1
$\begingroup$

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.

$\endgroup$
  • 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
3
+25
$\begingroup$

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.

New contributor
Francesco is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.