# Multiplicity of Cartier divisor on locally noetherian scheme is only non-zero at generic point

I'm following chapter 7 in Qing Liu's book 'Algebraic Geometry and Arithmetic Curves' about 'Divisors and applications to curves'.

My question concerns Definition 1.27:

Let $A$ be a Noetherian local ring of dimension 1. For any regular element $\ f \in A, \text{length}_A(A/fA)$ is finite. The map $f \mapsto \text{length}_A(A/fA)$ extends to a group homomorphism $\text{Frac}(A)^\times \rightarrow \mathbb{Z}$. Moreover, its kernel contains the invertible elements of $A$. Thus we obtain a group homomorphism $\text{mult}_A:\text{Frac}(A)^\times/A^\times \rightarrow \mathbb{Z}$.

This map will be used to define $\text{mult}_x(D)$ of a Cartier Divisor $D$ at a point $x \in X$:

Let $X$ be a locally Noetherian scheme. Let $D \in \text{Div}(X)$ be a Cartier divisor. For any point $x \in X$ of codimension 1, the stalk of $D$ at $x$ belongs to $$(\mathcal{K}_X^\times/\mathcal{O}_X^\times)_x = \text{Frac}(\mathcal{O}_{X,x})^\times/\mathcal{O}_{X,x}^\times,$$ thus we can define $$\text{mult}_x(D) := \text{mult}_{\mathcal{O}_{X,x}}(D).$$ Now let $U$ be an open everywhere dense (does this simply mean dense?) subset of $X$ such that $D_U = 0.$ Then: Any $x \in X$ of codimension 1 such that $\text{mult}_x(D) \neq 0$ is a generic point of $X-U$.

My Question: Why does the last statement hold?

The result seems to be very important and fundemanetal (hence the question), since following the statement, a Cartier divisor is determined on the generic points of $X$. Further, for assigning Weil divisors to Cartier divisors, we need that there are only finitely many prime divisors at which the Cartier divisor is non-zero - which is (with the above) a consequence if $X$ is assumed to be Noetherian.

I would be very grateful for any kind of help.

• I don't know the exact answer to your question right away, but I note one red flag: the highlighted text talks about generic points of $X-U$, but you talk about generic points of $X$. These are very different things! Jul 14 '16 at 14:51

Note that $D_U=0$ shows that multiplicity non-zero in particular implies $x \in X-U$, hence the closure of $x$ is contained in $X-U$ and has the same dimension by assumption. ($X-U$ having codimension at least $1$ is the everywhere dense condition)

Here is a rigorous proof using your definitions.

Let $x$ be a point with multiplicity non-zero. As I stated above, $D_U=0$ implies $x \in X-U$.

Let $y \in X-U$ with $x \in \overline {\{y\}}$. We have to show $x=y$, which is equivalent to $\overline {\{x\}}=\overline {\{y\}}$ (This equivalence essentially uses, that we work with a scheme. On arbitrary topological spaces, this is of course wrong).

Assume this does not hold, then we have $\overline {\{x\}} \subsetneq \overline {\{y\}}$. By the co-dimension $1$ assumption for $x$, we get that $\overline {\{y\}}$ is a maximal irreducible subset of $X$. Note that $\overline {\{y\}} \subset X-U$, since $y \in X-U$.

Now let $\mathcal Z$ be the set of maximal irreducible closed subsets of $X$. I claim that $S = \bigcup\limits_{Z \in \mathcal Z, Z \neq \overline {\{y\}}} Z$ is a closed subset. In particular it is a proper closed subset (it does not contain $y$), which contains $U$. This is a contradiction to the assumption that $U$ is dense.

To proof my claim, I will proof the following lemma:

Let $X$ be a locally noetherian scheme and $Z_i, i \in I$ a collection of maximal irreducible closed subsets. Then their union is a closed set. (Note that infinite unions are not closed in general)

Proof: Let $U$ be an open affine subset. By the locally noetherian assumption, $U$ is noetherian. For any $i$ $U \cap Z_i$ is an irreducible subset of $U$. By the noetherian property, the union $\bigcup_{i \in I} U \cap Z_i$ is actually a finite union, i.e. closed. Since we can cover $X$ be such $U$'s and we can test being closed on an open cover, the result follows.

You should note that the proof becomes essentially easier, when one assumes $X$ to be noetherian, because we have finitely many irreducible components then and thus can immediately derive that $\overline {\{x\}}$ is an irreducible component of $X-U$, which is equivalent to be a generic point.

• So, $U \subset X$ everywhere dense $\Leftrightarrow\ \text{codim}_X(X-U) \geq 1$? Jul 14 '16 at 16:05
• Only the forward direction is true (Think of the union of a curve and an isolated closed point. Take the open subset to be the curve). If all irreducible components have the same dimension, then the equivalence holds.
– MooS
Jul 14 '16 at 16:12
• Then, would you mind telling me what the definition of everywhere dense is? What is the difference to just dense? Because this would imply that for any $x \in X-U$ we have $\text{codim}_X \{x\}^- \geq 1$. This should further imply (how?) that if the codimension of such $x$ is exactly 1, that $\{x\}^- = X-U$, right? Jul 14 '16 at 16:15
• It is the same as dense. Meaning that $U$ does not leave out an irreducible component.
– MooS
Jul 14 '16 at 16:17
• Do we have to worry about the underlying space being not necessarily noetherian?
– Hoot
Jul 14 '16 at 16:55