## 53.13 Inseparable maps

Some remarks on the behaviour of the genus under inseparable maps.

Lemma 53.13.1. Let $k$ be a field. Let $f : X \to Y$ be a surjective morphism of curves over $k$. If $X$ is smooth over $k$ and $Y$ is normal, then $Y$ is smooth over $k$.

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. $\square$

Lemma 53.13.2. 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 the extension $k(X)/k(Y)$ of function fields is purely inseparable, then there exists a factorization

$X = X_0 \to X_1 \to \ldots \to X_ n = Y$

such that each $X_ i$ is a proper nonsingular curve and $X_ i \to X_{i + 1}$ is a degree $p$ morphism with $k(X_{i + 1}) \subset k(X_ i)$ inseparable.

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. $\square$

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. $\square$

Lemma 53.13.4. 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 purely inseparable, then there is a unique $n \geq 0$ and a unique isomorphism $Y = X^{(p^ n)}$ such that $f$ is the $n$-fold relative Frobenius of $X/k$.

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. $\square$

Lemma 53.13.5. Let $k$ be a field of characteristic $p > 0$. Let $f : X \to Y$ be a nonconstant morphism of proper nonsingular curves over $k$. Assume

1. $X$ is smooth,

2. $H^0(X, \mathcal{O}_ X) = k$,

3. $k(X)/k(Y)$ is purely inseparable.

Then $Y$ is smooth, $H^0(Y, \mathcal{O}_ Y) = k$, and the genus of $Y$ is equal to the genus of $X$.

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. $\square$

Example 53.13.6. This example will show that the genus can change under a purely inseparable morphism of nonsingular projective curves. Let $k$ be a field of characteristic $3$. Assume there exists an element $a \in k$ which is not a $3$rd power. For example $k = \mathbf{F}_3(a)$ would work. Let $X$ be the plane curve with homogeneous equation

$F = T_1^2T_0 - T_2^3 + aT_0^3$

as in Section 53.9. On the affine piece $D_+(T_0)$ using coordinates $x = T_1/T_0$ and $y = T_2/T_0$ we obtain $x^2 - y^3 + a = 0$ which defines a nonsingular affine curve. Moreover, the point at infinity $(0 : 1: 0)$ is a smooth point. Hence $X$ is a nonsingular projective curve of genus $1$ (Lemma 53.9.3). On the other hand, consider the morphism $f : X \to \mathbf{P}^1_ k$ which on $D_+(T_0)$ sends $(x, y)$ to $y \in \mathbf{A}^1_ k \subset \mathbf{P}^1_ k$. Then $f$ is a morphism of proper nonsingular curves over $k$ inducing an inseparable function field extension of degree $p$ but the genus of $X$ is $1$ and the genus of $\mathbf{P}^1_ k$ is $0$.

Proposition 53.13.7. Let $k$ be a field of characteristic $p > 0$. Let $f : X \to Y$ be a nonconstant morphism of proper smooth curves over $k$. Then we can factor $f$ as

$X \longrightarrow X^{(p^ n)} \longrightarrow Y$

where $X^{(p^ n)} \to Y$ is a nonconstant morphism of proper smooth curves inducing a separable field extension $k(X^{(p^ n)})/k(Y)$, we have

$X^{(p^ n)} = X \times _{\mathop{\mathrm{Spec}}(k), F_{\mathop{\mathrm{Spec}}(k)}^ n} \mathop{\mathrm{Spec}}(k),$

and $X \to X^{(p^ n)}$ is the $n$-fold relative frobenius of $X$.

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. $\square$

Lemma 53.13.8. Let $k$ be a field of characteristic $p > 0$. Let $X$ be a smooth proper curve over $k$. Let $(\mathcal{L}, V)$ be a $\mathfrak g^ r_ d$ with $r \geq 1$. Then one of the following two is true

1. there exists a $\mathfrak g^1_ d$ whose corresponding morphism $X \to \mathbf{P}^1_ k$ (Lemma 53.3.2) is generically étale (i.e., is as in Lemma 53.12.1), or

2. there exists a $\mathfrak g^ r_{d'}$ on $X^{(p)}$ where $d' \leq d/p$.

