# Degree of Frobenius

Let $k$ be an algebraically closed field of characteristic $p>0$ and $K/k$ be a function field, i.e. $K$ is finite over $k(t)$. Consider the field extension $K \subseteq K^{1/p}$. Why does it have degree $p$?

In the case $K = k(t)$ this is clear, because then $K^{1/p} = k(t^{1/p})$ and $t^{1/p}$ has degree $p$ over $K$, since $x^p - t$ is irreducible over $K$ and has $t^{1/p}$ as a root. But I don't know how to deal with the general case.

Remark: The equality $[K^{1/p} : K] = p$ is used in Hartshorne's book in the context of the Frobenius morphism of curves.

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Since, as you write, $[k(t)^{1/p}:k(t)]=p$, the result follows from $[K:k(t)]=[K^{1/p}:k(t)^{1/p}]$ by multiplicativity of degrees. $[K:k(t)]=[K^{1/p}:k(t)^{1/p}]$ holds since the map $x\mapsto x^p$ is an isomorphism $K^{1/p}\to K$ which restricts to an isomorphism $k(t)^{1/p}\to k(t)$ - i.e. the extensions $K^{1/p}\supset k(t)^{1/p}$ and $K\supset k(t)$ are isomorphic.