Algebra Prelim Solutions, Fall 2005

1. Let G be a group of order 18. The number of Sylow 3-subgroups is congruent to 1 mod 3 and divides 18/9 = 2. Therefore G has a unique Sylow 3-subgroup H of order 9, and H <G. Also G has an element x of order 2. Then GH X| < x >. Since groups of order p2 are prime, H is an abelian group. For a positive integer n, let Cn denote the cyclic group of order n. We have three cases to consider.
(a)
The conjugation action of x on H is trivial, that is xhx-1 = h for all hH and we have GH× < x >. There are two isomorphism classes for H, namely C9 and C3×C3. It follows that there are two isomorphism classes for G in this case.

(b)
The conjugation action of x on H is nontrivial and HC9. There is exactly one automorphism of H of order two, namely h |--> h-1 for hH. It follows that there is exactly one isomorphism class for G in this case.

(c)
The conjugation action of x on H is nontrivial and HC3×C3. Either x acts by inversion on H and this gives us one isomorphism class for G. Otherwise we may write H = A×B where A, BC3, x centralizes A and x acts by inversion on B. This yields a second isomorphism class for G.
We conclude that there are 2 + 1 + 2 = 5 isomorphism classes for a group of order 18.

2. First suppose that A is Noetherian. Then A[X, Y] is Noetherian by Hilbert's basis theorem. Since factor rings of Noetherian rings are Noetherian, we see that A[X, Y]/(X2 -Y2) is also Noetherian.

Conversely suppose A[X, Y]/(X2 -Y2) is Noetherian. Since (X, Y)⊃(X2 -Y2), we see that A[X, Y]/(X, Y) is also Noetherian. But A[X, Y]/(X, Y)≌A, because the homomorphism

X |--> 0, Y |--> 0 : A[X, Y] -> A

is surjective with kernel (X, Y), and the result follows.

3. Let 0≠uF2 and let S denote the stabilizer in GLn(F) of the one-dimensional subspace Fu. We need to prove that S is not simple. Set D = {diag(f, f ) | 0≠fF}, where diag(f, f ) indicates the invertible matrix in GL2(F) which has f's on the main diagonal and zeros elsewhere. Then D is a central subgroup of S and since | F|≥3, it is not 1. Let v be an element of F2 which is not in Fu. Then {u, v} is a basis of F2 and so we can define a linear isomorphism of F2 by u |--> u, v |--> u + v. This yields an element of S\D. Thus D is a normal subgroup of S which is neither 1 nor D, and we conclude that S is not simple.

4. Clearly M cannot be free of rank 0. Nor can M be free of rank at least 2, because if a, bM were part of a free R-basis for M, we would have 0≠abaRbR, which would mean that {a, b} was not linearly independent over R. Therefore the only possibility of M being free is that it is free of rank 1. This means we can write 2R + XR = cR for some cR. There are several methods to show that this is not possible; we present one of them.

Since 2∈cR, we see that c is a polynomial of degree zero and thus c = ±1 or ±2. Without loss of generality, we may assume that c = 1 or 2. Since XcR, we may write X = cf for some polynomial f∈ℤ[X]. By considering the leading coefficient (degree 1) of f, we see that c = 1 and we deduce that there exist g, hR such that 2g + Xh = 1. This is not possible because the left hand side has constant coefficient ∈2ℤ and in particular cannot be 1. It follows that M is not a free R-module.

5. Since σa -aF for all σ∈G, we see that (∑σ∈Gσa) - | G| aF. Now τ∑σ∈Gσa = ∑σ∈Gσa for all τ∈G. Since K/F is a Galois extension with Galois group G, it follows that σ∈GσaF and we deduce that | G| aF. We conclude that aF because F has characteristic zero.

6. Set m = √n. A finite dimensional simple algebra over an algebraically closed field is isomorphic to a full matrix ring over the field. In this situation, this means S is isomorphic to Mm(ℂ), the m×m matrices over ℂ. Let eij ( 1≤i, j < m) denote the matrix units of Mm(ℂ), so eij has 1 in the (i, j)th position and zeros elsewhere. Then Seii is the ith column of S and we see that S = Se11Se22⊕...⊕Semm. All that remains to prove is that Seii is irreducible for all i. Without loss of generality, we may assume that i = 1. Suppose M is a nonzero R-submodule of Se11. The general element α of Se11 is of the form iaiei1. If this is a nonzero element of M, then ai≠ 0 for some i, 1≤im, and we deduce that e11 = ai-1e1iα∈M. Thus M = Se11 and the result follows.

7. The map f |--> f⊗1 : -> FL is an algebra monomorphism with image ⊗1. Furthermore, if 1,...,λn} is a basis for L over F, then {1⊗λ1,..., 1⊗λn} is a basis for FL over ⊗1. It follows that FL is a field extension of degree n over and since is algebraically closed, we deduce that n = 1. Therefore L = F as required.

Peter Linnell 2012-08-04