Field of prime characteristic over two indeterminates

Question

Let $F$ have prime characteristic $p$ and let $E = F(Y,Z)$, where $Y, Z$ are indeterminates. Let $L=F(Y^{p} , Z^{p})$ $\subseteq E$.

a. Show that $\alpha^{p} \in L$ for all $\alpha \in E$.

b. Show that $E\ne L[\alpha]$ for any $\alpha \in E$.

I can see that $\mid E:L \mid$= $p^{2}$. But I don't know where to go from there. Any help will be appreciated. Thanks

Answer 1

a) Every $\alpha \in E$ is of the form $f(Y,Z)/g(Y,Z)$ for $f, g \in F[Y,Z]$. Then $\alpha^p = f^p/g^p$, and it suffices to show that $f^p \in L$ for all $f \in F[Y,Z]$. Writing $f = \sum_{i,j} a_{ij}Y^iZ^j$ and applying Frobenius gives $f^p = \sum_{i,j} a_{ij}^p (Y^p)^i(Z^p)^j \in F[Y^p,Z^p] \subseteq L$.

b) If $E = L(\alpha)$ for some $\alpha \in E$, then $[E : L] = [L(\alpha) : L]$ is the degree of the minimal polynomial of $\alpha$ over $L$. By (a), $\alpha^p \in L$, say $\alpha^p =: t \in L$. Then $\alpha$ is a root of $x^p - t$, so the minimal polynomial of $\alpha$ over $L$ has $\deg \le p$. But $[E : L] = [E : F(Y, Z^p)][F(Y, Z^p) : L] = p^2$.