$\operatorname{Gal}(k(t^{1/p},u^{1/p}) / k(t, u))$ contains only identity automorphism.

Question

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.

Answer 1

This is always the case for purely inseparable extensions in characteristic $p$. Your $t'$ is a root of $X^p-t=0$. Therefore, so is $\sigma(t)$. But this equation factors as $(X-t')^p$. Therefore $(\sigma(t)-t')^p=0$, and so $\sigma(t)=t$. Likewise $\sigma(u)=u$.

Answer 2

As you wrote we have $(\sigma(t'))^p=t$. This means $\sigma(t')$ is a root of the polynomial $x^p-t$ which belongs to $L$. But then note that since the characteristic is $p$ we have $x^p-t=x^p-(t')^p=(x-t')^p$. This means that $t'$ is the only root of this polynomial in $L$, and hence $\sigma(t')=t'$, we have no choice.