Purely inseparable extensions vs subextensions

Question

For prime $p$, suppose we have the base field $k = \mathbb{F}_p(x^p,y^p)$ and its extension $\ell = \mathbb{F}_p(x,y)$.

How could I prove that for some purely inseparable extension $K/k$ of degree $p$, $K\cong L$ where $\ell/L/k$?

Answer 1

$$(\Bbb{F}_p(x,y))^p = {\Bbb{F}_p}^ p(x^p,y^p)=\Bbb{F}_p(x^p,y^p)$$

That is to say $$\Bbb{F}_p(x,y)=(\Bbb{F}_p(x^p,y^p))^{1/p}$$

Given a field $F$ of characteristic $p$ then $F^{1/p}$ is the unique field extension generated by all the purely inseparable elements of degree $p$.

The latter fact follows from that $F^{1/p}$ is a field (isomorphic to $p$) and that for $a$ purely inseparable of degree $p$ over $F$ with minimal polynomial $f\in F[x]$ then $f$ has no other root ie. $f = (x-a)^p=x^p-a^p$ so that $a\in F^{1/p}$.