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The exercise is stated as follows:

Let $X$ be a projective smooth surface over an algebraically closed field $k$. If $H$ is an ample divisor on $X$, and $d \in \mathbf{Z}$, show that the set of effective divisors $D$ with $D . H=d$, modulo numerical equivalence, is a finite set. [Hint: Use the adjunction formula, the fact that $p_{a}$ of an irreducible curve is $\geqslant 0$, and the fact that the intersection pairing is negative definite on $H^{\perp}$ in Num $\left.X .\right]$

Fix a divisor $D$ satisfies the condition in the exercise, then my idea is to consider $D'\in div(X)$ such that: $D'.H=0$ and $D'+D$ is effective, then $D'\in H^{\perp}$. So it's equivalent to prove such $D'$ is finite. But then I don't know hot to apply the hint. To apply the adjunction formula, maybe I need to find a nonsingular curve? But D' is not effective, so what does it have to do with nonsingular curve? Could you provide some help? Thanks!

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  • $\begingroup$ Did you ever figure this one out? I've been kicking this around for two months now since you pointed out my answer was insufficient and it's eating at me. $\endgroup$ Commented Jul 30, 2022 at 3:55
  • $\begingroup$ I don’t know how to use this hint, but I think one could prove it along the lines of, working mod numerical equivalence, with $\mathbb Q$ divisors, the ample divisors form a cone, so the effective divisors are contained in the dual cone, so for any fixed ample divisor $H$, the intersection of the $H\cdot \_ = d$ affine hyperplane with the effective divisor lattice is contained in the finite volume slice of this hyperplane intersect the effective $\mathbb Q$ cone. Thus by lattice point considerations, this set is finite. $\endgroup$
    – Chris H
    Commented Aug 1, 2022 at 12:11
  • $\begingroup$ @ChrisH that sounds both interesting and promising - how do you know that the slice is finite volume, though? $\endgroup$ Commented Aug 2, 2022 at 1:59
  • $\begingroup$ Yea I think my approach is actually too simplistic, I don't think the slice ends up being finite volume. I think the finiteness doesn't follow from purely euclidean arguments, and one probably needs to use the canonical divisor to get the needed extra inequalities to get boundedness, but I'll keep thinking about it. $\endgroup$
    – Chris H
    Commented Aug 2, 2022 at 2:59
  • $\begingroup$ @HankScorpio I failed, and I also hope someone can give a clear answer... $\endgroup$
    – Richard
    Commented Aug 2, 2022 at 13:14

1 Answer 1

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The key is to use the adjunction formula on the prime divisors making up $D$ (you can think of this because Hartshorne specifically mentions irreducible curves, and attempts to adjust things so that you get a nice linear system by varying certain coefficients will almost certainly disrupt your control of the numerical quantities in the problem). Then the finite-dimensionality plus the inequalities and negative-definiteness give you the answer.

If $D=\sum_{i=1}^n b_iD_i$ for $D_i$ distinct prime divisors, adjunction says $2p_a(D_i) = D_i^2+D_i.K+2 \geq 0$, or $D_i^2 \geq -D_i.K-2$. Therefore $$ D^2 \geq \sum_{i=1}^n b_i^2 D_i^2 \geq \sum_{i=1}^n -b_i^2(D_i.K+2)$$ as $D_i.D_j\geq 0$ since they're distinct irreducible curves. Then since $d=D.H \geq b_i$ and $\sum_{i=1}^n b_i^2 \leq d^2$, $$D^2 \geq -2d^2 + \sum_{i=1}^n -b_idD_i.K = -dD.K - 2d^2.$$ So if $D,E$ are two effective divisors with $D.H=E.H=d$, then $$(D-E)^2 \geq (D-E).(-dK-2E) - 4d^2 - 2dE.K-2E^2.$$ Now, as $\operatorname{Num}(X)$ is finitely generated and the intersection form is negative definite on $H^\perp$, pairs of effective divisors $D,E$ with $D.H=E.H=d$ become lattice points in the finite-dimensional vector space $H^\perp\otimes\Bbb R$ satisfying $(D-E)^2 \geq (D-E).(-dK-2E) - 4d^2 - 2dE.K-2E^2$ by mapping $(D,E)\mapsto(D-E)$. By negative-definiteness, the set described by this inequality is an ellipsoid containing finitely many lattice points, and we're done.

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