January 1996 Algebra Prelim Solutions

1.
Since f is an epimorphism from G to H, the fundamental isomorphism theorem tells us that G/kerf H, so in particular | G/ker f| = | H|. Therefore | G|/| ker f| = | H|, hence | ker f| = | G|/| H| and we deduce that | kerf| = 33 112. Using the fundamental theorem for finitely generated abelian groups, we see that there are up to isomorphism six abelian groups of order | kerf| = 33 112, namely

 /27 X /121 /9 X /3 X /121 (/3)3 X /121 /27 X (/11)2 /9 X /3 X (/11)2 (/3)3 X (/11)2

2.
(i) Since A B is a subgroup of G whose order divides | A| = p4 and | B| = q5, we see that | A B| = 1 and hence A B = 1. Next if a A and b B, then a-1b-1ab = (a-1b-1a)b B, because B G. Similarly a-1b-1ab A and we deduce that a-1b-1ab A B = 1. Therefore a-1b-1ab = 1, consequently ab = ba for all a A and b B. We can now define a map : A X B - > G by (a, b) = ab. Then

((a1, b1)(a2, b2)) = (a1a2, b1b2) = a1a2b1b2 = (a1b1)(a2b2)

(because a2 commutes with b1), hence ((a1, b1)(a2, b2)) = (a1, b1)(a2, b2) and we deduce that is a homomorphism. If (a, b) ker , then ab = 1 and so a = b-1. Thus a = b-1 A B = 1 and we deduce that a = b = 1. Therefore ker = 1 and so is a monomorphism. Since | G| = | A| X | B| we conclude that is also onto, consequently is an isomorphism and the result follows.

(ii) Since A is a p-group, it has a normal subgroup P of order p. Similarly B has a normal subgroup Q of order q. Since P X Q is a normal subgroup of A X B of order pq, we see that (P X Q) is a normal subgroup of G of order pq, and so we may set N = P X Q to satisfy the requirements of the problem.

3.
Let G be a simple group of order 22 3 112. The number of Sylow 11-subgroups is congruent to 1 modulo 11 and divides 12, so the possibilities are 1 and 12. If there is 1 Sylow 11-subgroup, then it would have to be normal, which is not possible because G is simple. Therefore there are 12 Sylow 11-subgroups. If N is the normalizer of a Sylow 11-subgroup, then [G : N] is the number of Sylow 11-subgroups, so [G : N] = 12. By considering the permutation representation of G on the left cosets of N in G and using the fact that G is simple, we see that there is a monomorphism of G into A12, the alternating group of degree 12. This means that G is isomorphic to a subgroup of A12. This is not possible because 121 divides | G|, but 121 does not divide | A12|. We now have a contradiction and we deduce that no such G exists, as required.

4.
Since ( + )3 - 9( + ) = 2 , we see that [ + ] and we deduce that [ + ] = [,]. Now [[] : ] = 2, and [[,] : []] = 1 or 2, because satisfies x2 - 3, a degree 2 polynomial over . Therefore [[ + ] : ] = 4 or 2, depending on whether or not [].

Suppose []. Then we may write = a + b where a, b . Clearly a, b 0. Squaring we obtain 3 = a2 + 2ab + 2b2 and we deduce that is rational, which is not so. Therefore [] and consequently [ + ] = 4. Note we also have that {1,} is a -basis for [], and {1,} is a []-basis for [,]. Recall that if ei is an F-basis for E over F and fj is an E-basis for K, then eifj is an F-basis for K. It follows that {1,,,} is a -basis for [,].

5.
Let P be a prime ideal of D and suppose P was not maximal. Then there would exist M D such that MD and M properly containing P. Since D is a PID, we may write P = pD and M = mD for some m, p D with p 0. Then p = mx for some x D because M contains P. Since P is a prime ideal, we must have m or x P. We cannot have m P because M properly contains P. Therefore we must have x P and then we may write x = py for some y D. This yields p = mpy and since D is a domain, we see that 1 = my. This shows that mD = D, which contradicts the fact that M is a proper ideal of D and the first part of the problem is proven.

Suppose now that f : D - > K is a ring epimorphism onto the integral domain K with kerf 0. Then D/kerf K, so ker f is a prime ideal because D/ker f is an integral domain. Using the first part of the problem, we see that ker f is a maximal ideal of D. Therefore D/ker f is a field and it follows that K is a field as required.

For the last part a counterexample is R = /6 . Let P be the prime ideal 2/6 , Q the prime ideal 3/6 , and S = R P. Note that the ideals of R are precisely 0, R, P and Q, and the prime ideals of R are precisely P and Q. Then S-1P = 0 because s = 6 + 3 S and sp = 0 for all p P, and S-1Q = RP because Q S . Since all ideals of RP are of the form S-1I for some I R, we see that 0 and RP are the only ideals of RP and it follows that RP is a field. Similarly RQ is a field. Since R is not an integral domain, we have now established that R is a counterexample.

6.
Suppose is a prime ideal of R and R has a nonzero nilpotent element. Then we may assume that R has a nonzero element such that = 0. If S = R , then we may write the nilpotent element as r/s where r R and s S. Since (r/s)2 = 0, we see that r2t = 0 for some t S, and rt 0 because r/s 0. Also (rt)2 = r2tt = 0, so rt is a nonzero nilpotent element of R.

Suppose r is a nonzero nilpotent element of R. It remains to prove that R has a nonzero nilpotent element for some prime ideal of R. We may assume that r2 = 0. Let I = {s R | rs = 0} , an ideal of R which does not contain 1. By Zorn's lemma, there is a maximal ideal of R containing I; of course will also be a prime ideal. Then the image r/1 in R is nonzero because rt 0 for all t R . Since (r/1)2 = r2/12 = 0/12 = 0, we see that r/1 is a nonzero nilpotent element of R. This completes the proof.

7.
The submodule of M generated by A and B is A + B; this is the set {a + b | a A and b B} . Thus we need to prove that A B A + B. We define a map : A B - > M by (a, b) = a + b. Clearly this is an R-module homomorphism of A B onto A + B. If (a, b) ker , then a + b = 0, consequently a = - b. This shows that a and - b are both in A B = 0. Therefore a = b = 0 and hence ker = 0. It follows that is an isomorphism and so A B A + B as required.

8.
(i) Since there is a one-one correspondence between the subgroups of Gal(E/F) (the Galois group of E over F) and the proper subfields of E containing F, we need to show that S6 has at least 35 proper subgroups. One way to do this is to note that S6 has 144 5-cycles which gives 36 subgroups of order 5.

(ii) The subfield L required is Fix(A6), the subset of E which is fixed pointwise by all elements of the alternating subgroup A6 of S6. Since A6 is a normal subgroup of S6, we see that L is a Galois extension of F, and that Gal (E/L) A6. Since A6 is a simple group, there is no subfield between E and L which is Galois over L.

(iii) The dimension of L over F is | S6/A6| = 2.

Peter Linnell
1999-01-16