## 55.11 A formula for the genus

There is one more restriction on the combinatorial structure coming from a proper regular model.

Lemma 55.11.1. In Situation 55.9.3 suppose we have an effective Cartier divisors $D, D' \subset X$ such that $D' = D + C_ i$ for some $i \in \{ 1, \ldots , n\}$ and $D' \subset X_ k$. Then

$\chi (X_ k, \mathcal{O}_{D'}) - \chi (X_ k, \mathcal{O}_ D) = \chi (X_ k, \mathcal{O}_ X(-D)|_{C_ i}) = -(D \cdot C_ i) + \chi (C_ i, \mathcal{O}_{C_ i})$

Proof. The second equality follows from the definition of the bilinear form $(\ \cdot \ )$ in (55.9.6.1) and Lemma 55.9.6. To see the first equality we distinguish two cases. Namely, if $C_ i \not\subset D$, then $D'$ is the scheme theoretic union of $D$ and $C_ i$ (by Divisors, Lemma 31.13.10) and we get a short exact sequence

$0 \to \mathcal{O}_{D'} \to \mathcal{O}_ D \times \mathcal{O}_{C_ i} \to \mathcal{O}_{D \cap C_ i} \to 0$

by Morphisms, Lemma 29.4.6. Since we also have an exact sequence

$0 \to \mathcal{O}_ X(-D)|_{C_ i} \to \mathcal{O}_{C_ i} \to \mathcal{O}_{D \cap C_ i} \to 0$

(Divisors, Remark 31.14.11) we conclude that the claim holds by additivity of euler characteristics (Varieties, Lemma 33.33.2). On the other hand, if $C_ i \subset D$ then we get an exact sequence

$0 \to \mathcal{O}_ X(-D)|_{C_ i} \to \mathcal{O}_{D'} \to \mathcal{O}_ D \to 0$

by Divisors, Lemma 31.14.3 and we immediately see the lemma holds. $\square$

Lemma 55.11.2. In Situation 55.9.3 we have

$g_ C = 1 + \sum \nolimits _{i = 1, \ldots , n} m_ i\left([\kappa _ i : k] (g_ i - 1) - \frac{1}{2}(C_ i \cdot C_ i)\right)$

where $\kappa _ i = H^0(C_ i, \mathcal{O}_{C_ i})$, $g_ i$ is the genus of $C_ i$, and $g_ C$ is the genus of $C$.

Proof. Our basic tool will be Derived Categories of Schemes, Lemma 36.32.2 which shows that

$1 - g_ C = \chi (C, \mathcal{O}_ C) = \chi (X_ k, \mathcal{O}_{X_ k})$

Choose a sequence of effective Cartier divisors

$X_ k = D_ m \supset D_{m - 1} \supset \ldots \supset D_1 \supset D_0 = \emptyset$

such that $D_{j + 1} = D_ j + C_{i_ j}$ for each $j$. (It is clear that we can choose such a sequence by decreasing one nonzero multiplicity of $D_{j + 1}$ one step at a time.) Applying Lemma 55.11.1 starting with $\chi (\mathcal{O}_{D_0}) = 0$ we get

\begin{align*} 1 - g_ C & = \chi (X_ k, \mathcal{O}_{X_ k}) \\ & = \sum \nolimits _ j \left(-(D_ j \cdot C_{i_ j}) + \chi (C_{i_ j}, \mathcal{O}_{C_{i_ j}})\right) \\ & = - \sum \nolimits _ j (C_{i_1} + C_{i_2} + \ldots + C_{i_{j - 1}} \cdot C_{i_ j}) + \sum \nolimits _ j \chi (C_{i_ j}, \mathcal{O}_{C_{i_ j}}) \\ & = -\frac{1}{2}\sum \nolimits _{j \not= j'} (C_{i_{j'}} \cdot C_{i_ j}) + \sum m_ i \chi (C_ i, \mathcal{O}_{C_ i}) \\ & = \frac{1}{2} \sum m_ i(C_ i \cdot C_ i) + \sum m_ i \chi (C_ i, \mathcal{O}_{C_ i}) \end{align*}

Perhaps the last equality deserves some explanation. Namely, since $\sum _ j C_{i_ j} = \sum m_ i C_ i$ we have $(\sum _ j C_{i_ j} \cdot \sum _ j C_{i_ j}) = 0$ by Lemma 55.9.7. Thus we see that

