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 2 A 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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