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For a commutative ring $A$, let $R(A)$ be the set of regular elements in $A$ and $\textrm{Frac}(A)=R(A)^{-1}A$. It is known that if $A$ is Noetherian local with $\textrm{dim}(A)=1$, then $\textrm{length}(A/fA)<\infty$ for any $f\in R(A)$. Then we have a map $R(A) \to \mathbb{Z}$ and it can be extended to a group homomorphism $\textrm{Frac}(A)^{*}\to \mathbb{Z}$. Moreover, we have a map $\textrm{mult}_A:\textrm{Frac}(A)^{*}/A^{*}\to \mathbb{Z}$.

Let $X$ be a locally Noetherian scheme. Let $U$ be an open dense subset of $X$ (i.e. $\bar{U}=X$) and $D$ be a Cartier divisor on $X$ such that $D|_{U}=0$.

Claim 1: Any $x\in X$ of codimension 1 such that $\textrm{mult}_{\mathcal{O}_{X,x}}(D_x)\neq 0$ is a generic point of $X\setminus U$. Claim 2: Any affine open subset of $X$, there are only a finite number of points $x$ of codimension 1 such that $\textrm{mult}_{\mathcal{O}_{X,x}}(D_x)\neq 0$.

The author of my reference states Claim 2 comes from Claim 1, but I cannot understand how it works. Actually, I cannot prove the Claim 1. How can I use the conditions codimension 1 and density?

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  • $\begingroup$ @Matt For a codimension 1 point $x$, the dimension of local ring that corresponds to $x$ is 1. We don't have to assume $dim(X)=2$, do we? $\endgroup$ – User0829 Jul 16 '13 at 0:27
  • $\begingroup$ Sorry. Of course. I had something backwards in my head. $\endgroup$ – Matt Jul 16 '13 at 14:08
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Claim 1: If the multiplicity of $D$ at $x$ is non-zero, then $x\in X\setminus U$. As $U$ is dense, any point of $X\setminus U$ has codimension $>0$ in $X$. So any point of $X\setminus U$ of codimension $1$ in $X$ is a generic point of $X\setminus U$. This is purely topological.

Claim 2: can suppose $X$ is affine, hence noetherian. Then $X\setminus U$ has only finitely many generic points.

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