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

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.

• What is your definition of a Galois extension and inseparable closure? Is it not true, that $K$ is separable over $I$, if a inseparable closure exists (that is what I would call an inseparable closure) – kesa Feb 8 at 13:37
• Isn't it always the case that if $E$ is a field and $G$ is a finite group of automorphisms of $E$, then $E/E^G$ is Galois with Galois group $G$? – Jyrki Lahtonen Feb 12 at 4:33
• To me what Jyrki wrote is the most useful definition of "Galois" (it is the one which gives the minimal polynomials : for $a \in E$, $f(x)=\prod_{b\in G.a} (x-b) \in E^G[x]$) and the problem is to show given $K/I$ separable and normal that $I = K^G, G = Aut(K/I)$, which is a consequence of : separable implies $a \in K- I$ has a conjugate $b \ne a$ and normal implies there is $\sigma \in Aut(K/I)$ such that $b = \sigma(a)$, thus $a$ isn't in $K^G$ – reuns Feb 16 at 12:47

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.