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 ?
intersection of two affine open sets of a scheme
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algebraic-geometry
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0it is section 14.3.1. I think that the affine open sets containing a fi$x$ed point, and distinguished inclusions, $f$orm a filtered set, not for all affine open sets and distinguished inclusions. – 2011-07-18
1 Answers
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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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0Thank you, the convention that the empty set is an affine scheme is necessary! – 2011-07-19