Algebra Prelim Solutions, Summer 2002

1. (a)
H acts on the set of conjugates of Q according to the formula gQg-1 | - > hgQg-1h-1 for h H and g G. Note that gQg-1 < G for all g, so OQ is a set of subgroups. Let S = {h H | hQh-1 = Q}, the stabilizer of Q in H. Then |OQ|| S| = | H|. Now h S if and only if h H /\ NG(Q) = 1, hence | S| = 1 and the result follows.

(b)
Let P be a Sylow p-subgroup of G. We apply the above with H = P. Since P /\ NG(Q) = P /\ Q = 1, we see that OQ has | P| = pm subgroups. Furthermore all the subgroups of OQ have prime order q, so any two of them must intersect in 1. Now each nonidentity element of a subgroup in OQ has order q, consequently each subgroup of OQ yields q - 1 elements of order q and we deduce that G has at least (q - 1)pm elements of order q. Therefore G has at most pqm - (q - 1)pm = pm elements of order a power of p. Since | P| = pm and every element of a Sylow p-subgroup has order a power of p, we conclude that P is the only Sylow p-subgroup of G. Therefore P is normal in G and we are finished.

2. Let F = Q(,) and N = Gal(K/F). Then F K and is the splitting field over Q for (x2 - 2)(x2 - 3). Therefore N is a normal subgroup in Gal(K/Q) of index [F : Q]. Now satisfies x2 - 2 and Q. Therefore [Q() : Q] = 2.

Next we show that Q(). Suppose Q(). Then we could write = a + b with a, b Q, because every element of Q() can be written in this form. Squaring, we obtain 3 = a2 + 2b2 + 2ab. Since Q, we deduce that a or b = 0. But a = 0 yields Q, while b = 0 yields Q, neither of which is true. We conclude that Q(). Since satisfies x2 - 3, we deduce that [F : Q()] = 2. Therefore

[F : Q] = [Q(,) : Q()][Q() : Q] = 2*2 = 4.

Thus N is a normal subgroup of index 4 in Gal(K/Q).

Suppose K. Since satisfies x8 - 2 and x8 - 2 is irreducible over Q by Eisenstein's criterion for the prime 2, we see that 8 divides [K : Q], consequently 8 divides | Gal(K/Q)|. But | Gal(K/Q)| = 4| N|, so this is not possible if | N| is odd.

3. (a)
If X and Y are right R-modules and q : X - > Y is an R-module homomorphism, then there is a unique group homomorphism q 1 : X C - > Y C such that q(x c) = (qx) c for all x X and c C.

First we apply this to the maps

 (a, b) | - > a : A B - - > A (a, b) | - > b : A B - - > B.

We obtain a group homomorphism

q : (A B) C - - > A C B C

such that q((a, b) c) = (a c, b c). Next we apply it to the maps

 a | - > (a, 0) c : A C - - > (A B) C a | - > (0, b) c : B C - - > (A B) C.

We obtain a group homomorphism

f : (A C) (B C) - - > (A B) C

such that f(a c, b d )= (a, 0) c + (0, b) d.

Finally we show that fq is the identity on (A B) C, and that qf is the identity on (A C) (B C). We have

fq(a, b) c = f(a c, b c) = (a, b) c.

Since (A B) C is generated as an abelian group by elements of the form (a, b) c, we see that fq is the identity. Also

qf(a c, b d )= q(a, 0) c + q(0, b) d = (a c, b d ).

Since (A C) (B C) is generated as an abelian group by elements of the form (a c, b d ), we deduce that qf is the identity. It now follows that (A B) C @ (A C) (B C).

(b)
Since M is a finitely generated Z-module, we may express it as a finite direct sum of cyclic Z-modules, say M = Z/aiZ, where we may assume that ai =/= ±1 for all i. Then by the first part, we see that

M M @ Z/aiZ Z/ajZ.

