Let $k$ be a field of characteristic $p$, and let $F = k(t, u)$ and $L$ be the splitting field of $(x^p - t)(x^p - u) \in F[x]$. I want to show that $\operatorname{Gal}(L/F)$ contains only identity automorphism.

Proof:

Let's take $\sigma \in \operatorname{Gal}(L/F)$. Let $t' = t^{1/p}$ and $u' = u^{1/p}$.

Suppose $\sigma(t') = g(t', u')$ for some $g(x, y) \in F[x,y]$. Then I think I need to show that $g(t', u') = t'$. I tried:

$t
= \sigma(t)
= (\sigma(t'))^p
= g(t', u')^p
= \left( \sum c\_{i,j} {t'}^i {u'}^j \right)^p
= \sum c\_{i,j}^p {t}^i {u}^j
$ by bringing the exponent $p$ inside.

This implies that $c\_{i,j}^p = 0$ for all $(i,j)$ except $(1, 0)$, hence $g(t', u') = c\_{i,j} t'$, where $c\_{i,j}^p = 1$. (Im not sure if this part is valid, how can I show rigorously?)

Then I am stuck here. If I can show that $c\_{i,j} = 1$, then $\sigma(t') = t'$. Similarly, $\sigma(u') = u'$. And hence $\sigma$ is identity. But I don't see how to proceed.