The Stacks project

Proof. Let $y \in Y$. Pick $x \in X$ mapping to $y$. By Varieties, Lemma 33.25.9 it suffices to show that $f$ is flat at $x$. This follows from Lemma 53.2.3. $\square$

Proof. This follows from Theorem 53.2.6 and the fact that a finite purely inseparable extension of fields can always be gotten as a sequence of (inseparable) extensions of degree $p$, see Fields, Lemma 9.14.5. $\square$

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. $\square$

Proof. The $n$-fold relative Frobenius of $X/k$ is defined in Varieties, Remark 33.36.11. The lemma follows by combining Lemmas 53.13.3 and 53.13.2. $\square$

Proof. By Lemma 53.13.4 we see that $Y = X^{(p^ n)}$ is the base change of $X$ by $F_{\mathop{\mathrm{Spec}}(k)}^ n$. Thus $Y$ is smooth and the result on the cohomology and genus follows from Lemma 53.8.2. $\square$

Proof. By Fields, Lemma 9.14.6 there is a subextension $k(X)/E/k(Y)$ such that $k(X)/E$ is purely inseparable and $E/k(Y)$ is separable. By Theorem 53.2.6 this corresponds to a factorization $X \to Z \to Y$ of $f$ with $Z$ a nonsingular proper curve. Apply Lemma 53.13.4 to the morphism $X \to Z$ to conclude. $\square$

Proof. Pick two $k$-linearly independent elements $s, t \in V$. Then $f = s/t$ is the rational function defining the morphism $X \to \mathbf{P}^1_ k$ corresponding to the linear series $(\mathcal{L}, ks + kt)$. If this morphism is not generically étale, then $f \in k(X^{(p)})$ by Proposition 53.13.7. Now choose a basis $s_0, \ldots , s_ r$ of $V$ and let $\mathcal{L}' \subset \mathcal{L}$ be the invertible sheaf generated by $s_0, \ldots , s_ r$. Set $f_ i = s_ i/s_0$ in $k(X)$. If for each pair $(s_0, s_ i)$ we have $f_ i \in k(X^{(p)})$, then the morphism

factors through $X^{(p)}$ as this is true over the affine open $D_+(T_0)$ and we can extend the morphism over the affine part to the whole of the smooth curve $X^{(p)}$ by Lemma 53.2.2. Introducing notation, say we have the factorization

of $\varphi $. Then $\mathcal{N} = \psi ^*\mathcal{O}_{\mathbf{P}^1_ k}(1)$ is an invertible $\mathcal{O}_{X^{(p)}}$-module with $\mathcal{L}' = F_{X/k}^*\mathcal{N}$ and with $\psi ^*T_0, \ldots , \psi ^*T_ r$ $k$-linearly independent (as they pullback to $s_0, \ldots , s_ r$ on $X$). Finally, we have

as desired. Here we used Varieties, Lemmas 33.44.12, 33.44.11, and 33.36.10. $\square$

Proof. Set $\omega = \Omega _{X/k}$. By Lemma 53.8.4 this has degree $2g - 2$ and has $g$ global sections. Thus we have a $\mathfrak g^{g - 1}_{2g - 2}$. By the trivial Lemma 53.3.3 there exists a $\mathfrak g^1_{2g - 2}$ and by Lemma 53.3.4 we obtain a morphism

of some degree $d \leq 2g - 2$. Since $\varphi $ is flat (Lemma 53.2.3) and finite (Lemma 53.2.4) it is finite locally free of degree $d$ (Morphisms, Lemma 29.48.2). Pick any rational point $t \in \mathbf{P}^1_ k$ and any point $x \in X$ with $\varphi (x) = t$. Then

for example by Morphisms, Lemmas 29.57.3 and 29.57.2. Thus if $k$ is perfect (for example has characteristic zero or is finite) then the lemma is proved. Thus we reduce to the case discussed in the next paragraph.

Assume that $k$ is an infinite field of characteristic $p > 0$. As above we will use that $X$ has a $\mathfrak g^{g - 1}_{2g - 2}$. The smooth proper curve $X^{(p)}$ has the same genus as $X$. Hence its genus is $> 0$. We conclude that $X^{(p)}$ does not have a $\mathfrak g^{g - 1}_ d$ for any $d \leq g - 1$ by Lemma 53.3.5. Applying Lemma 53.13.8 to our $\mathfrak g^{g - 1}_{2g - 2}$ (and noting that $2g - 2/p \leq g - 1$) we conclude that possibility (2) does not occur. Hence we obtain a morphism

which is generically étale (in the sense of the lemma) and has degree $\leq 2g - 2$. Let $U \subset X$ be the nonempty open subscheme where $\varphi $ is étale. Then $\varphi (U) \subset \mathbf{P}^1_ k$ is a nonempty Zariski open and we can pick a $k$-rational point $t \in \varphi (U)$ as $k$ is infinite. Let $u \in U$ be a point with $\varphi (u) = t$. Then $\kappa (u)/\kappa (t)$ is separable (Morphisms, Lemma 29.36.7), $\kappa (t) = k$, and $[\kappa (u) : k] \leq 2g - 2$ as before. $\square$

Proof. After shrinking $X$ we may assume there is an étale morphism $\pi : X \to \mathbf{A}^1_ k$, see Morphisms, Lemma 29.36.20. Then we can consider the diagram of local rings

The horizontal arrows have ramification index $1$ as they correspond to étale morphisms. Moreover, the extension $\kappa (x)/\kappa (\pi (x))$ is separable hence $\kappa (x)$ and $\kappa (\pi (x))$ have the same inseparable degree over $k$. By multiplicativity of ramification indices it suffices to prove the result when $x$ is a point of the affine line.

Assume $X = \mathbf{A}^1_ k$. In this case, the local ring of $X$ at $x$ looks like

where $P$ is an irreducible monic polynomial over $k$. Then $P(t) = Q(t^ q)$ for some separable polynomial $Q \in k[t]$, see Fields, Lemma 9.12.1. Observe that $\kappa (x) = k[t]/(P)$ has inseparable degree $q$ over $k$. On the other hand, over $\overline{k}$ we can factor $Q(t) = \prod (t - \alpha _ i)$ with $\alpha _ i$ pairwise distinct. Write $\alpha _ i = \beta _ i^ q$ for some unique $\beta _ i \in \overline{k}$. Then our point $\overline{x}$ corresponds to one of the $\beta _ i$ and we conclude because the ramification index of

is indeed equal to $q$ as the uniformizer $P$ maps to $(t - \beta _ i)^ q$ times a unit. $\square$