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I'm trying to find a counter-example to following statement: if $K$ is the splitting field of $g\in F[x],$ then the extension $K/F$ is Galois.

I know the statement is true if $g$ is separable, so I need $F$ to be infinite of characteristic $p>0$. My thoughts were something like $K/\mathbb{F}_p(t)$, where $t$ is transcendental over $\mathbb{F}_p$, and $K$ is the splitting field of $g(x)=x^p-t$.

I know that $g$ is irreducible and not separable (it has a repeated root $s\in K$, $g(x)=x^p-t=x^p-s^p=(x-s)^p$), and if I could show that $g$ was the minimal polynomial of $s$ over $\mathbb{F}_p(t)$ then I think I'd be done ($K/F$ is Galois if and only if for every $\alpha\in K$, the minimal polynomial of $\alpha$ over $F$ is separable and $K$ contains its splitting field).

I'm pretty certain that this is the counter-example that I'm looking for, but I'm just struggling to fully prove to myself that it works. I know that the minimal polynomial of $s$ must be $(x-s)^q$ for some $0<q\le p$ since it must divide $g$, but not sure how to argue that $(x-s)^q\notin \mathbb{F}_p(t)[x]$ for $0<q<p$.

Thanks, Warren

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    $\begingroup$ You can save yourself a little trouble by taking $p = 2$, in which case all you need to do is show $\sqrt{t} \not\in \Bbb F_2(t)$. To see this suppose we had $u \in \Bbb F_2(t)$ with $u^2 = t$. Writing $u = \dfrac{p(t)}{q(t)}$ we have $[q(t)]^2t = [p(t)]^2$ where one polynomial has odd degree, and the other has even degree. $\endgroup$ Aug 1, 2014 at 19:22

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Hint: it turns out that $\mathbf{F}_p(s)$ is a rational function field as well, and the field extension identifies $\mathbf{F}_p(t)$ with the subfield $\mathbf{F}_p(s^p) \subseteq \mathbf{F}_p(s)$.

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  • $\begingroup$ Thanks! I was certain in my head, but just wanted something more concrete to justify myself, and that hint did the trick. $\endgroup$ Apr 14, 2013 at 13:48

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