abstract algebra - Splitting field that isn't a Galois extension

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

Thanks,
Warren

Answer

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 $\mathbf{F}_p(s^p) \subseteq \mathbf{F}_p(s)$.