MATH 601: Abstract Algebra II 6th Homework Partial Solutions

Jonathan Rosenberg

assignment due Wednesday, March 28, 2001

1 Hungerford Problems

V, Section 7, Exercise 6.b. Show that if K is a field of characteristic p and there exists a cyclic extension of degree p of K, then there exists a cyclic extension of degree p^{n} of K, for all n ≥ 1. Solution. We prove this by induction on n, the case of n = 1 being trivial. Assume the result is true for n−1, so there is a field E cyclic over K of degree p^{n−1}. Choose a generator σ of Gal(E/K). Since E is separable over K, T_{K}^{E} : E → K is a non-trivial K-linear map, and since dim_{K} K = 1, T_{K}^{E} is therefore surjective. So there is an element v ∈ E with T_{K}^{E}(v) = 1. In more concrete terms,

v + σ(v) + ⋯ + σ^{p^{n−1}−1}(v) = 1.

Clearly v ∉ K, since otherwise we’d have T_{K}^{E}(v) = [K : E]v = p^{n−1}v = 0 (since we’re in characteristic p), so in particular v^{p} ≠ v. Since σ commutes with the endomorphism x ↦ x^{p},

T_{K}^{E}(v^{p}) = v^{p} + σ(v^{p}) + ⋯ + σ^{p^{n−1}−1}(v^{p}) = (v + σ(v) + ⋯ + σ^{p^{n−1}−1}(v))^{p} = 1^{p} = 1.

So T_{K}^{E}(v^{p} − v) = 1 − 1 = 0 and by the additive analogue of Hilbert’s Theorem 90, there is an element u ∈ E with σ(u) − u = v^{p} − v. We claim x^{p} − x − u ∈ E[x] is irreducible. For if not, we know from Hungerford Corollary V.7.9 that w^{p} − w − u = 0 for some w ∈ E. Then v^{p} − v = σ(u) − u = σ(w^{p} − w) − (w^{p} − w) and so

(v − σ(w) + w)^{p} = v − σ(w) + w,

i.e., v − σ(w) + w lies in the fixed field of x ↦ x^{p}, which is 𝔽_{p} ⊆ K. So for some j ∈ 𝔽_{p}, σ(w) = v + w + j. By iteration, we get σ^{2}(w) = σ(v + w) + j = σ(v) + v + w + 2j, and so

σ^{p^{n−1}}(w) = (v + σ(v) + ⋯ + σ^{p^{n−1}−1}(v)) + w + p^{n−1}j = T_{K}^{E}(v) + w = w + 1.

Since σ had order p^{n−1}, this is a contradiction. So x^{p} − x − u ∈ E[x] is irreducible. Let w be a root; then E(w) is cyclic over E of order p, and of degree [E(w) : E][E : K] = p·^{n−1} = p^{n} over K. We claim E(w) is cyclic over K of order p^{n}. Indeed, if we define σ̃ to be σ on E and to send w ↦ w + v, then the calculation above shows that σ̃ is an automorphism of E(w) extending σ on E and with the property that σ^{p^{n−1}}(w) = w + 1, so that σ^{p^{n−1}} generates Gal(E(w)/E). So σ̃ has order p^{n} and generates Gal(E(w)/K). This proves that E(w) is Galois over K with cyclic Galois group. □

2 Additional Exercises

2. Let K be a finite field with q elements, and let L be a finite extension field with [L : K] = r. (Thus L has qr elements.) Recall that the multiplicative groups L^{×} and K^{×} are cyclic, and that G = Gal(L/K) is also cyclic. Compute the norm map N_{K}^{L} : L^{×} → K^{×} explicitly, and show that it is surjective. Show that your calculation agrees with the prediction of Hilbert’s Theorem 90. Solution. Recall that G is cyclic of order r, with generator the Frobenius automorphism σ : x ↦ x^{q}. For an element x ∈ L^{×}, the norm is given by

N_{K}^{L}(x) = ∏_{i=0}^{r−1} σ^{i}(x) = ∏_{i=0}^{r−1} x^{q^{i}} = x^{∑_{i=0}^{r−1} q^{i}} = x^{(q^{r}−1)/(q−1)}.

