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 Zp3×Zq, Zp2×Zp×Zq, and Zp×Zp×Zp×Zq.

    The first group above is generated by one element, while the third requires 3 elements. Therefore GZp2×Zp×ZqZp2×Zpq, because Zp×ZqZpq, 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 PQ⊕ker f. This proves the result, because direct summands of projective modules are projective.

  6. First observe that Q $ \otimes_{R}^{}$ QQ 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, bR 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}^{}$ VV. 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 Gi <G and | Gi+1/Gi| = 11 for all i. Now let Ki be the fixed subfield of Gi in K, for i = 0, 1,..., 4. Then Ki is a Galois extension of F for all i, because Gi <G. Since [Ki : Ki-1] = | Gi-1/Gi| = 11, the result is proven.

  8. Suppose R/I is a projective R-module. Then we may write R = IJ 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 IM, we must have J = R. But then JI and thus I + J is not a direct sum. We now have a contradiction and the result follows.

Peter Linnell 2007-12-09