If $E_\text{sep}/F$ is normal, then must $E/F$ be normal?

Question

This question has been asked in Math.StackExchange (see here) for more than a week and I even put a bounty on it. But still it hasn't been correctly answered (the current answer there was written by myself and it turned out to be false). Hence I decided to ask it here.

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Let $E/F$ be an algebraic field extension, denote by $E_\text{sep}$ the separable closure of $F$ in $E$. We know if $E/F$ is normal then $E_\text{sep}/F$ is normal, see tag 0EXK. The reverse should be false and I want an example. Note that $E/E_\text{sep}$ is purely inseparable and thus always normal.

Here is what I thought: Let $F=\mathbb{F}_p(t)$ and $m(x)$ be a degree $n$ irreducible separable polynomial with distinct roots $\alpha_1,\dots,\alpha_n$, e.g. $m(x)=x^n-t$ with $p\nmid n$ (but this example doesn't work, see the comment by François Gatine in the question on Math.StackExchange), then $F(\alpha_1,\dots,\alpha_n)/F$ is normal and separable. Let $E=F(\alpha_1,\dots,\alpha_n,\alpha_1^{1/p})$ so $E/F(\alpha_1,\dots,\alpha_n)$ is purely inseparable of degree $p$ and thus $E_\text{sep}=F(\alpha_1,\dots,\alpha_n)$.

Then the normal closure of $E/F$ is $F(\alpha_1^{1/p},\dots,\alpha_n^{1/p})$ which should generally be strictly bigger than $F(\alpha_1,\dots,\alpha_n,\alpha_1^{1/p})$, implying that $E/F$ is not normal. But I have difficulties showing this for any explicit example.

Since $x^n-t$ doesn't work, I'm now thinking about $m(x)=x^3+bx+c\in F[x]$ with $p>3$ and the discriminant $D=-4b^3-27c^2\notin F^2$, so $m(x)$ is irreducible and separable. Let $\alpha=\alpha_1,\alpha_2,\alpha_3$ be the roots of $m$ in $\overline{F}$, then $F(\alpha,\sqrt{D})=F(\alpha_1,\alpha_2,\alpha_3)$ is a splitting field of $m$ over $F$, see this calculation. Similarly it's not hard to see that the normal closure of $F(\alpha,\sqrt{D},\alpha^{1/p})$ is $F(\alpha_1^{1/p},\alpha_2^{1/p},\alpha_3^{1/p})=F(\alpha^{1/p},D^{1/2p})$. It remains to show that $D^{1/2p}\notin F(\alpha^{1/p},\sqrt{D})$. Maybe we can use trace or norm to set up a contradiction.

Answer 1

Let $K=M(X,Y)$ be the function field in two variables, for a field $M$ of characteristic $p>0$. Consider the Galois homomorphism $\gamma: f(X,Y)\mapsto f(Y,X)$ and let $F=K^\gamma$ the fixed field. Let $E=M(X^{1/p},Y)$, then $E/F$ is an extension of degree $2p$. By characteristic $p$, every element $b\in E$ satisfies $b^p\in K$ and therefore the extension $E/K$ is purely inseparable. Hence $E_\mathrm{sep}=K$. The element $\alpha=X^{1/p}Y$ of $E$ has minimal polynomial $g(Z)=Z^p-XY^p$ over $E_\mathrm{sep}$ and therefore $f(Z)=(Z^p-XY^p)(Z^p-X^pY)$ over $F$. Then $f$ has zeros $X^{1/p}Y$ and $XY^{1/p}$ the latter of which does not lie in $E$, therefore $E/F$ is not normal.