Exercise 7.3.4

Proof. In Example 7.3.6, $k$ has characteristic $p$, and the extension $L$ of $F=k(t,u)$ is the splitting field of $f=(x^p-t)(x^p-u)\in F[x]$.

We showed in Exercise 5.4.4 that $F\subset L$ is purely inseparable, and $L=F(\alpha ,\beta )$, where ${\alpha }^p=t,{\beta }^p=u$. Moreover the intermediate fields $F\subset F(\alpha +\mathit{\lambda \beta })\subset L$ are distinct.

Now we show that $\mathrm{Gal}(L∕F)=\{1_L\}$.

$\alpha$ is a root $x^p-t\in F[x]$, thus $\sigma (\alpha )$ is also a root. Since $x^p-t={(x-\alpha )}^p$ has the only root $\alpha$, $\sigma (\alpha )=\alpha$.

Similarly $\beta$ is the only root of $x^p-u={(x-\beta )}^p$, thus $\sigma (\beta )=\beta$.

Moreover $L=F(\alpha ,\beta )$, so an element $\sigma \in \mathrm{Gal}(L∕F)$ is uniquely determined by the images of $\alpha ,\beta$, therefore $\sigma =1_L$.

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