## 55.9 The geometry of a regular model

In this section we describe the geometry of a proper regular model $X$ of a smooth projective curve $C$ over $K$ with $H^0(C, \mathcal{O}_ C) = K$.

Lemma 55.9.1. Let $X$ be a regular model of a smooth curve $C$ over $K$.

1. the special fibre $X_ k$ is an effective Cartier divisor on $X$,

2. each irreducible component $C_ i$ of $X_ k$ is an effective Cartier divisor on $X$,

3. $X_ k = \sum m_ i C_ i$ (sum of effective Cartier divisors) where $m_ i$ is the multiplicity of $C_ i$ in $X_ k$,

4. $\mathcal{O}_ X(X_ k) \cong \mathcal{O}_ X$.

Proof. Recall that $R$ is a discrete valuation ring with uniformizer $\pi$ and residue field $k = R/(\pi )$. Because $X \to \mathop{\mathrm{Spec}}(R)$ is flat, the element $\pi$ is a nonzerodivisor affine locally on $X$ (see More on Algebra, Lemma 15.22.11). Thus if $U = \mathop{\mathrm{Spec}}(A) \subset X$ is an affine open, then

$X_ K \cap U = U_ k = \mathop{\mathrm{Spec}}(A \otimes _ R k) = \mathop{\mathrm{Spec}}(A/\pi A)$

and $\pi$ is a nonzerodivisor in $A$. Hence $X_ k = V(\pi )$ is an effective Cartier divisor by Divisors, Lemma 31.13.2. Hence (1) is true.

The discussion above shows that the pair $(\mathcal{O}_ X(X_ k), 1)$ is isomorphic to the pair $(\mathcal{O}_ X, \pi )$ which proves (4).

By Divisors, Lemma 31.15.11 there exist pairwise distinct integral effective Cartier divisors $D_ i \subset X$ and integers $a_ i \geq 0$ such that $X_ k = \sum a_ i D_ i$. We can throw out those divisors $D_ i$ such that $a_ i = 0$. Then it is clear (from the definition of addition of effective Cartier divisors) that $X_ k = \bigcup D_ i$ set theoretically. Thus $C_ i = D_ i$ are the irreducible components of $X_ k$ which proves (2). Let $\xi _ i$ be the generic point of $C_ i$. Then $\mathcal{O}_{X, \xi _ i}$ is a discrete valuation ring (Divisors, Lemma 31.15.4). The uniformizer $\pi _ i \in \mathcal{O}_{X, \xi _ i}$ is a local equation for $C_ i$ and the image of $\pi$ is a local equation for $X_ k$. Since $X_ k = \sum a_ i C_ i$ we see that $\pi$ and $\pi _ i^{a_ i}$ generate the same ideal in $\mathcal{O}_{X, \xi _ i}$. On the other hand, the multiplicity of $C_ i$ in $X_ k$ is

$m_ i = \text{length}_{\mathcal{O}_{C_ i, \xi _ i}} \mathcal{O}_{X_ k, \xi _ i} = \text{length}_{\mathcal{O}_{C_ i, \xi _ i}} \mathcal{O}_{X, \xi _ i}/(\pi ) = \text{length}_{\mathcal{O}_{C_ i, \xi _ i}} \mathcal{O}_{X, \xi _ i}/(\pi _ i^{a_ i}) = a_ i$

See Chow Homology, Definition 42.9.2. Thus $a_ i = m_ i$ and (3) is proved. $\square$

Lemma 55.9.2. Let $X$ be a regular model of a smooth curve $C$ over $K$. Then

1. $X \to \mathop{\mathrm{Spec}}(R)$ is a Gorenstein morphism of relative dimension $1$,

2. each of the irreducible components $C_ i$ of $X_ k$ is Gorenstein.

