Let $M$ be a variety and $Y\subset M$ a subscheme. Is the scheme-theoretic complement $M\setminus Y$ of $Y$ in $M$ equal to the scheme-theoretic complement $M\setminus Y_{\text{red}}$, where $Y_{\text{red}}$ denotes the underlying reduced scheme upon which $Y$ is supported? I'm thinking the answer should be "yes", since complements if I remember correctly are defined in terms of localizing at a prime, and if the subscheme $Y$ is non-reduced I don't think it's ideal can be prime. But if not, then how would one describe the scheme $M\setminus Y$, say when $M=\mathbb{A}_k^2$ and $Y$ is a fat point, e.g. when $Y$ is the subscheme corresponding to the ideal $(x,y)^2$?
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1$\begingroup$ Same difference. $\endgroup$– Allen KnutsonDec 13, 2013 at 4:38
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$\begingroup$ What's the definition of the scheme-theoretic complement for a non-reduced non-affine scheme anyway? $\endgroup$– Ryan ReichDec 13, 2013 at 6:51
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1$\begingroup$ I'm not sure. My question stems from considering the the Grothendieck group of schemes over Spec($k$) for some field $k$. This group doesn't see any scheme structure (and so is isomorphic to the Grothendieck group of varieties over $k$) due to the fact that the complement of a subscheme is the same as the complement of its support. So I was wondering if there could be a modification to the notion of subscheme complement which would make the Grothendieck group of schemes actually sensitive to scheme structure. $\endgroup$– dezignDec 13, 2013 at 7:04
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1$\begingroup$ I think you want something like constructible functions on $M$. Ripping out a thin point $p$ from $M$ would give you the function $1$ away from $p$, and $0$ at $p$. Ripping out a double point would give you the function $1$ away from $p$, and $-1$ at $p$. It's much more cycley than schemey, though. $\endgroup$– Allen KnutsonDec 13, 2013 at 7:22
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2$\begingroup$ How would you define the scheme-theoretic complement of $\mathbb{A}^1 \setminus \{0\}$ embedded in $\mathbb{A}^2$ as the $x$-axis minus origin? As far as I can tell, that's just a constructible set, not a scheme, but you don't seem to be removing that possibility in the text of the question. $\endgroup$– S. Carnahan ♦Dec 13, 2013 at 7:51
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1 Answer
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If $Y\subseteq X$ is a closed subscheme, then the topological space underlying $Y$ is a closed subset of $X$. Its complement $X\setminus Y$ is an open subset of $X$ and the restriction of $\mathcal{O}_X$ to it makes it into a scheme. It has the universal property Allen Knutson described in his comment. It is clear from this definition that $X\setminus Y = X\setminus Y_{red}$.
answered Dec 13, 2013 at 9:35