If $K/F$ is normal extension, $I$ is inseparable closure of $F$ in $K$, prove $K/I$ is Galois.

Question

Let $K/F$ be a normal extension and $I$ be the inseparable closure of $F$ in $K.$

Let $G=\text{Aut}_F(K),$ i.e., $F$-automorphisms of $K$, and similarly define $H=\text{Aut}_I(K).$

Now I have already shown that $I=K^{\mathrm{Aut}_F(K)},$ i.e., $G=H.$ How can I conclude that $K/I$ is Galois?

I need some help. Thanks.

Answer 1

It is a standard property that also $K/I$ is normal (because the minimal polynomial over $I$ divides the minimal polynomial over $F$), so it is enough to prove that $K/I$ is separable.

Let $\alpha \in K$ and consider the set $r_{\alpha}=\{\sigma(\alpha) \mid \sigma \in {\rm Aut}_F(K)\}=\{\alpha_1, \dots, \alpha_n\}$. The polynomial $f(x)= \prod_{i=1}^n(x-\alpha_i)$ is in $I[x]$, since $I=K^{{\rm Aut}_F(K)}$ and every $\sigma \in {\rm Aut}_F(K)$ only permutes the elements in $r_{\alpha}$. This means that $\alpha$ is separable over $I$ because $f(x)$ is separable and, then, $K/I$ is separable.