From the previous post, we know that (in characteristic ), if is finite, then (by the primitive element theorem to get some to satisfy hypotheses of previous explanation) and
(so is a splitting field over/Galois over ) equals the degree of the “splitting part” of min poly of .
So naturally, we wonder if we can say anything starting from the perspective of groups, i.e. “can be replaced by an arbitrary (finite) group of automorphisms (of ) ?”
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 is a finite splitting/Galois extension of .
First observe (and this is the reason is “not too unnatural from scratch”) that acts on (since automorphisms of must be bijective), so for any , we have (product over distinct elements of the orbit of ), since its coefficients are invariant under . In fact, this is the minimal polynomial of over , since otherwise some subset of the orbit would have to be fixed under .
We immediately get finiteness of , or else we could get an infinite chain of intermediate fields between , giving elements of larger than by the primitive element theorem, contradiction.
Now let and write by the primitive element theorem, so we can apply the previous explanation’s results. In fact, by the previous paragraphs the minimal polynomial splits completely in , so we already have Galois. But now equals the size of the orbit , which is at most (in fact, although we don’t need this, we can get exactly by the orbit-stabilizer theorem, since the stabilizer is trivial, since generates everything). Yet (since fixes by definition), so we conclude that and
strengthening our previous results.
1. Fundamental theorem of Galois theory
OK now this. For a Galois extension (although our proof might only work for characteristic ), we get a correspondence for subgroups and for intermediate fields.
But we just proved that for finite subgroups , so it remains to show that is surjective.
Here it’s important that is Galois over , and hence over any intermediate field . Indeed, we then have , since (alternatively, if fixes something not in , say for some of degree at most , then for all conjugates , so , so is identically constant, which is clearly bad).
(Another way of thinking about the correspondence: for every intermediate field , we associate some set of conjugates (subset of the ) who together form an irreducible polynomial in .)
1.1. Counterexample when is not Galois extension
has three intermediate fields . But 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 stuff above). See here.