Lemma 55.11.9 (0CEA)—The Stacks project

Lemma 55.11.9. In Situation 55.9.3 assume $X_ k$ has a $k$-rational point $x$ which is a smooth point of $X_ k \to \mathop{\mathrm{Spec}}(k)$. Then

\[ \dim _ k H^1((X_ k)_{red}, \mathcal{O}_{(X_ k)_{red}}) \geq g_{top} + g_{geom}(X_ k/k) \]

where $g_{geom}$ is as in Algebraic Curves, Section 53.18 and $g_{top}$ is the topological genus (Definition 55.3.11) of the numerical type associated to $X_ k$ (Definition 55.11.4).

Proof. We are going to prove the inequality

\[ \dim _ k H^1(D, \mathcal{O}_ D) \geq g_{top}(D) + g_{geom}(D/k) \]

for all connected reduced effective Cartier divisors $D \subset (X_ k)_{red}$ containing $x$ by induction on the number of irreducible components of $D$. Here $g_{top}(D) = 1 - m + e$ where $m$ is the number of irreducible components of $D$ and $e$ is the number of unordered pairs of components of $D$ which meet.

Base case: $D$ has one irreducible component. Then $D = C_ i$ is the unique irreducible component containing $x$. In this case $\dim _ k H^1(D, \mathcal{O}_ D) = g_ i$ and $g_{top}(D) = 0$. Since $C_ i$ has a $k$-rational smooth point it is geometrically integral (Varieties, Lemma 33.25.10). It follows that $g_ i$ is the genus of $C_{i, \overline{k}}$ (Algebraic Curves, Lemma 53.8.2). It also follows that $g_{geom}(D/k)$ is the genus of the normalization $C_{i, \overline{k}}^\nu $ of $C_{i, \overline{k}}$. Applying Algebraic Curves, Lemma 53.18.4 to the normalization morphism $C_{i, \overline{k}}^\nu \to C_{i, \overline{k}}$ we get

55.11.9.1

\begin{equation} \label{models-equation-genus-change-special-component} \text{genus of }C_{i, \overline{k}} \geq \text{genus of }C_{i, \overline{k}}^\nu \end{equation}

Combining the above we conclude that $\dim _ k H^1(D, \mathcal{O}_ D) \geq g_{top}(D) + g_{geom}(D/k)$ in this case.

Induction step. Suppose we have $D$ with more than $1$ irreducible component. Then we can write $D = C_ i + D'$ where $x \in D'$ and $D'$ is still connected. This is an exercise in graph theory we leave to the reader (hint: let $C_ i$ be the component of $D$ which is farthest from $x$). We compute how the invariants change. As $x \in D'$ we have $H^0(D, \mathcal{O}_ D) = H^0(D', \mathcal{O}_{D'}) = k$. Looking at the short exact sequence of sheaves

\[ 0 \to \mathcal{O}_ D \to \mathcal{O}_{C_ i} \oplus \mathcal{O}_{D'} \to \mathcal{O}_{C_ i \cap D'} \to 0 \]

(Morphisms, Lemma 29.4.6) and using additivity of euler characteristics we find

\begin{align*} \dim _ k H^1(D, \mathcal{O}_ D) - \dim _ k H^1(D', \mathcal{O}_{D'}) & = -\chi (\mathcal{O}_{C_ i}) + \chi (\mathcal{O}_{C_ i \cap D'}) \\ & = w_ i(g_ i - 1) + \sum \nolimits _{C_ j \subset D'} a_{ij} \end{align*}

Here as in Lemma 55.11.3 we set $w_ i = [\kappa _ i : k]$, $\kappa _ i = H^0(C_ i, \mathcal{O}_{C_ i})$, $g_ i$ is the genus of $C_ i$, and $a_{ij} = (C_ i \cdot C_ j)$. We have

\[ g_{top}(D) - g_{top}(D') = -1 + \sum \nolimits _{C_ j \subset D'\text{ meeting }C_ i} 1 \]

We have

\[ g_{geom}(D/k) - g_{geom}(D'/k) = g_{geom}(C_ i/k) \]

by Algebraic Curves, Lemma 53.18.1. Combining these with our induction hypothesis, we conclude that it suffices to show that

\[ w_ i g_ i - g_{geom}(C_ i/k) + \sum \nolimits _{C_ j \subset D'\text{ meets } C_ i} (a_{ij} - 1) - (w_ i - 1) \]

is nonnegative. In fact, we have

55.11.9.2

\begin{equation} \label{models-equation-genus-change} w_ i g_ i \geq [\kappa _ i : k]_ s g_ i \geq g_{geom}(C_ i/k) \end{equation}

The second inequality by Algebraic Curves, Lemma 53.18.5. On the other hand, since $w_ i$ divides $a_{ij}$ (Varieties, Lemma 33.44.10) it is clear that

55.11.9.3

\begin{equation} \label{models-equation-change-intersections} \sum \nolimits _{C_ j \subset D'\text{ meets } C_ i} (a_{ij} - 1) - (w_ i - 1) \geq 0 \end{equation}

because there is at least one $C_ j \subset D'$ which meets $C_ i$. $\square$

The Stacks project

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