# Closed immersions are affine morphisms

I want to prove that if a morphism of schemes $$f:X\to Y$$ is a closed immersion, then $$f$$ is affine.

Here's where I'm at.

If $$Y=\bigcup_iV_i$$ is an affine open cover, then $$f(X)\cap V_i$$ is an affine scheme, since it is a closed subset of the affine scheme $$V_i$$. Furthermore, the homeomorphism $$X\stackrel{\sim}{\to} f(X)$$ induces the homeomorphism $$f^{-1}(V_i)\stackrel{\sim}{\to}f(X)\cap V_i$$.

That still doesn't prove that $$f^{-1}(V_i)$$ is isomorphic to $$f(X)\cap V_i$$ as schemes, and I'm not sure how to prove it formally.

If $$I(V):=\ker(f_V^\#)$$, then $$f$$ being a closed immersion implies $$\mathcal{O}_Y/I\simeq f_*\mathcal{O}_X$$, where $$\mathcal{O}_Y/I$$ is the sheafification of $$V\mapsto \mathcal{O}_Y(V)/I(V)$$.

In particular $$(\mathcal{O}_Y/I)\big|_{V_i}\simeq f_*\mathcal{O}_X\big|_{V_i}$$, which looks like I'm almost there. But I don't know relate the structural sheaf $$\mathcal{O}_{f(X)\cap V_i}=\mathcal{O}_Y\big|_{f(X)\cap V_i}$$ of the affine $$f(X)\cap V_i$$ with the sheaf $$(\mathcal{O}_Y/I)\big|_{V_i}$$.

How do I do that?

• What is your definition of a closed immersion? Jun 12, 2020 at 10:09
• @user45878, a closed immersion is a morphism of schemes $(f,f^\#):(X,\mathcal{O}_X)\to(Y,\mathcal{O}_Y)$ such that $f(X)$ is closed in $Y$, $f$ induces a homeomorphism $X\stackrel{\sim}{\to}f(X)$ and $f^\#$ is an epimorphism Jun 12, 2020 at 18:32
• This is handled by the answer to your previous question here, particularly the sixth paragraph. Jun 14, 2020 at 0:06

You have to show that for any $$U=SpecB$$, open affine subescheme of $$X$$, the restriction of your closed immersion $$f^{-1}(U) \rightarrow U$$, is a closed immersion too, then you have that $$f^{-1}(U)$$ is $$SpecB/I$$, for some $$I$$ ideal of $$B$$.