called pintegers. We now outline a procedure for constructing the padic numbers formally.
DEFINITION The padic valuation on Q p is defined by vp 1, F j is a splitting field for xn  bl over F j_l . FlF is a cyclotomic (Galois) extel'}sion, and by (6.7.2), each F /F j _ l , i = 2, ... , r is a cyclic (Galois) extension. We now do some further preparation. Suppose that K is a splitting field for f over F, and that the Galois group of KIF is solvable, with Gal(KlF)
= Ho
~ HI ~ ... ~ Hr
=1
with each Hj_lHj abelian. By the fundamental theorem, we have the corresponding sequence of fixed fields F
= Ko
~ KI ~ ... ~ Kr
=K
with K/Kj_1 Galois and Gal(K/Kj_l ) isomorphic to Hj_lHj. Let us adjoin a primitive nth root of unity co to each K j , so that we have fields F j = K j ( co) with F ~ Fo ~ Fl ~ ... ~ Fr'
We take n = I Gal(KlF)I. Since F j can be obtained from F j _ 1 by adjoining everything in K j \ K j _ l , we have
F.I = F.1 IK.I = KF. I I 1the composite of Fj _ 1 and Kj , i = 1, ... , r. In the diamond diagram, at the top of the diamond we have F j, on the left K j , on the right F j _ 1, and on the bottom KI n F j _ 1 ~ K j _ 1 • We conclude that F /F j_ 1 is Galois, with a Galois group isomorphic to a subgroup ofGal(K/ K j_I ). Since Gal(K/Kj _ l ) ~ Hj_IIHI, it follows that Gal(F/Fj _ 1) is abelian. Moreover, the exponent of this Galois group divides the order of H o' which coincides with the size of Gal (KIF). (This explains our choice of n.)
Galois Theory
200 GALOIS' SOLVABILITY THEOREM
Let K be a splitting field for/over F, where F has characteristic O. Then/is solvable by radicals if and only if the Galois group of KIF is solvable. Proof Ifjis solvable by radicals, we have F = Fo ~ FI ~ ." ~ Fr = N
where NIF is Galois, N contains a splitting field K for / over F, and each F IFj_ 1 is Galois with an abelian Galois group. By the fundamental theorem, the corresponding sequence of subgroups is 1 = Hr :::! H r_1 :::! ... :::! Ho = G = Gal(NIF)
with each Hj_lHj abelian. Thus G is solvable, and since Gal(KlF)
[map Gal(NIF ) solvable.
~
~
Gal(NIF )/Gal(NIK)
Gal(KlF) by restriction; the kernel is Gal(NIK)], Gal( KIF) is
Conversely, assume that Gal(KlF) is solvable. We have F ~ Fo ~ FI ~ ... ~ Fr
where K ~ F r, each F j contains a primitive nth root of unity, with n = I Gal(KlF) I, and Gal(FIFi\) is abelian with exponent dividing n for all i = 1, ... , r. Thus each F IFj_ 1 is a Kummer extension whose Galois group has an exponent dividing n. For the case i = 1, each F IFj _ 1 is a radical extension. By transitivity, Fr is a radical extension of F. Since K ~ Fr,/is solvable by radicals. Example
Let.f{X) = Xi  10.x4 + 2 over the rationals. The Galois group ofjis 8 5, which is not solvable. Thus / is not solvable by radicals. There is a~fundamental idea that needs to be emphasized. The significance of Galois' solvability theorem is not simply that there are some examples of bad polynomials. The key point is there is no general method for solving a polynomial equation over the rationals by radicals, if the degree of the polynomial is 5 or more. Problems
In the exercises, we will sketch another classical problem,that of constructions with ruler and compass. In Euclidean geometry, we start with two points (0, 0) and (1, 0), and we are allowed the following constructions. 1. 2.
Given two points P and Q, we can draw a line joining them; Given a point P and a line L, we can draw a line through P parallel to L;
201
Galois Theory 3. 4.
5.
Given a point P and a line L, we can draw a line through P perpendicular to L; Given two points P and Q, we can draw a circle with center at P passing through Q; Let A, and similarly B, be a line or a circle. We can generate new points, called constructible points, by forming the intersection of A and B. If(c, 0) (equivalently (0, c)) is a constructible point, we call c a constructible number. It follows from (2) and (3) that (a, b) is a constructible point iff a and b are constructible numbers. It can be ·shown that every rational number is constructible, and that the constructible numbers form a field. Now in (5), the intersection of A and B can be found by ordinary arithmetic plus at worst the extraction of a square root. Conversely, the square roof of any nonnegative constructible number can be constructed.
Therefore c is constructible iff there are real fields Q E
= Fo :::; FI ... :::; Fr such that c
Fr and each [F j : Fj_tl is 1 or 2. Thus if c is constructible, then c is algebraic over Q and
[Q(c) : Q] is a power of2.
1.
(Trisecting the angle) If it is possible to trisect any angle with ruler and compass, then in particular a 60 degree angle can be trisected, so that a = cos 20 0 is constructible. Using the identity e i38 = cos 38 + i sin 38
= (cos 8 + i sin 8)3,
reach a contradiction.
2.
"",
(Duplicating the cube) Show that it is impossible to construct,with ruler and compass, a cube whose volume is exactly 2. (The side of such a cube would be ifi)
3.
(Squaring the circle) Show that if it were possible to construct a square with area 1t, then 1t would be algebraic over .Q. (It is known that 1t is transcendental over Q.) To construct a regular ngon,that is, a regular polygon with n sides, n ~ 3, we must be able to construct an angle of 27t1n; equivalently,cos 27t1n must be a constructible number. Let ro = e i2wn , a primitive nth root of unity.
4.
Show that [Q(ro) : Q(cos 27t1n)] = 2.
5.
Show that if a regular ngon is constructible,then the Euler phi function a l )· Show that in this case, V is cyclic but not simple. 5.
Suppose that M is a finitedimensional vector space over an algebraically closed field F, and in addition M is a module over a ring R containing F as a subring. If M is a simple Rmodule andfis an Rmodule homomorphism, in particular an Flinear transformation, on M, show that f is mUltiplication by some fixed scalar A E F. This result is frequently given as a third part of Schur's lemma.
6.
Let I be a left ideal of the ring R, so that RlI is an Rmodule but not necessarily a ring. Criticize the following statement: "Obviously", I annihilates RlI. SIMPLE AND SEMISIMPLE RINGS
DEFINITION
Since a ring is a module over itself, it is natural to call a ring R semisimple if it is semisimple as an Rmodule. Our aim is to determine, if possible, how semisimple rings are assembled from simpler components. Aplausible Idea is that the components are rings that are simple as modules over themselves. But this turns out to be too restrictive, since the components would have to be division rings. When we refer to a simple left ideal I of R, we will always mean that I is simple as a left Rmodule. We say that the ring R is simple if R is semisimple and all simple left ideals of R are isomorphic. [The definition of simple ring varies in the literature. An advantage of our choice is that we avoid an awkward situation in which a ring is simple but not semisimple.J Our goal is to show that the building blocks for semisimple rings are rings of matrices over a field, or more generally, over a division ring. The next two results give some properties of modules over semisimple rings. Proposition
If R is a semisimple ring, then every nonzero Rmodule Mis semisimple.
Noncommutative Algebra
254
Proof M is a quotient of a free Rmodule F. Since F is a direct sum of copies of R, and R is semisimple by hypothesis, it follows that F is semisimple. Mis semisimple. Proposition
Let Ibe a simple left ideal in the semisimple ring R, and let Mbe a simple Rmodule. Denote by 1M the Rsubmodule of M consisting of all finite linear combinations rj E I, xi E M. Then either 1M = M and I is isomorphic to M, or 1M = o.
I:
/iXi'
Proof If 1M "* 0, then since M is simple, 1M = M. Thus for some x E M we have Ix "* 0, and again by simplicity of M, we have Ix = M. Map I onto Mby r ~ rx, and note that the kernel cannot be I because Ix "* O. Since I is simple, the kernel must be 0, so I ~ M.
BEGINNING THE DECOMPOSITION Let R be a semisimple ring. We regard two simple left ideals of R as equivalent if they are isomorphic (as Rmodules), and we choose a representative Ii' i E T from each equivalence class. We define the basic building blocks of R as Bi = the sum of all left ideals of R that are isomorphic to Ii.
We have a long list of properties of the Bi to establish, and for the sake of economy we will just number the statements and omit the words "Lemma" and "Proof' in each case. We will also omit the end of proof symbol, except at the very end. If i"*}, then BiBj =
R= If r
E
o.
Apply with I replaced by Bi and M by Bj"
~iETBi
R, then (r) is a left ideal, which is a sum of simple left ideats.
Each Bi is a twosided ideal. we have
Thus RBi = BiR
= Bj"
R has only finitely many isomorphism classes of simple left ideals II' ... , P.
We can write the identity 1 of R as a finite sum of elements e i the notation if necessary, let 1 =
2::::=le
i .
Ifr
EB where} ¢ j
i = 1, ... , t, so r = r 1 = O. Thus Bj = 0 for} ¢ {I, ... , f}.
E
Bi' i
E
T. Adjusting
{I, ... , f}, then rei = 0 for all
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R = a/I "'~ e,", with e/. E B /.. R is the sum , B/". Thus 1 has a unique representation as D,=I of the Bj" If b I + ... + bt = 0, with b j E B j, then
o= e j(b I + ... + bt) = e /b 1 + .. e jb t = e jb j = (e I + ... + et)b j = 1b j = bI' Therefore the sum is direct. If b j E B j, then ejb j = b j = bl j. Thus e j is the identity on B j and B j = Re j = ejR. The first assertion follows from the computation along with a similar computation with e j multiplying on the right instead of the left. Now B j ~ Re j because b j = b/e j, and Rei ~ B j by and the fact that e j E B j. The proof that B j = el is similar. Each B j is a simple ring. By the computation B j is a ring (with identity e). Let J be a simple left ideal of B j • RJ = B/ = J, so J is a left ideal of R, necessarily simple. Thus J is isomorphic to some ~, and we must have j = i. [Otherwise, J would appear in the sums defining both B j and B .. ] Therefore B j has only one isomorphism class of simple left ideals. Now B j is a sum ~f simple left ideals of R, and a subset of B j that is a left ideal of R must be a left ideal of Bj" Consequently, B j is semisimple and the result follows. If M is a simple Rmodule, then M is isomorphic to SOine Ii" Thus there are only finitely many isomorphism classes of simple Rmodules. In particular, if R is a simple ring, then all simple Rmodules are isomorphic.
;=1
;=1
where the J are simple left ideals. Therefore t
t
M=RM= L:B;M=L:L:{JM:J~I;} ;=1
;=1
JM = 0 or J ~ M. The former cannot hold for all J, since M"* O. Thus lv! some i. If R is a simple ring, then there is only one i, and the result follows.
~
I j for
Let M be a nonzero Rmodule, so that Mis semisimple. Define M j as the sum of all simple submodules of Mthat are isomorphic to Ij' so that, M= L:;=IM;. Then t
M=.tB B;M and B#= e#= Mi, i = 1, ... , t. 1=1
By definition of Bj>
B~= L:{JMj :J~I;}
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Noncommututive Algebra
where the.r s are simple left ideals. If N is any simple module involved in the definition of ~, then IN is 0 or N, and IN = N implies that N ~ J ~ Ii" But all such N are isomorphic to~, and therefore Bf1.i = 0, i =I= j. Thus M.=RM.= I
~BjMi=BiMi j
I
and B.M=
~BjMi=BiMi J
I
Consequently, M j = BjM = ejRM = ef/, and all that remains is to show that the sum of the Mi is direct. Let xl + ... + x t = 0, Xi E M j. Then O=ej(x l
since ere}
+ ... +xt)=efi
Bf1.i = 0 for i
E
efi = (e}
=I= j.
Finally,
+ ... + et)xj = Xj'
Asemisimple ring R is ringisomorphic to a direct product of simple rings. For if aj' b i
E
Bj' then
(a l + .... + at)(b l + ... + bt) = alb l + ... + atbl' Problems
In Problems 1 and 2, let M be a semisimple module, so that M is the direct sum of simple modules M;, i E 1. We are going to show that M is a finite direct sum of simple modules if and only if M is finitely generated. 1.
Suppose that xl' ... , Xn generate M. It will follow that M is the direct sum of finitely many of the Mi" How would you determine which M;' s are involved?
2.
Conversely, assume that M is a finite sum of simple modules. Show that Mis finitely generated.
3.
Aleft ideal I is said to be minimal if I =I= 0 and I has no proper subideal except O. Show that the ring R is semisimple if and only if R is a direct sum of minimal left ideals.
4.
Is Z semisimple?
5.
Is Z n semisimple?
6.
Suppose that R is a ring with the property that every nonzero Rmodule is semisimple. Show that every Rmodule M is projective, that is, every exact sequence 0 ~ A ~ B ~ M ~ 0 splits. Moreover, M is injective, that is, every exact sequence 0 ~ M ~ A ~ B ~ 0 splits.
Noncommutative Algebra
7.
257
For any ring R, show that the following londitions are equivalent. a. R is semisimple; b. Every nonzero Rmodule is semisimple; c. Every Rmodule is projective; d. Every Rmodule is injective.
PROPERTIES OF SIMPLE RINGS, MATRIX RINGS, AND ENDOMORPHISMS To reach the Wedderburn structure theorem, we must look at simple rings in more detail, and supplement what we already know about matrix rings and rings of endomorph isms.