$0 = \sum \nolimits _{j \not= j'} (C_{i_{j'}} \cdot C_{i_ j}) + \sum m_ i(C_ i \cdot C_ i)$

by splitting this product into “nondiagonal” and “diagonal” terms. Note that $\kappa _ i$ is a field finite over $k$ by Varieties, Lemma 33.26.2. Hence the genus of $C_ i$ is defined and we have $\chi (C_ i, \mathcal{O}_{C_ i}) = [\kappa _ i : k](1 - g_ i)$. Putting everything together and rearranging terms we get

$g_ C = - \frac{1}{2}\sum m_ i(C_ i \cdot C_ i) + \sum m_ i[\kappa _ i : k](g_ i - 1) + 1$

which is what the lemma says too. $\square$

Lemma 55.11.3. In Situation 55.9.3 with $\kappa _ i = H^0(C_ i, \mathcal{O}_{C_ i})$ and $g_ i$ the genus of $C_ i$ the data

$n, m_ i, (C_ i \cdot C_ j), [\kappa _ i : k], g_ i$

is a numerical type of genus equal to the genus of $C$.

Proof. (In the proof of Lemma 55.11.2 we have seen that the quantities used in the statement of the lemma are well defined.) We have to verify the conditions (1) – (5) of Definition 55.3.1.

Condition (1) is immediate.

Condition (2). Symmetry of the matrix $(C_ i \cdot C_ j)$ follows from Equation (55.9.6.1) and Lemma 55.9.6. Nonnegativity of $(C_ i \cdot C_ j)$ for $i \not= j$ is part (1) of Lemma 55.9.7.

Condition (3) is part (3) of Lemma 55.9.7.

Condition (4) is part (2) of Lemma 55.9.7.

Condition (5) follows from the fact that $(C_ i \cdot C_ j)$ is the degree of an invertible module on $C_ i$ which is divisible by $[\kappa _ i : k]$, see Varieties, Lemma 33.44.10.

The genus formula proved in Lemma 55.11.2 tells us that the numerical type has the genus as stated, see Definition 55.3.4. $\square$

Now we match minimality of the model with minimality of the type.

Lemma 55.11.5. In Situation 55.9.3. The following are equivalent

1. $X$ is a minimal model, and

2. the numerical type associated to $X$ is minimal.

Proof. If the numerical type is minimal, then there is no $i$ with $g_ i = 0$ and $(C_ i \cdot C_ i) = -[\kappa _ i: k]$, see Definition 55.3.12. Certainly, this implies that none of the curves $C_ i$ are exceptional curves of the first kind.

Conversely, suppose that the numerical type is not minimal. Then there exists an $i$ such that $g_ i = 0$ and $(C_ i \cdot C_ i) = -[\kappa _ i: k]$. We claim this implies that $C_ i$ is an exceptional curve of the first kind. Namely, the invertible sheaf $\mathcal{O}_ X(-C_ i)|_{C_ i}$ has degree $-(C_ i \cdot C_ i) = [\kappa _ i : k]$ when $C_ i$ is viewed as a proper curve over $k$, hence has degree $1$ when $C_ i$ is viewed as a proper curve over $\kappa _ i$. Applying Algebraic Curves, Proposition 53.10.4 we conclude that $C_ i \cong \mathbf{P}^1_{\kappa _ i}$ as schemes over $\kappa _ i$. Since the Picard group of $\mathbf{P}^1$ over a field is $\mathbf{Z}$, we see that the normal sheaf of $C_ i$ in $X$ is isomorphic to $\mathcal{O}_{\mathbf{P}_{\kappa _ i}}(-1)$ and the proof is complete. $\square$

Remark 55.11.6. Not every numerical type comes from a model for the silly reason that there exist numerical types whose genus is negative. There exist a minimal numerical types of positive genus which are not the numerical type associated to a model (over some dvr) of a smooth projective geometrically irreducible curve (over the fraction field of the dvr). A simple example is $n = 1$, $m_1 = 1$, $a_{11} = 0$, $w_1 = 6$, $g_1 = 1$. Namely, in this case the special fibre $X_ k$ would not be geometrically connected because it would live over an extension $\kappa$ of $k$ of degree $6$. This is a contradiction with the fact that the generic fibre is geometrically connected (see More on Morphisms, Lemma 37.52.6). Similarly, $n = 2$, $m_1 = m_2 = 1$, $-a_{11} = -a_{22} = a_{12} = a_{21} = 6$, $w_1 = w_2 = 6$, $g_1 = g_2 = 1$ would be an example for the same reason (details omitted). But if the gcd of the $w_ i$ is $1$ we do not have an example.

