For any divisor \(D\) on the surface \(X\), we define its cohomology class \(c(D) \in H^1\left(X, \Omega_X\right)\) by using the isomorphism \(\Pic X \cong H^1\left(X, \mathcal{O}_X^*\right)\) of (III, Ex. 4.5) and the sheaf homomorphism \(d \log : \mathcal{O}^* \rightarrow \Omega_X\) (III, Ex. 7.4c). Thus we obtain a group homomorphism \(c: \Pic X \rightarrow H^1\left(X, \Omega_X\right)\). On the other hand, \(H^1(X, \Omega)\) is dual to itself by Serre duality (III, 7.13), so we have a nondegenerate bilinear map \[ \langle\quad, \quad\rangle: H^1(X, \Omega) \times H^1(X, \Omega) \rightarrow k . \]
- Prove that this is compatible with the intersection pairing, in the following sense: for any two divisors \(D, E\) on \(X\), we have \[ \langle c(D), c(E)\rangle=(D . E) \cdot 1 \] in \(k\).
Hint: Reduce to the case where \(D\) and \(E\) are nonsingular curves meeting transversally. Then consider the analogous map \(c: \Pic D \rightarrow H^1\left(D, \Omega_D\right)\), and the fact (III, Ex. 7.4) that \(c(\text{point})\) goes to 1 under the natural isomorphism of \(H^1\left(D, \Omega_D\right)\) with \(k\).
- If \(\operatorname{char} k=0\), use the fact that \(H^1\left(X, \Omega_X\right)\) is a finite-dimensional vector space to show that \(\Num X\) is a finitely generated free abelian group.
Solution
Write \[ V=H^1(X,\Omega_X). \] The Serre-duality pairing on \(V\) is the cup product followed by wedge product and the trace \[ H^1(X,\Omega_X)\times H^1(X,\Omega_X) \longrightarrow H^2(X,\omega_X)\xrightarrow{\operatorname{tr}_X}k. \]
1. If \(j:D\hookrightarrow X\) is a nonsingular curve, then for every\(\alpha\in V\),\[ \langle c(D),\alpha\rangle = \operatorname{tr}_D(j^*\alpha). \]
Exercise III.7.4 identifies the logarithmic class \(c(D)\) of the divisor with its normalized codimension-one cohomology class; this identification is proved on P-AGH374COHOMCLASS, part (d). For that Gysin class, compatibility of Serre duality with the closed immersion \(j\) gives \[ \langle c(D),\alpha\rangle_X = \langle 1,j^*\alpha\rangle_D = \operatorname{tr}_D(j^*\alpha). \] This is the claimed identity.
2. If \(D,E\) are nonsingular curves meeting transversally, then\[ \boxed{\langle c(D),c(E)\rangle=(D\cdot E)\cdot1}. \]
Naturality of \(d\log\) under restriction gives \[ j^*c(E)=c\qty(\OO_X(E)|_D). \] Because \(D\) and \(E\) meet transversally, \[ \OO_X(E)|_D \cong \OO_D\!\left(\sum_{P\in D\cap E}P\right), \] and the number of points in the sum is \(D\cdot E\).
On a nonsingular projective curve, Exercise III.7.4 identifies \(H^1(D,\Omega_D)\) with \(k\) by the trace and sends the logarithmic class of a point to \(1\). Since \(c\) is a group homomorphism, \[ \operatorname{tr}_D\qty(j^*c(E)) = (D\cdot E)\cdot1. \] Step 1 now gives the displayed equality.
3. Every divisor class on \(X\) is a difference of classes of nonsingularvery ample curves, and two such representatives can be chosen to meettransversally.
Fix a very ample divisor \(H\). For an arbitrary divisor \(A\), ampleness implies that for \(n\gg0\) the sheaf \[ \OO_X(A+(n-1)H) \] is globally generated. By P-AGH275AMPLEPROPS, part (d), \[ \OO_X(A+nH) = \OO_X(H)\tensor\OO_X(A+(n-1)H) \] is then very ample. Thus \[ A\sim A_1-A_2 \] with \(A_1\) and \(A_2\) very ample.
Choose general members of the two very ample linear systems. Bertini’s theorem makes them nonsingular. Given finitely many such curves, choosing them successively and generally also makes every pair needed below meet transversally. Replacing a divisor by a linearly equivalent one changes neither its logarithmic class nor its intersection numbers.
4. For arbitrary divisors \(D,E\) on \(X\),\[ \boxed{\langle c(D),c(E)\rangle=(D\cdot E)\cdot1}. \]
By step 3, write, up to linear equivalence, \[ D=D_1-D_2, \qquad E=E_1-E_2, \] where the four curves are nonsingular and the required pairs meet transversally. Both \(c\) and the intersection pairing are additive in each divisor. Bilinearity of the Serre pairing and step 2 therefore give \[ \begin{aligned} \langle c(D),c(E)\rangle &=\sum_{i,j=1}^2(-1)^{i+j} \langle c(D_i),c(E_j)\rangle\\ &=\sum_{i,j=1}^2(-1)^{i+j}(D_i\cdot E_j)\cdot1\\ &=(D\cdot E)\cdot1. \end{aligned} \] This proves part (a).
5. Assume \(\operatorname{char}k=0\). There exist divisors\(E_1,\ldots,E_r\) such that\[ c(E_1),\ldots,c(E_r) \]form a \(k\)-basis of the vector subspace of \(V\) spanned by \(c(\Pic X)\).
The vector space \(V\) is finite-dimensional. Hence the subspace \[ W=\operatorname{span}_k c(\Pic X)\subseteq V \] has finite dimension, say \(r\). Since \(W\) is spanned by elements of the image of \(c\), a basis can be extracted from that spanning set. Choose divisors \(E_1,\ldots,E_r\) representing the corresponding line bundles.
6. The homomorphism\[ \Phi:\Num X\longrightarrow\ZZ^r, \qquad [D]\longmapsto \bigl(D\cdot E_1,\ldots,D\cdot E_r\bigr) \]is injective.
Intersection numbers depend only on numerical equivalence, so \(\Phi\) is well-defined and additive.
Suppose \(\Phi([D])=0\). Step 4 gives \[ \langle c(D),c(E_i)\rangle=0 \qquad(1\le i\le r). \] By step 5, the classes \(c(E_i)\) span every class \(c(E)\) with \(E\in\Div X\). Hence \[ \langle c(D),c(E)\rangle=0 \] for every divisor \(E\). Applying step 4 again yields \[ (D\cdot E)\cdot1=0\in k \] for every \(E\). Since \(\operatorname{char}k=0\), the canonical map \(\ZZ\to k\) is injective, so \[ D\cdot E=0 \] as an integer for every divisor \(E\). Thus \(D\) is numerically equivalent to zero, and \([D]=0\) in \(\Num X\). Therefore \(\Phi\) is injective.
7. The group \(\Num X\) is finitely generated and free abelian.
By step 6, \(\Num X\) is isomorphic to a subgroup of the finitely generated free abelian group \(\ZZ^r\). Every subgroup of a finitely generated free abelian group is itself finitely generated free abelian. Hence \[ \boxed{\Num X\text{ is a finitely generated free abelian group}.} \] This proves part (b).
8. Q.E.D.
Steps <1>1–<1>4 prove part (a), and steps <1>5–<1>7 prove part (b).