# (non Galois) correspondence

If $L/K$ is a field extension we note $\textrm{Aut}_K (L)$ the group of field automorphisms of $L$ that are fixing each element of $K$.

Fix an extension $L/K$ and note $G:=\textrm{Aut}_K (L)$. If $H$ is a subgroup of $G$ then the set $L^H$ of elements of $L$ fixed by every element of $H$ is a field and we get a map $\varphi$ from the set $X$ of subgroups of $G$ to the set $Y$ of subextensions of $L/K$ sending a subgroup $H$ to the field $L^H$. Inversely if $M/K$ is a subextension of $L/K$ then $\textrm{Aut}_M (L)$ is subgroup of $\textrm{Aut}_K (L)$, and we get a map $\psi$ from $Y$ to $X$ sending a subextension $M$ to the subgroup $\textrm{Aut}_M (L)$.

I am interested in how we can, in general, describe $\varphi\circ\psi$ and $\psi\circ\varphi$. If $L/K$ is of finite degree, Artin's result shows that $\varphi\circ\psi = \textrm{Id}_Y$, but how can we describe $L^{\textrm{Aut}_M (L)}$ ? Is it the Galois closure of $M$ in $L$ ? What can we say if $L/K$ is not of finite degree anymore ?

Remark. Note that I am not interested in Galois correspondence per se, as I am not supposing $L/K$ Galois nor restricting $\psi$ and $\varphi$ to Galois subextensions and normal subgroups respectively. Nevertheless, I am interested in what we could say in the Galois case also.

Proposition. Let $L/K$ be an extension, $M/K$ be a subextension such that $L/M$ is normal (which is automatically the case if $L/K$ is normal) and $M' := L^{\textrm{Aut}_M (L)}$. Then $M'/M$ is a purely inseparable extension. If $\textrm{Aut}_M (L)$ is finite $M'$ is the purely inseparable closure of $M$ in $L$.
Proof. Take $x\in M'$, suppose that $x$ is separable over $M$, and let $P$ be the minimal polynomial of $x$ over $M$. As $L/M$ is normal $P$ splits over $L$. Now suppose that $x\not\in M$, so that the degree of $P$ is $\geq 2$ and as $P$ splits and has distinct roots (because $x$ is separable) we can find $y\in L$ with $y\not=x$ and $P(y)=0$. As $L/M$ is normal, we can find $g\in \textrm{Aut}_M (L)$ such that $g(x) = y \not = x$. But $x\in M' = L^{\textrm{Aut}_M (L)}$ must be fixed by $g$ and we have a contradiction. If $\textrm{Aut}_M (L)$ is finite Artin's lemma ensures that $L/M'$ is Galois and therefore separable, so that $M''/M'$ is separable. This shows that $M'$ is the purely inseparable closure of $M$ in $L$. $\square$
Moral. $L^{\textrm{Aut}_M (L)} / M$ being purely inseparable has few chances to be Galois, let alone to be the Galois closure of $M$ in $L$.
• What can we say without $L/M$ being normal ? I really don't know. And in inifinite dimension ? – EricFlorentNoube Mar 28 '15 at 21:13