In characteristic $p$, the field extension $k(X,Y)$ over $k(X^p,Y^p)$ has degree $p^2$

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 $K=k(X^p,Y^p)$. Prove that $[L:K]=p^2$

Help me a hint to prove it.

Thanks a lot.

Answer 1

The polynomial $f = T^p - X^p\in k(X^p,Y^p)[T]$ has the zero $X$. Thus, the minimal polynomial of $X$ over $k(X^p,Y^p)$ is a monic non-constant divisor of $f$.

By Frobenius, $f = (T - X)^p$, so over any splitting field of $f$, the monic non-constant factors have the form $f = (T - X)^i = \sum_{j = 0}^i (-1)^{i-j} \binom{i}{j} X^{i-j} T^j$ with $i\in\{1,\ldots p\}$. Its constant term is $(-1)^i X^i$, which for $i < p$ is not contained in $k(X^p,Y^p)$. This shows that no proper divisor of $f$ is in $k(X^p,Y^p)[T]$ and therefore, $f$ is the minimal polynomial of $X$ over $k(X^p,Y^p)$. So $$[k(X,Y^p) : k(X^p,Y^p)] = \deg(f) = p.$$

Similarly, we show that $$[k(X,Y) : k(X,Y^p)] = p.$$

Finally $$[k(X,Y) : k(X^p,Y^p)] = [k(X,Y) : k(X,Y^p)] \cdot [k(X,Y^p) : k(X^p,Y^p)] = p\cdot p = p^2. $$

Answer 2

Construct $k(X,Y)$ as an algebraic extension of $k(X^p, Y^p)$.

It may help keep things straight in your mind if you do a change of variable: e.g. call $X^p = U$ and $Y^p = V$.