Algebra Prelim Solutions, January 2012

1. Let n = | G|. Then G has an element x of order n. However if H is any proper subgroup of G, then every element of H has order strictly less than n. Thus x cannot be in any proper subgroup of G and the result follows.

2. Suppose G be a simple group of order 6435. Then the number of Sylow 5-subgroups is congruent to 1 mod 5 and divides 9·11·13. Furthermore this number is not 1 because G is not simple. Therefore this number must be 11 and it follows that G has a subgroup of index 11. Since G is simple, we deduce that G is isomorphic to a subgroup of S11 (even A11). This is not possible because 13, and hence 6435, does not divide 11! = | S11|. We conclude that there is no simple group of order 6435 as required.

3. Define θ : M2(Q)×M2(Q) -> M2(Q) by θ(A, B) = AB. It is easily checked that θ is an M2(Z)-balanced map. Therefore θ induces a group homomorphism

φ : M2(Q)⊗M2(Z)M2(Q) -> M2(Q).

It is easy to see that this map is a (M2(Q), M2(Q))-bimodule map. It remains to prove that φ is bijective, and we do this by producing the inverse map. Define ψ : M2(Q) -> M2(Q)⊗M2(Z)M2(Q) by ψ(A) = A⊗1. It is clear that φψ is the identity, so it remains to prove that φψ is the identity. Since φ and ψ are both group homomorphisms, it will be sufficient to show that ψφ is the identity on simple tensors, that is ψφ(AB) = AB. Therefore we need to prove that AB⊗1 = AB.

Choose a positive integer n such that Bn∈M2(Z). Then

AB⊗1 = A/n(Bn)⊗1 = A/nBn = (A/n)nB = AB,

and the result is proven.

4. By the structure theorem for finitely generated modules over a PID, M is a direct sum of modules of the form R and R/pn where p is a prime in R and n is a positive integer. If M is nonzero, then it must contain a summand which is either isomorphic to R or R/pnR, where p is a prime in R. Since M is injective and R is a domain, rM = M for all rR\ 0, in particular pM = M for primes p in R. Thus M cannot contain a summand isomorphic to R/pnR. On the other hand if M contains a summand isomorphic to R, let p be a prime in R, which exists because R is not a field. Since pRR, we see that pMM, a contradiction and the result follows.

5. (a)
Obviously Qp)⊆Q2p), because ζ2p2 = ζp. On the other hand ζ2p = -ζp, hence Q2p)⊆Qp) and the result follows.

(b)
Set f (x) = 1 + x2 + ... + x2p-2. Since f (x)(1 -x2) = 1 -x2p and ζp2p≠±1, we see that ζp and ζ2p both satisfy f (x). Thus the minimal polynomial of both these divides f (x). Now ζp satisfies g(x) : = 1 + x + ... + xp-1. By making the substitution y = x + 1, we see that g(x) is irreducible in Z[x] by Eisenstein for the prime p. Since deg(g) = p - 1≥1, it is also irreducible in Q[x]. It follows that g(x) is the minimal polynomial of ζp over Q. Also by considering the automorphism of Q[x] induced by x |--> -x, we see that g(-x) is the minimal polynomial of ζ2p over Q, so g(-x) divides f (x). It follows that f (x) = g(x)g(-x), the product of two irreducible polynomials.

6. Set f (x) = x5 - 5x - 1. Then f'(x) = 5x4 -5 = 5(x2 + 1)(x - 1)(x + 1). Thus f (x) has a maximum at -1, a minimum at 1. Since f (1) > 0 and f (- 1) < 0, we find that f has exactly 3 real roots and 2 complex roots. We want to prove that f is irreducible (as a polynomial in Q[x]). By Gauss's lemma, if f is not irreducible, then we may write f = gh where g, hZ[x], deg g, deg h≥1, and g, h are monic. Neither of g, h has degree one, because ±1 is not a root of f. Therefore we may without loss of generality assume that deg g = 3 and deg h = 2, say g = x3 + ax2 + bx + c and h = dx + e, where a, b, c, d, eZ. By equating coefficients, we find that a + d = 0, ad + bc + e = 0, ae + bd + c = 0, be + cd = 1, ce = 1. Thus c, e = 1 or c, e = -1, and we find that a2±a + 1 = 0. This last equation has no root in Z and we conclude that f is irreducible.

Let G denote the Galois group of f over Q. We consider G as a subgroup of S5 (by permuting the 5 roots of f). Since f is irreducible and 5 is prime, we see that G contains a 5-cycle. Also G contains a transposition, namely complex conjugation. Since S5 is generated by a 5-cycle and a transposition, we deduce that GS5.

7. (a)
We may write the general element of Q[x, y] as f0 + f1y + f2y2 + ... + fnyn, where n is a positive integer and fiQ[x] for all i. Then modulo the ideal (x3 -y2), we may replace y2 with x3 everywhere and we see that every element of Q[x, y] can be written in the form f + gy + (x3 -y2)h, where f, gQ[x] and hQ[x, y]. The result follows.

(b)
Define a ring homomorphism θ : Q[x, y] -> Q[t] by θ(x) = t2, θ(y) = t3 and θ(q) = q for qQ. Then imθ = Q[t2, t3]. Also if h∈kerθ, write h = k + f + yg, where k∈(x3 -y2), and f, gQ[x]. Then θ(h) = f (t2) + t3g(t2). Since f (t2) is a polynomial involving only even powers of t and t3g(t2) is a polynomial involving only odd powers of t, we see that θ(h) can only be zero if f, g = 0. It follows that kerθ = (x3 -y2) and the result now follows from the fundamental homomorphism theorem. Note that we have also proven that if h(x2, x3) = 0, then h∈(x3 -y2).

(c)
Note that t2 and t3 are irreducible in Q[t2, t3] (use unique factorization in Q[t]). Since t6 = (t2)3 = (t3)2, two different ways of factoring t6, we see that Q[t2, t3] is not a UFD.

(d)
Note that Z(x3 -y2) = {(t2, t3) | tQ}. Indeed (t2, t3)∈Z(x3, y2), because (t2)3 - (t3)2 = 0. On the other hand if (p, q)∈Z(x3 -y2), write t = q/p (t = 0 if p = 0). Since p3 = q2, we see that p = t2 and q = t3. Now suppose f is a polynomial vanishing on V. Then f (t2, t3) = 0 for all tQ. Since Q is infinite, we see that f (x2, x3) = 0 and it follows from (b) that f∈(x3 -y2). It follows that the coordinate ring Q[V] of V is Q[x, y]/(x3 -y2)≌Q[t2, t3] by (b).

(e)
Since Q is an infinite field, A1 has coordinate ring Q[x], a UFD. But Q[V] is not a UFD by (c), in particular Q[A1] is not isomorphic to Q[V]. Since isomorphic affine algebraic sets have isomorphic coordinate rings, we deduce that V is not isomorphic to A1.

Peter Linnell 2012-01-14