Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I hope this question is not completely trivial:

Suppose $V$ is an irreducible projective variety and $U\subset V$ is a Zariski open subset isomorphic to an affine variety. Is it true that $V\setminus U$ is a Cartier divisor in $V$? If not, what conditions should we impose on $V$? (I guess if $V$ is smooth, then everything is fine?)

share|cite|improve this question
Dear aglearner, You probably know this, but just to be clear --- the complement will always be pure of codimension one. Regards, –  Emerton Jan 24 '13 at 22:09
P.S. Sorry, I just saw that this was already posted in an answer. –  Emerton Jan 24 '13 at 23:21
thank you Matthew! –  aglearner Jan 24 '13 at 23:52

2 Answers 2

up vote 1 down vote accepted

aglearner, In his article on abelian varieties Bryden Cais proves the result you mention. You can find the statement/proof on the top of page 4. (

Specifically, he proves that if $X$ is separated, normal, noetherian and $U \subset X$ is a nonempty affine open subset the complement has pure codimension 1. Thus, with it's reduced-induced structure it is a Weil divisor.

share|cite|improve this answer
Thank you LMN, I will have a look (though my question was on Cartier divisors, it follows from what you say, that on a factorial varieties the answer is positive as well) –  aglearner Jan 24 '13 at 15:20
I have to admit that I can not follow the proof completely (due to the lack of knowledge in schemes), but it is nice to know that such a statement holds. –  aglearner Jan 25 '13 at 10:05
agleearner, I can sketch the proof: (1: Thm, An arbitrary morphism from an affine scheme to a separated scheme is affine, see… for a sketch of proof). Now, let $U \subset X$ an affine open set, $Y = X - U$ and $y$ a generic point of a component of $Y$. The map $\phi: Spec \mathcal{O}_{X,y} \rightarrow X$ is affine by thm. above, hence $\phi^{-1}(U) = Spec \mathcal{O}_{X,y} - \{y\}$ is affine. Since $X$ is normal, it follows that the dimension of this local ring is $1$. –  LMN Jan 25 '13 at 14:24
This completes the proof (for the last statement, I'm using an exercises from Hartshorne; that if one removes a point of codimension $\ge 2$ from a normal affine scheme the result is not affine.) –  LMN Jan 25 '13 at 14:27

As you say, if $V$ is smooth, then everything is fine. However, if $V$ is singular, the complement may fail to be the support of any Cartier divisor. For instance, take $V$ to be the projective cone over a smooth plane cubic $X$, and take $V\setminus U$ to be the line over any point $x$ of $X$ such that for every integer $n> 0$, $\mathcal{O}_X(n\cdot \underline{x})$ is not isomorphic to $\mathcal{O}(n)|_X$.

share|cite|improve this answer
I think you should also take $V$ to be a quadric cone in $\mathbb{P}^3$ and $V \setminus U$ a line $\ell$. Then $U$ is affine (since $2 \ell$ is a Cartier divisor) but $\ell$ itself is not a Cartier divisor. Of course, in this case the complement is the support of a Cartier divisor, namely $2 \ell$. –  Francesco Polizzi Jan 24 '13 at 13:40
Thank you Jason and Francesco, these nice examples. –  aglearner Jan 24 '13 at 13:44
Sorry I cannot understand the example about the cubic curve, can anyone explain it a little bit? How to find a point with the required property and what the non-isomorphism implies? Thanks! –  mqx Oct 21 '14 at 16:54

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.