Proof. Since $X \to \mathop{\mathrm{Spec}}(R)$ is flat, to prove (1) it suffices to show that the fibres are Gorenstein (Duality for Schemes, Lemma 48.25.3). The generic fibre is a smooth curve, which is regular and hence Gorenstein (Duality for Schemes, Lemma 48.24.3). For the special fibre $X_ k$ we use that it is an effective Cartier divisor on a regular (hence Gorenstein) scheme and hence Gorenstein for example by Dualizing Complexes, Lemma 47.21.6. The curves $C_ i$ are Gorenstein by the same argument. $\square$

Situation 55.9.3. Let $R$ be a discrete valuation ring with fraction field $K$, residue field $k$, and uniformizer $\pi$. Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_ C) = K$. Let $X$ be a regular proper model of $C$. Let $C_1, \ldots , C_ n$ be the irreducible components of the special fibre $X_ k$. Write $X_ k = \sum m_ i C_ i$ as in Lemma 55.9.1.

Proof. Consequence of More on Morphisms, Lemma 37.53.6. $\square$

Lemma 55.9.5. In Situation 55.9.3 there is an exact sequence

$0 \to \mathbf{Z} \to \mathbf{Z}^{\oplus n} \to \mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (C) \to 0$

where the first map sends $1$ to $(m_1, \ldots , m_ n)$ and the second maps sends the $i$th basis vector to $\mathcal{O}_ X(C_ i)$.

Proof. Observe that $C \subset X$ is an open subscheme. The restriction map $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (C)$ is surjective by Divisors, Lemma 31.28.3. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module such that there is an isomorphism $s : \mathcal{O}_ C \to \mathcal{L}|_ C$. Then $s$ is a regular meromorphic section of $\mathcal{L}$ and we see that $\text{div}_\mathcal {L}(s) = \sum a_ i C_ i$ for some $a_ i \in \mathbf{Z}$ (Divisors, Definition 31.27.4). By Divisors, Lemma 31.27.6 (and the fact that $X$ is normal) we conclude that $\mathcal{L} = \mathcal{O}_ X(\sum a_ iC_ i)$. Finally, suppose that $\mathcal{O}_ X(\sum a_ i C_ i) \cong \mathcal{O}_ X$. Then there exists an element $g$ of the function field of $X$ with $\text{div}_ X(g) = \sum a_ i C_ i$. In particular the rational function $g$ has no zeros or poles on the generic fibre $C$ of $X$. Since $C$ is a normal scheme this implies $g \in H^0(C, \mathcal{O}_ C) = K$. Thus $g = \pi ^ a u$ for some $a \in \mathbf{Z}$ and $u \in R^*$. We conclude that $\text{div}_ X(g) = a \sum m_ i C_ i$ and the proof is complete. $\square$

In Situation 55.9.3 for every invertible $\mathcal{O}_ X$-module $\mathcal{L}$ and every $i$ we get an integer

$\deg (\mathcal{L}|_{C_ i}) = \chi (C_ i, \mathcal{L}|_{C_ i}) - \chi (C_ i, \mathcal{O}_{C_ i})$

by taking the degree of the restriction of $\mathcal{L}$ to $C_ i$ relative to the ground field $k$1 as in Varieties, Section 33.44.

Lemma 55.9.6. In Situation 55.9.3 given $\mathcal{L}$ an invertible $\mathcal{O}_ X$-module and $a = (a_1, \ldots , a_ n) \in \mathbf{Z}^{\oplus n}$ we define

$\langle a, \mathcal{L} \rangle = \sum a_ i\deg (\mathcal{L}|_{C_ i})$

Then $\langle , \rangle$ is bilinear and for $b = (b_1, \ldots , b_ n) \in \mathbf{Z}^{\oplus n}$ we have

$\left\langle a, \mathcal{O}_ X(\sum b_ i C_ i) \right\rangle = \left\langle b, \mathcal{O}_ X(\sum a_ i C_ i) \right\rangle$

Proof. Bilinearity is immediate from the definition and Varieties, Lemma 33.44.7. To prove symmetry it suffices to assume $a$ and $b$ are standard basis vectors in $\mathbf{Z}^{\oplus n}$. Hence it suffices to prove that

$\deg (\mathcal{O}_ X(C_ j)|_{C_ i}) = \deg (\mathcal{O}_ X(C_ i)|_{C_ j})$

for all $1 \leq i, j \leq n$. If $i = j$ there is nothing to prove. If $i \not= j$, then the canonical section $1$ of $\mathcal{O}_ X(C_ j)$ restricts to a nonzero (hence regular) section of $\mathcal{O}_ X(C_ j)|_{C_ i}$ whose zero scheme is exactly $C_ i \cap C_ j$ (scheme theoretic intersection). In other words, $C_ i \cap C_ j$ is an effective Cartier divisor on $C_ i$ and

