12
$\begingroup$

This is a pretty basic question. Hartshorne defines "intersection multiplicity" for any two divisors on a surface. Fulton has an impressive framework of generalizing this in his book (my understanding of which is scant). But for whatever reason, in arithmetic texts one often sees intersection multiplicities defined only between a (general) divisor and a vertical divisor. What's really going on? Does this just make it easier to explain things, or is there an actual impediment to defining an intersection number between general divisors in the arithmetic setting? And if so, how does Fulton address it?

(just to be clear, when I say the "arithmetic setting" I mean in a scheme which has relative dimension 1 over a regular scheme of pure dimension 1)

Improve this question
$\endgroup$
1
  • $\begingroup$ Let O be the ring of int. of a numb field K and write S=Spec O. Let X be an integral regular flat projective S-scheme. You mention the intersection of a vertical divisor with a divisor. Now, take a horizontal divisor D on X. Then D surjects onto S. So D lies above all the primes of S. But now add formally the infinite points s on S (the embeddings of K into C). Then D should also lie above these points in a sense. When you intersect a horizontal divisor with D you expect to have some contribution at infinity. See Chapter 1.1 and 1.2 of de Jong's thesis math.leidenuniv.nl/~rdejong $\endgroup$
    – Ariyan Javanpeykar
    Commented Dec 11, 2010 at 16:40

4 Answers 4

Reset to default
13
$\begingroup$

When you intersect two divisor, you obtain a algebraic cycle of codimension 2. For a smooth surface, this is a collection of points that is well-defined up to rational equivalence. Now, if the surface is also proper ('compact'), you can count these points i.e. the number of points is the same for all representatives in the equivalence class.

The problem with arithmetical surfaces is that they are not compact! So you cannot apply the standard theory here. As far as I know, one usually tries to compactify arithmetic schemes using infinite points and Arakelov geometry. If one wants to avoid these matters, he has to put a restriction on divisors.

Improve this answer
$\endgroup$
6
$\begingroup$

This is more like a long comment on Mikhail's answer, which I beg to differ slightly from: Over a regular surface (all local rings are regular, so you don't need to work over fields) and if your divisors intersects properly (in other words, in codimension $2$), one can always define intersection multiplicity. This can be achieved by Serre's intersection formula. For a concise explanatio, see Section 3 of this survey paper by Henri Gillet.

The troubles arise when one does not have proper intersection (say if you intersect a divisor with itself). Traditionally this was resolved in two ways: Using moving lemma (to move the divisors so they intersect properly) or Fulton's approach via deformation to normal cone. Unfortunately both approaches need $X$ to be quasi-projective over a field.

If $X$ is not regular, one can still define intersection numbers, provided that one of the divisors have the so-called "exceptional support" (see Section 12,13 of Lipman's paper "Rational singularity with applications...")

Finally, I just want to mention a recent paper by Roberts-Spiroff on an algebraic approach to intersection of divisors (the authors prove the commutativity of the product without using Fulton's machinery, and so it should work more generally).

Improve this answer
$\endgroup$
4
  • $\begingroup$ These are interesting comments. Yet they do not seem to cancel my answer: if the divisors do not intersect properly, you have to move them somehow, and if the surface is not 'compact', moving can easily change the multiplicity. $\endgroup$
    – Mikhail Bondarko
    Commented Apr 17, 2010 at 21:05
  • $\begingroup$ Dear Mikhail, I did not mean to say what you wrote is not correct (sorry if my language was misleading). There are 2 differences, and perhaps they are merely about language: if we assume the divisors intersect in codimension 2, then one can always count without the compactness. Also, one can work over regular, not only smooth surfaces. $\endgroup$
    – Hailong Dao
    Commented Apr 17, 2010 at 23:09
  • $\begingroup$ I certainly do not (and did not) object against your answer; I just wanted to clarify its relation with my one. $\endgroup$
    – Mikhail Bondarko
    Commented Apr 18, 2010 at 10:52
  • $\begingroup$ @Mikhail: thanks for your response and patience. $\endgroup$
    – Hailong Dao
    Commented Apr 18, 2010 at 17:28
1
$\begingroup$

Also, I strongly suspect that arithmetic varieties are not really special in this matter. You would probably get the same restrictions in the 'geometric' situation i.e. when your surface is smooth proper over a smooth non-proper curve. The main difference is that in the arithmetic situation it is much more difficult to compactify the surface in an appropriate sense.

Improve this answer
$\endgroup$
0
$\begingroup$

I have a question in relation to the above discussion. If there exists a dominant morphism between two fibered, proper surfaces, then does one have a projection formula without vertical divisors?

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .