# 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 $div(x)=Q$.