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Assume we have an affine scheme $A$ that comes with a closed immersion into a smooth scheme $i \colon A \hookrightarrow M$, which is not necessarily affine.

Does there exist an affine open subscheme $j \colon V \hookrightarrow M$ such that $A$ already embeds in $V$?

Expressed in Diagrams, does there exist an affine open subscheme $V$ of $M$ such that we have a factorization

     i
A (----> M
 \      /
  \    /
     V ?
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  • $\begingroup$ I think that the following is a counter-example. Take a smooth plane cubic and remove a point such that it is of infinite order in the group law determined by a flex point. $\endgroup$ – damiano May 11 '11 at 10:54
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Let me expand on my comment. Let $E$ be an elliptic curve and let $p$ be a point of $E$ of infinite order. Embed $E$ in $\mathbb{P}^2$ as a plane cubic using the linear system $3O$, where $O$ is the origin of the group law. Remove from $\mathbb{P}^2$ the image of the point $p$ and note that $E \setminus \{p\}$ is affine. The inclusion of $E\setminus\{p\}$ in $\mathbb{P}^2 \setminus \{p\}$ is the required counter-example. Indeed, it is a closed immersion and if there were an affine open subset of $\mathbb{P}^2 \setminus \{p\}$ containing $E\setminus\{p\}$, then there would be a plane curve intersecting $E$ in only the point $p$. This is impossible, since the point $p$ has infinite order.

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    $\begingroup$ Note that it is quite natural to expect the answer to the question to be "no": a theorem of Goodman roughly asserts that the complement of an affine open subset can be blown up to become the support of an ample divisor. In particular, the complement of an affine open subset is "trying to meet" all subvarieties. On the other hand, your requirement is that you want one missing a particular one. Trying to play these facts against one another, allows you to find easily a counter-example. $\endgroup$ – damiano May 11 '11 at 14:12
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There are also examples of smooth proper (nonprojective) varieties $V$ with a finite subset not contained in any affine open subset.

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