MathOverflow is a question and answer site for professional mathematicians. 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

Let $V$ be an affine variety over an algebraically closed field $k$ and $D \subset V$ a Cartier divisor which is normal and has an isolated singularity at $p \in D$.
Let $\mathcal{O}_V^*, \mathcal{O}_D^*$ be the sheaves of invertible functions on $V$ and $D$. Then I think that we have an exact sequence $0 \rightarrow K \rightarrow \mathcal{O}_V^* \rightarrow \mathcal{O}_D^* \rightarrow 0$.

Question (Edited) Is there an affine open neighbourhood of $p \in V' \subset V$ such that $H^0(V', \mathcal{O}_{V'}^*) \rightarrow H^0(D', \mathcal{O}_{D'}^*)$ is surjective where $D':= D \cap V'$? That is, can we lift a surjection of stalks to that on some open neighbourhood?

I think the Question is reduced to the following.

Question' Is the cokernel of $H^0(V, \mathcal{O}_{V}^*) \rightarrow H^0(D, \mathcal{O}_{D}^*)$ finitely generated as an abelian group?

share|cite|improve this question
Affine space ${\bf A}^n_k$ has no globally invertible functions other than elements of $k^*$. So if the answer to your question were affirmative, then the same would be true for any Cartier divisor in affine space with the properties you mention. That doesn't seem very likely. – Damian Rössler Aug 22 '12 at 12:05
Thank you for the comment. I edited my question and focused on neighbourhood. – tarosano Aug 22 '12 at 12:51
The sequence you quote is not exact. If $U$ is an open subset of $V$, then it is not true in general that an element of the form $1+x$, where $x\in I_D$, is invertible in $U$ (because $x$ might be equal to $-1$ somewhere in $U$). Therefore $K$ is not the kernel of the morphism $O_V^*\to O^*_D$ - but it contains it. – Damian Rössler Aug 22 '12 at 13:47
Thank you for pointing out my mistake. I edited it. – tarosano Aug 22 '12 at 14:28
tarosano, for finite generation in the revised question, it seems to follow from… – Karl Schwede Aug 22 '12 at 14:45
up vote 4 down vote accepted

Suppose that $\bar{f} \in H^0(D, O_D^*)$ and consider a corresponding $f \in H^0(V, O_V)$ (which may or may not be invertible).

Then for every point $x \in D \subseteq V$, we let $\bar{f}'$ denote the element in the stalk $O_{D,x}$ and $f'$ the element in the stalk $O_{V,x}$. Since $\bar{f}'$ is not in the maximal ideal of $O_{D,x}$, neither is $f'$ in the maximal ideal of $O_{V,x}$. Thus $f'$ is invertible in a neighborhood of $x \in V$. Since this holds for all points $x \in D$, the vanishing locus $V(f')$ of $f'$ is away from $D$. It follows that there exists a neighborhood of $D$ where $f'$ is a unit.

Now, I just learned from THIS QUESTION (that at least in the geometric setting you are interested in) the set of units of $H^0(D, O_D)$ is finitely generated modulo constants. Thus, choose generators ${\bar f_1}, \dots, \bar{f_n}$ of $H^0(D, O_D^*)$ modulo constants. Lifting these to $f_i \in H^0(V, O_V)$, we can find an open set $U \subseteq V$ containing $D$ such that the $f_i$ are invertible in $H^0(U, O_U)$. It follows that $H^0(U, O_U^*) \to H^0(D, O_D^*)$ is surjective.

Thus it seems we can get a slightly stronger statement than what you asked for.

Statement: $\text{ }$ There exists an open neighborhood $U \subseteq X$ containing $D$ such that $H^0(U, O_U^{*}) \to H^0(V, O_V^{*})$ is surjective.

EDIT: Perhaps in view of the newly revised question which appeared while I was typing this (the finite generation part), this is more information than required. But perhaps it will be useful to someone.

share|cite|improve this answer
Thank you very much for the helpful answer. – tarosano Aug 22 '12 at 15:35
Nice answer ...! – Damian Rössler Aug 22 '12 at 15:55

Edit: As Jason points out, the following answers the original question, but not the revised question.

Let $E$ be an elliptic curve in $\mathbb{P}^2_{\mathbb{C}}$ and $P\in E$ a point of order $2$. The tangent line $L$ to $E$ at $P$ meets $E$ at $P$ and the identity $O$. Now $E\setminus L$ is a divisor in $\mathbb{A}^2_{\mathbb{C}} = \mathbb{P}^2_{\mathbb{C}}\setminus L$ that carries a non-constant invertible function.

share|cite|improve this answer
This example was correct before the OP edited his question. However, it is no longer correct: the OP allows himself to replace $\mathbb{P}^2\setminus L$ by a smaller open, e.g., the complement of the union of both $L$ and the flex line at $O$. – Jason Starr Aug 22 '12 at 13:39
Yeah, I saw the edit after I posted the answer. I'll edit to reflect this. – Ramsey Aug 22 '12 at 13:48
Thank you for comments. I think that, if the cokernel of $H^0(V,\mathcal{O}_V^*) \rightarrow H^0(D,\mathcal{O}_D^*) $ is finitely generated abelian group, then we can find $V'$ with the required property. Is it finitely generated? – tarosano Aug 22 '12 at 14:30
@tarosano: It certainly is in my example. Any invertible function on $E\setminus L$ has divisor $2n[P] - 2n[O]$ on $E$ for some integer $n$, so is a constant multiple of $f^n$ where $f$ is a fixed rational function with divisor $2[P]-2[O]$. – Ramsey Aug 22 '12 at 14:38

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.