# 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!

It’s not unnatural to do this from scratch, but we can run the arguments from before more or less in reverse. We’ll (among other things) show ${K}$ is a finite splitting/Galois extension of ${K^H}$.

First observe (and this is the reason ${K^H}$ is “not too unnatural from scratch”) that ${H}$ acts on ${K}$ (since automorphisms of ${K}$ must be bijective), so for any ${k\in K}$, we have ${\prod_{z\in H(k)} (x - z) \in K^H[x]}$ (product over distinct elements of the orbit of ${k}$), since its coefficients are invariant under ${H}$. In fact, this is the minimal polynomial of ${k}$ over ${K^H}$, since otherwise some subset of the orbit would have to be fixed under ${H}$.

We immediately get finiteness of ${[K:K^H]}$, or else we could get an infinite chain of intermediate fields between ${K^H,K}$, giving elements of degree larger than ${\lvert{H}\rvert}$ by the primitive element theorem, contradiction.

Now let ${F = K^H}$ and write ${K = K^H(\alpha)}$ by the primitive element theorem, so we can apply the previous explanation’s results. In fact, by the previous paragraphs the minimal polynomial ${f \in F[x]}$ splits completely in ${K}$, so we already have ${K / F = K / K^H}$ Galois. But now ${\lvert{G(K/K^H)}\rvert = [K:K^H]}$ equals the size of the orbit ${H(\alpha)}$, which is at most ${\lvert{H}\rvert}$ (in fact, although we don’t need this, we can get exactly ${\lvert{H}\rvert}$ by the orbit-stabilizer theorem, since the stabilizer is trivial, since ${\alpha}$ generates everything). Yet ${H \subseteq G(K/K^H)}$ (since ${H}$ fixes ${K^H}$ by definition), so we conclude that ${H = G(K/K^H)}$ and

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

strengthening our previous results.

1. Fundamental theorem of Galois theory

OK now this. For ${K/F}$ a Galois extension (although our proof might only work for characteristic ${0}$), we get a correspondence ${H\rightarrow K^H}$ for subgroups and ${L\rightarrow G(K/L)}$ for intermediate fields.

But we just proved that ${H\rightarrow K^H \rightarrow G(K/K^H) = H}$ for finite subgroups ${H}$, so it remains to show that ${H\rightarrow K^H}$ is surjective.

Here it’s important that ${K}$ is Galois over ${F}$, and hence over any intermediate field ${L}$. Indeed, we then have ${L \rightarrow G(K/L) \rightarrow K^{G(K/L)} = L}$, since ${[K:K^{G(K/L)}] = \lvert{G(K/L)}\rvert = [K:L]}$ (alternatively, if ${G(K/L)}$ fixes something not in ${L}$, say ${p(\alpha)}$ for some ${p\in L[x]}$ of degree at most ${n-1}$, then ${p(\alpha) = p(z_i)}$ for all conjugates ${z_i}$, so ${f \mid p - p(\alpha)}$, so ${p}$ is identically constant, which is clearly bad).

(Another way of thinking about the ${L \rightarrow G(K/L)}$ correspondence: for every intermediate field ${L}$, we associate some set of conjugates ${C_L(\alpha)}$ (subset of the ${z_i}$) who together form an irreducible polynomial in ${L[x]}$.)

1.1. Counterexample when ${K}$ is not Galois extension

${K = \mathbb{Q}(2^{1/4})}$ has three intermediate fields ${\mathbb{Q},\mathbb{Q}(\sqrt2),K}$. But ${G(K/\mathbb{Q}) \cong G(K/\mathbb{Q}(\sqrt2)) \cong C_2}$ only has two subgroups.

Comment. For many purposes I think we can avoid Galois theory in the intermediate fields problem. For example, we can say a lot by considering the extension field generated by the coefficients of the minimal polynomial (it vaguely resembles the ${K^H}$ stuff above). See here.

1. Thoughts to digest later: Suppose $F\subset L\subset K$ is a chain with $K/F$ Galois (so $K/L$ still a splitting field, hence Galois) so that $L/F\subset K/F$ is separable, and write $L = F(\beta)$ by PET. Then $\sigma\in G(K/F)$ fixes $\beta$ iff $\sigma\in H := G(K/L)$. So the orbit $G(\beta)$ has size $|G|/|H| = [K:F]/[K:L] = [L:F]$. Thus $G(\beta)$ is a full subset of (i.e. equals) the set of $F$-conjugates of $\beta$.
Furthermore, we easily see $L \subseteq K^H$ and (since $H$ fixes $K^H$ by definition) $G(K/L) = H \subseteq G(K/K^H) \subseteq G(K/L)$. So $L = K^H$.