LEMMA Let R be any ring, regarded as a left module over itself. If h: R ~ M is an Rmodule homomorpbism, then for some x E M we have her) = rx for every r E R. Moreover, we may choose x = h(l), and the map h ~ h(l) is an isomorphism ofHomR(R, M) and M. This applies in particular when M = R, in which case h E EndR(R). Proof The point is that h is determined by what it does to the identity. Thus
her) = h(rl) = rh(l) so we may take x = h(l). If s E Rand hE HomiR, M), we take (sh)(r) This makes HomR(R, M) into a left Rmodule isomorphic to M.
= hers) = rsx.
Notice that although all modules are left Rmodules, h is given by multiplication on the right by x. Corollary
Let J and J be simple left ideals of the simple ring R. Then for some x =Jx.
E
R we have J
Proof By the definition of a simple ring, R is semisimple, so R = J EB L for some left ideal L. Again by the definition of a simple ring, J and J are isomorphic (as Rmodules). If 't : J ~ J is an isomorphism and 1t is the natural projection of Ron J, then 't 1t E EndR(R), so there exists x E R such that 't1t(r) = rx for every r E R. Allow r to range over J to conclude that J = Ix.
Asemisimple ring can be expressed as a direct sum of simple left ideals. If the ring is simple, only finitely many simple left ideals are needed.
LEMMA A simple ring R is a finite direct sum of simple left ideals.
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Noncommutative Algebra
Proof Let R = EB j~ where the ~ are simple left ideals. Changing notation if necessary, we have 1 = Y 1 + ... + Y m with Yj E ~,j = 1, ... , m. If x E R, then m
m
x=x = LXYi ELI)" 1
. 1
J=
. 1
J=
Therefore R is a finite sum of the I., and the sum is direct because the original decomposition of R is direct. } Corollary
If I is a simple left ideal of the simple ring R, then IR = R.
Proof If J is any simple left ideal of R, J ~ 1R. R is a finite (direct) sum of simple left ideals, so R ~ 1R. The reverse inclusion always holds, and the result follows. We now have some insight into the structure of simple rings. Proposition
If R is a simple ring, then the only twosided ideals of Rare 0 and R.
Proof Let J be a nonzero 2sided ideal of R. J is a semisimple left Rmodule, J is a sum of simple left ideals of J, hence of R. In particular, J contains a simple left ideal!. Since J is a right ideal, it follows that J = JR. we have J=JR ;21R = R soJ=R. In the literature, a simple ring is often defined as a ring R whose only twosided ideals are 0 and R, but then extra hypotheses must be added to force R to be semisimple. Corollary
Let 1be a simple left ideal of the simple ring R, and let Mbe a simple Rmodule. Then 1M = M and M is faithful.
Proof The first assertion follows from a computation that uses associativity of scalar multiplication in a module: M= RM= (I R)M= 1(RM) = 1M Now let b belong to the annihilator of M, so that bM = o. We must show that b = O. By a computation similar to (l) (using in addition the associativity of ring multiplication),
RbRM= RbM= RO =
= O.
But RbR is a twosided ideal of R, RbR = 0 or R. In the latter case, M = RM = RbRM 0, contradicting the assumption that M is simple. Therefore RbR = 0, in particular, b =
Ibi =
o.
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259
We are now ready to show that a simple ring is isomorphic to a ring of matrices. Let
R be a simple ring, and Va simple Rmodule. [V exists because R is a sum of simple left ideals, and V is unique up to isomorphism.] Let D = EndiV), a division ring by Schur's lemma. Then, V is a Dmodule, in other words, a vector space over D. V is a faithful Rmodule, and if we can prove that V is finitedimensional as a vector space over D, then R is ringisomorphic to EndD(V). Ifn is the dimension of Vover D, then EndD(V) ~ Mn(DO), the ring of n by n matrices with entries in the opposite ring DO.
THEOREM Let R be a simple ring, Va simple Rmodule, and D the endomorphism ring EndR(V). Then V is a finitedimensional vector space over D. If the dimension of this vector space is n, then
R
~
EndD(V)
~
Mn(DO).
Proof Assume that we have infinitely many linearly independent elements xI' X2, .... Let 1m be the left ideal {r E R: rXj = 0 for all i = 1, ... , m}. Then the 1m decrease as m increases, in fact they decrease strictly. [Given any m, letfbe a Dlinear transformation on V such thatj(x) = 0 for 1 ~ i ~ m andj(xm+ l ) 1= O. There exists r E R such thatj(x) = rx j , i = 1, ... , m + 1. But then rX I = .... = rXm = 0, rXm+I1= 0, so r E 1m \ 1m+I'] Write 1m = J m EB 1m+I' [Since R is semisimple, so are all left ideals.] Iterating this process, we construct a left ideal J I EB J 2 EB ... , and again,
R = J o EB J I EB J 2 EB .... Therefore 1 is a finite sum of elements Yj
E
Jj, i
= 0, 1, ... , t. But then
R = J o EB J I EB ... EB J t and it follows that J t+ 1 must be 0, a contradiction. Problems
Problems 15 are the key steps in showing that a ring R is simple if and only if R is Artinian and has no twosided ideals except 0 and R. Thus if a simple ring is defined as one with no nontrivial twosided ideals, then the addition of the Artinian condition gives our definition of simple ring; in particular, it forces the ring to be semisimple. The result that an Artinian ring with no nontrivial twosided ideals is isomorphic to a matrix ring over a division ring is sometimes called the WedderburnArtin theorem. In Problems 15, "simple" will always mean simple in our sense. 1.
If R is an Artinian ring, show that there exists a simple Rmodule.
2.
Let R be an Artinian ring with no nontrivial twosided ideals. Show that R has a faithful, simple Rmodule.
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Noncommutative Algebra
3.
Continuing Problem 1, if Vis a faithful, simple Rmodule, and D = EndR(V), show that V is a finitedimensional vector space over D.
4.
Continuing Problem 3, show that R is ringisomorphic to EndD(V), and therefore to a matrix ring M/DO) over a division ring.
In the next section, we will prove that a matrix ring over a division ring is simple; this concludes the proof that R is simple iff R is Artinian with no nontrivial twosided ideals. 5.
If an Rmodule M is a direct sum EB7=1 M j of finitely many simple modules, show that M has a composition series.
6.
Conversely, if M is semisimple and has a composition series, show that M is a finite direct sum of simple modules.
STRUCTURE OF SEMISIMPLE RINGS We have now done all the work needed for the fundamental theorem.
WEDDERBURN STRUCTURE THEOREM Let R be a semis imp Ie ring. 1.
R is ringisomorphic to a direct product of simple rings B 1, ... , B('
2.
There are t isomorphism classes of simple Rmodules. If VI' ... , Vt are representatives of these classes, let D j be the division ring EndiV)' Then Vj is a finitedimensional vector space over D j • If nj is the dimension of this vector space, then there is a ring isomorphism
Consequently, R is isomorphic to the direct product of matrix rings over division rings. Moreover, (3)
Bj~
= 0, i '* j; Byj = Vj'
THEOREM The ring MnCR) of all n by n matrices with entries in the division ring R is simple. Proof We have done most of the work in the exercises. Let C k be the set of matrices whose entries are 0 except perhaps in column k, k = 1 ... , n. Then Ck is a left ideal of Mn(R), and if any nonzero matrix in Ck belongs to a left ideal I, then Ck ~ 1. Thus each Ck
is a simple left ideal, and MnCR), the direct sum of C 1,
... ,
Cn' is semisimple.
Now let I be a nonzero simple left ideal. Anonzero matrix in I must have a nonzero . entry in some column, say column k. Define/: I ~ C k by j{A) = A k, the matrix obtained
Noncommutative Algebra
261
from A by replacing every entry except those in column k by O. Then/is an Mn(R)module homomorphism, since j{BA)
= (BA)k = BAk = BftA).
By construction, / is not identically 0, so by Schur's lemma, / is an isomorphism. Since the Ck are mutually isomorphic, all simple left ideals are isomorphic, proving that MnCR) is simple. INTRODUCTION TO GROUP REPRESENTATIONS
A major application of semisimple rings and modules occurs in group representation theory, and we will try to indicate the connection. Let k be any field, and let G be a finite group. We form the group algebra kG, which is a vector space over k with basis vectors corresponding to the elements of G. In general, if G = {xl' ... , xm}, the elements of kG are of the form alx I + ... + a~m' where the a i belong to k. Multiplication in kG is defined in the natural way; we set (a.xi)(~x)
= a~xixj
and extend by linearity. Then kG is a ring (with identity IkIG) that is also a vector space over k, and a(xy) = (a.x)y = x(ay), a E k, x, Y E G, so kG is indeed an algebra over k. [This construction can be carried out with an arbitrary ring R in place of k, and with an arbitrary (not necessarily finite) group G. The result is the group ring RG, a free Rmodule with basis G.] Now let Vbe an ndimensional vector space over k. We want to describe the situation in which "G acts linearly on V". We are familiar with group action, but we now add the condition that each g E G determines a linear transformation peg) on V. We will write p(g)(v) as simply gv or g(v), so that g(av + ~w) = ag(v) + ~g(w). Thus we can multiply vectors in V by scalars in G. Since elements of kG are linear combinations of elements of G with coefficients in k, we can multiply vectors in Vby scalars in kG. To summarize very compactly, V is a kGmodule.
Now since G acts on V, (hg)v = h(gv) and 1 v = v, g, h E G, v E V. Thus p(hg) = p(h)p(g), and each peg) is invertible since p(g)p(gY) = p(IG) = the identity on V. Therefore p is a homomorphism from G to GL(V), the group of invertible linear transformations on V. Multiplication in GL(V) corresponds to composition of functions. The homomorphism p is called a representation of G in V, and n, the dimension of V, is called the degree of the representation. If we like, we can replace GL(V) by the group of all nonsingular n by n matrices with entries in k. In this case, p is called a matrix representation.
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Noncommutative Algebra
The above process can be reversed. Given a representation p, we can define a linear action ofG on Vby gv = p(g)(v), and thereby make Va kGmodule. Thus representations can be identified with kGmodules. REGULAR REPRESENTATION
If G has order n, then kG is an ndimensional vector space over k with basis G. We take Vto be kG itself, with gv the product of g and v in kG. As an example, let G = {e, a, a 2 }, a cyclic group of order 3. Vis a 3dimensional vector space with basis e, a, a 2 , and the action of G on V is determined by ee = e, ea = a, ea2 = a 2,. ae = a, aa = a 2, aa2 = e', a 2e = a 2, a 2a = e, a 2a 2 = a.
Thus the matrices peg) associated with the elements g 1 0 [e] = 0
0
0
0 0
1 O,[a]= 1 0
0 0
0
o ,[a 0
2
]= 0
E
G are
1 0
0
1.
0 0
Role of Semisimplicity Suppose that p is a representation of G in V. Assume that the basis vectors of V can be decomposed into two subsets v(A) and v(B) such that matrix of every g E G has the form [g]
= [~ :].
(The elements of A and B will depend on the particular g, but the dimensions of A and B do not change.) The corresponding statement about V is that V= VA EB VB
where VA and VB are kGsubmodules of V. We can study the representation by analyzing its behavior on the simpler spaces VA and VB' Maschke's theorem, to be proved in the next section, says that under wide conditions on the field k, this decomposition process can be continued until we reach subspaces that have no nontrivial kGsubmodules. In other words, every kGmodule is semisimple. In particular, kG is a semisimple ring, and the Wedderburn structure theorem can be applied to get basic information about representations. We will need some properties of projection operators, and it is convenient to take care of this now. DEFINITION
Alinear transformation
1t
on a vector space V [or more generally, a module
Noncommutative Algebra
263
homomorphism] is called a projection of V (on 7t(V)) if 7t is idempotent, that is, 7t2 = 7t. We have already met the natural projection of a direct sum onto a component, but there are other possibilities. For example, let p be the projection of lR 2 = lR EB lR given by p(x, y)
x
y x+2 y) . Note that 7t must be the identity on 7t(V), since 7t(7t(v)) = 7t(v).
= (2'
If we choose subspaces carefully, we can regard any projection as natural. Proposition
If 7t is a projection on V, then V is the direct sum of the image of 7t and the kernel of 7t. Proof Since v = 7t(v) + (v  7t(v)) and 7t(v 7t(v)) = 0, V= imV + ker V. To show that the sum is direct, let v = 7t(w) E ker 7t. Then 0 = 7t(v) = 7t2(w) = 7t(w) = v, so im 7t n ker 7t = O. Example
For real numbers x and y, we have (x, y) = (x  cy)(1, 0) + y(c, 1), where c is any fixed real number. Thus lR 2 = lR (1, 0) EB lR (c, 1), and if we take p(x, y) = (x  cy, 0), thenp is a projection of lR 2 onto lR (1,0). By varying c we can change the complementary subspace lR (c, 1). Thus we have many distinct projections onto the same subspace lR (1, 0). Problems
1.
Show that the regular representation is faithful, that is, the homomorphism p is injective.
2.
Let G be a subgroup of Sn and let V be an ndimensional vector space over k with basis v(l), ... , v(n). Define the action of G on V by g(v(i)) = v(g(i)), i = 1, ... , n.
Show that the action is legal. (V is called a permutation module.) 3.
Continuing Problem 2, if n = 4, find the matrix of g = (1, 4, 3).
4.
Here is an example of how a representation can arise in practice. Place an equilateral triangle in the plane V, with the vertices at VI = (I, 0), v2 =
(~,~J3)
and v3
=
(~,~J3); note that VI + v2 + v3 = O. Let G = D6 be
the group of symmetries of the triangle, with g = counterclockwise rotation by 120 degrees and h = reflection about the horizontal axis. Each member of D6 is of the form giN, i = 0, 1, 2,j = 0, I, and induces a linear transformation on V . Thus we have a representation of G in V (the underlying field k can be taken as lR).
