# Closed embedding are affine morphisms

$$\newcommand{\O}{\mathscr O}\newcommand{\Spec}{\operatorname{Spec}}\newcommand{\V}{\mathbb V}$$ Ok, I have been through the ringer with this problem today and I think I am finally on the right track, but for some reason proving this has been more difficult than anticipated.

Let $$f:X\rightarrow Y$$ be a closed embedding of schemes, by which I mean the topological map $$f$$ is a homeomorphism onto a closed subset of $$Y$$, and $$f^\sharp:\O_Y\rightarrow f_*\O_X$$ is a surjective morphism of sheafs. I am trying to show that if $$U=\Spec A\subset Y$$ is affine then $$f^{-1}(U)\cong \Spec A/I$$ for some $$I\subset A$$.

Let $$I_{X/Y}$$ be the sheaf of ideals on $$Y$$ given by $$I_{X/Y}=\ker f^\sharp$$. If $$U=\Spec A\subset Y$$ is affine, then $$I_{X/Y}(U)$$ is an ideal $$I\subset A$$. If we let $$V=f^{-1}(U)$$ then we obtain a morphism of schemes $$f|_V:V\rightarrow U$$ which must be a closed embedding of schemes as $$f|_V$$ is a homeomorphism onto $$U\cap f(X)$$ and the surjection property is inherently local. Now a priori, we don't know that $$V$$ is affine, but we do now that every morphism from a general scheme to an affine one is induced by the morphism on its ring of global sections. In particular, if $$\psi:A\rightarrow \O_V(V)$$ is our short hand for the morphism $$(f|_V)^\sharp_U$$, then the topological map $$f|_V$$ is given by: $$f|_V(x)=\psi^{-1}(\pi^{-1}_x(\mathfrak m_x))$$ where $$\pi_x$$ is the ring homomorphism $$\O_V(V)\rightarrow( \O_V)_x$$ and $$\mathfrak m_x\subset (\O_V)_x$$ is the unique maximal ideal of the stalk. It is then clear that $$f|_V$$ has image in $$\V(I)\subset \Spec A=U$$ as $$I=\ker\psi$$ and $$0\in \mathfrak m_x$$ so $$\psi^{-1}(0)=\ker\psi=I\subset f|_V(x)$$.

Where my trouble is is showing that $$\V(I)\subset f|_V(V)$$. In particular, I think this where I should invoke the surjectivity of $$(f_V)^\sharp_{\mathfrak p}$$ but I am not sure how...given a prime ideal such that $$I\subset \mathfrak p$$ I need to find an $$x\in V$$ such that $$f|_V(x)=\mathfrak p$$ but I am not sure how to do this.

Any help would be greatly appreciated; once I can show this homeomorphism I think I will be golden because I have an idea for how to construct the sheaf isomorphism to $$\O_{\Spec A/I}$$.

• It's equivalent and easier to work with $Y$ being affine right off the bat Commented May 2 at 0:14
• @FShrike Sure, but it doesn't really change where I'm having issue as I reduced to that case quickly Commented May 2 at 0:16
• You can also say $Z$ is isomorphic as a ringed space to the ringed space $(V(\ker f\#),O/\ker f\#)$ and then it just has to be argued this is an affine scheme of the expected form. Iirc there's some subtlety with closed subsets not having a unique subscheme structure and needing to take radicals and reductions everywhere Commented May 2 at 0:18
• @FShrike that is harder to prove as you actually need $i^{-1}O/\ker f^\sharp$ otherwise your sheaf is on the wrong topological space. Commented May 2 at 0:22
• Ok I've made an answer, I've tried to align it with your ideas and highlight the main points. Hopefully it is more helpful; I did not just copy this from an alg. geo textbook :) Commented May 2 at 10:11

$$\newcommand{\spec}{\operatorname{Spec}}\newcommand{\O}{\mathscr{O}}\newcommand{\V}{\mathbb{V}}\newcommand{\J}{\mathcal{J}}\newcommand{\I}{\mathcal{I}}$$Seeing what you want to see unfortunately involves a slightly subtle point where we require $$Z$$ to be a scheme and use an affine basis. We can't argue purely with the theory of locally ringed spaces. I go about it slightly differently, determining $$Z$$ as the support $$\V(\J)$$ of an appropriate sheaf and viewing the calculation $$\J=\ker\varphi$$ as the main point; I think it is harder to first define $$\J=\ker\varphi$$ and then try to compute that $$\V(\J)=Z$$ and identify $$\O_Z$$ with the quotient by $$\J$$.

If $$\iota:(Z;\O_Z)\hookrightarrow(X;\O_X)$$ is a closed immersion of schemes we know $$\J=\ker\iota^\#$$ is the important thing and - which seems to be the part you didn't get yet - $$\V(\J)=Z$$, by the general nonsense at the bottom. Suppose $$X$$ is affine. Using your nice idea about the spec-global adjunction (valid for locally ringed spaces in general) we can write $$\iota$$ as the composite: $$\iota:Z\overset{\eta}{\longrightarrow}\spec B\overset{\spec\varphi}{\longrightarrow}\spec A$$Where $$A:=\O_X(X),B:=\O_Z(Z)$$, $$\varphi=\iota^\#_X$$ and $$\eta$$ is the adjunction unit.

