# Degree of $K(X^{1/p}, Y^{1/p})$ over $K(X, Y)$ in characteristic $p$

Let $K$ be a field of characteristic $p$ and $L=K(X,Y)$ where $X$ and $Y$ are variables (i.e. $L$ is the field of fractions of the polynomial ring $K[X,Y]$.

Let $\alpha,\beta\in\overline L$ such that $\alpha^p=X$ and $\beta^p=Y$. Let's consider the extension $L(\alpha,\beta)/L$. Clearly, $L(\alpha)/L$ and $L(\beta)/L$ are extensions of degree $p$. But what about the degree of $$L(\alpha,\beta)/L\text{ ?}$$ It seems natural that $T^p-X$ is still irreducible over $L(\beta)$, because we essentially don't add anything about $X$ when we go up to $L(\beta)$.

However, I haven't found an elegant/rigorous way to show that. What I actually did some time ago was writing $T^p-X$ as the product of two polynomials in $L(\beta)$ and observing that it led to something "strange", but it was quite messy and not totally rigorous.

Is there a nice way to prove that ?

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You know that $L(\alpha, \beta) = (L(\alpha))(\beta)$ and that $[L(\alpha, \beta): L] = [L(\alpha, \beta):L(\alpha)][L(\alpha):L]$ right? –  Eugene Jun 10 '12 at 20:05
As a remark, this is an example of a non-simple extension: each element has degree at most $p$ over $L$, and yet the degree of the extension is $p^2$. –  Dylan Moreland Jun 10 '12 at 20:10
@Eugene: Yes, of course. My question is only how to prove that $L(\alpha,\beta)$ still has degree $p$ over $L(\alpha)$. –  Klaus Jun 10 '12 at 20:11
@Klaus As Arturo already noted, $L(\alpha, \beta)$ has degree $p$ over $L(\alpha)$ as $T^p - Y$ is irreducible over $L(\alpha)$. –  Eugene Jun 10 '12 at 20:13
@Eugene This is precisely the part I'm not able to formulate in a rigorous way (maybe my question was not clear enough, sorry). Why is $T^p-Y$ still irreducible over $L(\alpha)$ ? –  Klaus Jun 10 '12 at 20:15

$T^p - Y$ is irreducible over $L(\alpha)$ by Eisenstein's Criterion, applied to the integral domain $K(\alpha)[Y]$ and the prime ideal $(Y)$.
Now my question is how you proved that $T^p - Y$ was irreducible over $L$ in the first place.
Consider $F=K(x)$; then $[F(\alpha):F]$ is of degree $p$. Now look at $F(\alpha)(y)$, which is an extension of $F(y)$ also of degree $p$. And $[F(\alpha)(\beta):F(\alpha)(y)] = p$ (same argument was to why $[F(\alpha):F]=p$). Now note that $F(\alpha)(\beta) = L(\alpha,\beta)$.
So $[L(\alpha,\beta):L] = [F(\alpha)(\beta):F(y)] = [F(\alpha,\beta):F(\alpha)(y)][F(\alpha)(y):F(y)] = p^2$.