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I am trying to teach myself Galois theory. Is it true that every for a field extension K->L, that every normal subgroup of Gal(L:K) is of the form Gal(L:M) for some intermediate field M, ie K->M->L?

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Yes, this is true whenever $L$ is a finite Galois extension of $K$. But this is really not an MO level question. – Robin Chapman Jul 23 '10 at 10:50
@RC: if the OP is asking about finite Galois extensions (which I admit is likely) then yes, this is completely standard. But the case of field extensions of infinite degree is more interesting, and I have written an answer which I hope will be of interest to (some) research mathematicians. – Pete L. Clark Jul 23 '10 at 11:19
@Shamim: Assuming the extension $K<L$ was Galois in the first place, your statement is actually true for every subgroup of Gal$(L/K)$, not just the normal ones. – Kevin Ventullo Jul 23 '10 at 12:51
@Kevin: closed subgroups. – Martin Brandenburg Jul 23 '10 at 12:54
I'm confused by the unusual notation used here for field extensions: which are meant to be subfields of others? Normal subgroups in the case of finite extensions correspond to extensions of the smallest field by an intermediate Galois extension. Anyway, elementary Galois theory is presented clearly in many books; teaching oneself without a book at hand seems pointless. – Jim Humphreys Jul 25 '10 at 14:24

Let $L/K$ be an arbitrary extension of fields, and put $G = \operatorname{Aut}(L/K)$.

For a subgroup $H$ of $G$, put $L^H = \{x \in L \ | \ \forall \sigma \in H, \ \sigma(x) = x\}$.

We say that a subgroup $H$ of $G$ is closed if $H = \operatorname{Aut}(L/L^H)$.

Theorem (Artin-Kaplansky): Every finite subgroup $H$ of $G$ is closed.

Moreover, a closed subgroup $H$ of $G$ is normal iff for all $\sigma \in G$, $\sigma L^H = L^H$. If $L/K$ is algebraic, this says that $L^H/K$ is normal.

Also, if $[L:K]$ is finite, then $|G| \leq [L:K]$. So for finite extensions, every subgroup is closed.

If $[L:K]$ is infinite, it need not be the case that every subgroup $H$ of $G$ is closed. In fact, if $L/K$ is algebraic and Galois of infinite degree, then there are always nonclosed subgroups, since any closed subgroup $H$ of $G$ is profinite, hence finite or uncountably infinite. But the group $G$ itself is uncountably infinite, so has countably infinite subgroups.

To see a particular example of a nonclosed subgroup which is moreover normal, take $K = \mathbb{F}_p$ and $L = \overline{\mathbb{F}_p}$, an algebraic closure. Then $\operatorname{Aut}(L/K) \cong \widehat{\mathbb{Z}}$, and the closed subgroups are those of the form $n \widehat{\mathbb{Z}}$: they are all open, and in particular uncountably infinite. Then the subgroup of $G$ generated (not topologically generated!) by the Frobenius automorphism $\operatorname{Fr}: x \mapsto x^p$ is isomorphic to $\mathbb{Z}$, and is a nonclosed normal subgroup whose closure is $\widehat{\mathbb{Z}}$.

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