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Noncommutative Algebra
With vI and V z taken as a basis for V, find the matrices [g] and associated with g and h. 5.
Continue from Problem 4, and switch to the standard basis e l = VI = (1, 0), e z I). Changing the basis produces an equivalent matrix representation. The matrix representing the element a EGis now of the form [aJ' = pI[a]P
= (0,
where the similarity matrix P is the same for every a E G (the key point). Find the matrix P corresponding to the switch from {vI' Vz } to {e I , ez }, and the matrices [g]' and '. 6.
Consider the dihedral group D8, generated by elements R (rotation) and F (reflection). We assign to R the 2 by 2 matrix A=
[~l ~I
and to F the 2 by 2 matrix B
~ [~ ~J
Show that the above assignment determines a matrix representation of Dg of degree 2. 7.
Is the representation of Problem 6 faithful?
A very accessible basic text on group representation theory is "Representations and Characters of Groups" by James and Liebeck.
MASCHKE'S THEOREM We can now prove the fundamental theorem on decomposition of representations. It is useful to isolate the key ideas in preliminary lemmas.
LEMMA Let G be a fini~e group, and k a field whose characteristic does not divide I G I (so that division by I G I is legal). Let Vbe a kGmodule, and 'ljJ a linear transformation on Vas a vector space over k. Define e : V » V by
Then not only is homomorphism.
e a linear transformation on the vector space V, but it is also a kG
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Noncommutative Algebra
Proof Since 'ljJ is a linear transformation and G acts linearly on V, a is linear. Now if hE G, then a(hv)
=
1
~
1 I::gI'ljJg(hv).
gEG
As g ranges over all of G, so does gh. Thus we can let x = gh, g) = hx), to obtain a(hv) = _I_I::hxI'ljJ(xv) = h9(v) IGlgEG
and the result follows.
LEMMA Suppose that 'ljJ is a projection of Von a subspace W that is also a kGsubmodule of V. Then a is also a projection of Von W. Proof If v
E
W, then g(v)
since 'ljJ is a projection on W.
= g(v) By definition of a we have a(v) = v. To prove that a 2 = a, E
W since W is a kGsubmodule of V. Thus 'ljJ g(v)
note that since 'ljJ maps V into the kGsubmodule W, it follows from the definition ofa that a also maps V into W. But a is the identity on W, so a 2(v)
= a(a(v» = a(v)
and a is a projection. Since a maps into Wand is the identity on W, a is a projection of Von W.
MASCHKE'S THEOREM Let G be a finite group, and k a field whose characteristic does not divide 1G a kGmodule, then V is semisimple.
I. If V is
Proof Let W be a kGsubmodule of V. Ignoring the group algebra for a moment, we can write V = W EB U as vector spaces over k. Let 'ljJ be the natural projection of Von W. a is a kGhomomorphism and also a projection of Von W. V = ima EB ker a = W EB ker a as kGmodules. V is semisimple. .,
We have been examining the decomposition of a semisimple module into a direct sum of simple modules. Suppose we start with an arbitrary module M, and ask whether M can be expressed as M) EB M2, where M) and M2 are nonzero submodules. If so, we can try to decompose M) and M 2, and so on. This process will often terminate in a finite number of steps.
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Noncommutative Algebra
DEFINITION The module M is decomposable if M = MI EB M 2, where MI and M2 are nonzero submodules. Otherwise, M is indecomposable. Proposition
Let M be a module with a composition series; equivalently, M is Noetherian and Artinian. Then M can be expressed as a finite direct sum EB7=1 M j of indecomposable submodules. Proof If the decomposition process does not terminate, infinite ascending and descending chains are produced, contradicting the hypothesis.
As the above argument shows, the hypothesis can be weakened to M Noetherian or Artinian. Usually stated along with a uniqueness assertion which uses the stronger hypothesis: If M has a composition series and M = EB7=1 M j = EB~l~' where the ~ and ~ are indecomposable submodules, then n = m and the M j are, up to isomorphism, just a rearrangement of the N j • The full result (existence plus uniqueness) is most often known as the KrullSchmidt Theorem. The uniqueness proof is quite long, and we will not need the result. Returning to semisimple rings, there is an asymmetry in the definition in that a ring is regarded as a left module over itself, so that submodules are left ideals. We can repeat the entire discussion using right ideals, so that we should distinguish between leftsemisimple and rightsemisimple rings. However, this turns out to be unnecessary.
THEOREM A ring R is leftsemisimple if and only if it is rightsemisimple. Proof If R is leftsemisimple, R is isomorphic to a direct product of matrix rings over division rings. But a matrix ring over a division ring is rightsimple by with left ideals replaced by right ideals. Therefore R is rightsemisimple. The reverse implication is symmetrical. • Problems
1.
Let V be the permutation module for G = 83, with basis example of a nontrivial kGsubmodule of V.
VI' V2,
v 3 . Give an
In Problems 24, we show that Maschke's theorem can fail if the characteristic of k divides the order of G. Let G = {I, a, ... , aP I } be a cyclic group of prime order p, and let Vbe a twodimensional vector space over the field Fp ' with basis VI' v2 . Take the matrix of a as
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Noncommutative Algebra
[a]
=
[~
:1
so that [a] =
[~ :1
and [aP] is the identity. 2.
Show that W, the onedimensional subspace spanned by v)' is a kGsubmodule ofV.
3.
Continuing Problem 2, show that W is the only onedimensional kGsubmodule ofV.
4.
Continuing Problem 3, show that V is not a semisimple kGmodule.
5.
Show that a semisimple module is Noetherian iff it is Artinian.
6.
Let M be a decomposable Rmodule, so thatM is the direct sum of nonzero submodules M) and M 2 • Show that EndiM) contains a nontrivial idempotent e (that is, e2 = e with e not the zero map and not the identity).
7.
Continuing from Problem 6, suppose conversely that EndiM) contains a nontrivial idempotent e. Show that M is decomposable. (Suggestion: use e to construct idempotents e) and e2 that are orthogonal, that is, e)e 2 = e2e) = 0.) THE JACOBSON RADICAL
There is a very useful device that will allow us to look deeper into the structure of rings. DEFINITION
The Jacobson radical J(R) of a ring R is the intersection of all maximal left ideals of R. More generally, the Jacobson radical J(M) = JR(M) of an Rmodule M is the intersection of all maximal submodules of M. ["Maximal submodule" will always mean "maximal proper submodule".] If Mhas no maximal submodule, take J(M) = M. If M is finitely generated, then every submodule N of M is contained in a maximal submodule, by Zorn's lemma. [If the union of a chain of proper submodules is M, then the union contains all the generators, hence some member of the chain contains all the generators, a contradiction.] Taking N = 0, we see that J(M) is a proper submodule of M. Since R is finitely generated (by IR), J(R) is always a proper left ideal. Semi simplicity of M imposes a severe constraint on J(M).
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Noncommutative Algebra
Proposition
If Mis semisimple, then J(M) "obstruction" to semisimplicity.
= O. Thus in a sense, the Jacobson radical is an
Proof Let N be any simple submodule of M. M = N E9 N' for some submodule N. Now MIN ~ N, which is simple, so by the correspondence theorem, N' is maximal. Thus J(M) ~ N, and therefore J(M) n N = O. Since M is a sum of simple modules, J(M) = J(M) nM= O·
Here is another description of the Jacobson radical. Proposition
J(R) is the intersection of all annihilators of simple Rmodules. Proof Simple modules are isomorphic to RlI for maximal left ideals I. If r annihilates all simple Rmodules, then for every maximal left ideal I, r annihilates RlI, in particular, r annihilates 1 + I. Thus r(1 + l) = I, that is, rEI. Consequently, r E J(R). Conversely, assume r E J(R). If Mis a simple Rmodule, choose any nonzero element x E M. The map!x : R ~ M given by!x(s) = sx is an epimorphism by simplicity of M. The kernel ofjx is the annihilator of x, denoted by annex). By the first isomorphism theorem, M ~
RI annex). By simplicity of M, annex) is a maximal left ideal, so by hypothesis, r E nXEm
annex) = ann(M). Thus r annihilates all simple Rmodules. Corollary
J(R) is a twosided ideal.
DEFINITION The element a E R is left quaSiregular (lqr) if 1  a has a left inverse, right quasiregular (rqr) if 1  a has a right inverse, and quaSiregular (qr) if 1  a is invertible. Note that if a is both lqr and rqr, it is qr, because if b(1  a) = (1  a)c = 1, then b=bl =b(1a)c= Ic=c.
LEMMA Let I be a left ideal of R. If every element of I is lqr, then every element of I is qr. Proof If a E I, then we have b(1 a) = I for some b E R. Let c = 1  b, so that (1 c)(1  a) = I  a  c + ca = 1. Thus c = ca  a = (c  I)a E I. By hypothesis, c is lqr, so I  c has a left inverse. But we know that (1  c) has a right inverse (1  a), so c is rqr. c is qr and I  c is the twosided inverse of 1  a.
Noncommutative Algebra
269
Proposition
The Jacobson radical J(R) is the largest twosided ideal consisting entirely of quasiregular elements. Proof First, we show that each a E J(R) is lqr, each a E J(R) is qr. If I  a has no left inverse, then R(I  a) is a proper left ideal, which is contained in a maximal left ideal!. But then a E I and I  a E I, and therefore I E I, a contradiction.
Now we show that every left ideal (hence every twosided ideal) I consisting entirely of quasiregular elements is contained in J(R). If a E I but a ¢ J(R), then for some maximal left ideal L we have a ¢ L. By maximality of L, we have 1+ L = R, so I = b + C for some bEl, eEL. But then b is quasiregular, so c = I  b has an inverse, and consequently I E L, a contradiction. Corollary
J(R) is the intersection of all maximal right ideals of R. Proof We can reproduce the entire discussion beginning with left and right ideals interchanged, and reach exactly the same conclusion, namely that the "right" Jacobson radical is the largest twosided ideal consisting entirely of quasiregular elements. It follows that the "left"and "right" Jacobson radicals are identical.
We can now use the Jacobson radical to sharpen our understanding of semisimple modules and rings. Theorem
If M is a nonzero Rmodule, the following conditions are equivalent: (1) Mis semisimple and has finite length, that is, has a composition series;
(2) Mis Artinian and J(M) = o. Proof (l) implie~ (2), so assume M Artinian with J(M) = o. The Artinian condition implies that the collection of all finite intersections of maximal submodules of M has a minimal element N. If S is any maximal submodule of M. then N (\ S is a finite intersection of maximal submodules, so by minimality of N, N (\ S = N, so N ~ S. Since J(M) is the intersection of all such S, the hypothesis that J(M) = 0 implies that N = O. Thus for some
positive integer n we have maximal submodules M I , Now M is isomorphic to a submodule of M' =
... ,
Mn such that
EB~I (MIMi).
n7=1 Mi = o.
To see this, map x
E
M to
+ M I , ... , x + Mn) and use the first isomorphism theorem. Since M' is a finite direct sum of simple modules, it is semisimple and has a composition series. The same is true for M.
(x
270
Noncommutative Algebra
Corollary
The ring R is semisimple if and only if R is Artinian and J(R) =
o.
Proof It suffices to show that if R is semisimple, then it has a composition series. But this follows because R is finitely generated, hence is a finite direct sum of simple modules. The Jacobson radical of an Artinian ring has some special properties.
DEFINITION An ideal (or left ideal or right ideal) I of the ring R is nil if each element x E I is nilpotent, that is, rn = 0 for some positive integer m; I is nilpotent if l" = 0 for some positive integer n. Every nilpotent ideal is nil, and the converse holds if R is Artinian, as we will prove.
LEMMA If I is a nil left ideal of R, then I
Proof If x ... +xml.
E
~
J(R).
I and rn = 0, then x is quasiregular; the inverse of 1  x is 1 + x + Xl +
Proposition
If R is Artinian, then J(R) is nilpotent. Thus J(R) is the largest nilpotent ideal of R, and every nil ideal of R is nilpotent.
Proof Let J = J(R). The sequence J;;2 J2 ;;2 ... stabilizes, so for some n we have J'l = J'l+l = ... , in particular, J'l = J2n. We claim that J'l = o. If not, then the collection of all left ideals Q of R such that J'lQ "* 0 is nonempty (it contains J'l), hence has a minimal element N. Choose x E N such that J'lx "* O. By minimality of N, J'lx = N. Thus there is an element e E J'l such that ex = x, that is, (1  e)x = O. But e E:: J'l ~ J; 1  e is invertible, and consequently x = 0, a contradiction. Problems
1.
Show that an Rmodule is M cyclic if and only if M is isomorphic to RlI for some left ideal I, and in this case we can take I to be ann(M) , the annihilator ofM.
2.
Show that the Jacobson radical of an Rmodule M is the intersection of all kernels of homomorphisms from M to simple Rmodules.
3.
If I
4.
IfJis an Rmodule homomorphism from Mto N, show thatj(J(M»
5.
Assume R commutative, so that J(R) is the intersection of all maximal ideals of R. If a E R, show that a E J(R) if and only if 1 + ab is a unit for every b E R.
= J(R), show that J(RI!) = O. ~
J(N).
271
Noncommutative Algebra
If N is a submodule of the Jacobson radical of the Rmodule M, show that J(M)/N = J(M/N).
6.
