Algebra Prelim Solutions, December 2007

1. (a)

   By using the elementary divisor decomposition, up to isomorphism, there are three abelian groups of order p³q, namely Z_p³×Z_q, Z_p²×Z_p×Z_q, and Z_p×Z_p×Z_p×Z_q.

   (b)

   The first group above is generated by one element, while the third requires 3 elements. Therefore G≌Z_p²×Z_p×Z_q≌Z_p²×Z_pq, because Z_p×Z_q≌Z_pq, as required.

2. We have [k(α) : k] = deg f and [K : k(α)][k(α);k] = [K : k]. Thus deg f | [K : k] and the result follows.

3. By Gauss's lemma, f is irreducible in Q[x]. Since Q[x] is a PID, this tells us that fQ[x] is a maximal ideal of Q[x]. The result follows.

4. A₄ is a normal subgroups of S₄ and V : = {(1),(12)(34),(13)(24),(14)(23)} is a normal subgroup of A₄ (even normal in S₄). Since | S₄/A₄| = 2, | A₄|/V = 3, | V| = 4, the groups S₄/A₄, A₄/V and V are all abelian, because groups of order 2,3 or 4 are abelian. This proves that S₄ is solvable.

5. We have a short exact sequence 0 --> ker f --> P -- ^f--> Q --> 0. Since Q is projective, the sequence splits, so P≌Q⊕ker f. This proves the result, because direct summands of projective modules are projective.

6. First observe that Q $\otimes_{R}^{}$ Q≌Q as Q-modules. To do this, define f : Q×Q --> Q by f (p, q) = pq. Clearly this is R-bilinear, so induces an R-map g : Q $\otimes_{R}^{}$ Q --> Q satisfying g(p $\otimes$ q) = pq. Also we can define a Q-map h : Q --> Q $\otimes_{R}^{}$ Q by h(q) = q $\otimes$ 1. Since gh(q) = g(q $\otimes$ 1) = q, we see that gh is the identity on Q. Now consider hg(p $\otimes$ q) = pq $\otimes$ 1. Write q = a/b where a, b∈R with b≠ 0. Then pq $\otimes$ 1 = pa/b $\otimes$ 1 = p/b $\otimes$ ab/b = pb/b $\otimes$ a/b = p $\otimes$ q and it follows that hg is the identity on P $\otimes$ Q, because we only need to check that hg is the identity on the ``simple tensors". Thus h is one-to-one and onto, and our observation is established.

   Now observe that Q $\otimes_{Q}^{}$ V≌V. Indeed we can define a Q-bilinear map θ : Q×V --> V by θ(q, v) = qv, and this induces a Q-map φ : Q $\otimes_{Q}^{}$ V --> V satisfying φ(q $\otimes$ v) = qv. Also we can define a Q-map ψ : V --> Q $\otimes_{Q}^{}$ V by ψ(v) = 1 $\otimes$ v. Then φψ(v) = φ(1 $\otimes$ v) = v, so φψ is the identity on V. Since ψφ(q $\otimes$ v) = ψ(qv) = 1 $\otimes$ qv = q $\otimes$ v and ψφ is the identity on Q $\otimes_{Q}^{}$ V provided it is the identity on the simple tensors, we see that ψφ is the identity on Q $\otimes_{Q}^{}$ V, and the result follows.

   Note that this proof does not use the hypothesis that V is finite dimensional.

7. Let G denote the Galois group of F over K. Since G is a p-group for the prime p = 11, it has a sequence of normal subgroups 1 = G₄ <| G₃ <| G₂ <| G₁ <| G₀ = G, such that G_i <| G and | G_i+1/G_i| = 11 for all i. Now let K_i be the fixed subfield of G_i in K, for i = 0, 1,..., 4. Then K_i is a Galois extension of F for all i, because G_i <| G. Since [K_i : K_i-1] = | G_i-1/G_i| = 11, the result is proven.

8. Suppose R/I is a projective R-module. Then we may write R = I⊕J for some R-submodule J of R. Of course R-submodules of R are the same as ideals, so J is an ideal of R. Since M is the unique maximal ideal of R and I⊆M, we must have J = R. But then J⊇I and thus I + J is not a direct sum. We now have a contradiction and the result follows.