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In page 299 of Ravi Vakil's lecture "Foundations of algebraic geometry" , there is a statement: For a scheme X, the category of affine open sets, and distinguished inclusions, forms a filtered set. Given two affine open sets U and V of the scheme X, if the intersection of U and V is empty, how can we find an upper bound of U and V ?

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Is the union of the $U,V$ off the mark here? A lower bound would be the open set. Am I missing something? –  Eivind Dahl Jul 18 '11 at 12:04
    
Could you please give a reference in terms of sections rather than pages, because of the changes due to successive online versions ? –  Georges Elencwajg Jul 18 '11 at 16:15
    
it is section 14.3.1. I think that the affine open sets containing a fixed point, and distinguished inclusions, form a filtered set, not for all affine open sets and distinguished inclusions. –  Yubin Jul 18 '11 at 16:46
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First of all, note that this doesn't come up often. If $X$ is irreducible, and $U$ and $V$ are nonempty opens of $X$, then $U \cap V$ is nonempty.

That said, if you want Ravi's statement to be correct in full generality, then you should consider the empty set to be an affine scheme. Specifically, it should be $\mathrm{Spec}$ of the zero ring. I don't recall what Ravi's conventions for the zero ring are, but in my opinion the correct conventions are that the zero ring is a ring, is not a field, and is a $k$-algebra for every $k$.

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Thank you, the convention that the empty set is an affine scheme is necessary! –  Yubin Jul 19 '11 at 5:02
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