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

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    That $y$ is taken in $\mathbf{F}_p((x))$ seems like a strange detail.2011-07-22

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