Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm looking for an example of the following. Let $X$ be a smooth quasiprojective variety over $\mathbb{C}$ and let $\overline{X}$ be a compactification of $X$. We then have a map $Pic(\overline{X}) \rightarrow Pic(X)$. I want an example where this map is not surjective. It will be surjective if, for example, $\overline{X}$ is smooth, but I don't think it should be surjective in general. I'd also be interested in conditions under which it will be surjective.


EDIT : Here are a couple of thoughts. Since $X$ is smooth, every line bundle on it comes from a Weil divisor. The closure in $\overline{X}$ of a Weil divisor in $X$ is another Weil divisor. The only thing that could go wrong is that this new Weil divisor might not come from a line bundle. Thus we are looking for Weil divisors that don't come from Cartier divisors, but I don't know enough examples of this to get what I'm looking for.

share|cite|improve this question
up vote 3 down vote accepted

Your argument is absolutely correct, as long as $\bar{X} \setminus X$ is of codimension at least 2; this ensures that the only possible Weil divisor on $\bar{X}$ which restricts to your chosen $D$ on $X$ is just the closure of $D$ in $\bar{X}$.

So you need to think of your favourite example of a projective variety with a non-Cartier divisor, which is probably the quadric cone $x^2 + y^2 = z^2$ in $\mathbb{P}^3$. Any line through the vertex of the cone is a Weil divisor which is not locally principal. Take $X$ to be the cone with the vertex removed, and you have your example. In this case $X$ is isomorphic to $(\textrm{conic}) \times \mathbb{A}^1$, so $\textrm{Pic}\; X \cong \mathbb{Z}$. On the other hand, any two lines passing through the vertex together form a plane section, which is a Cartier divisor. So the image of $\textrm{Pic}\; \bar{X} \to \textrm{Pic}\; X$ is $2\mathbb{Z} \subset \mathbb{Z}$. (This example is described in many algebraic geometry textbooks, including Hartshorne, but I'm at home at the moment and can't give you a reference. If you haven't seen it before, it's worth spending a little while seeing in detail what's going on here.)

One condition which implies surjectivity is that your variety be locally factorial (since then every Weil divisor is Cartier).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.