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$(X,\mathcal O_X)$ is a scheme, and $Y \subset X$ is a closed subset. So on $Y$ we can define a canonical structure sheaf, i.e. the restriction structure sheaf as $(Y,\mathcal O_X|_Y)$. I guess it is not hard to see this is a closed subscheme of $(X,\mathcal O_X)$. However, we know that,(Hartshorne Ex 3.2.6) there are many different closed subschemes with underlying space $Y$ but different structure sheaves. In that Ex 3.2.6, it defines reduced induced closed subscheme which is "smaller" (universal) than any other closed subschemes with space $Y$.

So my question is for those two specific constructions of closed subschemes, are they the same or if not, related in some way? For example, if $X=Spec \, A$, then we know that the reduced induced closed subscheme $Y=Spec \, A/I$, where $I$ is the largest ideal for which $V(I)=Y$ (Also in Ex 3.2.6), then what will be the ideal $I$ corresponding to the restricted subscheme $Y, \mathcal O_X|_Y$??

Thank you so much for any help!

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  • $\begingroup$ How are you defining $\mathcal{O}_X|_Y$? $\endgroup$
    – KReiser
    Nov 16, 2020 at 2:58
  • $\begingroup$ Sorry, I think that's defined in any textbook right? it is the inverse image sheaf of the inclusion map from $Y$ to $X$. $\endgroup$
    – jingyey
    Nov 16, 2020 at 3:04
  • $\begingroup$ I ask because that does not even give a scheme - taking $X=\Bbb A^1_k$ and $Y$ to be the origin, you get a point with the constant sheaf $k[t]_{(t)}$, and $\operatorname{Spec} k[t]_{(t)}$ has two points. $\endgroup$
    – KReiser
    Nov 16, 2020 at 3:06
  • $\begingroup$ @KReiser Oh sorry! I don't understand your example, but I see your point that the restricted sheaf doesn't always give you a scheme. Can you explain that example a little bit more? $\endgroup$
    – jingyey
    Nov 16, 2020 at 3:15
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    $\begingroup$ Sure - a scheme is a locally ringed space where every point has an affine neighborhood isomorphic to an affine scheme (this is a definition). In particular, any scheme which has underlying space a point must have structure sheaf which is a dimension-zero local ring. I showed that your proposed definition applied to the case where $X=\operatorname{Spec} k[t]$ and $Y$ is the closed point $(t)\in X$ gives something not of this form - that is, your claim that $(Y,\mathcal{O}_X|_Y)$ is a scheme is false. $\endgroup$
    – KReiser
    Nov 16, 2020 at 3:24

1 Answer 1

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Here's a recap of the discussion from the comments and chat: the locally-ringed space $(Y,\mathcal{O}_X|_Y)$ is not in general a scheme, because it is not the case that every point has an open neighborhood isomorphic as a locally ringed space to $\operatorname{Spec} A$ for some ring $A$ (this is the defining property for schemes among locally ringed spaces). In particular, taking $X=\operatorname{Spec} k[t]$ for $k$ a field and $Y$ to be the set consisting of the single closed point $(t)\in X$, we have that $(Y,\mathcal{O}_X|_Y)$ is the locally ringed space which is a point with the structure sheaf the constant sheaf with value $k[t]_{(t)}$. If this were to be a scheme, then it must be an affine scheme, and it would need to be isomorphic to the affine scheme $\operatorname{Spec} k[t]_{(t)}$. But this is impossible, because this has two points while $Y$ is a singleton.

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