Is $F\hat{E_s}/E$ normal, where $F/E$ is a finite field extension?

Question

Let $F/E\,$ be a finite field extension, let $E_s$ be the separable closure of $E$ in $F$ and let $\hat{E_s}$ be its normal closure. Then $\hat{E_s}/E$ is Galois and $F\hat{E_s}/\hat{E_s}$ is purely inseparable (where $F\hat{E_s}$ is the compositum of $F$ and $\hat{E_s}$). Is it true that $F\hat{E_s}/E$ is normal?

Answer 1

• If $F$ is algebraic over $\Bbb{F}_p(t)$ then yes, this is because there is at most one purely inseparable extension of each degree over $\hat{E}_s$:

  if $L/\hat{E}_s$ is purely inseparable of degree $p$ then $L=\hat{E}_s(a^{1/p})$ for some $a\in \hat{E}_s$. For any $z\in \hat{E}_s,\not \in \overline{\Bbb{F}}_p$ let $f$ be $a$'s $\Bbb{F}_p(z)$-minimal polynomial, the $\Bbb{F}_p(z)$-minimal polynomial of $a^{1/p}$ is $f(x^p) = g(x)^p$ where $g\in \Bbb{F}_p(z^{1/p})[x]$ and $g(a^{1/p})=0$, from which you get that $$[\Bbb{F}_p(z^{1/p},a^{1/p}):\Bbb{F}_p(z)] = \deg(g)p=p\deg(f)= [\Bbb{F}_p(z,a^{1/p}):\Bbb{F}_p(z)]$$ ie. $z^{1/p}\in \Bbb{F}_p(z,a^{1/p})\subset L$ and $L=(\hat{E}_s)^{1/p}$. Repeating the same argument you get that $F\hat{E}_s=(\hat{E}_s)^{1/p^k}$ which is obviously a normal extension of $E$.

• In general no, try with $F=\Bbb{F}_p(x,y^p),E=\Bbb{F}_p(x^p+y^p,x^py^p)$

  Then $\hat{E}_s=E_s=\Bbb{F}_p(x^p,y^p)\subset F$ and $(x^p,y^p)\to (y^p,x^p)$ is in $Gal(\hat{E}_s/E)$ but it doesn't extend to an automorphism of $F/E$.