Let be a smooth, projective surface over the algebraically closed field . Previous posts have set up an intersection theory

on with very convenient formal properties. We also described a historically important use of this machinery: the Weil bound on points on a smooth curve over a finite field. The purpose of this post is to prove an entirely numerical criterion for ampleness of a line bundle on a surface, due to Nakai and Moishezon.

Let be a very ample divisor on . Then we have:

- for all curves (i.e., strictly effective divisors) on . In fact, if defines an imbedding , then the degree of under this imbedding is .
- As a special case of this, . In fact, must be effective.

Since a power of an ample divisor is very ample, the same is true for an ample divisor.

The purpose of this post is to prove the converse:

Theorem 1 (Nakai-Moishezon)Let be a smooth projective surface as above. If is a divisor on (not necessarily effective!) satisfying and for all curves on , then is ample. In particular, ampleness depends only on the numerical equivalence class of .

Once again, the source for this material is Hartshorne’s *Algebraic geometry. *The goal is to get to some computations and examples as soon as possible.

The strategy behind the proof of the Nakai-Moisezhon criterion is a three-step process: first to show that some multiple of is an effective divisor, second to show that some multiple of is generated by global sections, and third to analyze the map to projective space that defines.

**1. is effective, **

The first step is to show that for is an effective divisor. In order to do that, we’ll use the Riemann-Roch theorem, in the form for surfaces

Since , this expression tends to as . It follows by Serre duality, as we saw in the proof of the Hodge index theorem, that

where

In particular, for , either or is effective; but since intersected with any ample divisor is positive, it follows that can’t be effective for . Thus

and is effective for . (This is the only place where we needed to use that .)

**2. is generated by global sections, **

Replacing by a large multiple , we may assume that is thus effective, and is associated to a curve on the surface . We want to show that some power of the associated line bundle is generated by global sections.

We have a global section of which is nonzero away from the curve (namely, “”) and an exact sequence

We will show that, after tensoring with a high power of , we get a surjection

and that is generated by global sections. Together, these will prove the claim of this section.

First of all, is ample. It suffices for this to check that for each irreducible, reduced component of , that is ample. It suffices to check this after pulling back to the normalization of . But we know that

which proves ampleness. In general, we use the formula

valid for a reduced, irreducible (not necessarily smooth!) curve on with normalization ; this is evident when the divisor doesn’t intersect the singularities of and follows in general by additivity.

Anyway, we conclude from this that is ample, so it suffices to show that (1) is a surjection. So let’s prove that.

Applying cohomology to the short exact sequence of sheaves

we get an exact sequence

Since the last group is zero for by ampleness, it follows that we have a sequence of surjections for

which must eventually become isomorphisms by finite-dimensionality. Using the long exact sequence again, we find that induces a surjection in for .

This proves the claim of this section.

**3. Completion of the proof**

In the previous sections, we considered a divisor on a surface satisfying the conditions , for all curves . We showed that, at least after replacing by some multiple, that could be assumed effective and, even better, generated by its global sections. Thus defines a map

for some , with the property that is the line bundle associated to .

Unfortunately, this map need not be a closed immersion. But we do know that for any curve on , or is ample. In particular, it is nontrivial, so no curve is mapped to a point under . It follows that the fibers of are finite; Zariski’s main theorem implies that is in fact a finite morphism.

Now we use the following fact, applied to and on , to conclude.

Lemma 2A finite morphism preserves ampleness. In other words, if is a finite morphism of finite type -schemes, and is a line bundle on which is ample, then is ample.

*Proof:* Let’s check this in the case that are proper (which is all that we need). In this case, we need to show that for any coherent sheaf on ,

Since is finite, it is equivalent to show that

But we have an isomorphism

from the projection formula, so that

using ampleness of on .

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