In particular N_{K}^{L} : L^{×} → K^{×} is a homomorphism. If x_{0} is a generator of L^{×}, then x_{0} has order q^{r} − 1 and (N_{K}^{L}(x_{0}))^{j} = 1 exactly when j(q^{r} − 1)/(q − 1) is divisible by q^{r} − 1, i.e., when (q − 1) | j. That shows N_{K}^{L}(x_{0}) has order exactly q − 1 and is thus a generator of K^{×}, so N_{K}^{L} : L^{×} → K^{×} is surjective. Then by the isomorphism theorems, K^{×} ≅ L^{×}/ ker N_{K}^{L}, so ker N_{K}^{L} has order |L^{×}|/|K^{×}| = (q^{r} − 1)/(q − 1). Now Hilbert’s Theorem 90 claims that the kernel of the normal map should consist of elements of the form σ(y)/y. Since σ(y)/y = y^{q}/y = y^{q−1} and (q − 1) | |L^{×}|, ker N_{K}^{L} should have order |L^{×}|/(q − 1) = (q^{r} − 1)/(q − 1), which is just what we showed. □

3. Let K be the splitting field over ℚ(ω), ω a primitive cube root of unity, of the polynomial x^{3} − 3x + 1. Show that K is a cyclic extension of ℚ(ω) of degree 3, and use the Lagrange resolvant method to show it’s obtained by adjoining a cube root of something. Again write down the norm map K^{×} → ℚ(ω)^{×} explicitly and verify the conclusion of Hilbert’s Theorem 90 for this case. Solution. Let α be a root of x^{3} − 3x + 1. We claim ℚ(ω)(α) = ℚ(ω, α) is a splitting field of x^{3} − 3x + 1 over ℚ(ω). First observe since x^{3} − 3x + 1 has no integral roots, x^{3} − 3x + 1 is irreducible over ℚ by Gauss’s Lemma (any factorization over ℚ would give a linear factor over ℤ, and thus an integral root), and hence α has degree 3 over ℚ. Since [ℚ(ω) : ℚ] = 2, α cannot lie in ℚ(ω) and ω cannot lie in ℚ(α), hence [ℚ(ω, α) : ℚ] = [ℚ(ω, α) : ℚ(α)][ℚ(α) : ℚ] = [ℚ(ω, α) : ℚ(ω)][ℚ(ω) : ℚ] = 3 · 2 and [ℚ(ω, α) : ℚ(ω)] = 3. Next observe that the discriminant of x^{3} − 3x + 1 is −27 − 4 · (−3)^{3} = (−27) · (1 − 4) = 81, which is a perfect square, and hence ℚ(ω, α) is a splitting field of x^{3} − 3x + 1 over ℚ(ω). Since [ℚ(ω, α) : ℚ(ω)] = 3, the Galois group G is cyclic of order 3. Let σ be a generator of G, and consider the Lagrange resolvant ζ = α + σ(α)ω + σ^{2}(α)ω^{2}. Then σ(ζ) = σ(α) + σ^{2}(α)ω + αω^{2} = ω^{2}ζ, so ζ ∉ ℚ(ω) (since it is not fixed by σ) while ζ^{3} ∈ ℚ(ω) (since σ(ζ^{3}) = (σ(ζ))^{3} = (ω^{2}ζ)^{3} = ζ^{3}). The method developed in class shows (with appropriate sign choices) that ζ is a cube root of 27ω, so ζ is 3 times a primitive 9th root of 1.

Now write K as ℚ(ω)[ζ] = ℚ(ω) ⊕ ℚ(ω)ζ ⊕ ℚ(ω)ζ^{2}. The generator σ of the Galois group may be chosen to send ζ ↦ ζω, so

N(x + yζ + zζ^{2}) = (x + yζ + zζ^{2})(x + yζω + zζ^{2}ω^{2})(x + yζω^{2} + zζ^{2}ω)
= x^{3} + 27ωy^{3} + 3^{6}ω^{2}z^{3} − (81/2)xyzω.

Thus the kernel of the norm map consists of those x + yζ + zζ^{2}, x, y, z ∈ ℚ(ω), for which x^{3} + 27ωy^{3} + 3^{6}ω^{2}z^{3} − (81/2)xyzω = 1. Hilbert’s Theorem 90 says this is the same as elements of the form

(x + yζω + zζ^{2}ω^{2})/(x + yζ + zζ^{2}).

□