Therefore it will be sufficient to prove Z/aiZ Z/aiZ =/=  0 for all i (of course even for just one i will be sufficient). However we can define a bilinear map

q : Z/aiZ X Z/aiZ - > Z/aiZ

by q(x, y) = xy. This induces a Z-module homomorphism Z/aiZ Z/aiZ - > Z/aiZ, which is obviously onto. We conclude that Z/aiZ Z/aiZ =/=  0, as required.

4. Note that Ann(m) is an ideal of R. Since R is Noetherian, we may choose w M such that Ann(w) is maximal (that is Ann(w) is as large as possible, but not R). Suppose Ann(w) is not prime. Then there exists a, b R\Ann(w) such that ab Ann(w), i.e. abw = 0. But then a Ann(bw) and Ann(w) Ann(bw). Furthermore Ann(bw) =/= R because bw =/=  0, so the maximality of Ann(w) has been contradicted and the result follows.

5. Suppose R is a field. Then an R-module is the same thing as an R-vector space, and since every vector space has a basis this means that every R-module is free; in particular every R-module is projective.

Conversely suppose every R-module is projective. Since R is an integral domain, to prove R is a field we only need show that every nonzero element of R is invertible. Suppose to the contrary that x is a nonzero element of R which is not invertible. Then R/Rx is a nonzero R-module, so it has a nonzero element u. Note that xu = 0. Consider the exact sequence

0 - - > Rx - - > R - - > R/xR - - > 0.

Since R/xR is projective, the sequence splits, in particular R/xR is isomorphic to a submodule of R. Now R is an integral domain, so xv =/=  0 for all nonzero v R and we deduce that xu =/=  0. We now have a contradiction and the result follows.

6. Since ZN = ZN + 1, we see that Z(G/ZN) = 1. Therefore K N.

Now suppose L is a normal subgroup of G such that Z(G/L) = 1 and L does not contain ZN. Then there is a nonnegative integer n such that

Zn L,    Zn + 1 L.

Choose x Zn + 1\L. Then xL =/= 1 in G/L. Also xgx-1g-1 Zn + 1 for all g G, because xZn + 1 Z(G/Zn + 1). Therefore xgx-1g-1 L and we deduce that xL Z(G/L). This is a contradiction, and so the result is proven.

7. Let I be the set of matrices in M2(Q[x]/(x2 - 1)) of the form

(
 a(x + 1) + (x2 - 1) 0 b(x + 1) + (x2 - 1) 0
)

with a, b Q. Note that if f Q[x], then (x - 1) divides f (x) - f (1), consequently f (x)(x + 1) + (x2 - 1) = f (1)(x + 1) + (x2 - 1) in Q[x]/(x2 - 1).

Now we verify that I is a left ideal of Q[x]/(x2 - 1). Clearly I is an abelian group under addition. Since

(
 f (x) + (x2 - 1) g(x) + (x2 - 1) h(x) + (x2 - 1) k(x) + (x2 - 1)
) (
 a(x + 1) + (x2 - 1) 0 b(x + 1) + (x2 - 1) 0
) = (
 af (1)(x + 1) + (x2 - 1) 0 bh(1)(x + 1) + (x2 - 1) 0
)

we see that I is closed under left multiplication by elements of Q[x]/(x2 - 1), and it now follows that I is a left ideal.

Finally we need to show that I is a minimal ideal. Obviously I =/=  0 (note x + 1(x2 - 1)). Suppose J is a nonzero left ideal contained in I. We need to show that J = I. By multiplying on the left by the matrix

(
 0 1 + (x2 - 1) 1 + (x2 - 1) 0
)

if necessary, we may assume that I contains a matrix of the form

(
 a(x + 1) + (x2 - 1) 0 b(x + 1) + (x2 - 1) 0
)

with a =/=  0. Then by multiplying on the left by

(
 c/a + (x2 - 1) 0 d /a + (x2 - 1) 0
)

we see that J must be the whole of I and the result follows.

Peter Linnell
2002-08-17