Let $X$ be a projective, noetherian $k$-scheme for an algebraically closed field $k$ of characteristic zero. Let $Y_1,...,Y_r$ be locally closed subschemes (open subschemes of closed subschemes) of $X$. Does the scheme structure on $X$ necessarily induce a scheme structure on $Y_1 \cup ... \cup Y_r$?
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2 Answers
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You just discovered constructible sets!
It is really easy to give counter-examples to your suggestion (as Ja ok already has), but here is a general idea: Take your favorite locally closed but neither open nor closed subscheme of your favorite irreducible scheme. Then prove (as a homework) that its complement is a union of locally closed subschemes (hint: an open and a closed), but itself is not locally closed (and hence not a subscheme).
answered Aug 22, 2017 at 2:49
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$\begingroup$ One motivation for introducing this notion is Chevalley's theorem, that the image of a morphism is a constructible set (example: consider $(x,y)\mapsto (x,xy)$). $\endgroup$ Commented Sep 20, 2017 at 14:09
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No. Take the Union of the origin and the complement of the coordinate axes inside X = A^2.
This is not a subscheme because the origin doesn't have an affine open neighborhood.