THEOREMS OF HOPKINSLEVITZKI AND NAKAYAMA We know that a Noetherian ring need not be Artinian, and an Artinian module need not be Noetherian. But the latter situation can never arise for rings, because ofthe following result.
THEOREM Let R be an Artinian ring, and M a finitely generated Rmodule. Then Mis both Artinian and Noetherian. In particular, with M = R, an Artinian ring is Noetherian. Proof Mis Artinian. Let J be the Jacobson radical of R. The Jacobson radical of RI J is zero, and since RlJ is Artinian, it is semisimple. Now consider the sequence Mo
= M,
Ml =JM, M2 =J2M, .... J is nilpotent, so Mn = 0 for some n. Since JMi = M i+ l , J annihilates M/Mi+ l , M/Mi+ 1 is an RlJmodule.
We claim that each M/Mi+ l has a composition series. We can assume that M/Mi+ l "* 0, otherwise there is nothing to prove. M/Mi+ 1 is semisimple, M/Mi+1 is Artinian. [Note that submodules of M/Mi+ l are the same, whether we use scalars from R or from RlJ.] M/Mi+ l is Noetherian, hence has a composition series. Now intuitively, we can combine .,!he composition series for theM/Mi+ 1 to produce a composition series for M, proving that M is Noetherian. Formally, M n 1 ~ Mn_lMn has a composition series. Since MniMn1 has a composition series, so does M n 2. Iterate this process until we reach M. We now proceed to a result that has many applications in both commutative and noncommutative algebra.
NAKAYAMA'S LEMMA Let M be a finitely generated Rmodule, and I a twosided ideal of R. If I ~ J(R) and 1M= M, then M= O. Proof Assume M"* 0, and let Xl'
(Then
n
... , Xn
generate M, where n is as small as possible.
2: 1 and the Xi are nonzero.) Since xn
EM = 1M, we can write xn = 2::1 b,y,
for
some bi E I andYi EM. ButYj can be expressed in terms of the generators aSYi= 2:~=laijXi with
aij E
R. Thus
272
Noncommutative Algebra
where cj = .2::::1bjaij . Since I is a right ideal, cj
E
I ~ J(R). (We need Ito be a left
ideal to make 1M a legal submodule of M.) The above equation can be written as
and I  cn is invertible. If n > 1, then xn is a linear combination of the other x/ s, contradicting the minimality ofn. Thus n = 1, in which case (1 cl)x l = 0, so xl = 0, again a contradiction. There is another version of Nakayama's lemma, which we prove after a preliminary result.
LEMMA Let N be a submodule of the Rmodule M, I a left ideal of R. Then M only if MIN = 1(M/N). Proof Assume M = N + 1M, and let X + N z
E
1M. Write z = x
.2::::=1 bj Wj ,
ai
E
I,
wi E
E
= N + 1M if and
lWN. Then x = y + Z for some YEN and
M. It follows that
+ N = al(w l + N) + ... + at(wt + N)
Conversely, assume MIN = I(MlN), and let x
E
E
1(M/N).
M. Then
t
x+N=
with a·I
E
I and w,'
.2::: a j(Wj ;=1
E
+N)
M. Consequently, x  " , I a/,wj L..,.,=1
E
N, so X
E
N + 1M.
NAKAYAMA'S LEMMA Let N be a submodule of the Rmodule M, with MIN finitely generated over R. [This will be satisfied if M is finitely generated over R.] If I is a twosided ideal contained in J(R), and M = N + 1M, then M = N. Proof I(MlN)
= MIN,
MIN = 0, hence M
= N.
Here is an application of Nakayama's lemma. Proposition
Let R be a commutative local ring with maximal ideal J. Let Mbe a finitely generated Rmodule, and let V = MlJM. Then:
Noncommutative Algebra
273
V is a finitedimensional vector space over the residue field k = R/J.
(i)
(ii) If {xl + JM, ... , xn + JM} is a basis for V over k, then {xl' ... , xn} is a minimal set of generators for M. (iii) Any two minimal generating sets for M have the same cardinality.
Proof (i) Since J annihilates MlJM, that V is a kmodule, that is, a vector space over k. Since M is finitely generated over R, V is finitedimensional over k.
(ii) Let N = L:~=1 Rx; . Since the xi + JM generate V = MlJM, we have M = N + JM. M = N, so the xi generate M. If a proper subset of the Xi were to generate M, then the corresponding subset of the Xi + JM would generate V, contradicting the assumption that V
is ndimensional. (iii) Agenerating set S for M with more than n elements determines a spanning set for V, which must contain a basis with exactly n elements. By (ii), S cannot be minimal. Problems
= (at+ 1) for some
1.
Let a be a nonzero element of the integral domain R. If (at) positive integer t, show that a is invertible.
2.
Continuing Problem 1, show that every Artinian integral domain is a field.
3.
If R is a commutative Artinian ring, show that every prime ideal of R is maximal.
4.
Let R be a commutative Artinian ring. If S is the collection of all finite intersections of maximal ideals of R, then S is not empty, hence contains a minimal element I = II n 12 n ... n In' with the ~ maximal. Show that if P is any maximal ideal of R, then P must be one of the ~. Thus R has only finitely many maximal ideals.
5.
Viewed as a left Rmodule, R = Z [X] EB Z [X]Y . Show that R is leftNoetherian.
Geometry of Algebra
VARIETIES DEFINITION
We will be working in k[X\, ... , X n], the ring of polynomials in n variables over the field k. (Any application of the Nullstellensatz requires that k be algebraically closed, but we Vi~;i1 not make this assumption until it becomes necessary.) The set An = An(k) of all ntuples with components in k is called affine nspace. If S is a set of polynomials in k[X\, ... , X n], then the zeroset of S, that is, the set V = V (S) of all x E An such thatf(x) = 0 for every f E S, is called a variety. (The term "affine variety" is more precise, but we will use the short form because we will not be discussing projective varieties.) Thus a variety is the solution set of simultaneous polynomial equations. If I is the ideal generated by S, then I consists of all finite linear combinations
2.::::: gJ;
with gj E k[X\, ... , Xn] and.!; E S. It follows that V(S) = V (1), so every variety is the variety of some idfal. We now prove that we can make An into a topological space by taking varieties as the closed sets.
Proposition 1.
If Va
=
V (Ia) for all a
E
T, then
nV;y = v(U1Q). Thus an arbitrary
intersection of varieties is a variety. 2.
If ~ = V(~.),j = 1, ... , r, then U~=IVj = V({f\
:1,.:Jj E~,
1 ~ j ~ r}). Thus
a finite union of varieties is a variety. 3.
An
= V (0) and = V(1), so the entire space and the empty set are varieties.
Geometry of Algebra
275
Consequently, there is a topology on An, called the Zariski topology, such that the closed sets and the varieties coincide. Proof (l) If x X E
E
An, then x
E
nV;, iff every polynomial in every la vanishes at x iff
V(U1a). 2.
x E U~=l Vj iff for some}, every fj
E
~ vanishes at x iff x
E
V ({fl ... .t;.
:fj
E
~ for
all)} ). 3. The zero polynomial vanishes everywhere and a nonzero constant polynomial vanishes nowhere. Note that condition (2) can also be expressed as
We have seen that every subset of k[XI' ... , X n], in particular every ideal, determines a variety. We can reverse this process as follows. If X is an arbitrary subset of An, we define the ideal ofX as 1(X) = {f E k[X1, fvanishes on X}. By definition we have: 4.
If X
~
Ythen 1(X)
;;;2
I(Y); if S ~ Tthen V (S)
;;;2
... ,
Xn] :
V (1).
Now if S is any set of polynomials, define IV (S) as I(V (S)), the ideal oftlie zeroset of S; we are simply omitting parentheses for convenience. Similarly, if X is any subset of An, we can define V leX), IV leX), V IV (S), and so on. From the definition we have:
5.
lV(S);;;2 S; V 1(X)
;;;2
x.
[Iff E S thenfvanishes on V (S), hencef E IV (S). Ifx 1(X) vanishes at x, so x belongs to the zeroset of 1(X).]
E
Xthen every polynomial in
Ifwe keep applying V's and rs alternately, the sequence stabilizes very quickly: 6.
V IV (S) = V (S); IV 1(X) = leX).
[In each case, apply (4) and (5) to show that the left side is a subset of the right side. If x E V (S) andf E IV (S) thenj(.;\,) = 0, so X E V IV (S). Iff E 1(X) and x E V 1(X) then x belongs to the zeroset of leX), soj(x) = O. Thusfvanishes on V leX), sof E IV 1(X).] Since every polynomial vanishes on the empty set (vacuously), we have:
7.
1( 1, letf= arx{ + .. , + aIXI + a o where the a i are polynomials in X 2, .. " Xn and ar:;t: O. By the induction hypothesis, there is a point (x 2, .. " xn ) at which ar does not vanish. Fixing this point, we can regardfas a polynomial in Xl' which cannot possibly vanish at all xl E k. Thusf ~ I(An) , To prove (9), note that the right side is contained in the left side because Xi  a i is 0 when X; = a i · Also, the result holds for n = 1 by the remainder theorem. Thus assume n > 1 and letf= brX{ + .. , + biXI + bo E I({x}), where the bi are polynomials inX2' .. " Xn and br:;t: O. By the division algorithm, we have f= (Xl  aI)g(XI, .. " Xn) + h(X2, .. " Xn) and h must vanish at (a 2, .. " an)' By the induction hypothesis, hE (X2  a2, .. " X n all ), hencef E (Xl  aI' X 2  a 2. ... , Xn  an)' Problems
A variety is said to be reducible if it can be expressed as the union of two proper snbvarieties; otherwise the variety is irreducible. In Problems 14, we are going to show th~tt a variety V is irreducible if a!1d only if I( V) is a prime ideal.
2. 3.
Assume that I(V) is not prime, and letf[2 E I(V) withfI' f2 ~ I(V). If X E V, show that x ~ V ([1) implies x E V (12) (and similarly, x ~ V ([2) implies x E V ([1))' Show that V is reducible. Show that if Vand Ware varieties with V, c W, then I(V) :::) I(W).
4.
Now assume that V
5. 6.
choose J; E ideal. Show that any variety is the union of finitely many irreducible subvarieties. Show that the decomposition of Problem 5 is unique, assuming that we discard any subvariety that is contained in another one, Assume that k is algebraically closed, Suppose that An is covered by open sets AnI V (1) in the Zariski topology. Let I is the ideal generated by the I" so that I
1.
7.
= VI UV2' with VI' V2, c V. By Problem 3, we can I( Vi) with J; ~ I( V). Show that f [2 E I( V), so I( V) is not a prime
= L:1i , the set of all finite sums 8,
Xii
+ .. ,+ Xi
y
with
Show that An is compact in the Zariski topology.
Xi)
Eli) . Show that 1
E
1.
Geometry of Algebra
277
HILBERT BASIS THEOREM If S is a set of polynomials in k[XI , """' Xn]' we have defined the variety V (S) as the zeroset of S, and we know that V(S) = Vel), where 1 is the ideal generated by S. Thus any set of simultaneous polynomial equations defines a variety. In general, infinitely many equations may be involved, but as Hilbert proved, an infinite collection of equations can always be replaced by a finite collection. The reason is that every ideal of k[XI, """' Xn] has a finite set of generators, in other words, k[XI' """' Xn] is a Noetherian ring. The field k is, in particular, a PID, so k is Noetherian. The key step is to show that if R is a Noetherian ring, then the polynomial ring in n variables over R is also Noetherian.
HILBERT BASIS THEOREM If R is a Noetherian ring, then R[XI' """' Xn] is also Noetherian. Proof By induction, we can assume n = 1. Let 1 be an ideal of R[X], and let J be the ideal of all leading coefficients of polynomials in 1. (The leading coefficient of 5X2  3X + 17 is 5; the leading coefficient of the zero polynomial is 0.) By hypothesis, J is finitely generated, say by aI' """' an' Leth be a polynomial in 1 whose leading coefficient is ai' and let d j be the degree ofh . Let 1* consist of all polynomials in 1 of degree at most d = max {dj : 1 ~ i ~ n}. Then 1* is an Rsubmodule of the free Rmodule M of all polynomials b o+ blX + """ + b~, b j E R. Now a finitely generated free Rmodule is a finite direct sum of copies of R, hence M, and therefore 1*, is Noetherian. Thus 1* can be generated by finitely many polynomials gl' """' gm" Take 10 to be the ideal of R[X] generated by fl' """' Irf gl' """' gm' We will show that 10 = I, proving that 1 is finitely generated.
First observe thath Eland ~ E 1* ~ I, so 10 ~ 1. Thus we must show that each h E 1 belongs to 10 , Case 1 : deg h ~ d
Then hE 1*, so h is a linear combination of the gj (with coefficients in R ~ R[X]), so hE 10 , Case 2 : deg h
=r
> d
Let a be the leading coefficient of h. Since a E J, we have a = L~=lc;a; with the c j E R. Let n
q = h Lc;X
r

;=1
The coefficient of X' in q is n
a Lc;a;=O ;=1
d
;
fi E I.
278
Geometry of Algebra
so that deg q < r. We can iterate this degreereduction process until the resulting polynomial has degree d or less, and therefore belongs to 10 , But then h is a finite linear combination of the 1; and g/ Corollary
Every variety is the intersection of finitely many hypersurfaces (zerosets of single polynomials). Proof Let V= Vel) be a variety. Ihas finitely many generators.fi, ... ,!,.. But then V= n;=I V (h).
