A finite field extension that is not simple

Question

Let $F=\mathbb{F}_p(X,Y)$ be the field of rational functions in variables $X,Y$ over the finite field of $p$ elements. Let $K=\mathbb{F}_p(X^p,Y^p)$ be a subfield. Note that for any $f\in F$, $f^p\in K$. Deduce from this that $F/K$ is not a simple extension.

Answer 1

First of all $\mathbb{F}_p(X^p,Y^p)$ is a function field, which means that we may treat $X^p,Y^p$ as variables. The polynomial $(T-X)^p = T^p-X^p$ is irreducible over $\mathbb{F}_p(Y^p)[X^p]$ by Eisenstein's criterion, but then also over the field of fractions $\mathbb{F}_p(X^p,Y^p)$. Hence, $\mathbb{F}_p(X,Y^p)$ has degree $p$ over $\mathbb{F}_p(X^p,Y^p)$. Similarily one proves that $\mathbb{F}_p(X,Y)$ has degree $p$ over $\mathbb{F}_p(X,Y^p)$. Hence, $F=\mathbb{F}_p(X,Y)$ has degree $p^2$ over $K=\mathbb{F}_p(X^p,Y^p)$. But since $F^p \subseteq K$, every element of $F$ has degree $\leq p$ over $K$, so that it cannot generate $F/K$.