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Can it happen that an affine open subset of an affine variety is not the complement of a hypersurface? I think this must be able to happen because if not, I imagine I would have encountered a proposition to that effect in one of the books I am studying. However, I so far don't have an example: in all the toy examples I am thinking of, the minute I construct an open subset so that its complement isn't a hypersurface, it also stops being affine.

By the way, I am interested in this question due to a homework assignment (hence the homework tag) but the question itself is not the one I was assigned. I am not including the assigned question out of fear somebody might answer it or inadvertently give me a bigger hint than I want.

Thanks in advance.

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    What you say is possible for affine schemes (i.e., an affine open subset need not be a distinguished open), but I have not seen an easy example of this. I am very interested to know if there is an easy example that does not involve using more sophisticated machinery like Riemann-Roch.2012-11-02
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    There are several nice examples at http://mathoverflow.net/questions/7153/open-affine-subscheme-of-affine-scheme-which-is-not-principal .2012-11-02
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    @QiaochuYuan - in reference to Hailong Dao's example at the MathOverflow link, why does $Yf+Xg=1$ imply that $U$ is affine?2012-11-02
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    @Ben: I don't know. Admittedly I only understand the elliptic curve example.2012-11-02
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    Dear @BenBlum-Smith: I think he is using exercise 2.17(b) from Chapter 2 of Hartshorne.2012-11-02
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    @Rankeya - thanks.2012-11-13

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