Algebra Prelim Solutions, Winter 2003

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

    Of course a consequence of this is that x4 + 3 remains irreducible over Q(i). Let q $ \in$ Gal(K/Q(i)) satisfy q(w$ \sqrt[4]{3}$) = w3$ \sqrt[4]{3}$. Then q(w3$ \sqrt[4]{3}$) = w5$ \sqrt[4]{3}$, q(w5$ \sqrt[4]{3}$) = w7$ \sqrt[4]{3}$ and q(w7$ \sqrt[4]{3}$) = w$ \sqrt[4]{3}$, in particular q has order 4. Furthermore gqg(i) = i and gqg(w$ \sqrt[4]{3}$) = w7$ \sqrt[4]{3}$, 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 $ \oplus$ 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 $ \in$ 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 $ \in$ 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 g$ \notin$K. 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 $ \supseteq$ IP for all n>P, so we need to show the reverse inclusion. Let x $ \in$ In. Since (0 XB) + In = (0 XB) + IP, we may write x = b + i where b $ \in$ 0 X B and i $ \in$ IP. Since x, i $ \in$ In, we see that b $ \in$ (0 XB) /\ In and hence b $ \in$ (0 XB) /\ IP, because (0 XB) /\ In = (0 XB) /\ IP. This shows that x $ \in$ 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 $ \in$ I and p $ \in$ P, then a(ip) = iap $ \in$ IM = 0. Therefore IP $ \subseteq$ 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 $ \oplus$ Q @ F for some R-modules Q, F with F free. Then P/IP $ \oplus$ 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 $ \in$ k and i, j $ \in$ I, then (a + i)(b + j) = ab + (aj + ib + ij) $ \in$ k + I, because aj, ib, ij $ \in$ I by using I <k[x].

    (b)
    Note that if i $ \in$ I, then i $ \otimes$ a = 1 $ \otimes$ ia = 0 for all a $ \in$ 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 $ \otimes$ a (where b $ \in$ k[x] and a $ \in$ k[x]/I) is a k-linear sum of elements of the form xr $ \otimes$ (I + xs), where r, s < n. Since every element of k[x] $ \otimes_{k+I}^{}$ k[x]/I is a sum of simple tensors, we deduce that k[x] $ \otimes_{k+I}^{}$ k[x]/I is generated as a k-module by elements of the form xr $ \otimes$ (I + xs) where r, s < n and the result follows.





Peter Linnell
2003-01-11