In characteristic $p$, the field extension $k(X,Y)$ over $k(X^p,Y^p)$ is not simple

Question

Let $k$ be a field with characteristic $p>0$, $L=k(X,Y)$ be the field of rational fractions of two variables over $k$. Let $K=k(X^p,Y^p)$. Then $[L:K]=p^2$ (for a proof see In characteristic $p$, the field extension $k(X,Y)$ over $k(X^p,Y^p)$ has degree $p^2$). Show that $L^{p}\subseteq K$ and $L$ is not a simple extension over $K$.

Thank in advance.

Answer 1

Obviously $k^p \subseteq k\subseteq K$ and furthermore clearly $X^p\in K$ and $Y^p\in K$. Since the characteristic of our fields is $p$, the Frobenius mapping $x\mapsto x^p$ is a field homomorphism. This implies $L^p \subseteq K$.

Assume that $L = K(a)$ with $a\in L$. Since $L^p\subseteq K$, in particular $a^p\in K$. So $a$ is a zero of $f = T^p - a^p\in K[T]$. Thus $[L : K] \leq \deg(f) = p$. Contradiction.