Algebra Prelim Solutions, August 2009

1. Let sS and let H denote the stabilizer of s in G. Since G acts transitively on S, we have | G| = pn| H|, hence pn | | G|/| PH| and we deduce that pn | | P|/| PH, because p | | G|/| P|. Therefore pn divides the size of the orbit of s under P, because PH is the stabilizer of s in P. Thus we must have the orbit of s under P is the whole of S and the result is proven.

2. Let G be a simple group of order 448. The number of Sylow 2-subgroups of G is congruent to 1 mod 2 and divides 7, and cannot be 1 because G is not simple. Therefore G has exactly 7 Sylow 2-subgroups and because G is simple, we deduce that G is isomorphic to a subgroup of A7. This is not possible because 448 does not divide | A7|, so the result is proven.

3. (a)
If x2 + 1 was not irreducible, then it would have a root in Z/3Z. This is not the case, because x2 = 0 or 1 mod 3.

(b)
We have an epimorphism Z/3Z[x] --> > Z[i]/3Z[i] determined by x |--> i whose kernel contains x2 + 1. Thus from part (a), we see that Z[i]/3Z[i] is a field and hence 3 is a prime in Z[i]. We can now apply Eisenstein's criterion for the prime 3. Since 3 divides 3 and -9, but 32 does not divide 12 in Z[i], the result is proven.

4. By the structure theorem for finitely generated modules over a PID, there is an R-submodule K of M containing N such that M/K is a torsion module and K/N is a free module, so there exists 0≠rR such that MrK. Since K/N is free, there exists a submodule L of K such that L + N = K and LN = 0. The result follows.

5. Let bB. Since f is onto, there exists aA such that f (a) = b. Now set k(b) = g(a). If we had instead chosen a'A such that f (a') = b, then

jg(a') = hf (a') = h(b) = hf (a) = jg(a)

and we deduce that g(a') = g(a) because j is one-to-one; in other words, the definition of k does not depend on the choice of a. Next we need to show that k is an R-module homomorphism. Suppose b, b'B and choose a, a'A such that f (a) = b and f (a') = b'. Then f (a + a') = b + b'. Thus k(b + b') = g(a + a') = g(a) + g(a') = k(b) + k(b'). Also if rR, then f (ar) = br, consequently k(br) = g(ar) = g(a)r = k(b)r and we have shown that k is an R-module homomorphism. Clearly kf = g. Furthermore jkf = jg = hf and since f is onto, we deduce that jk = h. Finally k is unique because j is one-to-one.

6. Solving x4 -2x2 + 9 = 0, we find that x2 = 1±2√2i and we deduce that the roots of x4 -2x2 + 9 are ±√2±i. It follows that the splitting field is Q[i,√2]. Since this has degree 4 over Q, we see that the Galois group has order 4. The automorphisms induced by i |--> - i, √2 |--> √2 and i |--> i, √2 |--> - √2 both have order 2 and we conclude that the Galois group is isomorphic to Z/2Z×Z/2Z.

7. We can define an R-bilinear map R/I×R/J -> R/(I + J) by (r + I, s + J) |--> rs. This induces an R-module map θ : R/IRR/J -> R/(I + J) satisfying θ((r + I)⊗(s + J)) = rs + I + J. Now define φ : R -> R/IRR/J by φ(r) = (r + I)⊗R(1 + J). Then φ is an R-module map and clearly I⊆kerφ. Also if jJ, then φ(j) = (j + I)⊗(1 + J) = (1 + I)⊗(j + J) = 0. It follows that I + J⊆kerφ and we deduce that φ induces an R-module map ψ : R/(I + J) -> R/IRR/J such that ψ(r + I + J) = (r + I)⊗(1 + J). Note that θψ(r + I + J) = θ((r + I)⊗(1 + J)) = r + I + J so θψ is the identity map. Finally ψθ(r + I)⊗(s + J) = ψ(rs + I + J) = (rs + I)⊗(1 + J) = (r + I)⊗(s + J) and we conclude that ψθ is also the identity map. This shows that θ and ψ are isomorphisms, and the result is proven.

Peter Linnell 2011-07-21