# Galois theory basics, part 2

From the previous post, we know that (in characteristic ${0}$), if ${[K:F]}$ is finite, then (by the primitive element theorem to get some ${\alpha}$ to satisfy hypotheses of previous explanation) ${G(K/F) = G(K/K^{G(K/F)})}$ and

$\displaystyle [K:K^{G(K/F)}] = \lvert{G(K/F)}\rvert = \lvert{G(K/K^{G(K/F)})}\rvert$

(so ${K}$ is a splitting field over/Galois over ${K^{G(K/F)}}$) equals the degree of the “splitting part” of min poly of ${\alpha}$.

So naturally, we wonder if we can say anything starting from the perspective of groups, i.e. “can ${G(K/F)}$ be replaced by an arbitrary (finite) group of automorphisms (of ${K}$) ${H}$?”

Perhaps not surprisingly, yes!

# Galois theory basics, part 1

I was recently looking at algebraic proofs of FTA, and decided to review some Galois theory. So here’s some notes/an attempt to motivate Galois theory (perhaps up to or a bit before fundamental theorem of Galois theory), from the perspective of someone who much prefers field theory/polynomials to group theory/automorphisms (although maybe I should just stop being stubborn).

(Random side note: in the context of abstract algebra, polynomials are naturally motivated by “minimal (free) ring extensions”: if we have a commutative ring ${R}$, then ${R[x]}$ is the smallest ring containing ${R}$ with the additional element ${x}$ satisfying no relations. On the other hand, any constraints/extra relations would be polynomials, so at least for monic polynomials ${f}$ we get the quotient construction ${R[x]/(f)}$.)

Suppose ${K = F(\alpha)}$ is a singly-generated field extension (by primitive element theorem this is broader than it seems). If ${f}$ is the minimal polynomial of ${\alpha}$ of degree ${d}$, then let’s look at how it splits/factors in ${K[x]}$. If ${f}$ has some set of roots ${r\le n}$ of roots lying in ${K}$ (the “splitting part”/linear factors of ${f}$ in ${K[x]}$), say ${z_1 = \alpha, z_2, \ldots, z_r}$.

Generally, the main object of Galois theory seems to be splitting fields (or if we’re lazy, algebraic closures), but I’m still fighting through the material myself so I don’t fully appreciate/communicate this myself. Perhaps the point is just that it’s much easier to work concretely with roots (in algebraic closures) than directly with irreducible polynomials. (For example, we can then work with symmetric sums of the roots, and generally draw lots of intuition from number fields.)

We’ll work in characteristic ${0}$ for convenience, so e.g. the ${z_i}$ are pairwise distinct.

1. “Symmetry” of the ${z_i}$: crux of Galois theory, and introducing Galois groups

The key is that ${K = F(z_1) = F(z_2) = \cdots = F(z_r)}$ (recall ${\alpha = z_1}$ by definition), since the ${z_i}$ share minimal polynomials ${f \in F[x]}$.

To make this symmetry precise, we phrase things in terms of ${F}$-automorphisms of ${K}$; each ${F}$-automorphism fixes coefficients of ${f}$, hence is uniquely determined by sending ${\alpha \rightarrow z_i}$. Thus they form a Galois group ${G := G(K/F)}$ of size ${r}$, since we easily check the automorphisms to be bijections of ${K}$.