$\deg (\mathcal{O}_ X(C_ j)|_{C_ i}) = \deg (C_ i \cap C_ j)$

by Varieties, Lemma 33.44.9. By symmetry we obtain the same (!) formula for the other side and the proof is complete. $\square$

In Situation 55.9.3 it is often convenient to think of $\mathbf{Z}^{\oplus n}$ as the free abelian group on the set $\{ C_1, \ldots , C_ n\}$. We will indicate an element of this group as $\sum a_ i C_ i$; here we think of this as a formal sum although equivalently we may (and we sometimes do) think of such a sum as a Weil divisor on $X$ supported on the special fibre $X_ k$. Now Lemma 55.9.6 allows us to define a symmetric bilinear form $(\ \cdot \ )$ on this free abelian group by the rule

55.9.6.1
\begin{equation} \label{models-equation-form} \left(\sum a_ i C_ i \cdot \sum b_ j C_ j\right) = \left\langle a, \mathcal{O}_ X(\sum b_ j C_ j) \right\rangle = \left\langle b, \mathcal{O}_ X(\sum a_ i C_ i) \right\rangle \end{equation}

We will prove some properties of this bilinear form.

Lemma 55.9.7. In Situation 55.9.3 the symmetric bilinear form (55.9.6.1) has the following properties

1. $(C_ i \cdot C_ j) \geq 0$ if $i \not= j$ with equality if and only if $C_ i \cap C_ j = \emptyset$,

2. $(\sum m_ i C_ i \cdot C_ j) = 0$,

3. there is no nonempty proper subset $I \subset \{ 1, \ldots , n\}$ such that $(C_ i \cdot C_ j) = 0$ for $i \in I$, $j \not\in I$.

4. $(\sum a_ i C_ i \cdot \sum a_ i C_ i) \leq 0$ with equality if and only if there exists a $q \in \mathbf{Q}$ such that $a_ i = qm_ i$ for $i = 1, \ldots , n$,

Proof. In the proof of Lemma 55.9.6 we saw that $(C_ i \cdot C_ j) = \deg (C_ i \cap C_ j)$ if $i \not= j$. This is $\geq 0$ and $> 0$ if and only if $C_ i \cap C_ j \not= \emptyset$. This proves (1).

Proof of (2). This is true because by Lemma 55.9.1 the invertible sheaf associated to $\sum m_ i C_ i$ is trivial and the trivial sheaf has degree zero.

Proof of (3). This is expressing the fact that $X_ k$ is connected (Lemma 55.9.4) via the description of the intersection products given in the proof of (1).

Part (4) follows from (1), (2), and (3) by Lemma 55.2.3. $\square$

Lemma 55.9.8. In Situation 55.9.3 set $d = \gcd (m_1, \ldots , m_ n)$ and let $D = \sum (m_ i/d)C_ i$ as an effective Cartier divisor. Then $\mathcal{O}_ X(D)$ has order dividing $d$ in $\mathop{\mathrm{Pic}}\nolimits (X)$ and $\mathcal{C}_{D/X}$ an invertible $\mathcal{O}_ D$-module of order dividing $d$ in $\mathop{\mathrm{Pic}}\nolimits (D)$.

Proof. We have

$\mathcal{O}_ X(D)^{\otimes d} = \mathcal{O}_ X(dD) = \mathcal{O}_ X(X_ k) = \mathcal{O}_ X$

by Lemma 55.9.1. We conclude as $\mathcal{C}_{D/X}$ is the pullback of $\mathcal{O}_ X(-D)$. $\square$

Lemma 55.9.9. In Situation 55.9.3 let $d = \gcd (m_1, \ldots , m_ n)$. Let $D = \sum (m_ i/d) C_ i$ as an effective Cartier divisor. Then there exists a sequence of effective Cartier divisors

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

such that $Z_ j = Z_{j - 1} + C_{i_ j}$ for some $i_ j \in \{ 1, \ldots , n\}$ for $j = 1, \ldots , m$ and such that $H^0(Z_ j, \mathcal{O}_{Z_ j})$ is a field finite over $k$ for $j = 0, \ldots m$.