Lemma 55.11.7. In Situation 55.9.3 assume $C$ has a $K$-rational point. Then

1. $X_ k$ has a $k$-rational point $x$ which is a smooth point of $X_ k$ over $k$,

2. if $x \in C_ i$, then $H^0(C_ i, \mathcal{O}_{C_ i}) = k$ and $m_ i = 1$, and

3. $H^0(X_ k, \mathcal{O}_{X_ k}) = k$ and $X_ k$ has genus equal to the genus of $C$.

Proof. Since $X \to \mathop{\mathrm{Spec}}(R)$ is proper, the $K$-rational point extends to a morphism $a : \mathop{\mathrm{Spec}}(R) \to X$ by the valuative criterion of properness (Morphisms, Lemma 29.42.1). Let $x \in X$ be the image under $a$ of the closed point of $\mathop{\mathrm{Spec}}(R)$. Then $a$ corresponds to an $R$-algebra homomorphism $\psi : \mathcal{O}_{X, x} \to R$ (see Schemes, Section 26.13). It follows that $\pi \not\in \mathfrak m_ x^2$ (since the image of $\pi$ in $R$ is not in $\mathfrak m_ R^2$). Hence $\mathcal{O}_{X_ k, x} = \mathcal{O}_{X, x}/\pi \mathcal{O}_{X, x}$ is regular (Algebra, Lemma 10.106.3). Then $X_ k \to \mathop{\mathrm{Spec}}(k)$ is smooth at $x$ by Algebra, Lemma 10.140.5. It follows that $x$ is contained in a unique irreducible component $C_ i$ of $X_ k$, that $\mathcal{O}_{C_ i, x} = \mathcal{O}_{X_ k, x}$, and that $m_ i = 1$. The fact that $C_ i$ has a $k$-rational point implies that the field $\kappa _ i = H^0(C_ i, \mathcal{O}_{C_ i})$ (Varieties, Lemma 33.26.2) is equal to $k$. This proves (1). We have $H^0(X_ k, \mathcal{O}_{X_ k}) = k$ because $H^0(X_ k, \mathcal{O}_{X_ k})$ is a field extension of $k$ (Lemma 55.9.9) which maps to $H^0(C_ i, \mathcal{O}_{C_ i}) = k$. The genus equality follows from Lemma 55.9.10. $\square$

Lemma 55.11.8. In Situation 55.9.3 assume $X$ is a minimal model, $\gcd (m_1, \ldots , m_ n) = 1$, and $H^0((X_ k)_{red}, \mathcal{O}) = k$. Then the map

$H^1(X_ k, \mathcal{O}_{X_ k}) \to H^1((X_ k)_{red}, \mathcal{O}_{(X_ k)_{red}})$

is surjective and has a nontrivial kernel as soon as $(X_ k)_{red} \not= X_ k$.

Proof. By vanishing of cohomology in degrees $\geq 2$ over $X_ k$ (Cohomology, Proposition 20.20.7) any surjection of abelian sheaves on $X_ k$ induces a surjection on $H^1$. Consider the sequence

$(X_ k)_{red} = Z_0 \subset Z_1 \subset \ldots \subset Z_ m = X_ k$

of Lemma 55.9.9. Since the field maps $H^0(Z_ j, \mathcal{O}_{Z_ j}) \to H^0((X_ k)_{red}, \mathcal{O}_{(X_ k)_{red}}) = k$ are injective we conclude that $H^0(Z_ j, \mathcal{O}_{Z_ j}) = k$ for $j = 0, \ldots , m$. It follows that $H^0(X_ k, \mathcal{O}_{X_ k}) \to H^0(Z_{m - 1}, \mathcal{O}_{Z_{m - 1}})$ is surjective. Let $C = C_{i_ m}$. Then $X_ k = Z_{m - 1} + C$. Let $\mathcal{L} = \mathcal{O}_ X(-Z_{m - 1})|_ C$. Then $\mathcal{L}$ is an invertible $\mathcal{O}_ C$-module. As in the proof of Lemma 55.9.9 there is an exact sequence

