The Stacks project

Lemma 53.13.3. Let $k$ be a field of characteristic $p > 0$. Let $f : X \to Y$ be a nonconstant morphism of proper nonsingular curves over $k$. If $X$ is smooth and $k(Y) \subset k(X)$ is inseparable of degree $p$, then there is a unique isomorphism $Y = X^{(p)}$ such that $f$ is $F_{X/k}$.

Proof. The relative frobenius morphism $F_{X/k} : X \to X^{(p)}$ is constructed in Varieties, Section 33.36. Observe that $X^{(p)}$ is a smooth proper curve over $k$ as a base change of $X$. The morphism $F_{X/k}$ has degree $p$ by Varieties, Lemma 33.36.10. Thus $k(X^{(p)})$ and $k(Y)$ are both subfields of $k(X)$ with $[k(X) : k(Y)] = [k(X) : k(X^{(p)})] = p$. To prove the lemma it suffices to show that $k(Y) = k(X^{(p)})$ inside $k(X)$. See Theorem 53.2.6.

Write $K = k(X)$. Consider the map $\text{d} : K \to \Omega _{K/k}$. It follows from Lemma 53.12.1 that both $k(Y)$ is contained in the kernel of $\text{d}$. By Varieties, Lemma 33.36.7 we see that $k(X^{(p)})$ is in the kernel of $\text{d}$. Since $X$ is a smooth curve we know that $\Omega _{K/k}$ is a vector space of dimension $1$ over $K$. Then More on Algebra, Lemma 15.46.2. implies that $\mathop{\mathrm{Ker}}(\text{d}) = kK^ p$ and that $[K : kK^ p] = p$. Thus $k(Y) = kK^ p = k(X^{(p)})$ for reasons of degree. $\square$

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