Proof. The reduction $D_{red} = (X_ k)_{red} = \sum C_ i$ is connected (Lemma 55.9.4) and proper over $k$. Hence $H^0(D_{red}, \mathcal{O})$ is a field and a finite extension of $k$ by Varieties, Lemma 33.9.3. Thus the result for $Z_0 = D_{red} = (X_ k)_{red}$ is true. Suppose that we have already constructed

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

with $Z_ j = Z_{j - 1} + C_{i_ j}$ for some $i_ j \in \{ 1, \ldots , n\}$ for $j = 1, \ldots , t$ and such that $H^0(Z_ j, \mathcal{O}_{Z_ j})$ is a field finite over $k$ for $j = 0, \ldots , t$. Write $Z_ t = \sum a_ i C_ i$ with $1 \leq a_ i \leq m_ i/d$. If $a_ i = m_ i/d$ for all $i$, then $Z_ t = D$ and the lemma is proved. If not, then $a_ i < m_ i/d$ for some $i$ and it follows that $(Z_ t \cdot Z_ t) < 0$ by Lemma 55.9.7. This means that $(D - Z_ t \cdot Z_ t) > 0$ because $(D \cdot Z_ t) = 0$ by the lemma. Thus we can find an $i$ with $a_ i < m_ i/d$ such that $(C_ i \cdot Z_ t) > 0$. Set $Z_{t + 1} = Z_ t + C_ i$ and $i_{t + 1} = i$. Consider the short exact sequence

$0 \to \mathcal{O}_ X(-Z_ t)|_{C_ i} \to \mathcal{O}_{Z_{t + 1}} \to \mathcal{O}_{Z_ t} \to 0$

of Divisors, Lemma 31.14.3. By our choice of $i$ we see that $\mathcal{O}_ X(-Z_ t)|_{C_ i}$ is an invertible sheaf of negative degree on the proper curve $C_ i$, hence it has no nonzero global sections (Varieties, Lemma 33.44.12). We conclude that $H^0(\mathcal{O}_{Z_{t + 1}}) \subset H^0(\mathcal{O}_{Z_ t})$ is a field (this is clear but also follows from Algebra, Lemma 10.36.18) and a finite extension of $k$. Thus we have extended the sequence. Since the process must stop, for example because $t \leq \sum (m_ i/d - 1)$, this finishes the proof. $\square$

Lemma 55.9.10. In Situation 55.9.3 let $d = \gcd (m_1, \ldots , m_ n)$. Let $D = \sum (m_ i/d) C_ i$ as an effective Cartier divisor on $X$. Then

$1 - g_ C = d [\kappa : k] (1 - g_ D)$

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

Proof. By Lemma 55.9.9 we see that $\kappa$ is a field and a finite extension of $k$. Since also $H^0(C, \mathcal{O}_ C) = K$ we see that the genus of $C$ and $D$ are defined (see Algebraic Curves, Definition 53.8.1) and we have $g_ C = \dim _ K H^1(C, \mathcal{O}_ C)$ and $g_ D = \dim _\kappa H^1(D, \mathcal{O}_ D)$. By Derived Categories of Schemes, Lemma 36.32.2 we have

$1 - g_ C = \chi (C, \mathcal{O}_ C) = \chi (X_ k, \mathcal{O}_{X_ k}) = \dim _ k H^0(X_ k, \mathcal{O}_{X_ k}) - \dim _ k H^1(X_ k, \mathcal{O}_{X_ k})$

We claim that

$\chi (X_ k, \mathcal{O}_{X_ k}) = d \chi (D, \mathcal{O}_ D)$

This will prove the lemma because

$\chi (D, \mathcal{O}_ D) = \dim _ k H^0(D, \mathcal{O}_ D) - \dim _ k H^1(D, \mathcal{O}_ D) = [\kappa : k](1 - g_ D)$

Observe that $X_ k = dD$ as an effective Cartier divisor. To prove the claim we prove by induction on $1 \leq r \leq d$ that $\chi (rD, \mathcal{O}_{rD}) = r \chi (D, \mathcal{O}_ D)$. The base case $r = 1$ is trivial. If $1 \leq r < d$, then we consider the short exact sequence

