Selected Solutions to Assignment 3

Question 3

We first want to show that f(x) is separable, we take its derivative which is simply f′(x) = −1 (we note that the field has characteristic p). Then we have that gcd(f, f′) = 1 implying that f is separable (lemma 7.2.2 in Goren’s notes). This thus implies that K/F is Galois.

We now look at GL_{2}(F_{p}), and note that |GL_{2}(F_{p})| = p(p − 1)^{2}(p + 1). We intend to show that Gal(K/F) ⊆ GL_{2}(F_{p}). The crucial remark is that the set R of roots of f(x) is not just a set, but is also equipped with a structure of a vector space over F_{p}, namely, it is closed under addition as well as multiplication by scalars in F_{p}. (This unusual feature of R arises from the fact that f(x) is a linear polynomial, i.e., an F - linear combination of monomials of the form x^{p^{j}}.) Furthermore, the Galois automorphisms of K over F = F_{p}(t) act on R not just as permutations on this set of p^{2} elements (which would merely imply that the Galois group is contained in the symmetric group S_{p^{2}} on p^{2} elements) but actually operates on R as F_{p}-linear transformations. All of this implies that

Gal(K/F) < GL_{2}(F_{p}),

implying that Gal(K/F) ⊆ GL_{2}(F_{p}) and one concludes that [K : F]|p(p − 1)^{2}(p + 1).

Question 4

We know that the Galois group in question here is (ℤ/7ℤ)^{×} = (ℤ/6ℤ). The subgroups are isomorphic to (ℤ/2ℤ), (ℤ/3ℤ), and the identity. Each subgroup corresponds to a subfield. We’ll look at (ℤ/3ℤ) here. We take its elements as id, σ_{2}, σ_{4}, then an element that fixes the field is a := ζ + ζ^{2} + ζ^{4}, whose Galois conjugate is b := ζ^{6} + ζ^{5} + ζ^{3}. Thus we get that one of the subfields is ℚ(ζ + ζ^{2} + ζ^{4}). An easy calculation shows that a + b = −1 and ab = 2 – the fact that these expressions are rational was expected, given that they are preserved under the full group of symmetries of Q(ζ). It follows that a and b satisfy the polynomial x^{2} + x + 2, therefore

a, b = \frac{−1 ± \sqrt{−7}}{2}.

It is rather interesting that the field of seventh roots of unity contains a square root of −7.

Question 5

Let E be the splitting field of f(x) over F. By assumption, Gal(E/F) acts as the full permutation group S_{n} on the roots of f(x), and hence Gal(E/K) acts as S_{n−1} on the n − 1 roots of g(x) ∈ K[x]. Since this action is transitive, it follows that g(x) is irreducible in K[x].

Question 6

We want to show that F (α^{2}) ⊆ F (α) and F (α) ⊆ F (α^{2}). The first assertion is obvious, since obviously α · α = α^{2}. The other way a little trickier but not too hard. We note that α solves x^{2} − α^{2}, and hence we know that

[F (α) : F (α^{2})] ≤ 2

So then we get that

[F (α) : F] = [F (α) : F (α^{2})][F (α^{2}) : F]

But then since F satisfies an irreducible polynomial of odd degree over F, we know that [F (α) : F (α^{2})] is odd, implying that [F (α) : F (α^{2})] is also odd. Since then [F (α) : F (α^{2})] ≤ 2, and it must be odd, we know now that [F (α) : F (α^{2})] =, implying that F (α) ⊆ F (α^{2}), and we are done.

Question 9 Let us view σ as an F -linear transformation acting on K. Since σ^{p} = 1 its eigenvalues consist of p-th roots of unity. Since these roots of unity are contained in F, it follows that σ is diagonalisable. Furthermore, σ ≠ 1, hence there is a non-trivial p-th root of unity, ζ, which is an eigenvalue for σ. Let b ∈ K be an associated eigenvector. Since σ(b) = ζb ≠ b, it generates a non-trivial extension of F, hence it generates all of K since K is of prime degree over F. Furthermore, a = b^{p} is fixed by σ, since σ(b^{p}) = σ(b)^{p} = (ζb)^{p} = b^{p}. It follows that K = F (b) = F (a^{1/p}).

Question 10a)

Noting that the polynomials are irreducible (Einstein Criterion), we start by doing

F (u^{1/p}, v^{1/p}) ≅ (\frac{F [x]}{(x^{p} − u)})[y]/(y^{p} − v)

Then we have that

[F (u^{1/p}, v^{1/p}) : F] = [F (u^{1/p}, v^{1/p}) : F (v^{1/p}][F (v^{1/p} : F] = deg(x^{p}−u)deg(y^{p}−v) = p^{2}.

Question 10b)

Let σ ∈ Aut(K/F). Since x^{p} − u = (x − u^{1/p})^{p} is the minimal polynomial of u^{1/p} over F, the automoprhism σ must send u^{1/p} to another root of the same polynomial, but there is only one such root, hence σ fixes u^{1/p}. By the same reasoning it also fixes v^{1/p} and it follows that σ = 1. Hence Aut(K/F) = {1}. This means that K is not Galois over F.

Question 10c)

With α ∈ F, define a new field

F_{α} = {a + +b(u^{1/p} + αv^{1/p})|a, b ∈ F }

We clearly have that F ⊆ F_{α} ⊆ K. We know further that

p^{2} = [K : F] = [K : F_{α}][F_{α} : F]

implying that [F_{α} : F] ∈ {1, p, p^{2}}. Since (a + bu^{1/p} + αbv^{1/p})^{p} = a^{p} + b^{p}u + αb^{p}v ∈ F (this holds due to the field having characteristic p), which then implies that [F_{α} : F] ≠ p^{2}. But then clearly F_{α} ≠ F, then we must have [F_{α} : F] = p. Thus for different values of α ∈ F, we get different fields F_{α} that have degree p over F, and there are an infinite number of them since F is infinite.