We would love to say $$\J$$ is isomorphic with the sheaf of ideals associated to $$\ker\varphi:=\I$$. Then $$\V(\J)=\V(\I)\cong\spec A/\I$$ (using two 'different' notions of $$\V$$ and $$\cong$$ is a homeomorphism of spaces) and under this homeomorphism we could easily identify $$j^\ast(\O_X/\J)\cong\O_{\spec A/\I}$$ since localisation of rings commutes with quotients. Thus $$Z\cong\spec A/\I$$ as a scheme, underneath $$X$$, would follow as desired. Make sure you can check $$\V(\widetilde{\I})$$ - in the sense of the vanishing set of a sheaf of ideals, defined in the bottom section - really is equal to $$\V(\I)=\{\mathfrak{p}\in\spec A:\mathfrak{p}\supseteq\I\}$$.

To determine $$\J\cong\widetilde{\I}$$ we just need to check (using the nature of $$\eta,\spec\varphi$$) that $$\I_f$$ is the kernel of $$A_f\to B_{\varphi(f)}=\O_Z(Z)_{\varphi(f)}\to\O_Z(Z_{\varphi(f)})$$ for all $$f\in A$$. This is $$\iota^\#=\eta^\#(\spec\varphi)^\#$$'s action on the principal affine open $$D(f)$$. The exact sequence of $$A$$-modules $$0\to\I\to A\to B$$ certainly tells you after localisation that $$\I_f$$ is the kernel of $$A_f\to B_{\varphi(f)}$$; thus, the game is to determine that $$\O_Z(Z)_{\varphi(f)}\to\O_Z(Z_{\varphi(f)})$$ is necessarily injective.

Well, it sure would be an injective map (isomorphism/equality) if $$Z$$ were affine so here is where we have to use the fact $$Z$$ locally looks like an affine scheme:

$$X$$ is compact, so so is $$Z$$. $$Z$$ thus admits a finite cover by affine opens $$U_1,\cdots,U_n$$ (more precisely we require $$(U_\bullet;\O_Z|_{U_\bullet})$$ is affine). This is the only place where we use the assumption $$Z$$ is a scheme rather than an arbitrary locally ringed space. We have a commutative diagram, with $$g:=\varphi(f)$$: $$\require{AMScd}\begin{CD}\O_Z(Z)_g@>>>\O_Z(Z_g)\\@VVV@VVV\\\bigoplus_{k=1}^n\O_Z(U_k)_g@>>\cong>\bigoplus_{k=1}^n\O_Z(U_k\cap Z_g)\end{CD}$$Where the vertical maps are injective by the separatedness property of a sheaf and exactness of localisation (which commutes with finite direct products) and the bottom map is an isomorphism because the $$U_\bullet$$ are affine.

It follows the top map is injective, as desired, so $$\I_f\hookrightarrow A_f$$ really is identifiable with $$\J=\ker\iota^\#$$. For we have just constructed an isomorphism between the two on the basis of principal affine opens $$(D(f))_{f\in A}$$, a basis for $$X$$, so by general sheaf theory these sheaves must be isomorphic and we win.

A generic lemma that has nothing to do with schemes per se, just locally ringed spaces: Say $$(X;\O_X)$$ is a locally ringed space and $$\J$$ is a sheaf of ideals on $$X$$. We first claim $$(\V(J);j^\ast(\O_X/\J))$$ is a locally ringed space, where $$\V(\J):=\{x\in X:\J_{X,x}\neq\O_{X,x}\}$$, where $$j:\V(J)\hookrightarrow X$$. Firstly we must convince ourselves $$j^\ast$$ of a sheaf of rings is again a sheaf of rings in such a way that the unit/counit of adjunction are (local) ring homomorphisms. This is fairly clear. The reason it is a locally ringed sheaf is because each stalk is isomorphic with $$\O_{X,x}/\J_{X,x}$$ and if $$A$$ is a local ring and $$I\subset A$$ a proper ideal (this is the only reason we need to restrict to $$\V$$) then $$A/I$$ is also a local ring. Now, say $$\iota:(Z;\O_Z)\hookrightarrow(X;\O_X)$$ is a closed immersion of locally ringed spaces. If $$\J=\ker\iota^\#$$ we would want to identify $$(Z;\O_Z)\cong(\V(\J);j^\ast(\O_X/\J))$$ underneath $$X$$. First note for $$x\in X\setminus Z$$ that $$(\iota_\ast\O_Z)_x=0$$ by closedness, thus $$\J_x=\O_{X,x}$$. Now if $$z\in Z$$ then we are given, by the surjectivity condition, that: $$0\to\J_z\to\O_{X,z}\to\O_{Z,z}\to0$$Is exact; as $$\O_Z$$ is a locally ringed sheaf $$\O_{Z,z}\neq0$$ and it follows $$\J_z$$ is a proper ideal of $$\O_{X,z}$$. Therefore $$\V(\J)=Z$$ as an exact equality of subspaces of $$X$$ and $$j=\iota$$ on the point-set level. Moreover: $$0\to j^\ast\J\to j^\ast\O_X\to\O_Z\to0$$Is an exact sequences of sheaves on $$Z$$ and $$j^\ast$$ preserves quotients so $$\O_Z\cong j^\ast(\O_X/\J)$$ compatibly w.r.t $$X$$, as desired.