The Stacks project

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. $\square$


Comments (7)

Comment #2367 by Leonardo Zapponi on

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

Indeed, setting , the curve obtained by base extension is not smooth over (and smooth morphisms are stable under base extension).

Equivalently, on the affine open , the element generates a prime ideal (since the quotient is a field) and the morphism 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 in the example is not smooth over whereas in the lemma it is required that be smooth over . So the lemma does not apply to the situation of the example. Am I misunderstanding your comment?

Comment #2369 by Junyan Xu on

So 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 is not smooth over .

Indeed, setting , the curve obtained by base extension is not smooth over (and smooth morphisms are stable under base extension).

Equivalently, on the affine open , the element generates a prime ideal (since the quotient is a field) and the morphism 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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