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(A⊗B) = A⊗B. Therefore we need to prove that AB⊗1 = A⊗B.

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

   AB⊗1 = A/n(Bn)⊗1 = A/n⊗Bn = (A/n)n⊗B = A⊗B,
   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 r∈R\ 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 pR≠R, we see that pM≠M, a contradiction and the result follows.

5. (a)

   Obviously Q(ζ_p)⊆Q(ζ_2p), because ζ_2p² = ζ_p. On the other hand ζ_2p = -ζ_p, hence Q(ζ_2p)⊆Q(ζ_p) and the result follows.

   (b)

   Set f (x) = 1 + x² + ... + x^2p-2. Since f (x)(1 -x²) = 1 -x^2p and ζ_p,ζ_2p≠±1, we see that ζ_p and ζ_2p both satisfy f (x). Thus the minimal polynomial of both these divides f (x). Now ζ_p satisfies g(x) : = 1 + x + ... + x^p-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 ζ_p over Q. Also by considering the automorphism of Q[x] induced by x |--> -x, we see that g(-x) is the minimal polynomial of ζ_2p 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, h∈Z[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, e∈Z. 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 G≌S₅.

7. (a)

   We may write the general element of Q[x, y] as f₀ + f₁y + f₂y² + ... + f_nyⁿ, where n is a positive integer and f_i∈Q[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, g∈Q[x] and h∈Q[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 q∈Q. Then imθ = Q[t², t³]. Also if h∈kerθ, write h = k + f + yg, where k∈(x³ -y²), and f, g∈Q[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) | t∈Q}. 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 t∈Q. 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.