Algebra Prelim Solutions, August 2011

1. First we write 380 as a product of prime powers, namely 22*5*19. Suppose by way of contradiction G is a simple group of order 380. The number of Sylow 19-subgroups is congruent to 1 mod 19 and divides 20, hence is 1 or 20. But 1 is ruled out because then G would have a normal subgroup of order 19, which would contradict the hypothesis that G is simple. Therefore G has 20 Sylow 19-subgroups. Next we consider the Sylow 5-subgroups. The number is congruent to 1 mod 5 and divides 4*19. Thus there are 1 or 76 Sylow 5-subgroups.

Now we count elements. If P and Q are distinct Sylow 19-subgroups, then PQP and PQ < P. Since | PQ| divides | P| = 19 by Lagrange's theorem, we deduce that PQ = 1. It follows that G has at least 20*18 = 360 elements of order 19. Similarly two distinct Sylow 5-subgroups intersect trivially and we deduce that G has at least 76*4 = 304 elements of order 5. We conclude that G has at least 360 + 304 = 664 > 380 elements, which is a contradiction. Therefore there is no simple group of order 380.

2. Let ω = F7, so ω≠1 = ω3. We have (f - g)(f - ωg)(f - ω2) = h3. Since f, g are coprime, we see that f - g, f - ωg, f - ω2g are pairwise coprime. Now use the fact that k[x1,..., xn] is a UFD; remember that the units of k[x1,..., xn] are precisely the nonzero elements of k. Write h = up1r1...pmrm where 0≠uk, m is a nonnegative integer, pi is prime for all i, and ri is a positive integer for all i. Since f - g, f - ωg, f - ω2g are pairwise coprime, we see that if pi divides one of these three polynomials, then pi doesn't divide the other two polynomials, and it follows that pi3ri is the precise power of pi which divides this polynomial. We deduce that each of f - g, f - ωg, f - ω2g is of the form uq3 for some unit u and some polynomial q, and the result follows.

3. We use the structure theorem for finitely generated modules over a PID, elementary divisor form. We may write M = i∈I(R/piR)eiiCi, where eiN, I is a finite subset of N, and the Ci are modules of the form R or R/qhR, where hN and q is a prime which is not associate to p. This expresses M in a unique way as a direct sum of indecomposable R-modules. The hypothesis that pm = 0≠m implies Rm is not a direct summand of M tells us that 1∉I. Similarly we may write N = i∈J(R/piR)fiiDi, where fiN, J is a finite subset of N not containing 1, and the Di are modules of the form R or R/qfR, where fN and q is a prime which is not associate to p. Note that pCiCi and pDiDi for all i. Also p(R/piR)≌R/pi-1R≠ 0 for i≥2. Thus pMi∈IR/pi-1RiCi and pNi∈JR/pi-1RiDi, and these expressions are direct sums of indecomposable modules. Since pMpN, the uniqueness statement in the structure theorem for modules over a PID yields I = J and after renumbering if necessary, CiDi for all i. The result follows.

4. Let G denote the Galois group of K over Q. Then | G| = 27 and there is a one-to-one correspondence between subfields of K and subgroups of G which is reverse including. Also [Q(α) : Q] = 9 because f is irreducible with degree 9. Therefore Q(α) corresponds to a subgroup H of order 3. To find a subfield of Q(α) which has degree 3 over Q, we need to find a subgroup of order 9 which contains H. Since G is a nontrivial finite 3-group, it contains a central subgroup Z of order 3. If Z is not contained in H, then HZ = 1, hence

| HZ|/3 = | HZ|/| Z| = | HZ/Z| = | H/HZ| = | H| = 3

and we see that HZ is a subgroup of order 9 containing H. On the other hand if ZH, then H = Z and hence H <G. Thus G/H is a group of order 9, and hence has a subgroup of order 3, which by the subgroup correspondence theorem we may write as K/H, where K is a subgroup of G containing H. The order of K is 3| H| = 9, which finishes the proof.

5. We note that given a, bA, there exists nN and cA such that pna = 0 and pnc = b. This shows that for a, bA,

ab = apnc = pnac = 0⊗c = 0.

Since AA is generated as an abelian group by simple tensors" ab, we deduce that every element of AZA is zero, in other words AZA = 0.

6. The minimal polynomial divides the characteristic polynomial, so is x, x2 or x3. Also since the minimal polynomial factors into linear factors over k, the Jordan canonical form for A is defined over k.

If the minimal polynomial is x, then A = 0, so we could take B = 0, since then B2 = 0 = A.

If the minimal polynomial is x2, then the invariant factors of A are x, x2. Consider the matrix

C : = (
 0 0 1 0 0 0 0 0 0
)

Then C≠ 0 and C2 = 0, so the invariant factors of C are x, x2 and therefore C is similar to A. Thus it will be sufficient to find a matrix B such that B2 = C; here we could take B to be

(
 0 1 0 0 0 1 0 0 0
)

Then B2 = C as required.

Finally suppose A has one invariant factor, which will necessarily be x3. Then the Jordan canonical form of A is

(
 0 1 0 0 0 1 0 0 0
)

Suppose there is a matrix B such that B2 = A. Then B6 = A3 = 0. Therefore the minimal polynomial of B divides x6 and since B is a 3 by 3 matrix, we deduce that the minimal polynomial of B divides x3. Therefore B3 = 0 and we conclude that A2 = B4 = 0. But

A2 = (
 0 0 1 0 0 0 0 0 0
)

which is nonzero, and the result follows.

7. Let K denote the field of fractions of R and let I denote the ideal of K[x1,..., xn] generated by S. Then Z(S) = {(r1,..., rn)∈Rn | f (r1,..., rn) = 0} for all fI. By Hilbert's basis theorem, there is a finite subset T of S which generates the ideal I. Then Z(S) = Z(T).

Peter Linnell 2011-08-19