Algebra Prelim Solutions, Winter 2003

1. We have f (x) = (x + 1)(x4 + 3); since -1 Q, the splitting field for x4 + 3 is also K. Let w = (1 + i)/, a primitive 8th root of 1. Then w4 = - 1 and we see that the four roots of x4 + 3 are wr for r = 1, 3, 5, 7. Therefore K = Q(w,w3,w5,w7). Since x4 + 3 is irreducible by Eisenstein for the prime 3, we see that [Q(w) : Q] = 4. Let g denote complex conjugation. Since x4 + 3 is a polynomial with real coefficients, we see that g Gal(K/Q). Thus K, because = w + g(w). Now satisfies x4 - 12, which is irreducible by Eisenstein for the prime 3. Therefore [Q() : Q] = 4. Note that Q() =/= Q(w), because the former is contained in R while the latter is not. We deduce that K =/= Q(w). Also i = w3/w, which shows that i K. Since w2r + 1 = irw, we conclude that K = Q(i,w). Therefore [K : Q(w)] = 2 and hence [K : Q] = 8.

Of course a consequence of this is that x4 + 3 remains irreducible over Q(i). Let q Gal(K/Q(i)) satisfy q(w) = w3. Then q(w3) = w5, q(w5) = w7 and q(w7) = w, in particular q has order 4. Furthermore gqg(i) = i and gqg(w) = w7, which shows that gqg = q-1. We now see that Gal(K/Q) = {qrgs | r = 0, 1, 2, 3, s = 0, 1} and is isomorphic to the dihedral group of order 8.

2. This is false. Consider the group Z4 Z2, where Zn denotes the integers modulo n. Then (2, 0) and (0, 1) both have order 2 (when we write (2, 0), the 2 means 2 modulo 4). Suppose q is an automorphism such that q(2, 0) = (0, 1). Then 2(q(1, 0)) = q(2, 0) = (0, 1). On the other hand 2(q(1, 0)) is of the form 2(a, b) = (2a, 0), and so cannot be equal to (0, 1). Thus we have a contradiction and we conclude that there is no such q.

3. Let n denote the number of Sylow 2-subgroups. Since 2002 = 2*1001, we see that a Sylow 2-subgroup has order 2 and n | 1001. Therefore each Sylow 2-subgroup has exactly one element of order 2 and n is odd. Also any element of order 2 is in exactly one Sylow 2-subgroup, consequently the number of elements of order 2 is n. Since the number of elements in the set {h H | h2 = e} is n + 1 (the +1" for the identity), we conclude that this number is even.

4. Suppose G is nonabelian, so there exist a, b G such that ab =/= ba. Set g = aba-1b-1, so g =/= 1. By hypothesis there exists K <G such that G/K is abelian and gK. But KaKb(Ka)-1(Kb)-1 = Kg =/= 1, which shows that KaKb =/= KbKa and hence G/K is nonabelian, which is a contradiction. The result follows.

5. Certainly if A and B are commutative rings, then A X B is also a commutative ring. We need to show that if A and B are in addition Noetherian, then so is A X B. Suppose I1, I2,... is an ascending chain of ideals in A X B. Then (0 XB) /\ I1,(0 XB) /\ I2,... is an ascending chain of ideals in 0 X B. But 0 XB @ B and B is Noetherian, hence there exists a positive integer M such that 0 XB /\ In = 0 XB /\ IM for all n>M. Also (A XB)/(0 XB@ A as rings, so (A XB)/(0 X B) is Noetherian. Therefore the ascending chain of ideals (0 XB) + I1,(0 XB) + I2,... of (A XB)/(0 X B) becomes stationary, that is there is a positive integer N such that (0 XB) + In = (0 XB) + IN for all n>N. Let P be the maximum of M and N. We claim that In = IP for all n>P. Obviously In IP for all n>P, so we need to show the reverse inclusion. Let x In. Since (0 XB) + In = (0 XB) + IP, we may write x = b + i where b 0 X B and i IP. Since x, i In, we see that b (0 XB) /\ In and hence b (0 XB) /\ IP, because (0 XB) /\ In = (0 XB) /\ IP. This shows that x IP and hence In = IP for n>P.

6. Obviously P/IP is an R/I-module; we need to prove that it is projective. Suppose we are given an R/I-epimorphism m : M - - > > N of R/I-modules and an R/I-map q : P/IP - > N. We need an R/I-map b : P/IP - > M such that q = mb. Let p : P - > P/IP denote the natural epimorphism. We can also view M and N as R-modules, and then m is also an R-map. Since P is a projective R-module, certainly there exists an R-map a : P - > M such that ma = qp. If i I and p P, then a(ip) = iap IM = 0. Therefore IP kera and we deduce that a induces an R/I-map b : P/IP - > M satisfying bp = a. Then mbp = ma = qp and since p is onto, we conclude that mb = q.

Sketch of alternate proof. Since P is projective, we may write P Q @ F for some R-modules Q, F with F free. Then P/IP Q/IQ @ F/IF and since F/IF is a free R/IR-module, we see that P/IP is a projective R/IR-module.

7. (a)
Certainly k + I is a subgroup of k[x] under addition; we need to show that it is closed under multiplication. However if a, b k and i, j I, then (a + i)(b + j) = ab + (aj + ib + ij) k + I, because aj, ib, ij I by using I <k[x].

(b)
Note that if i I, then i a = 1 ia = 0 for all a k[x]/I. Since I is a nonzero ideal of k[x], we know that k[x]/I is finite dimensional over k. Thus there is a positive integer n such that {I + 1, I + x,..., I + xn} spans k[x]/I as a k-module. It follows that any simple tensor b a (where b k[x] and a k[x]/I) is a k-linear sum of elements of the form xr (I + xs), where r, s < n. Since every element of k[x] k[x]/I is a sum of simple tensors, we deduce that k[x] k[x]/I is generated as a k-module by elements of the form xr (I + xs) where r, s < n and the result follows.

Peter Linnell
2003-01-11