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.

  • 1
    $\begingroup$ 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) $\endgroup$ – kesa Feb 8 at 13:37
  • $\begingroup$ 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 $G$? $\endgroup$ – Jyrki Lahtonen Feb 12 at 4:33
  • $\begingroup$ 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 $K^G$ $\endgroup$ – 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.

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