# Intersection of affine opens is affine for separated schemes of an affine scheme

Let $$X$$ be a scheme over $$S=Spec A$$ and let $$U, V$$ be affine opens in $$X$$. Let $$X$$ be separated over $$A$$ and let $$\delta$$ be the diagonal morphism $$X \rightarrow X \times_S X$$. Then the $$U \cap V$$ is also affine. The proof I've seen goes like this:

• $$\delta$$ restricts to a map $$U \cap V \rightarrow U \times_S V$$.
• The restriction of $$\delta$$ is still a closed immersion (it is a homeomorphism onto a closed set with surjective stalk maps, since they are the same stalk maps as $$\delta$$).
• Therefore, $$Z = (U \times_S V) \cap \delta(X)$$ is a closed set.
• At this point, the proof I saw says that $$Z$$ is affine (since closed sets in affine schemes are affine) and that $$\delta: U \cap V \rightarrow Z$$ is an isomorphism, so $$U \cap V$$ is also affine.

I know that $$Z$$ is homeomorphic to $$U \cap V$$. The problem I have is that we haven't defined a scheme structure on $$Z$$. But since $$Z$$ is a closed subset of an affine scheme, it has the natural structure from Spec of the quotient ring. So let's define this for $$Z$$. At this point, how do we restrict the morphism $$\delta$$, which currently maps to $$U \times V$$, to $$Z$$ (I know the topological map of points restricts to $$Z$$, but how about as a morphism of schemes)? And if we do this, how do we show it is an isomorphism of schemes?

Another approach would be the following:

Let $$U=Spec B$$, $$V=Spec C$$. So $$U \times V = Spec B \otimes_A C$$, and $$Spec R$$ an affine open in $$U \cap V$$. Then we have a ring homomorphism $$\phi: B \otimes_A C \rightarrow R$$ coming from the map $$\delta$$ to $$U\times V$$. Define $$J$$ to be the kernel of this map. Then define $$I'$$ to be the intersection of all such $$J$$'s where we range over all $$R$$'s. So we can map $$(B \otimes_A C)/I' \rightarrow (B \otimes_A C)/J \rightarrow R$$ and hence we have a morphism $$U \cap V \rightarrow Spec (B \otimes_A C)/I'$$. Is this an isomorphism? Does $$I'=I$$, where $$Z=Spec (B \otimes_A C)/I$$?

• It's better to observe that $U \cap V$ can be identified with $(U \times_S V) \times_{X \times_S X} X$ as schemes. Commented Jan 22, 2023 at 14:27

A reference to your first proof is https://stacks.math.columbia.edu/tag/01KP (Its a bit more general, but the difference is really just going to affines). There it references https://stacks.math.columbia.edu/tag/01IN which is basically the answer to your question as $$U\cap V =\Delta^{-1}(Z) = \Delta^{-1}(U\times_SV)$$ and $$Z=Spec(B\otimes_A C)$$, is affine.
To formulate it in my own words: You got $$\Delta:X\to X\times_SX$$, the diagonal and $$U = Spec(B),V=Spec(C) \subset X$$ affine opens of $$X$$. Because $$X$$ is separated $$\Delta$$ is a closed immersion. Now $$U\times_S V = Spec(B\otimes_A C)\subset\Delta X$$ is affine. So we get a closed immersion from $$U\cap V = \Delta^{-1}(U\times_SV)$$ to $$U\times_S V$$ and any closed subscheme of an affine scheme is affine. Therefor we get $$U\cap V = Spec((B\otimes_A C)/J)$$, where $$J$$ is the ideal such that $$J^{\sim}$$ is the kernel of this closed immersion.
So to conclude we don't get an isomorphism of $$U\cap V$$ and $$U\times_S V$$ but we only need that $$\Delta(U\cap V)$$ is closed in $$U\times_S V$$ for our claim.