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I have some elementary questions related to divisors. I will state them below.

Question 1: Let $X$ a Noetherian scheme be regular in codimension 1 (every local ring of dimension $1$ is regular). Then in Ravi FOAG's (Jun 2013 version) he defines the following. Let $\mathcal{L}$ be an invertible sheaf, $s$ a rational section. Then $\operatorname{div}(s) := \sum_Y \nu_Y(s) [Y]$ where the sum is over all irreducible codimension $1$ subsets of $X$. What is this $\nu_Y(s)$? Do I simply take any $y\in Y$ and calculate the valuation of $s$ in $\mathcal{O}_{X,y}$?

<p><strong>Question 2:</strong> Suppose $A$ is a domain and $x \in A$ such that $(x)$ a prime ideal. Then is the divisor $[(x)]$ principal, equal to $\operatorname{div}(x)$? Some might say this is pretty much the definition, but I have $\operatorname{div} (x) := \sum_{ Y} \nu_Y(x)[Y]$ where $Y$ is a codimension $1$ closed subset of $\operatorname{Spec} A$. It is not clear to me from this definition that $\operatorname{div}(x) = [(x)]$.</p>
  • For Question 1, $\nu_Y$ is the valuation on the local ring $O_{\eta,X}$ where $\eta$ is the generic point of $Y$. (That local ring has dimension 1, so it is a DVR.) For Question 2: yes! If $Y$ is different from ${x=0}$, then $\nu_Y(x)=0$, so $[Y]$ doesn't appear in the sum. –  Oct 17 '13 at 11:08
  • Dear @Asal: what you write in your comment about Question 1 does not solve the OP's question. – Georges Elencwajg Oct 17 '13 at 11:31
  • Dear @GeorgesElencwajg: can you elaborate a little bit? He asks "What is this $\nu_Y(s)$?" I thought the question was about the definition of $\nu_Y$, but maybe I am incorrect. –  Oct 17 '13 at 11:41
  • Dear @Asal, sincere apologies for having been too concise and cryptic. The problem is that $v_Y(s)$ does not make sense because $s$ is not a rational function but a section of a line bundle. I'll write an answer developing this difficulty and its solution, since comments tend to make me be too enigmatic! – Georges Elencwajg Oct 17 '13 at 11:53
  • Dear @GeorgesElencwajg: I see your point now, and I agree it needs a little elucidation. Thanks for keeping me on my toes! –  Oct 17 '13 at 11:57
  • @user38268: I am indeed saying that; this is almost the definition of "codimension 1". –  Oct 17 '13 at 12:02
  • Dear user 38268 : for the sake clarity, in question 2 you should repeat the assumption that $A$ is noetherian and that $Spec(A)$ regular in codimension $1$. – Georges Elencwajg Oct 17 '13 at 12:39
  • @AsalBeagDubh I'm very stupid. $(x)$ is already a prime ideal and so if $0 \subset (x) \subset \mathfrak{p}$ where $\mathfrak{p}$ is our codimension $1$ prime we get a contradiction unless $(x) = \mathfrak{p}$. Facepalm...... –  Oct 17 '13 at 13:14
  • Dear @GeorgesElencwajg For (a), I don't want to assume that $A$ is regular in codimension $1$. This is because I am actually trying to show that if $(x)$ is prime then the map $\langle (x) \rangle \to \operatorname{Cl} \operatorname{Spec} A$ is the zero map. –  Oct 17 '13 at 13:16
  • Dear user38268: beware that if you do not have regularity in codimension one, the very notions of $v_Y$ and $div(x)$ become more problematic. – Georges Elencwajg Oct 17 '13 at 13:20
  • @GeorgesElencwajg Right. But the reason I ask question (2) is because I want to know why the class group of a Noetherian domain is zero, given the class group of $A_x$ is zero for some $x\in A$. That's why I need to know that the map $\langle (x) \rangle \to \text{Cl} \text{Spec} A$ is zero. –  Oct 17 '13 at 13:21
  • Dear user38268, may I be so bold as to encourage you to use your real name as username? I would find it much more friendly and I am sure many here feel the same. – Georges Elencwajg Oct 18 '13 at 07:09
  • Dear user38268: oh, I'm very glad to hear such good news! But still, wouldn't you like to use your real name or, at least, a pseudonym which is friendlier and easier to remember than user38268? Although I much hope you will adopt my suggestion, it goes without saying that I will strictly respect your decision and call you whatever you decide. – Georges Elencwajg Oct 18 '13 at 10:25
  • Dear @Grothendieck: with pleasure! – Georges Elencwajg Oct 18 '13 at 10:43
  • @GeorgesElencwajg I posted a new question here, would you like to answer it? Thanks. –  Oct 18 '13 at 13:19
  • Dear @Grothendieck, Martin has answered your question and I'll leave it at that since I don't feel I can give a very different answer. – Georges Elencwajg Oct 18 '13 at 13:49

