Can an irreducible curve in a surface be the divisor of a function

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 k. So S being projective means that the natural map S\to Spec(k) is a projective morphism, in that it factors through the following closed immersion i:

S\to^{i} \mathbb{P}^n_k\to Spec(k)

Here \mathbb{P}^n_k =Proj k[x_0,x_1,\ldots,x_n] and the map from this to Spec(k) is given by gluing maps from the usual affine cover D_{+}(x_i)\forall x_i\to Spec(k) where

D_+(f)=\{P\in Proj k[x_0,\ldots,x_n]| f\not\in P\} \cong Spec degree 0 homogeneous elements of k[x_0,\ldots,x_n]_{f}

To check : Thus D_{+}(x_i)\cong Spec k[\frac{x_0}{x_i},\ldots, \frac{x_n}{x_i}]

which has a natural map to Spec(k).

Now the ring of functions regular on S is k^* because S 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 f\in k(S)^*, the function field of S, then the divisor of f is div(f)=\sum v_{P}(f) where P runs over all codimension 1 points of S. This is the construction of a principal weil divisor.

In more detail, if S 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 k(S) called v_P.

Coming back to our problem, let C be an irreducible curve. If possible, let it be div(f). Thus we get div(f)=nP (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 n>0 (if n<0, just look at \frac{1}{f}. Of course if n=0, then it is not a curve !)

So this means f\in m_{P,S}, the maximal ideal of the local ring of S at P and in O_{Q,S}, 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 Spec(R).

Claim : R is also integrally closed.

Proof : R=\cap R_M where M runs over maximal ideals of R

(Look at a\in \cap R_M and look at I=\{x\in R|xa\in R\}, this is an ideal of R. If it is all of R, then 1\in I and hence a.1=a\in R. If it is a proper ideal, then let it be inside maximal ideal N. Since a\in R_N, a=b/c for c\not\in N and hence c\not\in I. But ca=b\in R and hence c has to be in I, a contradiction.)

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

Finally we have one more theorem :

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

Thus, our f will be in R since it is in each R_P.

This means that f is regular on each affine open set of S, and hence it is a regular function on S, and hence a constant in k. 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 S=\mathbb{A}^2=Spec k[x,y].

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 k[x,y] is a UFD.

But the Krull dimension of k[x,y] is 2, hence every height one prime looks like (t) where t is an irreducible polynomial in k[x,y].

Let us look at div(x). Since x is irreducible, x\in (t) implies (t)=(x), thus x is a unit in k[x,y]_P for every height one prime ideal P except Q=(x). At Q, x has valuation 1.

Thus div(x)=Q.

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