A quick question on transcendence I've seen the following claim in my notes, but I couldn't see why it's true:
Suppose that $y \in F_p((x))$ is transcendent over $F_p(x)$, denote $L:=F_p(x, y)$ and let $L^p$ be the field of $p$th powers of $L$.
My question is:

Why is $[L : L^p]=p^2$?

 A: Here are some hints; if you like, I'll say more. If you can show that $[L : L^p(x)]$ and $[L^p(x) : L^p]$ are both $p$, then the tower law gives you the degree of $L$ over $L^p$, and the standard proof of that law shows that maxymoo gave you a basis.
I think the interesting step is showing that if $k$ is a field of characteristic $p > 0$, and $t$ is transcendental over $k$, then $[k(t) : k(t^p)] = p$. Clearly $t$ satisfies the polynomial $f(X) = X^p - t^p$ over $k(t^p)$, so we must show that $f$ is irreducible. As $f(X) = (X - t)^p$ in the larger field, if we have a non-trivial factor of $f$ over $k(t^p)$ then it is associate to $(X - t)^r$ for some $1 \leqq r < p$. Assuming we have such a factor, show that $t \in k(t^p)$.
Thus it is enough to show that $t$ is not in $k(t^p)$. If it were, then we'd have $g(X), h(X) \in k[X]$ satisfying
$$
t = \frac{g(t^p)}{h(t^p)}.
$$
Why is this impossible?
Fun corollary: $L$ is not a simple extension of $L^p$.
A: i'm not sure i misunderstand the question, but is it because you can take $\{x^iy^j\}_{0\leqslant i,j \leqslant p-1}$ as your $L^p$-basis of $L$?
