Let \(X\) be a surface. Recall that we have defined an algebraic family of effective divisors on \(X\), parametrized by a nonsingular curve \(T\), to be an effective Cartier divisor \(D\) on \(X \times T\), flat over \(T\) (III, 9.8.5). In this case, for any two closed points \(0,1 \in T\), we say the corresponding divisors \(D_0, D_1\) on \(X\) are prealgebraically equivalent.
Two arbitrary divisors are prealgebraically equivalent if they are differences of prealgebraically equivalent effective divisors. Two divisors \(D, D^{\prime}\) are algebraically equivalent if there is a finite sequence \(D=D_0, D_1, \ldots, D_n=D^{\prime}\) with \(D_i\) and \(D_{i+1}\) prealgebraically equivalent for each \(i\).
Show that the divisors algebraically equivalent to 0 form a subgroup of \(\Div X\).
Show that linearly equivalent divisors are algebraically equivalent.
Hint: If \((f)\) is a principal divisor on \(X\), consider the principal divisor \((t f-u)\) on \(X \times \PP^1\), where \(t, u\) are the homogeneous coordinates on \(\PP^1\).
- Show that algebraically equivalent divisors are numerically equivalent.
Hint: Use (III, 9.9) to show that for any very ample \(H\), if \(D\) and \(D^{\prime}\) are algebraically equivalent, then \(D . H=D^{\prime} . H\).
Note. The theorem of Néron and Severi states that the group of divisors modulo algebraic equivalence, called the Néron-Severi group, is a finitely generated abelian group. Over \(\CC\) this can be proved easily by transcendental methods (App. B, §5) or as in (Ex. 1.8) below. Over a field of arbitrary characteristic, see Lang and Néron [1] for a proof, and Hartshorne [6] for further discussion. Since \(\Num X\) is a quotient of the Néron-Severi group, it is also finitely generated, and hence free, since it is torsion-free by construction.
Solution
Write \(D\sim_{\mathrm{pa}}D'\) for prealgebraic equivalence and \(D\sim_{\mathrm{alg}}D'\) for algebraic equivalence.
1. Prealgebraic equivalence is symmetric, is preserved by negation, and ispreserved by translating both divisors by the same divisor.
For effective divisors, symmetry follows by interchanging the two parameter points in the same flat family.
Suppose arbitrary divisors \(D,D'\) are prealgebraically equivalent. By definition there are effective divisors \(A,B,A',B'\) such that \[ D=A-B, \qquad D'=A'-B', \] with \[ A\sim_{\mathrm{pa}}A', \qquad B\sim_{\mathrm{pa}}B'. \] Then \[ -D=B-A, \qquad -D'=B'-A', \] so the same definition gives \[ -D\sim_{\mathrm{pa}}-D'. \]
Let \(E=E_+-E_-\) with \(E_+,E_-\) effective. If a flat family of effective Cartier divisors connects \(A\) to \(A'\), adding the constant divisor \(E_+\times T\) gives a flat family connecting \(A+E_+\) to \(A'+E_+\); locally its equation is the product of the family equation with the fixed Cartier equation, and the fibrewise Cartier criterion of Hartshorne III.9.8.5 applies. The same statement holds for \(B+E_-\) and \(B'+E_-\). Hence \[ D+E=(A+E_+)-(B+E_-) \sim_{\mathrm{pa}} (A'+E_+)-(B'+E_-)=D'+E. \]
2. The set\[ \boxed{\{D\in\Div X:D\sim_{\mathrm{alg}}0\}} \]is a subgroup of \(\Div X\).
It contains \(0\). Suppose \[ 0=D_0\sim_{\mathrm{pa}}D_1\sim_{\mathrm{pa}}\cdots \sim_{\mathrm{pa}}D_r=D. \] Negating each step and using step 1 gives an algebraic-equivalence chain from \(0\) to \(-D\). Thus the set is closed under inverses.
Now suppose \(D\sim_{\mathrm{alg}}0\) and \(E\sim_{\mathrm{alg}}0\). Choose a chain from \(0\) to \(D\) and a chain from \(0\) to \(E\). By the translation property in step 1, translating the second chain by \(D\) gives a chain from \(D\) to \(D+E\). Concatenating the two chains gives \[ 0\sim_{\mathrm{alg}}D+E. \] Hence the displayed set is closed under addition and inverses, proving part (a).
