Let $p$ be a prime number, let $F_p=\mathbb Z/p\mathbb Z$ and consider the rational function fields $F=F_p (X^p,Y^p)$ and $K=F_p (X,Y)$. Is $K$ a simple extension over $F$ i.e. is $K=F(a)$ for some $a\in K$ ?
I think I can show that $[F_p(X,Y):F_p(X^p,Y)]=[F_p(X^p,Y):F_p(X^p,Y^p)]=p$. But I don't know whether it is useful here or not.
Please help.
Your fact is indeed useful. From this you should be able to show that $[F_p(X,Y):F_p(X^p,Y^p)] = p^2$. Now, take some $a \in F_p(X,Y)$. You can see that, by applying $Frob_p$ the Frobenius morphism, that $a^p \in F_p(X^p,Y^p)$. So now, we have that $[F_p(X^p, Y^p)(a): F_p(X^p,Y^p)] \leq p$. This shows that there is no generating element so it is not simple.