Over a perfect field $k$ it is well known that an algebraic field extension $k \subseteq E$ is normal if and only if $\mathrm{Aut}_{E}(\bar{k})$ is a normal subgroup of $\mathrm{Aut}_{k}(\bar{k})$, as in the infinite Galois correspondence closed normal subgroups of $\mathrm{Aut}_k(\bar{k})$ correspond exactly to normal field extensions of $k$. Here $\bar{k}$ denotes an algebraic closure of $k$. Is the same true when $k$ is not perfect?

The closest I have gotten to providing an answer is the following: The group $\mathrm{Aut}_k(\bar{k})$ acts on the set of $k$-algebra homomorphisms $E \to \bar{k}$ by postcomposition. The stabiliser of the inclusion $E \subseteq \bar{k}$ is given by $\mathrm{Aut}_E(\bar{k})$; as by assumption $\mathrm{Aut}_E(\bar{k})$ is a normal subgroup of $\mathrm{Aut}_k(\bar{k})$ we see that it is the stabiliser of all $k$-algebra homomorphisms $E \to \bar{k}$, and therefore any conjugate of $E$ must lie within the purely inseparable closure of $E$, as this field coincides with the fixed field of $\mathrm{Aut}_E(\bar{k})$.


The statement is wrong. I will first give a moral reason, and then a concrete counter example.

Moral reason: The extension $k \subseteq E$ can be divided into $k \subseteq E_s \subseteq E$, where $E_s$ is the separable closure of $k$ in $E$; it is then well known, that $E_s \subseteq E$ is purely inseparable. We thus have $\mathrm{Aut}_{E_s}(\overline{k}) = \mathrm{Aut}_E(\overline{k})$, so that $E_s \subseteq E$ is normal.
The group $\mathrm{Aut}_k(\overline{k})$ acts on the $k$-algebra homomorphisms $E_s \to \overline{k}$ by postcomposition; as $\mathrm{Aut}_{E_s}(\overline{k}) = \mathrm{Aut}_E(\overline{k})$ is normal in $\mathrm{Aut}_k(\overline{k})$ it is the stabiliser of all $k$-algebra homomorphisms $E_s \to \overline{k}$, i.e. the images of all homomorphisms $E_s \to \overline{k}$ is fixed by $\mathrm{Aut}_{E_s}(\overline{k})$, and thus by the (possibly infinite) Galois correspondence the images of these homomorphisms must coincide, so that $k \hookrightarrow E_s$ is normal.
Assume now that $E_s \subseteq E$ is generated by the $p$-th root of some element $\alpha \in E_s \setminus E$, then the minimal polynomial of $\alpha^{\frac{1}{p}}$ over $k$ is of the form $f(X^p)$ for some separable polynomial which splits in $E_s$. If $k \subseteq E$ were normal, then $E$ would contain all $p$-th roots of all conjugates of $\alpha$, but there is no reason for this to happen.

Counter example: For $p> 2$ consider the irreducible (by Eisenstein) polynomial $f(X) = X^{2p} + uvX^p + u$ over $\mathbb{F}_p(u,v)$, then by adjoining a root $\alpha$ of $f(X)$ to $\mathbb{F}_p(u,v)$, we obtain a degree $2p$ extension $\mathbb{F}_p(u,v) \subseteq E$ such that the separable closure of $\mathbb{F}_p(u,v)$ in $E$ is $\mathbb{F}_p(u,v)(\alpha^{\frac{1}{p}})$, a field obtained by adjoining the root $\alpha^{\frac{1}{p}}$ of the separable polynomial $X^2 + uvX + u$. As this polynomial has degree $2$, the extensions $E \subseteq E_s$ is normal, thus, by the above discussion, $\mathrm{Aut}_E(\overline{k})$ is normal in $\mathrm{Aut}_k(\overline{k})$.
The inseparable closure $E_i$ of $k$ in $E$ is $k$ (this is exercise V.5.3, Algèbre, Bourbaki). If $E$ were normal, then $E$ would be $E_sE_i$ (Proposition V.10.11.13, Algèbre, Bourbaki), but this is clearly false.


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