1 Answers1

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Question 1
There is a subtle trap here: we certainly have a group morphism $v_Y:K(X)^*\to \mathbb Z $ defined on rational functions but $s$ is not a rational function and the notation $v_Y(s)$ does not a priori make sense !
The key to this riddle is to choose a trivialization $\mathcal L|U \stackrel {\cong} {\to} \mathcal O_U$ on some open subset $U\subset X$.
The rational section $s|U\in \Gamma_{\text {rat }} (U,\mathcal L)$ will be transformed into a rational function $s'\in K(U)=K(X)$ and we define $v_Y(s):=v_Y(s')$.
This definition trivially does not depend on the trivialization.
[For ease of notation I have assumed the scheme $X$ integral. If it isn't, this construction must be adapted.]

The simplest example
Consider the point $p=(0:1)\in \mathbb P^1_{z:w}$ and the line bundle $\mathcal L=\mathcal O(p).$
(Recall that $\mathcal O(p)$ consists of rational functions having at most a pole of order one at $p$)
What is the divisor $div(s)$ associated to the section $s=1\in \Gamma(\mathbb P^1,\mathcal O(p))$ ?
Despite appearances it is not the zero divisor!
A trivialization $\mathcal L|U \stackrel {\cong} {\to} \mathcal O_U$ of our line bundle around $p$ over $U=\mathbb A^1_t$ is given by multiplication by $t=\frac zw$
So the constant function $1|U$ is transformed into the coordinate function $s'=t\in \Gamma(\mathbb A^1,\mathcal O)$ and since $v_p(t)=1$, we get finally $$div(s)=1\cdot p$$

Question 2
Write $\mathfrak p:=(x)\in \text {Spec} A$
(i) Since $x$ generates the maximal ideal $\mathfrak pA_\mathfrak p$ of the DVR $A_\mathfrak p$ we already have $$ v_\mathfrak p (x)=1 $$ (ii) Now consider some other prime $\mathfrak q\neq \mathfrak p$ of height one in $ \text {Spec} A$ .
We have $x\in A\subset A_\mathfrak q$ but I claim that $x\notin \mathfrak q A_\mathfrak q$.
Indeed if that were the case we would have $x\in \mathfrak qA_\mathfrak q \cap A=\mathfrak q$ and thus $(x)=\mathfrak p\subset \mathfrak q$.
This is absurd since different height one primes cannot be included in one another.
So in reality $x\in A_\mathfrak q\setminus \mathfrak q A_\mathfrak q $ and thus $$v_\mathfrak q(x)=0$$

From (i) and (ii) we conclude that, as desired, $$\text {div}(x)=1\cdot \mathfrak p=1\cdot (x) $$

Georges Elencwajg
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  • Dear Georges, thanks for your answer. So upon choosing the trivialization, $\nu_Y(s')$ is as Asal has defined it to be yes? Namely, $\nu_Y(s')$ is the valuation of the image of $s'$ in $\mathcal{O}_{X,\eta}$ where $\eta$ is the generic point of $Y$.

    Also, would you like to answer question 2?

     –  Oct 17 '13 at 12:02
  • Dear user38268: yes $v_Y(s')$ is exactly what @Asal and you say. And I might try to answer Question 2 (this is not a promise!) But in any case let me say that it is a very, very good question and that few would have noticed that there is a genuine difficulty there. – Georges Elencwajg Oct 17 '13 at 12:27
  • @user38268: OK, I have kept my non-promise and answered question 2 ! – Georges Elencwajg Oct 17 '13 at 13:12
  • Sorry for asking question (2). It was a stupid question and I was not thinking about the definition of height $1$ when I asked it! –  Oct 17 '13 at 13:20
  • "This definition trivially does not depend on the trivialization." That's when we realize we are funny people. – Patrick Da Silva Mar 28 '15 at 00:14