FORMAL POWER SERIES The argument used to prove the Hilbert basis theorem can be adapted to show that if R is Noetherian, then the ring R[[X]] of formal power series is Noetherian. We cannot simply reproduce the proof because an infinite series has no term of highest degree, but we can look at the lowest degree term. Iff= a,xr + ar+IXr+1 + ... , where r is a nonnegative integer and a r 7= 0, let us say thatfhas degree r and leading coefficient ar' (Iff= 0, take the degree to be infinite and the leading coefficient to be 0.) If 1 is an ideal of R[[X]], we must show that 1 is finitely generated. We will inductively construct a sequence ofelements1; E R[[X]] as follows. Letfl have minimal degree among elements of 1. Suppose that we have chosenfl' ... ,1;, where 1; has degree d j and leading coefficient a r We then selectfi+1 satisfying the following three requirements: 1. 2.
1;+ I belongs to 1; a j +1 does not belong to (aI' ... , a), the ideal of R generated by the aj'i = l, ... , i;
3. Among all elements satisfying the first two conditions,1;+1 has minimal degree. The second condition forces the procedure to terminate in a finite number of steps; otherwise there would be an infinite ascending chain (a l ) c (aI' a2) c (aI' a2, a 3) c .... If stabilization occurs at step k, we will show that 1 is generated by fl' ... , fie Let g = aJ(!i + ... be an element of 1 of degree d and leading coefficient a. Then a ... , k) (Problem 1).
E
(aI'
Case 1 : d 2: dk. Since di :::s; di+1 for all i (Problem 2), we have d 2: d j for i = 1, ... , k.
Now a
= I:::=lc;Oa; with the c iO
E
k
g =
o
I:: c;oX i=1
d d  ,
h
R. Define
Geometry of Algebra
279
so that go has degree d and leading coefficient a, and consequently g  go has degree greater than d. Having defined go' ... , gr greater than d + r, say
E(fl' ... , he) such that g  L:;=I/;
has degree
r • g Lgi=bX d +r +1 + .... i=O
(The argument is the same if the degree is greater than d + r + 1.) Now b (Problem 1 again), so
E
(ai' ... , ak )
k
b = L:Ci,r+1 ai i=1
with ci' r+1
E
R. We define k
_ '"' c. Xd+r+1dj I". gr+1  L..J l,r+1 Ji i=1
so that g

,",r+1
L..J i=O gi has degree greater than d + r + I. Thus N
g
N
k
= L:gr = L:~CirXd+rd, /; r=O
r=O i=1
and it follows upon reversing the order of summation that g E (fj, ... , he). (The reversal is legal because the inner summation is finite. For a given nonnegative integer j, there are onJy :f:in.:iteJy m any t:elm s of the:fDnn bXi.) Case 2 : d < dk . As above, a
E
(ai' ... , ak ), so there is a smallest m between 1 and k
such that a E (ai' ... , am)' It follows that d 2:: dm. As in case 1 we have a = ~:ociai with c j E R. Define m
h = L:ciX d  d, /;
E (fi, ... ,Jk) ~
I.
i=1
The leading coefficient of h is a, so the degree of g  h is greater than d. We replace g by g  h and repeat the procedure. After at most dk  d iterations, we produce an element g  ~hi in I of degree at least dk, with all hi E (fl' ... , he). By the analysis of case 1, g E (fl' ···,jk)·
NULLSTELLENSATZ: PRELIMINARIES We have observed that every variety V defines an ideal I(V) and every ideal I defines a variety V (1). Moreover, if I(VI ) = I(V2 ), then VI = V2 . But it is entirely possible for many ideals to define the same variety. For example, the ideals (j) and (f') need not coincide,
280
Geometry of Algebn:
but their zerosets are identical. Appearances to the contrary, the two statements in part are not symmetrical. Avariety V is, by definition, always expressible as V (8) for some collection S of polynomials, but an ideal I need not be of the special form I(X). Hilbert's Nullstellensatz says that if two ideals define the same variety, then, informally, the ideals are the same "up to powers". More precisely, if g belongs to one of the ideals, then g belongs to the other ideal for some positive integer r. Thus the only factor preventing a onetoone correspondence between ideals and varieties is that a polynomial can be raised to a power without affecting its zeroset. In this section we collect some results needed for the proof of the Nullstellensatz. We begin by showing that each point of An determines a maximal ideal.
LEMMA If a = (aI' ... , an)
E
An, then I
= (Xl
 aI' ... , Xn  an) is a maximal ideal.
Proof Suppose that I is properly contained in the ideal J, with IE J \ 1. Apply the division algorithm n times to get
1= AI(XI 
al)
+ AiX 2 
a2)
+ .,. + An(Xn 
an)
+b
where AlE k[XI , ... , X n], A2 E k[X2, ... , X n], ... , An E k[Xn], b E k. Note that b cannot be 0 sincel rt 1. Butl E J, so by solving the above equation for b we have b E J, hence 1 = (lIb)b E J. Consequently, J = k[XI , ... , X,J. The following definition will allow a precise statement of the Nullstellensatz. DEFINITION The radical of an ideal I (in any commutative ring R) is the set of all elements IE R such thatf E I for some positive integer r. A popular notation for the radical of I is binomial theorem,
if + g) r+sl
E
.JJ. IfI
I, and it follows that
rand gs belong to I, then by the
.JJ is an ideal.
LEMMA If I is any ideal of k[XI , .... X n], then
.JJ ~ IV (1).
Proof Iff E .JJ, thenf E I for some positive integer r. But thenf vanishes on V (1), hence so does f Therefore I E IV (1).
The Nullstellensatz states that IV (1) =
.JJ, and the hard part is to prove that IV (1) ~
.JJ. The technique is known as the "Rabinowitsch trick", and it is indeed very clever. Assume thatf ~ IV (1). We introduce a new variable Yand work in k[XI , ... , Xw Y]. If I is an ideal of k[XI, ... , X n], then by the Hilbert basis theorem, I is finitely generated, say byfi,
Geometry of Algebra
281
... , f m· LetI* be the ideal of k[XI, ... , X n, Y] generated by ii, ... ,fm' 1  Y f. [There is a slight ambiguity: by J;(XI , ... , Xn' Y) we meanJ;(XI' ... , X n), and similarly forf.] At an appropriate moment we will essentially set Yequal to 1/j and come back to the original problem.
LEMMA If (aI' ... , art an+ I ) is any point in An+I and (aI' ... , an) E V(1) (in other words, the!;, i = 1, ... , m, vanish at (aI' ... , an»' then (aI' ... , an' an+ l ) t/. V (1*). Proof. We are assuming thatf E JV(1), so thatfvanishes on the zeroset of {fI' ... ,fm }. In particular, fiaI, ... , an) = O. The value of 1 E Y f at (aI' ... , an' an+ l ) is therefore 1 an+J{a l , ... , an) = 1  an+I(O) = 1 = O. But 1  Y f  J*, so (aI' ... , an' an+l) does not belong to the zeroset of J*.
LEMMA If(a l , ... , an' an+ l ) is any point inAn+I and (aI' ... , an) V (1*). Consequently, V (1*) = $.
t/. V (1), then (aI' ... , an' an+l) t/.
Proof By hypothesis,J;(a l , ... , an' an+ I ) = 0 for some i, and since!; cannot belong to the zeroset of J*.
E
1*, (aI' ... , a n+ l )
At this point we are going to assume what is called the weak Nullstellensatz, namely that if J is a proper ideal of k[XI, ... , X n ], then V (1) is not empty.
LEMMA There are polynomials gl' ... , gm' h
E
k[XI' ... , Xn, Y] such that
m
1=
"'£ gJ; + h(l Yf). ;=1
This equation also holds in the rational function field k(XI, ... , X n, Y) consisting of quotients of polynomials in k[XI, ... , X n, Y]. Proof V (1*) = $, so by the weak Nullstellensatz, J* = k[XI, ... , X n, Y]. In particular, 1 E J*, and since J* is generated by fl' ... , fm' 1  Yf, there is an equation ofthe specified form. The equation holds in the rational function field because a polynomial is a rational function. RABINOWITSCH TRICK
The idea is to set Y= 1/f, so that (1) becomes
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Geometry of Algebra
Is this legal? First of all, ifJis the zero polynomial, then certainlyf E JI, so we can assumeJ* O. To justify replacing Yby Iff, consider the ring homomorphism from k[XI' ... , X,l' Y] to k(XI , ... , Xn) determined by ~ ~ Xi' i = 1, ... , n, Y ~ I1{XI , ... , X n). Applying this mapping to (1), we get (2). Now the right side of (2) is a sum of rational functions whose denominators are various powers off Iff is the highest power that appears, we can absorb all denominators by mUltiplying (2) by f. The result is an equation of the form m
fr
= L:hi(X\> ... ,Xn).!;(X\' .... ,Xn ) i=1
where the hi are polynomials in k[XI , ... , X n]. Consequently,Jr
E
1.
The final ingredient is a major result in its own right.
NOETHER NORMALIZATION LEMMA Let A be a finitely generated kalgebra, where k is a field. In other words, there are finitely many elements Xl' ... , Xn i~ A that generate A over k in the sense that every element of A is a polynomial in the Xi' Equivalently, A is a homomorphic image of the polynomial ring k[XI , ... , Xn] via the map determined by ~ ~ Xi' i = 1, ... , n. There exists a subset {YI' ... , Yr} of A such that the Yi are algebraically independent over k and A is integral over k[YI' ... , Yr ]. Proof Let {Xl' ... , x r } be a maximal algebraically independent subset of {XI' ... , x n}. If n = r we are finished, since we can take Yi = Xi for all i. Thus assume n > r, in which case Xl' ... , Xn are algebraically dependent over k. Thus there is a nonzero polynomialJ E k[Xl' ... , Xn] such thatj{x l, ... , xn) = O. We can assume n > 1, for ifn = 1 and r = 0, then A = k[xl]
and we can take {YI' ... , Y r } to be the empty set.
We first assume that k is infinite and give a proof by induction on n. (It is possible to go directly to the general case, but the argument is not as intricate for an infinite field.) Decomposefinto its homogeneous components (sums of monomials of the same degree). Say that g is the homogeneous component of maximum degree d. Then, regarding g as a polynomial in Xn whose coefficients are polynomials in the other Xi' we have, relabeling variables if necessary, g(XI, ... , X n_l , 1) * O. Since k is infinite, that there are elements ai' ... , an_I E k such that g(a l , ... , an_I' 1) * O. Set zi = Xi  afn' i = 1, ... , n  1, and plug into j{x l , ... , xn) = 0 to get an equation of the form g(a l , ... , an_I' 1) x~
+ terms of degree less than din xn = O.
Aconcrete example may clarify the idea. Ifj{xl' x2) a l x 2 , then the substitution yields
= g(xl'
x 2) =
xlxi
and XI = zi +
Geometry of Algebra
283
which indeed is g(a l , l)x~ plus terms of degree less than 5 in x 2• Divide by g(a l ,
... ,
anI' I) *" 0 to conclude thatxn is integral over B = k[zl' ... , zntl. By the induction hypothesis, there are elements YI' ... , Yr algebraically independent over k such that B is integral over k[YI' ... , Yr ]· But the Xj' i < n, are integral over B since Xj = Zj + a By transitivity, Xl' ... , Xn are integral over k[Yl' ... , Yr ]. A is integral over k[YI' ... , y r ].
rn .
Now we consider arbitrary k. As before, we produce a nonzero polynomialfsuch that = O. We assign a weight Wj = snj to the variable~, where s is a large positive integer. (It suffices to take s greater than the total degree off, that is, the sum of the degrees
j{x l , ... , xn)
of all monomials inf) If h = >Xfl .. X:/l is a monomial off, we define the weight of has w(h)
= 2::1 a jWj .The point is that if h ' = f1X~ .. X!/l,
then w(h) > w(h) iff h > h' in the
lexicographic ordering, that is, for some m we have a j = bj for i ::; m, and am+1 > bm+1' We take h to be the monomial of maximum weight. (If two monomials differ in the lexicographic ordering, they must have different weights.) Set Zj =
X::i
::;
1 ::; i ::; n  I, and plug into
j{x l , ... , xn) = 0 to get CX:'(h)
+ terms of lower degree in xn = O.
and w(h) = 3w I + 2w2 = 3w 1 + 2 since sl12 = sO = 1. Thus xn is integral over B = k[zl, ... , zntl, and an induction argument finishes the proof as in the first case. Corollary
Let B be a finitely generated kalgebra, where k is a field. If I is a maximal ideal of B, then BII is a finite extension of k.
Proof The field k can be embedded in BII via c ~ c + I, Ie
C E
k. [If eEl, e *" 0, then e
= 1 E I, a contradiction.] Since A = BII is also a finitely generated kalgebra, it follows
from that there is a subset {Yl' ... , Yr } of A with the Yj algebraically independent over k and A integral over k[Yl' ... , y r]. Now A is a field (because I is a maximal ideal), and therefore so is k[Yl' ... , Y,J But this will lead to a contradiction ifr ~ I, because lIYl ¢ k[Yl' ... , y r]. (If IIYl = g(yl' ... , Yr ) E k[Yl' ... , y r ], then YIg(yI' ... , Yr ) = 1, contradicting algebraic independence.) Thus r must be 0, so A is integral, hence algebraic, over the field k. Therefore A is generated over k by finitely many algebraic elements, A is a finite extension of k.
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Geometry of Algebra
Corollary
Let A be a finitely generated kalgebra, where k is a field. If A is itself a field, then A is a finite extension of k. Problems
I.
