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I'm trying to understand the connection between closed subschemes and closed immersions. More precisely, the notion of closed subscheme of a scheme $X$ I'm working with is that of a closed subspace $i:Z\hookrightarrow X$ and a sheaf of ideals $\mathcal{I}\subseteq\mathcal{O}_X$ such that $(Z,\mathcal{O}_Z:=i^{-1}\mathcal{O}_X/\mathcal{I})$ is a scheme, such that $Z=\operatorname{supp}\mathcal{O}_X/\mathcal{I}$. I'm trying to see that the inclusion $(i,i^\flat):(Z,\mathcal{O}_Z)\rightarrow(X,\mathcal{O}_X)$ is a closed immersion (by working through the definition rigorously, instead of accepting this rather as "intuitively true") so basically that $i^{\flat}:\mathcal{O}_X\rightarrow \mathcal{O}_Z$ is surjective. However, I even don't know how to define $i^{\flat}$ correctly. The thing I can think of now is the adjunction $$\operatorname{Hom}(i^{-1}(\mathcal{O}_X/\mathcal{I}),i^{-1}(\mathcal{O}_X/\mathcal{I}))\rightarrow\operatorname{Hom}(\mathcal{O}_X/\mathcal{I},i_*i^{-1}\mathcal{O}_X/\mathcal{I}),$$ so that I get a morphism $\eta:\mathcal{O}_X/\mathcal{I}\rightarrow i_*i^{-1}\mathcal{O}_X/\mathcal{I}$ of sheafs corresponding to the identity.

Question 1: As $i^{\flat}$ I would take $\mathcal{O}_X\rightarrow \mathcal{O}_X/\mathcal{I}\stackrel{\eta}{\rightarrow}i_*i^{-1}\mathcal{O}_X/\mathcal{I}=i_*\mathcal{O}_Z,$ is this correct?

Question 2: Why is this surjective? That the first morphism is surjective I see, but why is $\eta$?

Edit: I saw this related question, Why a closed subscheme give rise to a closed immersion, but it doesn't answer my question.

Thanks.
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  • Hint: surjectivity can be checked on stalks. Check that $i^{-1}$ preserves stalks for any $i$ and $i_*$ preserves stalks when $i$ is a closed immersion. – KReiser Apr 02 '21 at 18:46
  • @KReiser Thank you very much! I posted it as an answer, but now I think I would need to go into the proof of the adjunction formula to see that $\eta$ does indeed induce isomorphism on stalks, or am I mistaken? – Thanks. Apr 02 '21 at 20:53
  • Yes, you do need to check that, but it shouldn't be so onerous: the fact that $i$ is a closed immersion will again be helpful. – KReiser Apr 02 '21 at 21:07
  • @KReiser Now I'm a bit confused: doesn't I want to show that $i$ is closed immersion? – Thanks. Apr 02 '21 at 21:39
  • I meant topologically a closed immersion, sorry for being imprecise. – KReiser Apr 02 '21 at 21:42
  • @KReiser, so for a $V\subseteq X$ open, $\eta_V$ is given by $$(\mathcal{O}_X/\mathcal{I})(V)\rightarrow i^+(\mathcal{O}_X/\mathcal{I})(i^{-1}(V))\rightarrow i^{-1}(\mathcal{O}_X/\mathcal{I})(i^{-1}(V)),$$ where $^+$ means the inverse-image presheaf. Since the canonical map into the sheaffification induces isomorphisms on stalks, I just need to see that the first map induces isomorphism on stalks, right? – Thanks. Apr 02 '21 at 22:04
  • I think that seeing that it is surjective would be just taking representatives in the two respective limits, or am I mistaken? – Thanks. Apr 02 '21 at 22:06
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    Yes, that's what you do. – KReiser Apr 02 '21 at 22:08
  • Thank you very much! – Thanks. Apr 02 '21 at 22:12

1 Answers1

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I am trying to follow the hint from the comment. For a point $x\in X$, we can compute the stalk $$(i_*i^{-1}\mathcal{O}_X/\mathcal{I})_X)_x=0$$ if $x\notin Z$ since $Z$ is closed, and if $x\in Z$ then it is equal to $$(i^{-1}\mathcal{O}_X/\mathcal{I})_x=(\mathcal{O}_X/\mathcal{I})_x.$$ Oh, now I see where the condition with $\operatorname{supp}\mathcal{O}_X/\mathcal{I}=Z$ comes into play, because now we see that the stalks are the same. So we would even had an isomorphism $\mathcal{O}_X/\mathcal{I}\cong i_*i^{-1}\mathcal{O}_X/\mathcal{I}$, if we can show that the equality on stalks is induced by $\eta$.

Thanks.
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