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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?)

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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
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.

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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 $\mathcal{O}(n)|_X$.

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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

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