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.

One thought on “Galois theory basics, part 2”

  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.


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