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

$\varphi = \varphi _{(\mathcal{L}', (s_0, \ldots , s_ r)} : X \longrightarrow \mathbf{P}^ r_ k = \text{Proj}(k[T_0, \ldots , T_ r])$

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

$X \xrightarrow {F_{X/k}} X^{(p)} \xrightarrow {\psi } \mathbf{P}^ r_ k$

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

$d = \deg (\mathcal{L}) \geq \deg (\mathcal{L}') = \deg (F_{X/k}) \deg (\mathcal{N}) = p \deg (\mathcal{N})$

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

Lemma 53.13.9. Let $k$ be a field. Let $X$ be a smooth proper curve over $k$ with $H^0(X, \mathcal{O}_ X) = k$ and genus $g \geq 2$. Then there exists a closed point $x \in X$ with $\kappa (x)/k$ separable of degree $\leq 2g - 2$.

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

$\varphi : X \longrightarrow \mathbf{P}^1_ k$

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

$d \geq [\kappa (x) : \kappa (t)] = [\kappa (x) : k]$

for example by Morphisms, Lemmas 29.56.3 and 29.56.2. 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

$\varphi : X \longrightarrow \mathbf{P}^1_ k$

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. $\square$

The following lemma does not really belong in this section but we don't know a good place for it elsewhere.

Lemma 53.13.10. Let $X$ be a smooth curve over a field $k$. Let $\overline{x} \in X_{\overline{k}}$ be a closed point with image $x \in X$. The ramification index of $\mathcal{O}_{X, x} \subset \mathcal{O}_{X_{\overline{k}}, \overline{x}}$ is the inseparable degree of $\kappa (x)/k$.

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

$\xymatrix{ \mathcal{O}_{X_{\overline{k}}, \overline{x}} & \mathcal{O}_{\mathbf{A}^1_{\overline{k}}, \pi (\overline{x})} \ar[l] \\ \mathcal{O}_{X, x} \ar[u] & \mathcal{O}_{\mathbf{A}^1_ k, \pi (x)} \ar[l] \ar[u] }$

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

$\mathcal{O}_{X, x} = k[t]_{(P)}$

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

$k[t]_{(P)} \longrightarrow \overline{k}[t]_{(t - \beta _ i)}$

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

Comment #2367 by Leonardo Zapponi on

Example 46.11.6 contradicts the previous lemma. The problem comes from the fact that the curve $X$ is not smooth over $k$.

Indeed, setting $k'=k(a^{1/3})$, the curve $X'$ obtained by base extension is not smooth over $k'$ (and smooth morphisms are stable under base extension).

Equivalently, on the affine open $U=\mbox{Spec}(k[x,y]/(x^2-y^3+a)$, the element $x$ generates a prime ideal (since the quotient $A/xA$ is a field) and the morphism $X\to\mbox{Spec}(k)$ is not smooth at the corresponding (closed) point.

Comment #2368 by on

@#2367. No, Example 53.13.6 does not contradict Lemma 53.13.5 and there is no problem. In fact, as you yourself point out, the curve $X$ in the example is not smooth over $k$ whereas in the lemma it is required that $X$ be smooth over $k$. So the lemma does not apply to the situation of the example. Am I misunderstanding your comment?

Comment #2369 by Junyan Xu on

So $X$ should be singular, not nonsingular.

Comment #2371 by Junyan Xu on

OK, I found Tag 00TY and it resolves my prior misconception. So by nonsingular you mean that the local rings are regular, which is not the same as smooth over a non-perfect field. It may be a good idea to cross reference Tag 00TY.

Comment #2372 by Leonardo Zapponi on

Example 46.11.6 contradicts the previous lemma. The problem comes from the fact that the curve $X$ is not smooth over $k$.

Indeed, setting $k'=k(a^{1/3})$, the curve $X'$ obtained by base extension is not smooth over $k'$ (and smooth morphisms are stable under base extension).

Equivalently, on the affine open $U=\mbox{Spec}(k[x,y]/(x^2-y^3+a)$, the element $x$ generates a prime ideal (since the quotient $A/xA$ is a field) and the morphism $X\to\mbox{Spec}(k)$ is not smooth at the corresponding (closed) point.

Edit: Sorry, reading (too) quickly the example, I thought that the curve X was supposed to be smooth over k. As Johan and Junyan Xu point out, its regularity does not imply the smoothness over k (Tag 00TY being the standard example). I apologize for the misreading.

Comment #2373 by on

@Junyan, @Leonardo: Right! Maybe there should also be a reference to Definition 28.9.1 in Example 53.13.6.

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