Let S be a multiplicative subset of the ring R. If I is an ideal that is disjoint from S, then by Zorn's lemma, there is an ideal J that is maximal among ideals disjoint from S. Show that J must be prime.
2.
Show that the radical of the ideal I is the intersection of all prime ideals containing 1. [Iff E I ~ P, P prime, thenf E P. Conversely, assumef rt JI. With a clever choice of multiplicative set S, show that for s~me prime ideal P containing I, we havef rt P.]
3.
An algebraic curve is a variety defined by a nonconstant polynomial in two variables. Show (using the Nullstellensatz) that the polynomialsfand g define the same algebraic curve iff f divides some power of g and g divides some power off. Equivalently,fand g have the same irreducible factors.
4.
Show that the variety V defined over the complex numbers by the two polynomials y2  XZ and Z2  X2 Y is the union of the line L given by Y = Z = 0, X arbitrary, and the set W of all ([3, [4, (5), [E C.
5.
The twisted cubic is the variety V defined over the complex numbers by Y X2 and Z  X3. In parametric form, V = {(i, [2, (3) : [ E C}. Show that V is irreducible. [The same argument works for·any variety that can be parametrized over an infinite field.]
6.
Find parametrizations of the following algebraic curves over the complex numbers. (It is permissible for your parametrizations to fail to cover finitely many points of the curve.) a. The unit circle x 2 + y2
= I;
b. The cuspidal cubic y2 = ~; c. The nodal cubic y2 7.
= x2 + ~.
Letfbe an irreducible polynomial, and g an arbitrary polynomial, in k[x, y]. If f does not divide g, show that the system of simultaneous equations j{x, y) = g(x, y) = 0 has only finitely many solutions.
THE NULLSTELLENSATZ: EQUIVALENT VERSIONS AND PROOF We are now in position to establish the equivalence of several versions of the Nullstellensatz.
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Geometry of Algebra
THEOREM For any field k and any positive integer n, the following statements are equivalent. 1.
Maximal Ideal Theorem The maximal ideals of k[XI , ... , Xn] are the ideals of the form (Xl  aI' "" Xn  an)' aI' ,." an E k. Thus maximal ideals correspond to points.
2.
Weak Nullstellensatz If I is an ideal of k[XI , "" Xn] and V(1) = $, then 1= k[XI , "" Xnl. Equivalently, if I is a proper ideal, then V (1) is not empty. Nullstellensatz If I is an ideal of k[XI , "" X n], then
3.
IV (1)
4.
=
JI,
k is algebraically closed.
Proof (1) implies (2). Let I be a proper ideal, and let J be a maximal ideal containing 1. Part (4), V (.I) s;: V (1), so it suffices to show that V (.I) is not empty. By (1), J has the form (Xl  aI' .'" Xn  an)' But then a = (aI' "" an) E V (.I). [In fact V (.I) = {a}.]
We use the fact that the radical of an ideal I is the intersection of all prime ideals containing 1. Let I be a proper ideal of k[XI , "" Xn]. Then I is contained in a maximal, hence prime, ideal P. By the result just quoted, HI is also contained in P, hence JI is a proper ideal. But if V (1) = $, then, IV (1) = k[XI , "" X n], a contradiction. If I is a maxima! ideal, then by (2) there is a point a = (aI' "" an) E V (1), Thus every I vanishes at a, in other words, Is;: I( {a}). But (Xl  a l' "" Xn an) = I( {a}); to see this, decompose f E I( {a}). Therefore the maximal ideal I is contained in the maximal ideal (Xl  aI' ,.. , Xn  an)' and it follows that 1= (Xl  aI' "" Xn  an)' Let Ibe a maximal ideal of k[XI , "" X n], and let K = k[XI , "" Xn]/I, a field containing an isomorphic copy of kvia c ~ c + I, C E k, K is a finite extension of k, so by (4), K = k. But then ~ + 1= a j + I for some a j E k, i = 1, ... , n. Therefore ~ E a j is zero in k[XI , ,.. , Xn]/I, in other words, Xj  a j E 1. Consequently, I ~ (Xl  aI' .. " Xn  an)'
f
E
Let f be a nonconstant polynomial in k[Xtl with no root in k. We can regard f is a polynomial in n variables with no root in An' Let I be a maximal ideal containing the proper ideal (j). I is of the form (Xl  aI' .. " Xn  an) = I({a}) for some a = (aI' "" an) E An' Therefore f vanishes at a, a contradiction. Corollary
If the ideals I and J define the same variety and a polynomial g belongs to one of the ideals, then some power of g belongs to the other ideal.
E
Proof If V (1) = V (.I), then by the Nullstellensatz, J for some positive integer r.
I
JI = JJ . If gEl s;: JI ' then g
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Corollary
The maps V ~ I(V) and I ~ V (1) set up a onetoone correspondence between varieties and radical ideals (defined by I = Ji). Proof. V I(V) = V. By the Nullstellensatz, IV (1) = Ji
to prove that for any variety V, I(V) is a radical ideal. Iff on V, so! E I(V).
E
= I for radical ideals. It remains I(V), thenf, hence/, vanishes
Corollary
Let!I' ... , I,. g E k[XI , ... , X n], and assume that g vanishes wherever the ~ all vanish. Then there are polynomials hI' ... , hr E k[XI, ... , Xn] and a positive integer s such that if =
hJi + ..... + h/,..
Proof. Let I be the ideal generated by fi, ... ,!,.. Then V (1) is the set of points at which all~ vanish, so thatIV (1) is the set of polynomials that vanish wherever all~ vanish. Thus
g belongs to IV (1), which is
integer s, we have if
E
Ji
by the Nullstellensatz. Consequently, for some positive I, and the result follows.
Problems
I.
Let! be a polynomial in k[XI , into irreducibles is f
... ,
X n], and assume that the factorization of I
= ,hfIJ. ... frnr . Show that the decomposition of the variety
V if) into irreducible subvarieties is given by V if) =
U;=lU;)·
2.
Under the hypothesis of Problem 1, show that IV if) = (fl ... fr).
3.
Show that there is a onetoone correspondence between irreducible polynomials in k[XI, ... , Xn] and irreducible hypersurfaces in An(k), if polynomials that differ by a nonzero multiplicative constant are identified.
4.
For any collection of subsets Xi of An, show that I( CUi Xi)
5.
Show that every radical ideal I of k[XI , many prime ideals.
6.
In Problem 5, show that the decomposition is unique, subject to the condition that the prime ideals P are minimal, that is, there is no prime ideal Q with I ~ Q, ~P.
7.
Suppose that X is a variety in A2, defined by equations fi (x, y) = .... =!m(x, y) = 0, m ~ 2. Let g be the greatest common divisor of the~. If g is constant, show that X is a finite set (possibly empty).
... ,
= ni I(Xi)'
Xn] is the intersection of finitely
Geometry of Algebra
287
8.
Show that every variety in A2 except for A2 itself is the union of a finite set and an algebraic curve.
9.
Give an example of two distinct irreducible polynomials in k[X; Y] with the same zeroset, and explain why this cannot happen if k is algebraically closed.
10.
Give an explicit example of the failure of a version of the Nullstellensatz in a nonalgebraically closed field.
LOCALIZATION GEOMETRIC MOTIVATION Suppose that V is an irreducible variety, so that l(V) is a prime ideal. Ap olynomial g will belong to l(V) if and only if it vanishes on V. Ifwe are studying rational functions fig in the neighborhood of a point x E V, we must have g(x) = O. It is very convenient to have
;z£ leV) available as a legal object, even though g may vanish at some
every polYRomial g
points of V. The technical device that makes this possible is the construction of the ring of fractions SIR, the localization of R by S, where R = k[X1, ••• , Xn] and S is the multiplicative set R \ leV). We will now study the localization process in general.
NOTATION Let S be a multiplicative subset of the ring R, and Sl R the ring of fractions of R by S. Let h be the natural homomorphism of R into Sl R, given by h(a) = all. If X is any subset of R, define SIX= {xis: x EX; S E S}. We will be especially interested in such a set when X is an ideal. If land J are ideals of R, the product of land J, denoted by lJ, is defined as the set of all finite sums
I:: /
X/Yi ,xi E
l, Yi
E
J. It follows from the definition thatIJ is an ideal. The
sum of two ideals has already been defined.
LEMMA If 1 is an ideal of R, then Sl 1 is an ideal of Sl R. If J is another ideal of R, the
= Sl 1 + SIJ;
1.
SI(I + J)
2.
SI(IJ) = (SIl)(SIJ);
3.
SI(I n J)
4.
Sl 1 is a proper ideal iff S n 1 = .
= Sll n
SIJ;
Proof The definition of addition and mUltiplication in Sl R implies that Sl 1 is an ideal, and that in (1), (2) and (3), the left side is contained in the right side. The reverse inclusions in (1) and (2) follow from
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Geometry of Algebra
a b at + bs a b ab ,=s t st st st To prove (3), let als = bit where a E I, b E J, s, t  bs) = O. Then als = uatlust = ubslust E SI(I (1.1).
+=
E
S. There exists u
E
S such that u(at
Finally, if s E S (1 I then III = sis E SII, so SII= SIR. Conversely, if SII= SIR, then III = als for some a E I, s E S. There exists t E S such that t(s  a) = 0, so at = st E S(11.
Ideals in SI R must be of a special form.
LEMMA If J is an ideal of SIR and 1= h I(.1), then I is an ideal of Rand SII = J. E E
Proof I is an ideal by the basic properties of preimages of sets. Let als E SI I, with a I and s E S. Then all E J, so als = (all)(1/s) E J. Conversely, let als E J, with a E R, s S. Then h(a) = all = (als)(s/l) E J, so a E I and als E SI1. Prime ideals yield sharper results.
LEMMA If I is any ideal of R, then I ~ hI(SIl), with equality if I is prime and disjoint from S.
Proof If a E I, then h(a) = all E SI1. Thus assume that I is prime and disjoint from S, and let a E hI(SIl). Then h(a) = all E SII, so all = bls for some bEl, s E S. There exists t
E
S such that teas  b) = O. Thus as! = bt
is prime, we have a
E
E
I, with st
;i I since S (1 1= cj>. Since I
1.
LEMMA If I is a prime ideal of R disjoint from S, then SII is a prime ideal of SIR.
Proof Part (iv), SIJis a proper ideal. Let (als)(blt) = ablst E SII, with a, b E R, s, t E S. Then ablst = c/u for some eEl, u E S. There exists v E S such that v(abu  cst) = O. Thus abuv = cstv E I, and uv rt. I because S (1 1= cj>. Since I is prime, ab E I, hence a E I or b E 1. Therefore either als or bit belongs to SI1. The sequence of lemmas can be assembled to give a precise conclusion.
THEOREM There is a onetoone correspondence between prime ideals P of R that are disjoint from S and prime ideals Q of SI R, given by P + SI P and Q + hI(Q).
Geometry of Algebra
289
Proof S1(hI(Q)) = Q, and, hI(Slp) = P. Slp is a prime ideal, and hI(Q) is a prime ideal by the basic properties of preimages of sets. If hI(Q) meets S, part (iv), Q = S1(hI(Q)) = Sl R, a contradiction. Thus the maps P ~ Sl P and Q ~ hI(Q) are inverses of each other, and the result follows.
DEFINITION If P is a prime ideal of R, then S = RIP is a multiplicative set. In this case, we write R(P) for SIR, and call it the localization of R at P. (The usual notation is R p , but it's easier to read without subscripts.) If I is an ideal of R, we write I(P) for S11. We are going to show that R(P) is a local ring, that is, a ring with a unique maximal ideal. First we give some conditions equivalent to the definition of a local ring. Proposition For a ring R, the following conditions are equivalent. 1.
R is a local ring;
2.
There is a proper ideal I of R that contains all non units of R;
3.
The set of non units of R is an ideal.
Proof (I) implies (2). If a is a nonunit, then (a) is a proper ideal, hence is contained in the unique maximal ideal 1. (2) impli~s (2). If a and bare nonunits, so are a + band ra. If not: then I contains a unit, so 1= R, a contradiction. (3) implies (I). If I is the ideal of nonunits, then I is maximal, because any larger ideal J would have to contain a unit, so that J = R. If H is any proper ideal, then H cannot contain a unit, so H fS 1. Therefore I is the unique maximal ideal.
THEOREM R(P) is a local ring. Proof Let Q be a maximal ideal of R(P). Then Q is prime, Q = I(P) for some prime ideal I of R that is disjoint from S, in other words, contained in P. Thus Q = I(P) k PcP). If PCP) = R(P), then part (iv), P is not disjoint from S = RIP, which is impossible. Therefore PCP) is a proper ideal containing every maximal ideal, so it must be the unique maximal ideal. If R is an integral domain and S is the set of all nonzero elements of R, then Sl R is the quotient field of R. In this case, Sl R is a local ring, because any field is a local ring. ({ O} is the unique maximal ideal.) Alternatively, we can appeal to with P = {O}.
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LOCALIZATION OF MODULES If M is an Rmodule and S a multiplicative subset of R, we can essentially repeat the construction 1. Nutrition and Feeding to form the localization ~lM of M by S, and thereby divide elements of M by elements of S. If x, Y E M and s, t E S, we call (x, s) and (y, t) equivalent if for some U E S, u(tx  sy) = O. The equivalence class of (x, s) is denoted by xis, and addition is defined by .::.+y =!x+sy. s! sf E ~lR
Ifals
andxlt
ax s f
E ~1M,
we define
ax sf
In this way, ~lMbecomes an ~lRmodule. Exactly, if MandN are submodules ofa module L, then ~l(M + N) = ~lM + ~lN and ~l(M n N) = ~lM n ~1N.
