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(1) Let $X$ be a projective (integral) curve over $\mathbb{C}$ and let $P$ be a singular point of $X$. Is there always a Cartier divisor whose support is exactly $P$ (set-theoretically)?

The following may be two related questions:

(2) Let $f: Y \to X$ be the normalisation of the curve $X$. Then we have a pushforward functor for Weil divisors from $Y$ to $X$. Is there some analogous contruction for Cartier divisors?

(3) Is every Weil divisor on a singular curve $\mathbb{Q}$-Cartier (i.e. a mutiple of the divisor is Cartier)? A 2-dimensional analog seems to have been discussed in the following MO question:

Is every Weil divisor on an arithmetic surface Q-Cartier

I still wonder what is known in the 1-dimensional situation.

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2 Answers

up vote 7 down vote accepted

Ok, the first thing you have to specify is what does it mean for to talk about a Cartier divisor (or Weil divisors). There are a number of options here. The one I like best for Cartier divisors would be invertible subsheaves of $K(X)$, the fraction field of $X$. This coincides with the one in Fulton's intersection theory if I recall correctly.

Now, what do you mean by Weil divisors? One option is formal sums of points. This is ok, but there are problems (mentioned below). Another option is $O_X$-module subsheaves of $K(X)$ in general (if you weren't dealing with curves maybe we should require these to be S2). In higher dimension sometimes people require them to be Cartier in codimension 1. But this just gives us Cartier divisors...

Why is the naive notion of Weil divisor problematic? The map from Cartier divisors to Weil divisors is not necessarily injective for non-normal varieties. Let me give an example. Consider the curve singularity $k[x,y]/(xy + x^3 + y^3)$ and consider the Cartier divisors corresponding to the principal ideals $(x)$ and $(y)$ respectively. These are different subsheaves of $K(X)$ (they are different ideals). However, they both give Weil divisors $3P$ (I assume you compute the map from Cartier to Weil divisors by computing lengths at points).

This implies that there isn't a well defined notion of $O_X(nP)$. For example, if there are two Cartier divisors giving the same $nP$, which sheaf do you choose to represent $O_X(nP)$?

The subsheaf notion is also problematic since not all subsheaves are invertible. So, I think the best behaved notion of Weil is the one that requires divisors to be Cartier in codimension 1. Unfortunately, for curves, this is not interesting.

Now, let me answer your questions.

  1. Yes, depending on what you mean by Cartier divisor. Embed your curve in projective space. Take a hyperplane $H$ (not containing $X$) passing through $P$ and some other smooth points on $X$. $O_{\mathbb{P}^n}(H) \cdot O_X$ is certainly invertible. Let $D$ denote the effective Cartier divisor on $X$ such that $O_X(D)$ agrees with $O_{\mathbb{P}^n}(H) \cdot O_X$ away from $P$ and agrees with $O_X$ at $P$. Now simply consider $O_{\mathbb{P}^n}(H) \cdot O_X(-D)$. This agrees with $O_X$ away from $P$, and is not trivial at $P$, so I would say it is a Cartier divisor supported at $P$.

  2. I don't think so. This doesn't work in higher dimensions either. I assume you want some properties? Like preserving linear equivalence?

  3. Since $O_X(nP)$ is not well defined, then I don't think you can define the notion of $\mathbb{Q}$-Cartier. I guess you could say that there is some Cartier divisor that maps to $nP$ in the natural map from Cartier divisors to Weil divisors?

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Hi, Karl. I'm using the definitions of Cartier and Weil divisors in Hartshorne's book. So, a Cartier divisor is an invertible subsheaf of the function field $K(X)$ and a Weil divisor is a formal sum of points. (Let's stick to the case of curves.) The support of a Cartier divisor $D$ is by definition the set of points $x$ such that the stalks of $O_X(D)$ and $O_X$ at $x$ are different as submodules of $K(X)$. The support of a linearly equivalent Cartier divisor can be different from that of $D$. I hopes this makes the meaning of question (1) clear. –  Yong Hu Aug 1 '12 at 14:50
    
To Karl: (continued) For (3), I agree with what you said. For me, a Weil divisor $E$ is called $Q$-Cartier if there exists some multiple $nE$ which lies in the image of the cycle map from the group of Cartier divisors to that of Weil divisors (I'm not considering the linear classes!). I think your argument for (1) also gives a positive answer to (3). Thanks! –  Yong Hu Aug 1 '12 at 14:58
    
Great, let me know if I can clarify anything. The reason I gave so many different definitions of Weil divisors is that Hartshorne defined Weil divisors only for schemes that are R1 in his book. For the sheafy perspective, Harsthorne has written a couple papers Generalized divisors... where he studies notions of divisors on singular varieties. –  Karl Schwede Aug 1 '12 at 15:06
    
I understood that the definition of "Cartier divisor" is standard and applies to any scheme. (Rather, there are several definitions, but they are all equivalent.) Can you cite any references for non-equivalent definitions of "Cartier divisor"? –  Charles Staats Aug 2 '12 at 13:53
    
Charles, you are right. Certainly I've seen people sometimes talk about Cartier divisors that are also honest divisors (for example the codimension 1 points of the support are in the regular locus of the scheme), but this is really more of a specialization of the above definition. –  Karl Schwede Aug 2 '12 at 15:04
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Here is a part of an email I sent you because I couldn't log in MO.

The answer to (3) is yes: there is a Cartier divisor $D$ on $X$ such that the associated Weil divisor ($0$-cycle for a curve) is $[P]$. This idea is just to consider the normalization $f : Y\to X$. Let $Q\in Y$ be a point lying over $P$. Then $[Q]\sim Z$ (rational equivalence) for some $0$-cycle $Z$ on $Y$ with support disjoint from $f^{-1}(P)$ (use e.g. Riemann-Roch). By pushforward, $[P]=f_*[Q] \sim f_*Z$. By definition of rational equivalence, this means that $[P]=f_*Z+[\mathrm{div}(h)]$ for some rational function $h$ on $X$. As the support of $f_*Z$ is disjoint from $P$, this means that in an open neighborhood $V$ of $P$, $[P]$ coincides with $[\mathrm{div}(h)]$. Define a Cartier divisor $D$ by ${D|}\_{V}=\mathrm{div}(h)$ and $D|_{X\setminus { P}}=1$. Then $[P]$ equals to (the $0$-cycle associated to) $D$. The idea of taking normalization can be found, say, in a paper of Colliot-Thélène (Inv. Math. 2005, p. 599).

This works over any algebraically closed field. If the base field is not necessarily algebraically closed, let $Q_1, \cdots, Q_r$ be the points of $f^{-1}(P)$, let $n$ be the gcd of the $[k(Q_i): k(P)]$. Then the above argument shows that $n[P]$ is associated to a Cartier divisor on $X$. And $n$ is the smallest possible.

In higher dimension, as you already known, this conclusion is false. However the following weaker result holds. Let $X$ be a noetherian scheme, let $E$ be a prime cycle. Then for some positive integr $n$, $nE$ is the cycle associated to a Cartier divisor (on an integral closed subscheme of $X$ containing $E$) in an open neighborhood of the generic point $P$ of $E$. The integer $n$ can be controlled by some invariants related to the singularity at $P$ (can take $n=1$ is $P$ is a regular point of $X$, but this is not necessary as we saw in the case of curves). See Theorem 4.5, §5, and Theorem 6.5 in this paper.

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