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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. – Rankeya Nov 2 '12 at 6:44
There are several nice examples at… . – Qiaochu Yuan Nov 2 '12 at 6:46
@QiaochuYuan - in reference to Hailong Dao's example at the MathOverflow link, why does $Yf+Xg=1$ imply that $U$ is affine? – Ben Blum-Smith Nov 2 '12 at 15:57
@Ben: I don't know. Admittedly I only understand the elliptic curve example. – Qiaochu Yuan Nov 2 '12 at 18:18
Dear @BenBlum-Smith: I think he is using exercise 2.17(b) from Chapter 2 of Hartshorne. – Rankeya Nov 2 '12 at 21:22

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