Algebra Prelim Solutions, Winter 2003

1. We have f (x) = (x + 1)(x⁴ + 3); since -1∈ℚ, the splitting field for x⁴ + 3 is also K. Let ω = (1 + i)/√2, a primitive 8th root of 1. Then ω⁴ = -1 and we see that the four roots of x⁴ + 3 are ω^r 4√3 for r = 1, 3, 5, 7. Therefore K = ℚ(ω ⁴√3,ω^3 4√3,ω^5 4√3,ω^7 4√3). Since x⁴ + 3 is irreducible by Eisenstein for the prime 3, we see that [ℚ(ω ⁴√3) : ℚ] = 4. Let γ denote complex conjugation. Since x⁴ + 3 is a polynomial with real coefficients, we see that γ∈Gal(K/ℚ). Thus  ⁴√12∈K, because  ⁴√12 = ω ⁴√3 + γ(ω ⁴√3). Now  ⁴√12 satisfies x⁴ - 12, which is irreducible by Eisenstein for the prime 3. Therefore [ℚ( ⁴√12) : ℚ] = 4. Note that ℚ( ⁴√12)≠ℚ(ω ⁴√3), because the former is contained in ℝ while the latter is not. We deduce that K≠ℚ(ω ⁴√3). Also i = ω^3 4√3/ω ⁴√3, which shows that i∈K. Since ω^2r+1 4√3 = i^rω ⁴√3, we conclude that K = ℚ(i,ω ⁴√3). Therefore [K : ℚ(ω ⁴√3)] = 2 and hence [K : ℚ] = 8.

   Of course a consequence of this is that x⁴ + 3 remains irreducible over ℚ(i). Let θ∈Gal(K/ℚ(i)) satisfy θ(ω ⁴√3) = ω^3 4√3. Then θ(ω^3 4√3) = ω^5 4√3, θ(ω^5 4√3) = ω^7 4√3 and θ(ω^7 4√3) = ω ⁴√3, in particular θ has order 4. Furthermore γθγ(i) = i and γθγ(ω ⁴√3) = ω^7 4√3, which shows that γθγ = θ^-1. We now see that Gal(K/ℚ) = {θ^rγ^s | r = 0, 1, 2, 3, s = 0, 1} and is isomorphic to the dihedral group of order 8.

2. This is false. Consider the group ℤ₄⊕ℤ₂, where ℤ_n denotes the integers modulo n. Then (2, 0) and (0, 1) both have order 2 (when we write (2, 0), the 2 means 2 modulo 4). Suppose θ is an automorphism such that θ(2, 0) = (0, 1). Then 2(θ(1, 0)) = θ(2, 0) = (0, 1). On the other hand 2(θ(1, 0)) is of the form 2(a, b) = (2a, 0), and so cannot be equal to (0, 1). Thus we have a contradiction and we conclude that there is no such θ.

3. Let n denote the number of Sylow 2-subgroups. Since 2002 = 2*1001, we see that a Sylow 2-subgroup has order 2 and n | 1001. Therefore each Sylow 2-subgroup has exactly one element of order 2 and n is odd. Also any element of order 2 is in exactly one Sylow 2-subgroup, consequently the number of elements of order 2 is n. Since the number of elements in the set {h∈H | h² = e} is n + 1 (the ``+1" for the identity), we conclude that this number is even.

   However a better proof is to pair each h∈H with h^-1. If h²≠e, then {h, h^-1} has order 2, otherwise {h, h^-1} has order 1. It follows that the number of elements h∈H such that h²≠e has even order and since | H| is even, it follows that the number of elements h∈H such that h² = e is even, as required.

4. Suppose G is nonabelian, so there exist a, b∈G such that ab≠ba. Set g = aba^-1b^-1, so g≠1. By hypothesis there exists K⊲G such that G/K is abelian and g∉K. But KaKb(Ka)^-1(Kb)^-1 = Kg≠1, which shows that KaKb≠KbKa and hence G/K is nonabelian, which is a contradiction. The result follows.

5. Certainly if A and B are commutative rings, then A×B is also a commutative ring. We need to show that if A and B are in addition Noetherian, then so is A×B. Suppose I₁, I₂,... is an ascending chain of ideals in A×B. Then (0×B)∩I₁,(0×B)∩I₂,... is an ascending chain of ideals in 0×B. But 0×B≌B and B is Noetherian, hence there exists a positive integer M such that 0×B∩I_n = 0×B∩I_M for all n≥M. Also (A×B)/(0×B)≌A as rings, so (A×B)/(0×B) is Noetherian. Therefore the ascending chain of ideals (0×B) + I₁,(0×B) + I₂,... of (A×B)/(0×B) becomes stationary, that is there is a positive integer N such that (0×B) + I_n = (0×B) + I_N for all n≥N. Let P be the maximum of M and N. We claim that I_n = I_P for all n≥P. Obviously I_n⊇I_P for all n≥P, so we need to show the reverse inclusion. Let x∈I_n. Since (0×B) + I_n = (0×B) + I_P, we may write x = b + i where b∈0×B and i∈I_P. Since x, i∈I_n, we see that b∈(0×B)∩I_n and hence b∈(0×B)∩I_P, because (0×B)∩I_n = (0×B)∩I_P. This shows that x∈I_P and hence I_n = I_P for n≥P.

6. Obviously P/IP is an R/I-module; we need to prove that it is projective. Suppose we are given an R/I-epimorphism μ : M↠N of R/I-modules and an R/I-map θ : P/IP→N. We need an R/I-map β : P/IP→M such that θ = μβ. Let π : P→P/IP denote the natural epimorphism. We can also view M and N as R-modules, and then μ is also an R-map. Since P is a projective R-module, certainly there exists an R-map α : P→M such that μα = θπ. If i∈I and p∈P, then α(ip) = iαp∈IM = 0. Therefore IP⊆kerα and we deduce that α induces an R/I-map β : P/IP→M satisfying βπ = α. Then μβπ = μα = θπ and since π is onto, we conclude that μβ = θ.

   Sketch of alternate proof. Since P is projective, we may write P⊕Q≌F for some R-modules Q, F with F free. Then P/IP⊕Q/IQ≌F/IF and since F/IF is a free R/IR-module, we see that P/IP is a projective R/IR-module.

7. 1. Certainly k + I is a subgroup of k[x] under addition; we need to show that it is closed under multiplication. However if a, b∈k and i, j∈I, then (a + i)(b + j) = ab + (aj + ib + ij)∈k + I, because aj, ib, ij∈I by using I⊲k[x].

   2. Let R = k + I. We first prove that k[x] is finitely generated as an R-module. We may write I = (f ) where f is a monic polynomial in k[x]. Let d denote the degree of f and set M = R + Rx + ... + Rx^d, an R-submodule of k[x]. We prove by induction on n that xⁿ∈M for all n≥ 0. This is obviously true if n = 0, because 1∈R. It is also obviously true for d = 0 because then R = k[x]. Now suppose d, n > 0. Then by the division algorithm xⁿ = qf + r, where q, r∈k[x] and deg r < n. Then we must have deg q < n. Therefore by induction q, r∈M, and it follows that xⁿ∈M as required.

      Now k[x]⊗_Rk[x]/I is an R-module and also an R/I-module. Since k[x] is a finitely generated R-module, we see that k[x]/I is also a finitely generated R-module, and we deduce that k[x]⊗_Rk[x]/I is a finitely generated R/I-module. Since R/I = (k + I)/I≌k/k∩I = k/0≌k, we conclude that k[x]⊗_Rk[x]/I is a finitely generated k-module and the result follows.