Further properties will be given in the exercises. Problems
M, 1 + x is a
1.
Let M be a maximal ideal of R, and assume that for every x unit. Show that R is a local ring (with maximal ideal M).
2.
Show that if p is prime and n is a positive integer, then Z pn is a local ring
E
with maximal ideal (p). 3.
Let R be the ring of all n by n matrices with coefficients in a field F. If A is a nonzero element of Rand 1 is the identity matrix, is {l, A, A 2, ... } always a mUltiplicative set?
Let S be a mUltiplicative subset of the ring R. We are going to construct a mapping from Rmodules to ~l Rmodules, and another mapping from Rmodule homomorphisms to ~l Rmodule homomorphisms, as follows. If M is an Rmodllle, we let M ~ ~l M. Iff : M ~ N is an Rmodule homomorphism, we define ~1j: ~l M ~ ~l N by x s
I(x)
.+
S
Sincefis a homomorphism, so is ~l f 4.
If g : N
~
L and composition of functions is written as a product, show that
~l(gj) = ~l(g)~l(f), and if 1M is the idendty mappIfig on M, then ~1(1M) =
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291
IS1M. We say that Sl is a functor from the category of Rmodules to the category of Sl Rmodules.
5.
If
M~N~L is an exact sequence, show that
6.
7.
is exact. We say that Sl is an exact functor. Let R be the ring of rational functionsflg with/, g E k[Xl , ... , Xn] and g(a) "* 0, where a = (aI' ... , an) is a fixed point in An' Show that R is a local ring, and identify the unique maximal ideal. If M is an Rmodule and S is a multiplicative subset of R, denote SIMby Ms. If N is a submodule of M, show that (MlN)s ~ M/Ns'
PRIMARY DECOMPOSITION We have seen that every radical ideal in k[Xl , ... , Xn] can be expressed as an intersection of finitely many prime ideals. Anatural question is whether a similar result holds for arbitrary ideals. The answer is yes if we generalize from prime to primary ideals.
DEFINITION The ideal Q in the ring R is primary ,f Q is proper and whenever a product ab belongs to Q, either a E Q or bn E Q for some positive integer n. [The condition on b is equivalent to b E JQ.] An equivalent statement is that RlQ"* 0 and whenever (a + Q)(b + Q) = 0 in RlQ, either a + Q = 0 or (b + Q)n = 0 for some positive integer n. This says that if b + Q is a zerodivisor in RlQ, then it is nilpotent, that is, some power of b + Q is O.
It follows from the definition that every prime ideal is primary. Also, if Q is primary, then
JQ
is the smallest prime ideal containing Q. [Since
prime ideals containing Q, it suffices to show that
JQ
JQ
is the intersection of all
is prime. But if anb n E Q, then an
JQ. Note also that since Q is proper, it is contained in a maximal, hence prime, ideal, so JQ is also proper.] E
Q or bnm
E
Q for some m, so either a or b must belong to
If Q is primary and
JQ
= P,
we say that Q is Pprimary.
Examples
1.
In IE, the primary ideals are {O} and (]I), where p is prime. In IE 6' 2 and 3 are zerodivisors that are not nilpotent, and a similar situation will occur in IE m whenever more than one prime appears in the factorization of 111.
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Geometry of Algebra
2.
n
Let R = k[X where k is any field, and take Q = (X y3), the ideal generated by X and y3. This is a nice example of analysis in quotient rings. We are essentially setting X and y3 equal to zero, and this collapses the ring R down to polynomials ao + a 1y + a2 y2, with the ai ! k and arithmetic mod y3. Formally, RlQ is isomorphic to k[n/(y3). The zerodivisors in RlQ are of the form cY + dy2, c j k, and they are nilpotent. Thus Q is primary. If/! R, then the only way for /not to belong to the radical of Q is for the constant term of/to be nonzero. Thus
JQ
= (X Y).
Now we claim that Q cannot be a power of a prime ideal; this will be a consequence of the next result.
LEMMA If P is a prime ideal, then for every positive integer n, Proof Since P is a prime ideal containing pn, E
.jp;;
.jp;; = P.
~ P. If x E P, then xn E pn, so X
JP.
If Q = (K, y3) is a prime power pn, then its radical is P, so P must be (X, Y). But X E Q and X ¢ pn, n 2: 2; since Y belongs to P but not Q, we have reached a contradiction. After a preliminary definition, we will give a convenient sufficient condition for an ideal to be primary.
DEFINITION The nilradical N(R) ofa ring R is the set of nilpotent elements of R, that is, {x
E
R:
xn
= 0 for some positive integer n}. Thus N(R) is the radical of the zero ideal, which is the intersection of all prime ideals of R. Proposition
If the radical of the ideal Q is maximal, then Q is primary. Proof Since
JQ
is maximal, it must be the only prime ideal containing Q. By the
correspondence theorem and the fact that the preimage of a prime ideal is a prime ideal, RI Q has exactly one prime ideal, which must coincide with N(RlQ). Any element of RlQ that is not a unit generates a proper ideal, which is contained in a maximal ideal, which again must be N(RlQ). Thus every element of RlQ is either a unit or nilpotent. Since a zerodivisor cannot be a unit, every zerodivisor of RlQ is nilpotent, so Q is primary . . Corollary
If M is a maximal ideal, then M" is Mprimary for all n = 1, 2, ....
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Geometry of Algebra
Proposition If Q is a finite intersection of Pprimary ideals Qi , i
= 1, ... , n, then Q is Pprimary.
Prooj First note that the radical of a finite intersection of ideals is the intersection of the radicals. It follows that the radical of Q is P, and it remains to show that Q is primary. If ab E Q but a ¢ Q, then for some i we have a ¢ Qi. Since Q/ is Pprimary, b belongs to P
= JQi' But then some power of b belongs to Q.
We are going to show tha. in a Noetherian ring, every proper ideal I has a primary decomposition, that is, I can be expressed as a finite intersection of primary ideals.
LEMMA Call an ideal I irreducible if for any ideals J and K, I = J n K implies that I = J or I = K. If R is Noetherian, then every ideal of R is a finite intersection of irreducible ideals.
Prooj Suppose that the collection S of all ideals that cannot be so expressed is nonempty. Since R is Noetherian, S has a maximal element I, necessarily reducible. Let I = J n K, where I is properly contained in both J and K. By maximality of I, the ideals J and K are finite intersections of irreducible ideals, and consequently so is I, contradicting I E S. Ifwe can show that every irreducible proper ideal is primary, we then have the desired primary decomposition. Let us focus on the chain of reasoning we will follow. If I is an irreducible proper ideal of R, then by the correspondence theorem, 0 is an irreducible ideal of the Noetherian ring RlI. If we can show that 0 is primary in RlI, then again by the correspondence theorem, I is primary in R.
PRIMARY DECOMPOSITION THEOREM Every proper ideal in a Noetherian ring R has a primary decomposition. (We can drop the word "proper" if we regard R as the intersection of the empty collection of primary ideals.)
Prooj By the above discussion, it suffices to show that if 0 is an irreducible ideal of R, then it is primary. Let ab = 0 with a * O. Since R is Noetherian, the sequence of annihilators ann b ~ ann b2 ~ ann b 3 ~ ... stabilizes, so ann bn = ann bn+1 for some n. Ifwe can show that (a) n (b n ) = 0
we are finished, because a * 0 and the zero ideal is irreducible (by hypothesis). Thus let x = ca = db n for some c, dE R. Then bx = cab = db n+ 1 = 0 (because ab = 0), so d annihilates bn+ l , hence d annihilates bn . Thus x = db n = O.
Geometry of Algebra
294 Problems
1.
If II' ... , In are arbitrary ideals, show that
kyj~OFII 2.
Let I be the ideal (XY  Z2) in k[X, Y, Z], where k is any field, and let R = k[X, Y, Z]/1. If P is the ideal (X + 1, Z + /), show that P is prime.
3.
Let R
= k[X, YJ,
prime and 4.
pl
and let PI
= (X),
P 2 = (X, 1'),
Q = (X2'
1'). Show that PI is
and Q are P2primary.
Continuing Problem 4, let I
= (X2,
XY). Show that PIn
pl
and PIn Q are
both primary decompositions of 1. Notice that the radicals of the components of the primary decomposition (referred to as the primes associated with /) are PI and P 2 in both cases. [PI is prime, so ~
JQ
and P 2 is maximal, so P 2 =
JQ;]
JP 1 = PI; P 2
Uniqueness questions involving primary
decompositions are treated in detail in textbooks on commutative algebra. 5.
Let R = k[X, YJ and let In be the ideal (X3, XY, yn). Show that for every positive integer n, In is a primary ideal of R.
~ENSOR
PRODUCT OF MODULES OVER A COMMUTATIVE RING
MOTIVATION In many areas of algebra and its applications, it is useful to multiply, in a sensible way, an element x of an Rmodule M by an element Y of an Rmodule N. In group representation theory, M and N are free modules, in fact finitedimensional vector spaces, with bases {Xi} and {Yj}. Thus if we specify that multiplication is linear in each variable, then we need only specify products of Xi and Yj" We require that the these products, to be denoted by Xi ® Yj' form a basis for a new Rmodule T.
Iff: R
~
S is a ring homomorphism and M is an Smodule, then M becomes an Rmodule via rx = j(r)x, r E R, x E M. This is known as restriction of scalars. In algebraic topology and algebraic number theory, it is often desirable to reverse this process. If Mis an Rmodule, we want to extend the given multiplication rx, r E R, x E M, to multiplication of an arbitrary s E S by x E M. This is known as extension of scalars, and it becomes possible with the aid of the tensor product construction. The tensor product arises in algebraic geometry in the following way. Let M be the coordinate ring ofa variety Vin affine space Am, in other words, Mis the set of all polynomial
295
Geometry of Algebra
functions from V to the base field k. Let N be the coordinate ring of the variety W in An. Then the cartesian product V x W is a variety in Am+n, and its coordinate ring turns out to be the tensor product of M and N. Let's return to the first example above, where M and N are free modules with bases {xi} and {y.}. Suppose that f is a bilinear map from M x N to an Rmodule P. (In other words,fis Rlinear in each variable.) Information aboutf can be completely encoded into a function g of one variable, where g is an Rmodule homomorphism from Tto P. We take g(xi 0 y) =j(xi, yjJ and extend by linearity. Thusfis the composition of the bilinear map h from M x N to l' specified b~l (xi' y) ) Xi 0 Yj , followed by g. To summarize: Every bilinear mapping on M
x
N can be factored through T.
The Rmodule T is called the tensor product of M and N, and we write T = M 0 JIV. We are going to construct a tensor product of arbitrary modules over a commutative ring, and sketch the generalization to noncommutative rings.
DEFINITION Let M and N be arbitrary Rmodules, and let F be a free Rmodule with basis M x N. Let G be the submodule of F generated by the "relations" (x + x', y)  (x, y)  (x', y); (x, y + y)  (x, y)  (x, y); (rx, y)  rex, y); (x, ry)  rex, y)
where x, x'
E
M, y, y'
N, r
E
E
R. Define the tensor product of M and N (over R) as
T=M 0RN=F/G and denote the element (x, y) + G of T by x0Y. Thus the general element of T is a finite sum of the form
with xi E M and Yi necessarily unique.
E
N. It is important to note that the representation (1) is not
The relations force x 0 y to be linear in each variable, so that x 0 (y + y') rex 0 y)
=x
= rx
0 y + x 0 y', (x + x') 0 y
0 y
=x
=x
0 y + x' 0 y,
0 ry.
Now iffis a bilinear mapping from M x Nto the Rmodule P, thenfextends uniquely to a homomorphism from Fto P, also calledf Bilinearity means that the kernel offcontains G, so by the factor theorem, there is a unique Rhomomorphism g : T ) P such that g(x 0 y) = j(x, y) for all x E M,y E N. Ifwe compose the bilinear map h : (x, y) ) x 0 Y with g, the result is f Again, we say that Every bilinear mapping on AI x N can be factored through T.
296
Geometry of Algebra
We have emphasized this sentence, known as a universal mapping property (abbreviated UMP), it indicates how the tensor product is applied in practice. The detailed construction we have just gone through can now be forgotten. In fact any two Rmodules that satisfy the universal mapping property are isomorphic. The precise statement and proof of this result will be developed in the exercises. In a similar fashion, using multilinear rather than bilinear maps, we can define the tensor product of any finite number of Rmodules. [In physics and differential geometry, a tensor is a multilinear map on a product MI x .,. x M r , where each M j is either a finitedimensional vector space Vor its dual space V*. This suggests where the terminology "ten~or product" comes from.] ,
In the discussion to follow, M, Nand Pare Rmodules. The ring R is assumed fixed, and we will usually write @ rather than @ R'
Proposition M@N~N@M.
Proof Define a bilinearmappingf: M x N ~N @Mby.f(x,y)=y @ x. By the UMP, there is a homomorphism g : M @ N ~ N @ M such that g(x @ y) = y @ x. Similarly, there is a homomorphism g' : N @ M ~ M @ N with g'(y @ x) = x @ y. Thus g is an isomorphism (with inverse g).
Proposition M
@
(N
@
P)
~
(M
N)
@
P.
@
Proof DefineJ: M x N x P ~ (M @ N) @ P by.f(x, y, z) = (x @ y) ® z. The UMP produces g : M x (N @ P) ~ (M @ N) @ P with g(x, (y @ z») = (x @ y) @ z. [We are applying the UMP for each fixed x E M, and assembling the maps to produce g.] Since g is bilinear, the UMP yields h : M @ (N @ P) ~ (M @ N) @ P with hex @ (y @ z» = (x @ y) @ z. We can construct the inverse of h, so h is the desired isomorphism.
