Algebra Prelim Solutions, Fall 2007

  1. Let G be a simple group of order 168. The number of Sylow 7-subgroups of G is congruent to 1 mod 7 and divides 168/7 = 24. This number cannot be 1 because that would mean that G has exactly one Sylow 7-subgroup, consequently G would have a normal Sylow 7-subgroup and we would deduce that G is not simple, contrary to the hypothesis. It follows that G has exactly 8 Sylow 7-subgroups. Also by Lagrange's theorem, two distinct Sylow 7-subgroups must intersect in the identity. Since any element of order 7 is contained in a Sylow 7-subgroup and there are 6 elements of order 7 in each Sylow 7-subgroup, we deduce that there are 8*6 = 48 elements of order 7 in G.

  2. Note that ρ = i. The following are easy to check: Q(√2)≠Q√3. Thus Q(√2,√3) is a Galois extension of degree 4 over Q. Let K be the splitting field of x4 - 2 over Q and let L be the splitting field of x2 - 3 over Q. The roots of x4 - 2 are ± 4√2, ±i 4√2, hence K is a Galois extension of degree 8 over Q, and has maximal real subfield of degree 4 over Q, namely Q( 4√2). Since this subfield is not normal over Q, we deduce that Q(√2,√3) is not contained in K. Therefore KL = Q, and we deduce that KL(i) = Q(i). The Galois group of L/Q has order two and is therefore isomorphic Z/2Z. Also the Galois group of K/Q is a group of order 8 and not every subgroup is normal, because Q(4√2) is not normal over Q. and we deduce that this group is isomorphic to the dihedral group D8 of order 8. Finally Gal(K/Q(i))≌Z/4Z, being generated by the automorphism determined by 4√2 | --> i 4√2.

    The Galois group of (x4 -2)(x2 - 3) over Q is Gal(K/Q)×Gal(L/Q)≌D8×Z/2Z. The Galois group of (x4 -2)(x2 - 3) over Q(i) is

    Q(i) is Galois over Q because it is the splitting field of x2 + 1 over Q.

    Yes, because Gal(LK/Q(i)) has nontrivial normal subgroups.

  3. Since 0 --> A -- f--> B -- g--> C --> 0 is split exact, there exists h : B --> A such that hf = 1A, the identity map on A. Then (1D $ \otimes$ h)(1D $ \otimes$ f )= 1D $ \otimes$ hf = 1D $ \otimes$ 1A = 1. Thus if xD $ \otimes_{R}^{}$ A and (1D $ \otimes$ f )(x) = 0, then (1D $ \otimes$ f )(1D $ \otimes$ h)(x) = 0, consequently 1(x) = 0 and we conclude that x = 0, as required.

  4. Certainly S-1R is an integral domain, since it is a subring of the field of fractions of R, so we need to prove that every ideal of S-1R is principal. Let I <S-1R and let J = IR. Then J <R, so J = xR for some xR. Obviously xS-1RI, so it remains to prove that xS-1RI. However if yI, then syIR = J where sS and hence we may write sy = xr for some rR. Therefore y = s-1(sy) = xs-1rxS-1R and the result is proven.

  5. Let G be a group of order 24·112. The number of Sylow 11-subgroups is congruent to 1 mod 11 and divides 16, consequently there is exactly one Sylow 11-subgroup; call this Sylow 11-subgroup H. Then H <G. Now G/H and H are p-groups for p = 2 and 11 respectively, and p-groups are solvable (even nilpotent). However the property of being solvable is closed under extensions, that is H and G/H solvable implies G is solvable, which is the required result.

  6. (a)
    Apply Eisenstein's criterion for the prime 3.

    We know that f is irreducible (from (a)) and that g is irreducible (use Eisenstein for the prime 2). Since Q[x] is a PID, we see that (f ) and (g) are maximal ideals of Q[x]. Furthermore (f )≠(g), because f and g are not scalar multiples of each other. It now follows from the Chinese remainder theorem that Q[x]/(fg)≌Q[x]/(fQ[x]/(g), a product of two fields. The dimension over Q of these two fields are the degrees of the polynomials f and g, that is 4 and 2 respectively.

  7. (a)
    Since 1·0 = 0, we see that 0∈t(X). Next suppose that x, yt(X). Then there exist r, sR\ 0 such that rx = 0 = sy and we have (rs)(x + y) = 0. Since rs≠ 0 because R is an integral domain, we conclude that x + yt(X). Finally suppose that xt(X) and rR. Then there exists sR\ 0 such that sx = 0 and consequently s(rx) = 0. This shows that rxt(X) and we have established that t(X) is an R-submodule of X.

    Write T = t(X) and let xt(T); we want to prove that xT. Since xt(T), there exists sR\ 0 such that sxT, and then there exists tR\ 0 such that t(sx) = 0. It follows that (st)x = 0 and since st≠ 0 because R is an integral domain, we conclude that xT as required.

    Because t(X/t(X)) is cyclic, t(X/t(X))≌R/I for some I <R. But t(X/t(X)) = 0 by (b), hence I = 0 and we deduce that X/t(X)≌R. Since R is a projective R-module, 0 --> t(X) --> X --> X/t(X) --> 0 splits, in particular Xt(X)⊕R, as required.

Peter Linnell 2007-08-11