3. If \(D\) and \(D'\) are linearly equivalent, then\[ \boxed{D\sim_{\mathrm{pa}}D'}, \]and therefore \(D\sim_{\mathrm{alg}}D'\).
Choose an effective divisor \(G\) whose coefficients are large enough that \[ A=D+G, \qquad A'=D'+G \] are both effective. Since \(D\sim D'\), the effective divisors \(A,A'\) are linearly equivalent. Let \(\mcl\) be their common invertible sheaf and let \(s_0,s_1\in H^0(X,\mcl)\) be sections whose zero divisors are \(A,A'\). If the two sections are proportional then \(A=A'\) and there is nothing to prove. Otherwise, on \(X\times\PP^1\) the universal pencil section \[ t s_0+u s_1 \] of \[ p_X^*\mcl\tensor p_{\PP^1}^*\OO_{\PP^1}(1) \] is nonzero on every fibre over \(\PP^1\). Its zero scheme is therefore an effective Cartier divisor flat over \(\PP^1\) by Hartshorne III.9.8.5, and its fibres over \([1:0]\) and \([0:1]\) are \(A\) and \(A'\). Thus \[ A\sim_{\mathrm{pa}}A'. \] The constant family shows \(G\sim_{\mathrm{pa}}G\), so the definition for arbitrary divisors gives \[ D=A-G\sim_{\mathrm{pa}}A'-G=D'. \] This pencil is the homogeneous form of the source hint using \((tf-u)\). Thus part (b) follows.
4. Let \(H\) be a very ample divisor. If effective divisors \(A,A'\) areprealgebraically equivalent, then\[ A\cdot H=A'\cdot H. \]
Let \(\mathcal A\subseteq X\times T\) be a flat family of effective Cartier divisors with fibres \(A\) and \(A'\) at two closed points of the nonsingular curve \(T\). The divisor \(H\) embeds \[ X\hookrightarrow\PP^N, \] and hence gives a projective flat family \[ \mathcal A\hookrightarrow\PP^N\times T. \] By Hartshorne III.9.9, recorded in T-COHFLATCHI, all fibres have the same Hilbert polynomial. In particular, they have the same degree in \(\PP^N\). For a divisor on the surface \(X\), this degree is its intersection number with \(H\). Therefore \[ A\cdot H=A'\cdot H. \]
5. If arbitrary divisors \(D,D'\) are algebraically equivalent, then\[ D\cdot H=D'\cdot H \]for every very ample divisor \(H\).
First suppose \(D\sim_{\mathrm{pa}}D'\). Write \[ D=A-B, \qquad D'=A'-B' \] with \(A\sim_{\mathrm{pa}}A'\) and \(B\sim_{\mathrm{pa}}B'\) effective. Step 4 and bilinearity give \[ D\cdot H =A\cdot H-B\cdot H =A'\cdot H-B'\cdot H =D'\cdot H. \] If \(D\sim_{\mathrm{alg}}D'\), apply this equality to each consecutive pair in an algebraic-equivalence chain and use transitivity of equality.
6. Every divisor \(E\) on \(X\) is a difference of two very ample divisors.
Fix a very ample divisor \(H_0\). Since \(\OO_X(H_0)\) is ample, for all sufficiently large \(n\) the sheaf \[ \OO_X(E+(n-1)H_0) \] is generated by global sections. Part (d) of P-AGH275AMPLEPROPS then shows that \[ \OO_X(E+nH_0) = \OO_X(H_0)\tensor\OO_X(E+(n-1)H_0) \] is very ample. The divisor \(nH_0\) is also very ample. Hence \[ E=(E+nH_0)-nH_0 \] is a difference of very ample divisors.
7. Algebraic equivalence implies numerical equivalence:\[ \boxed{D\sim_{\mathrm{alg}}D'\Longrightarrow D\equiv_{\mathrm{num}}D'}. \]
Let \(E\) be any divisor. By step 6, write \[ E=H_1-H_2 \] with \(H_1,H_2\) very ample. Step 5 gives \[ D\cdot H_i=D'\cdot H_i \qquad(i=1,2). \] Therefore \[ \begin{aligned} D\cdot E &=D\cdot H_1-D\cdot H_2\\ &=D'\cdot H_1-D'\cdot H_2\\ &=D'\cdot E. \end{aligned} \] Since this holds for every divisor \(E\), the divisors \(D,D'\) are numerically equivalent. This proves part (c).
8. Q.E.D.
Step 2 proves part (a), step 3 proves part (b), and steps <1>4–<1>7 prove part (c).