A finite field extension that is not simple

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.

• I tried contradiction methods. So assuming $F=K(f)$, $f^p=g\in K$ then $f$ is not separable over $K$. That's all I could think of, and I don't think this is taking me anywhere. Commented Jan 12, 2013 at 23:51
• Maybe if you can prove that the degree of the extension is $p^2$ all is done. Try it!
– user26857
Commented Jan 13, 2013 at 0:01
• I couldn't show the set $\{X,Y,XY,X^2,Y,XY^2,...X^pY^p\}$ is a linearly independent set. I took that this was what you meant by $p^2$. But this is an exercise in a section on separability. Commented Jan 13, 2013 at 0:21
• Presumably the reason it's in a section on separability is that this section (or maybe the next section) contains the theorem of the primitive element, and the point of the exercise is to show that this theorem really needs the hypothesis of separability. Commented Jan 13, 2013 at 2:11

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 $F/K$.