I'm trying to prove that $x^p-t\in \mathbb{F}_{p}(t)[x]$ is not separable and finding the $\mathbb{F}_{p}(t)$-automorphism of $L:=\mathbb{F}_{p}(t)[x]/(x^p-t)$.
I have this:
By Gauss's Lemma, $x^p-t$ irreducible in $\mathbb{F}_{p}(t)[x]$ if and only if is irreducible in $\mathbb{F}_{p}[t][x]$. Now, the ideal $(t)$ is prime in $\mathbb{F}_{p}[t][x]$. Indeed, if $f(t)g(t)\in (t)$ then $f(t)g(t)=th(t)$ some $h(t)\in \mathbb{F}_{p}[t]$. If $f(t),g(t)\not\in (t)$ then $f(t)=tk(t)+r(t)$ with $deg(r(t))<deg(t)=1$ or $r(t)=0$. If $r(t)=0$ then $f(t)\in (t)$ trivially. If $f(t)=tk(t)+r$ with $r$ constant then $f(t)g(t)=tk(t)g(t)+rg(t)$. This implies $t\mid g(t)$, therefore $g(t)\in (t)$ a contradiction. Now, $-t\in (t)$ and $-t\in (t)^2$ therefore $x^p-t$ is irreducible in $\mathbb{F}_{p}[t]$ by Eisentein.
$D_x(x^p-t)=px^{p-1}=0$ because $char(\mathbb{F}_{p})=p$. Since $x^p-t$ irreducible by 1) then $x^p-t$ is not separable in $\mathbb{F}_{p}(t)[x]$.
$\mathbb{F}_{p}(t)$ is a field and $x^p-t$ irreducible then is maximal. Therefore $L$ is a field.
I would like to find out which a field is $L$ isomorphic to follow a proof similar to the one I wrote in these threads. (Finding the $\mathbb{Q}$-automorphisms of the splitting field of $x^p-2$ over $\mathbb{Q}$. and Finding the group of $\mathbb{Q}$-automorphisms of the field $\mathbb{Q}[x]/(x^{p-1}+x^{p-2}+\cdots +x+1)$
To what field is $L$ isomorphic? I think that $L\simeq \mathbb{F}_{p}(\sqrt[p]{t})$ because if $\alpha$ is a root of $x^p-t$ then $t=\alpha^p$. Now, $(x^p-t)=(x^p-\alpha^p)=(x-\alpha)^p$ (because $char(\mathbb{F}_{p}=p$) then $x^p-t$ has a unique root, this root is $\sqrt[p]{t}$ (at least as a symbol notation)
Is it correct to say that $L\simeq \mathbb{F}_{p}(\sqrt[p]{t})$? Thanks.