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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?

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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 mathoverflow.net/questions/57352/… –  Karl Schwede Aug 22 '12 at 14:45

2 Answers 2

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.

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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.

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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
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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

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