Proposition M
@
(N
@
P) < == (M
@
N)
@
(M
@
P).
Proof LetJbe an arbitrary bilinear mapping from M x (N EB P) to Q. Ifx E M,y E N, z E P, then.f(x, y + z) = .f(x, y) +.f(x, z). The UMP gives homomorphisms gl : M @ N ~ Q and g2 : M @ P ~ Q such that gl(x @ y) = .f(x, y) and g2(x @ z) = .f(x, z). The maps gl and g2 combine to give g: (M @ N) EB (M @ P) ~ Q such that g(x
@
y) + (x'
@
z»
= gl(x
@
y) + g2(x'
In particular, with x' = x, g(x
@
y) + (x
@
z» =.f(x, y + z),
@
z).
Geometry of Algebra
297
so if h : M x (N EB P)
~
M ® (N EB P) is defined by
h(x, Y + z) = (x ® y) + (x ® z), then f = gh. Thus (M ® N) EB (M ® P) satisfies the universal mapping property, hence must be isomorphic to the tensor product. Proposition
Regarding R as a module over itself, R ® R M
~
M.
Proof The map (r, x) ~ rx of R x M ~ M is bilinear, so there is a homomorphism g : R ® M ~ M such that g(r ® x) = rx. Define h : M ~ R ® Mby h(x) = 1 ® x. Then h(rx) = 1 ® rx = rl ® x = r ® x. Thus g is an isomorphism (with inverse h). Corollary
Let Rm be the direct sum of m copies of R, and M'" the direct sum of m copies of M. Then Rm ® M ~ M"'. Proof Rm ® M is isomorphic to the direct sum of m copies of R ® M, which is isomorphic to M"'. Proposition
Rm ® Rn ~ Rmn. Moreover, if {x 1, ... , xm} is a basis for Rm and {YI' ... , Y n} is a basis for Rn, then ® Yj' i = 1, ... , m,j = I, ... , n} is a basis for Rmn.
ri
TENSOR PRODUCT OF HOMOMORPHISMS LetJi : MI ~ NI andJi : M2 ~ N2 be Rmodule homomorphisms. The map (xI' x 2) ~ fl(x l ) ® fix2) of M1 x M2 into N1 ® N2 is bilinear, and induces a uniquef: MI ® M2 ~ NI ® N2 such that j{x 1 ® x 2) =Ji(x l ) ® f 2(x 2), XI E M I, x 2 E M2. We writef= fl ® f 2, and call it the tensor product ofJi andf2. Similarly, if gl : Nt ~'! P t and g2 : N2 ~ P2, then we can compose gl ® g2 withJt ® Ji, and (gt ® g2)(lt ® Ji)(x t ® x 2) = gJj(x t ) ® gj'ix 2)' hence (gl ® g2)
0
(It ® Ji) = (gt 0 J t) ® (g2 0 Ji).
When M) = Nt = V, a free Rmodule of rank m, and M2 = N2 = W, a free Rmodule of rank n, there is a very concrete interpretation of the tensor product of the endomorphisms J: V ~ Vandg: W ~ W. IfJis represented by the matrix A andgbythe matrixB, then the action ofJand g on basis elements is given by
GeometnJ of Algebra
298
j(vj) =
~aitv"
g(wl) = ;;=bklWk
where i andj range from I to m, and k and I range from I to n. Thus
if 0
g)(Vj 0 wI)
= j(v) 0
g(wl)
=
~a .. bkl I,k
Ij
(vi 0 w k )·
The mn by mn matrix representing the endomorphismJ 0 g: V 0 W ~ V 0 Wis denoted by A 0 B and called the tensor product or Kronecker product of A and B. It is given by
A 0 B=
The ordering of the basis of V 0 W is 0 w n· To determine the column of A 0 B corresponding to vj  " wi, locate the aijB block (i = I, ... , m;j fixed) and proceed to column I of B. As we move down this column, the indices i and k vary according to the above ordering of basis elements. If this road map is not clear, perhaps writing out the entire matrix for m = 2 and n = 3 will help. VI
0
wI' ... , VI
0 wn' ... , vm 0
wI' ... , vm
Problems
I.
If m and n are relatively prime, show that Z m 0
2.
If A is a finite abelian group and Q is the additive group of rationals, show
Z
Z n = O.
that A 0 Z Q = O. Generalize to a wider class of abelian groups A. 3.
The definition of M 0 R N via a universal mapping property is as follows. The tensor product is an Rmodule T along with a bilinear map h : M x N ~ T such that given any bilinear map f : M x N ~ P, there is a unique Rhomomorphism g : T ~ P such thatJ= gh. See the diagram below. h MxN
) T
~lg p
Now suppose that another Rmodule T', along with a bilinear mapping h : M x N ~ T, satisfies the universal mapping property. Using the above diagram with P = T' and J replaced by h " we get a unique homomorphism g : T ~ T' such that h ' = gh. Reversing the roles of T' and T', we get g' : T' ~ T such that h = g'h '.
299
Geometry of Algebra
Show that T and T' are isomorphic. 4.
Consider the element n x in Z @ Z n' where x is any element of Z n and we are tensoring over Z, i.e., R = z. Show that n @ x = o.
5.
Continuing Problem 5, take x ::j:. 0 and regard n Z n rather than Z @ Z n. Show that n @ X::j:. o.
6.
Let M, N, M', N' be arbitrary Rmodules, where R is a commutative ring. Show that the tensor product of homomorphisms induces a linear map from HomiM, M) @ R HomR(N, N) to HomiM @ R N, M' @ R N'). Let M be a free Rmodule of rank m, and N a free Rmodule of rank n. Show that there is an Rmodule isomorphism of EndR(M) @ R EndR(N) and EndR(M @ N).
7.
@
x as an element of n Z
@
General Tensor Products
We now consider tensor products of modules over noncommutative rings. Anatural question is "Why not simply repeat the construction for an arbitrary ring R?". But this construction forces rx
@
sy = rex
rx
@
sy = s(rx
@
= rs(x
@
y)
y) = sr(x
@
y)
sy)
and @
which cannot hold in general if R is noncommutative. Asolution is to modify the construction so that the tensor product T is only an abelian group. Later we can investigate conditions under which T has a module structure as well. DEFINITION
Let M be a right Rmodule and N a left Rmodule. (We often abbreviate this as MR and RN.) Letf: M x N ~ P, where P is an abelian group. The map f is biadditive if it is additive in each variable, that is,j(x + x', y) =j(x, y) +j(x', y) andj(x, y + y) =j(x, y) + j(x, y) for all x, x E M, y, YEN. The map fis Rbalanced ifj(xr, y) = j(x, ry) for all x E M, y EN, r E R. As before, the key idea is the universal mappingproperty: Every biadditive, Rbalanced map can be factored through the tensor product. CONSTRUCTION OF THE GENERAL TENSOR PRODUCT
If MR and ~, let F be the free abelian group with basis M x N. Let G be the subgroup of R generated by the relations (x + x', y)  (x, y)  (x', y); (x, y + y)  (x, y)  (x, y); (xr, y)  (x, ry)
300
Geometry of Algebra where x, x'
E
M, y, y'
E
N, r
E
R. Define the tensor product of M and N over R as
T=M ®RN=F/G (
and denote the element (x, y) + G of T by x ® y. Thus the general element of T is a finite sum of the form
The relations force the map h : (x, y) balanced, so that x ® (y + y ')
=x
® y
~
x ® y of M x N into Tto be biadditive and R
+ x ® y', (x + x') ® Y = x ® y + x' ® y,
xr ® y=x ® ry.
Ifjis a biadditive, Rbalanced mapping from M x Nto the abelian group P, thenf extends uniquely to an abelian group homomorphism from Fto P, also called! Sincefis biadditive and Rbalanced, the kernel ofj contains G, so by the factor theorem there is a unique abelian group homomorphism g : T ~ P such that g(x ® y) = j(x, y) for all x E M, YEN. Consequently, gh =fand we have the universal mapping property: Every biadditive, Rbalanced mapping on M
x
N can be factored through T.
As before, any two abelian groups that satisfy the universal mapping property are isomorphic.
BIMODULES Let Rand S be arbitrary rings. We say that M is an S  R bimodule if M is both a left Smodule and a right Rmodule, and in addition a compatibility condition is satisfied: (sx)r = s(xr) for all s E S, r E R. We often abbreviate this as sUR' Iff: R ~ S is a ring homomorphism, then S is a left Smodule, and also a right Rmodule by restriction of scalars. The compatibility condition is satisfied:
(sx)r = sxf{r) = s(xr). Therefore S is an S  R bimodule. Proposition
If sUR and ~1" then M ® R N is an S  T bimodule. Proof Fix s E S. The map (x, y) ~ sx ® y of M x N into M ® R N is biadditive and Rbalanced. The latter holds because by the compatibility condition in the bimodule property ofM, s(xr) ® Y
= (sx)r
® y
= sx
® ry.
301
Geometry of Algebra
Thus there is an abelian group endomorphism on M ® R N such that x ® y + sx ® y, and we use this to define scalar multiplication on the left by s. Asymmetrical argument yields scalar multiplication on the right by t. To check the compatibility condition, [s(x ® y)]t = (sx ® y)t = sx ® yt = s(x ® yt) = s[(x ® y)t]. Corollary
If sMR and ~, then M ® RN is a left Smodule. If MR and ~1" then M ® RN is a right Tmodule. Proof The point is that every module is, in particular, an abelian group, hence a Z module. Thus for the first statement, take T = Z, and for the second statement, take S= Z.
EXTENSIONS We can define the tensor product of any finite number of modules using multi additive maps (additive in each variable) that are balanced. For example, suppose that MR, RNS and SP. If!: M x N x P + G, where G is an abelian group, the condition of balance is j{xr, y, z) for all x shows that
E
= j{x, ry, z) andj{x, ys, z) = j{x, y, sz)
M, YEN, z
E
P, r
E
R, s
E
S. An argument similar to the earlier proof
(a) M ® RN®sP ~ (M ® RN) ®sP ~ M ® R (N ®sP),
If M is a right Rmodule, and Nand P are left Rmodules, then (b) M ® R (N 97 P) ~ (M ® R N) EB (M ® R P).
In fact the result can be extended to the direct sum of an arbitrary (not necessarily finite) number of left Rmodules. If M is a left Rmodule, we have (c) R ® R M
~
M and
(d) Rm ®M
~
M".
Let MI and M2 be right Rmodules, and let NI and N2 be left Rmodules. If!I : MI + NI and!2 : M2 + N2 are Rmodule homomorphisms, the tensor product!I ® fi can be defined. As before, the key property is
ifI for all
Xl E
® !2)(x l ® x 2) =!I(x I) ® !2(x2)
M I, x 2
E
M 2.
TENSOR PRODUCT OF ALGEBRAS If A and B are algebras over the commutative ring R, then the tensor product A ® R B becomes an Ralgebra if we define multiplication appropriately. Consider the map of A x B x A x B into A ® R B given by
302
Geometry of Algebra
(a, b, a', b)
~
aa' 0 bb', a,a'
E
A, b, b'
E
B.
The map is 4linear, so it factors through the tensor product to give an Rmodule homomorphism g : A 0 B 0 A 0 B ~ A 0 B such that g(a 0 b 0 a' 0 b)
Now let h : (A 0 B)
x
= aa' 0
bb'.
(A 0 B) '! A 0 B 0 A 0 B be the bilinear map given by
h(u, v) = u 0 v.
o
If we apply h followed by g, the result is a bilinear map f: (A 0 B) B with
x
(A 0 B)
~
A
j{a 0 b, a' 0 b')=aa' 0 bb',
and this defines our multiplication (a 0 b )(a' 0 b) on A 0 B. The multiplicative identity is 1A 0 1/3' and the distributive laws can be checked routinely. Thus A 0 RBis a ring that is also an Rmodule. To check the compatibility condition, note that if r E R, a, a' E A, b, bE B, then r[(a 0 b)(a' 0 b)]
= [rea 0
b)](a' 0 b)
= (a 0
b)[r(a' 0 b)];
all three of these expressions coincide with raa' 0 bb' = aa' 0 rbb '. Problems
We will use the tensor product to define the exterior algebra of an Rmodule M, where R is a commutative ring. If p is a positive integer, we form the tensor product M 0 R ... 0 # of M with itselfp times, denoted by M0 p. Let Nbe the submodule of M0 P generated by those elements XI 0 ... 0Xp' with Xi E Mfor all i, such that Xi = Xj for some i:t= j. The pth exterior power of M is defined as
"PM=M0P/N In most applications, M is a free Rmodule with a finite ~sis XI' ... , xn (with 1 ~ P ~ n), and we will only consider this case. To simplify notation, we write the element a 0 b 0 ... 0c + N of "PM as ab .... c. (The usual notation is a " b " .. ~ " c.)
the
1.
Let )Ii' .,., Yp E M. Show that if Yi and Yj are interchanged in the product YI ... Yp ' then the product is mUltiplied by  1.
2.
Show that the products
3.
Letf: Mp ~ Q be a multilinear map from MP to the Rmodule Q, and assume that f is alternating, that is, j{m l , ... , m~ = 0 if fl1i c= mj for some.i :t= j. Show that f can be factored through· "PM, in other words, there is a unique Rhomomorphism g: "PM ~ Q such that g(Y1 .... Yb) = j(yl' .... , Yp ).
xi)····xip ,
where i l < ....