$0 \to \mathcal{L} \to \mathcal{O}_{X_ k} \to \mathcal{O}_{Z_{m - 1}} \to 0$

of coherent sheaves on $X_ k$. We conclude that we get a short exact sequence

$0 \to H^1(C, \mathcal{L}) \to H^1(X_ k, \mathcal{O}_{X_ k}) \to H^1(Z_{m - 1}, \mathcal{O}_{Z_{m - 1}}) \to 0$

The degree of $\mathcal{L}$ on $C$ over $k$ is

$(C \cdot -Z_{m - 1}) = (C \cdot C - X_ k) = (C \cdot C)$

Set $\kappa = H^0(C, \mathcal{O}_ C)$ and $w = [\kappa : k]$. By definition of the degree of an invertible sheaf we see that

$\chi (C, \mathcal{L}) = \chi (C, \mathcal{O}_ C) + (C \cdot C) = w(1 - g_ C) + (C \cdot C)$

where $g_ C$ is the genus of $C$. This expression is $< 0$ as $X$ is minimal and hence $C$ is not an exceptional curve of the first kind (see proof of Lemma 55.11.5). Thus $\dim _ k H^1(C, \mathcal{L}) > 0$ which finishes the proof. $\square$

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

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

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

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

Lemma 55.11.10. If equality holds in Lemma 55.11.9 then

1. the unique irreducible component of $X_ k$ containing $x$ is a smooth projective geometrically irreducible curve over $k$,

2. if $C \subset X_ k$ is another irreducible component, then $\kappa = H^0(C, \mathcal{O}_ C)$ is a finite separable extension of $k$, $C$ has a $\kappa$-rational point, and $C$ is smooth over $\kappa$

Proof. Looking over the proof of Lemma 55.11.9 we see that in order to get equality, the inequalities (55.11.9.1), (55.11.9.2), and (55.11.9.3) have to be equalities.

Let $C_ i$ be the irreducible component containing $x$. Equality in (55.11.9.1) shows via Algebraic Curves, Lemma 53.18.4 that $C_{i, \overline{k}}^\nu \to C_{i, \overline{k}}$ is an isomorphism. Hence $C_{i, \overline{k}}$ is smooth and part (1) holds.

Next, let $C_ i \subset X_ k$ be another irreducible component. Then we may assume we have $D = D' + C_ i$ as in the induction step in the proof of Lemma 55.11.9. Equality in (55.11.9.2) immediately implies that $\kappa _ i/k$ is finite separable. Equality in (55.11.9.3) implies either $a_{ij} = 1$ for some $j$ or that there is a unique $C_ j \subset D'$ meeting $C_ i$ and $a_{ij} = w_ i$. In both cases we find that $C_ i$ has a $\kappa _ i$-rational point $c$ and $c = C_ i \cap C_ j$ scheme theoretically. Since $\mathcal{O}_{X, c}$ is a regular local ring, this implies that the local equations of $C_ i$ and $C_ j$ form a regular system of parameters in the local ring $\mathcal{O}_{X, c}$. Then $\mathcal{O}_{C_ i, c}$ is regular by (Algebra, Lemma 10.106.3). We conclude that $C_ i \to \mathop{\mathrm{Spec}}(\kappa _ i)$ is smooth at $c$ (Algebra, Lemma 10.140.5). It follows that $C_ i$ is geometrically integral over $\kappa _ i$ (Varieties, Lemma 33.25.10). To finish we have to show that $C_ i$ is smooth over $\kappa _ i$. Observe that

$C_{i, \overline{k}} = C_ i \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(\overline{k}) = \coprod \nolimits _{\kappa _ i \to \overline{k}} C_ i \times _{\mathop{\mathrm{Spec}}(\kappa _ i)} \mathop{\mathrm{Spec}}(\overline{k})$

where there are $[\kappa _ i : k]$-summands. Thus if $C_ i$ is not smooth over $\kappa _ i$, then each of these curves is not smooth, then these curves are not normal and the normalization morphism drops the genus (Algebraic Curves, Lemma 53.18.4) which is disallowed because it would drop the geometric genus of $C_ i/k$ contradicting $[\kappa _ i : k] g_ i = g_{geom}(C_ i/k)$. $\square$

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