(Black + Gray) and (Black + White) form the whole space. That is, in an irreducible space, any two non-empty open sets intersect

**Question** : Let S be a projective surface and C, an irreducible curve on it. Show that C can’t be the divisor of a function.

**Answer** : A surface is a two dimensional scheme. I guess we assume S is defined over a field . So S being projective means that the natural map is a projective morphism, in that it factors through the following closed immersion :

Here and the map from this to is given by gluing maps from the usual affine cover where

degree 0 homogeneous elements of

To check : Thus

which has a natural map to .

Now the ring of functions regular on S is because is projective. (I guess that for this you need that S is projective over a field, or atleast a nice enough affine scheme below, like an integral domain ??)

What is the divisor of a function ?

If , the function field of , then the divisor of is where P runs over all codimension 1 points of . This is the construction of a principal weil divisor.

In more detail, if is a *nice enough scheme* (noetherian, separated, all local rings of dimension one are regular), then at any codimension 1 point P (which will correspond to a height one prime ideal of the ring of any affine open subset containing it), the local ring will be a noetherian regular one dimensional local ring and hence a DVR. This DVR defines a valuation on called .

Coming back to our problem, let be an irreducible curve. If possible, let it be . Thus we get (for an irreducible curve can’t have more than 1 generic point in it. Note that since S is two dimensional, closure of any codimension 1 point is 1-dimensional)

We will show that this is not possible.

Note that we can assume wlog that (if , just look at . Of course if , then it is not a curve !)

So this means , the maximal ideal of the local ring of S at P and in , the local ring of S at every other codimension 1 point Q.

**Probably a stupid question : Is an integral projective scheme normal ? Ie, are the local rings integrally closed ?**

Let us assume for a moment that our scheme S is normal. That is, all the local rings (whether at codimension 1 points or not are integrally closed in the function field). Let us look at any affine open set .

Claim : is also integrally closed.

Proof : where M runs over maximal ideals of

(Look at and look at , this is an ideal of R. If it is all of R, then and hence . If it is a proper ideal, then let it be inside maximal ideal . Since , for and hence . But and hence has to be in , a contradiction.)

We have assumed every local ring is normal, so are all integrally closed in . If such that it is integral over , then it is clearly integral over each (a bigger ring than ) and hence because of normality, in each . Thus .

Finally we have one more theorem :

**Thm : If R is a noetherian integrally closed domain, then where P runs over height one prime ideals only.**

Thus, our will be in since it is in each .

This means that is regular on each affine open set of S, and hence it is a regular function on S, and hence a constant in . In which case, it can’t have positive valuation at any codimension 1 point (for it is a unit in every local ring which are all k-algebras)

Hence we are done, modulo the question about whether S is a normal surface or not.

**Counter example**

If we drop the assumption that S is projective, then this is false. For example, take .

What are the height one primes here ?

Well every prime ideal has an irreducible element in it. And the ideal generated by any irreducible element is a prime ideal because is a UFD.

But the Krull dimension of is 2, hence every height one prime looks like where is an irreducible polynomial in .

Let us look at . Since is irreducible, implies , thus is a unit in for every height one prime ideal except . At , x has valuation 1.

Thus .