May 7, 2014

http://coursework.tylerlogic.com/courses/upenn/math503/homework11
Given the Fundamental Theorem of Galois Theory, in order to find fields F ⊂ K ⊂ L such that K∕F and L∕K are finite Galois, but L∕F is not Galois, it suffices to find a finite Galois extension M∕F with Galois group G with subgroups H

In the last homework we saw that the splitting field of T^{4} - 3 over ℚ is finite Galois with Galois group D_{8}. In the
group D_{8} we have subgroups ⟨sr⟩≤⟨sr,r^{2}⟩≤ D_{8} where s is the mirror symmetry and r is the rotational
symmetry.

The fact that r(sr)r^{3} = rsr^{4} = rs = sr^{3} demonstrates that ⟨sr⟩ is not normal in D_{8}. Since ⟨sr,r^{2}⟩ = {1,r^{2},sr,sr^{3}}, then

Thus for F = ℚ, K = Ω^{⟨sr,r2⟩
} and L = Ω^{⟨sr⟩} where Ω = ℚ(,i) is the splitting field of T^{4} - 3 over ℚ, we will have our
desired scenario.

We can determine the fixed fields K and L as follows. Define s and r as the automorphisms

where θ = . This then sets

Because ⟨sr⟩ is a subgroup of size two, then L∕ ℚ must be an extension of degree four. Since

and ℚ(θ(i- 1)) ⊂ Ω is an extension of degree four, then we must have L = ℚ(θ(i- 1)). Similarly, ⟨sr,r^{2}⟩ is a subgroup of
size four, and ℚ(iθ^{2}) ⊂ ℚ(θ(i - 1)) ⊂ Ω has degree two. Then since

and

then we have K = ℚ(iθ^{2}).

Let L be a finite Galois extension of ℝ.

Proof. If K is a non-trivial finite extension of ℝ, then it is algebraic over ℝ. For any α ∈ K - ℝ, it’s minimal
polynomial m_{α}(x) ∈ ℝ[x] is nonlinear, irreducible. Since all odd degree polynomials in ℝ[x] have a real root, we
conclude deg(m_{α}(x)) is even. However, [K : ℝ] = deg(α) = deg(m_{α}(x)) and thus K has even degree. __

By the preceeding lemma, [L : ℝ] is even. Then so is the size of Gal(L∕ ℝ). So define H to be a 2-Sylow subgroup of
Gal(L∕ ℝ). Then the index of H in Gal(L∕ ℝ) is not divisible by two and therefore E = L^{H} is an extension of ℝ
with odd degree. However, according to Lemma 3.1, the only such extension is ℝ itself. Thus E = ℝ. This
furthermore implies that H = Gal(L∕ ℝ), which, since H is a 2-Sylow subgroup, also means Gal(L∕ ℝ) is a
2-group.

Proof. Any quadratic over ℂ has roots in ℂ provided by the quadratic formula. __

Let f(T) ∈ ℂ[T] and α be a root of f(T). Assume by way of contradiction that ℂ(α) is a nontrivial extension over ℂ. Then ℂ(α) is an extension of ℝ of even degree by Lemma 3.1. Thus, Gal(ℂ(α)∕ ℝ) would be a nontrivial 2-group with size greater than or equal to 4, by part (a) of this problem. But then Gal(ℂ(α)∕ ℝ) would have a subgroup of size 4 (Theorem 6.1 [DF04]: “p-groups have subgroups of all applicable sizes”), i.e. there would exist an extension K∕ ℂ with degree 2. However, this contradicts the Lemma 3.2, and hence ℂ(α) is trivial, implying α ∈ ℂ.