Algebra Prelim Solutions, Spring 1980

  1. We shall use the fundamental theorem for finitely generated abelian groups. We may write

    A = i=1n(Z/qiZ)ai,    B = i=1n(Z/qiZ)bi,    C = i=1n(Z/qiZ)ci

    where ai, bi, ci, n are nonnegative integers, and the qi are distinct prime powers. Then ABAC yields


    The fundamental theorem now shows that ai + bi = ai + ci and hence bi = ci for all i. It follows that BC as required.

  2. Suppose by way of contradiction G is a simple group of order 56. Sylow's theorem for the prime 7 shows that the number of Sylow 7-subgroups is congruent to 1 modulo 7 and divides 8, hence the number of Sylow 7-subgroups is 1 or 8. If there is one Sylow 7-subgroup, x then this subgroup must be normal in G which contradicts the hypothesis that G is simple, consequently G has 8 Sylow 7-subgroups.

    Let A, B be two distinct Sylow 7-subgroups. Then AB is a subgroup of A, so by Lagrange's theorem | AB| divides | A|, hence | AB| = 1 or 7. Now A and B both have order 7, so we cannot have | AB| = 7. Therefore we must have | AB| = 1. Since every nonidentity element of a Sylow 7-subgroup has order 7, we deduce that G has at least 6×8 elements of order 7.

    Now every Sylow 2-subgroup has order 8, and every element of a Sylow 2-subgroup has order a power of 2 by Lagrange's theorem, so if G has at least two Sylow 2-subgroups, then G has at least 9 elements of order a power of 2. Since we have already shown that G has at least 48 elements of order 7, we now have that G has at least 9 + 48 = 57 elements, which is not possible because | G| = 56. Therefore G has exactly one Sylow 2-subgroup, and so this Sylow subgroup must be normal which contradicts the hypothesis that G is simple.

  3. Since we are working over C, an algebraically closed field, we may use the Jordan canonical form. Here every matrix has a unique Jordan canonical form, and two canonical forms are in the same equivalence class of T if and only if they are equal. Since the eigenvalues of a matrix in T are 4,4,17,17,17, the Jordan canonical form of such a matrix must look like

     4 a 0 0 0
     0 4 0 0 0
     0 0 17 b 0
     0 0 0 17 c
     0 0 0 0 17

    where a, b, c are 1 or 0, and b = 0 if c = 0. Therefore there are 6 equivalence classes in T.

  4. There are many answers to this problem; perhaps the simplest example of a UFD which is not a PID is Z[X]. This is a UFD because Z is a UFD, and a polynomial ring over UFD is again a UFD. We now establish that Z[X] is not a PID by showing that the ideal (2, X) (the ideal generated by 2 and X) is not principal.

    Suppose on the contrary that (2, X) = f for some fZ[X]. Then there exist g, hZ[X] such that fg = 2 and fh = X. The equation fg = 2 shows that f must have degree 0, in other words fZ, and then fh = X shows that f = ±1. Thus we must have (2, X) = Z[X]. Now the general element of (2, X) is 2a + Xb where a, bZ[X]. The constant term of 2a must be divisible by 2, and the constant term of Xb must be zero, hence the constant term of 2a + Xb must be divisible by 2. In particular we cannot have 2a + Xb = 1, so 1 ∉ (2, X) and we have the required contradiction.

  5. Let G denote the Galois group of x3 - 10 over Q. By Eisenstein's criterion for the prime 2 (or otherwise), we see that x3 - 10 is irreducible over Q, hence GS3 or A3. Let ω = (-1 + i√3)/2, a primitive cube root of unity. Then the roots of x3 - 10 are 3√10, ω 3√10, and $ \bar{{\omega}}$ 3√10 ( $ \bar{{\omega}}$ = (-1 - i√3)/2), hence x3 - 10 has one real root and two complex roots, so complex conjugation is an element of G. Therefore G has an element of order 2, which rules out the possibility GA3. Therefore GS3.

    A splitting field of x3 - 10 is Q(i√3,3√10). The normal subfields of the splitting field correspond to normal subgroups of G. The normal subfields corresponding to the subgroups 1 and G are Q(i√3,3√10) and Q respectively. G has exactly one other normal subgroup, namely A3, so there is exactly one other normal subfield. Since subfields of degree two over Q are always normal, this other normal subfield must be Q(i√3).

  6. (a) How to prove this depends on how much field theory one is allowed to assume. Also the result is true without the hypothesis that f is separable. Here is one way to proceed. Suppose K1, K2 are fields, θ : K1 -> K2 is an isomorphism, gK1[X] is an irreducible polynomial, α1 is a root of g in a splitting field L1 for g, and α2 is a root of the irreducible polynomial θg in a splitting field L2 for θg (recall if g = a0 + a1X + ... + anXn with aiK1, then θg = θa0 + θa1X + ... + θanXnK2[X]). Then θ extends to an isomorphism φ : K11) -> K22) such that φ(α1) = φ(α2). Using induction on the degree of g, we deduce that φ in turn extends to an isomorphism of L1 onto L2. This is what is required.

    (b) Note that x4 - 2 is indeed irreducible over Q, by Eisenstein's criterion for the prime 2. The roots of x4 - 2 are ± 4√2 and ±i 4√2. Consider the permutation of the roots 4√2 -> 4√2, - 4√2 -> i 4√2 -> -i 4√2 -> - 4√2. This cannot be induced by an element θ of the Galois group of x4 - 2 over Q, because if θ 4√2 = 4√2, then we must have θ(- 4√2) = - 4√2.

  7. Since HomA(M, k) = 0, we must have M = MM. Then Nakayama's lemma yields the result.

Peter Linnell 2011-08-15