$0 \to \mathcal{O}_ X(rD)|_ D \to \mathcal{O}_{(r + 1)D} \to \mathcal{O}_{rD} \to 0$

of Divisors, Lemma 31.14.3. By additivity of Euler characteristics (Varieties, Lemma 33.33.2) it suffices to prove that $\chi (D, \mathcal{O}_ X(rD)|_ D) = \chi (D, \mathcal{O}_ D)$. This is true because $\mathcal{O}_ X(rD)|_ D$ is a torsion element of $\mathop{\mathrm{Pic}}\nolimits (D)$ (Lemma 55.9.8) and because the degree of a line bundle is additive (Varieties, Lemma 33.44.7) hence zero for torsion invertible sheaves. $\square$

Lemma 55.9.11. In Situation 55.9.3 given a pair of indices $i, j$ such that $C_ i$ and $C_ j$ are exceptional curves of the first kind and $C_ i \cap C_ j \not= \emptyset$, then $n = 2$, $m_1 = m_2 = 1$, $C_1 \cong \mathbf{P}^1_ k$, $C_2 \cong \mathbf{P}^1_ k$, $C_1$ and $C_2$ meet in a $k$-rational point, and $C$ has genus $0$.

Proof. Choose isomorphisms $C_ i = \mathbf{P}^1_{\kappa _ i}$ and $C_ j = \mathbf{P}^1_{\kappa _ j}$. The scheme $C_ i \cap C_ j$ is a nonempty effective Cartier divisor in both $C_ i$ and $C_ j$. Hence

$(C_ i \cdot C_ j) = \deg (C_ i \cap C_ j) \geq \max ([\kappa _ i: k], [\kappa _ j : k])$

The first equality was shown in the proof of Lemma 55.9.6. On the other hand, the self intersection $(C_ i \cdot C_ i)$ is equal to the degree of $\mathcal{O}_ X(C_ i)$ on $C_ i$ which is $-[\kappa _ i : k]$ as $C_ i$ is an exceptional curve of the first kind. Similarly for $C_ j$. By Lemma 55.9.7

$0 \geq (C_ i + C_ j)^2 = -[\kappa _ i : k] + 2(C_ i \cdot C_ j) - [\kappa _ j : k]$

This implies that $[\kappa _ i : k] = \deg (C_ i \cap C_ j) = [\kappa _ j : k]$ and that we have $(C_ i + C_ j)^2 = 0$. Looking at the lemma again we conclude that $n = 2$, $\{ 1, 2\} = \{ i, j\}$, and $m_1 = m_2$. Moreover, the scheme theoretic intersection $C_ i \cap C_ j$ consists of a single point $p$ with residue field $\kappa$ and $\kappa _ i \to \kappa \leftarrow \kappa _ j$ are isomorphisms. Let $D = C_1 + C_2$ as effective Cartier divisor on $X$. Observe that $D$ is the scheme theoretic union of $C_1$ and $C_2$ (Divisors, Lemma 31.13.10) hence we have a short exact sequence

$0 \to \mathcal{O}_ D \to \mathcal{O}_{C_1} \oplus \mathcal{O}_{C_2} \to \mathcal{O}_ p \to 0$

by Morphisms, Lemma 29.4.6. Since we know the cohomology of $C_ i \cong \mathbf{P}^1_\kappa$ (Cohomology of Schemes, Lemma 30.8.1) we conclude from the long exact cohomology sequence that $H^0(D, \mathcal{O}_ D) = \kappa$ and $H^1(D, \mathcal{O}_ D) = 0$. By Lemma 55.9.10 we conclude

$1 - g_ C = d[\kappa : k](1 - 0)$

where $d = m_1 = m_2$. It follows that $g_ C = 0$ and $d = m_1 = m_2 = 1$ and $\kappa = k$. $\square$

 Observe that it may happen that the field $\kappa _ i = H^0(C_ i, \mathcal{O}_{C_ i})$ is strictly bigger than $k$. In this case every invertible module on $C_ i$ has degree (as defined above) divisible by $[\kappa _ i : k]$.

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