# Proving that the disjoint union of two affine schemes is isomorphic to an affine scheme

An "Easy Exercise" from Vakil's text:

Show that the disjoint union of a finite number of affine schemes is also an affine scheme.

To me it's not so easy.

Let's do it for two affine schemes $$(Spec A_i,\mathscr O_{Spec A_i})$$ where $$i=1,2$$.

We need to prove that there is an isomorphism of two ringed spaces: $$(Spec A_1\coprod Spec A_2, \mathscr O)\simeq (Spec_{A_1\times A_2},\mathscr O_{Spec(A_1\times A_2)})$$ where $$\mathscr O$$ is the sheaf that is constructed from the following sheaf on a base $$O$$ by the procedure described in Vakil's Theorem 2.5.1: $$O(D(f_i))=(A_i)_{f_i}$$.

Suppose we know somehow that $$Spec A_1\coprod Spec A_2$$ and $$Spec(A_1\times A_2)$$ are homeomorphic via $$\pi: Spec A_1\coprod Spec A_2\to Spec(A_1\times A_2)\\P_1\mapsto P_1\times A_2 \text{ if P_1 is a prime ideal of A_1}\\P_2\mapsto A_1\times P_2\text{ if P_2 is a prime ideal of A_2}$$ Then the only thing that remains to be proved is that there is a natural isomorphism of functors $$\mathscr O_{Spec A_1\times A_2} \simeq \pi^\ast\mathscr O$$, and this is where problems arise.

We need to define a natural transformation $$\alpha$$ whose components are isomorphisms in the category of rings. Is there a way to reduce this to defining only $$\alpha_{D(f)}$$ (the components of $$\alpha$$ at distinguished open sets)? (If we invoke the construction of $$\mathscr O$$ from Theorem 2.5.1 via tuples of compatible germs, I feel this will become very messy, but it's supposed to be an "easy exercise".)

Even if it's enough to define only $$\alpha_{D(f)}$$, I'm having trouble doing this. Each $$\alpha_{D(f)}$$ should be a ring isomorphism $$\alpha_{D(f)}:(A_1\times A_2)_f\to \mathscr O(\pi^{-1}D(f))=O(\pi^{-1}(D(f)))$$

I'm not sure how to simplify $$O(\pi^{-1}(D(f)))$$ and how to define $$\alpha_{D(f)}$$. And if it's not enough to define $$\alpha_{D(f)}$$, how to define $$\alpha_U$$ in general?

• For "not so easy": I remember thinking about one of those "easy" exercises for eternity whereafter it turned out that it requires somehow knowing that field extensions are faithfully flat (it was still not a one-liner). ¯\_(ツ)_/¯ Don't be underwhelmed by this. Commented Sep 18, 2021 at 21:33
• Previously discussed here - does this resolve your issues? Commented Sep 18, 2021 at 22:03
• Using the fact that $\mathsf{Aff} \simeq \mathsf{CRing}^{op}$ via global sections and taking spectrum, it is indeed a oneliner. The fact itself however is not. Commented Sep 19, 2021 at 11:07

Answer: There is a "direct" approach: Let $$X,Y$$ be affine schemes and let the disjoint union (as topological spaces) have the following open sets: A set $$U⊆X∪Y$$ is open iff $$U:=U_1∪U_2$$ where $$U_1⊆X,U_2⊆Y$$ are open sets. By definition: For any open set $$U_1∪U_2$$ let $$\mathcal{O}_{X∪Y}(U_1∪U_2):=\mathcal{O}_X(U_1)⊕\mathcal{O}_Y(U_2)$$. It follows $$(X∪Y,\mathcal{O}_{X∪Y})$$ is a locally ringed space. If $$U:=X⊆X∪Y$$ it follows $$(U,(\mathcal{O}_{X∪Y})_U)≅(X,\mathcal{O}_X)$$ and similar for $$Y$$. Hence $$X∪Y$$ has an open cover of affine schemes and is therefore a scheme. If $$X:=Spec(A),Y:=Spec(B)$$ there is a canonical map of schemes $$ϕ:X∪Y→Spec(A⊕B)$$ induced by the canonical map $$Id∈Hom_{rings}(A⊕B,Γ(X∪Y,\mathcal{O}_{X∪Y}))$$(HH.Ex.II.2.4). The map $$\phi$$ is checked to be an isomorphism. This is because the ring $$Γ(X∪Y,\mathcal{O}_{X∪Y}):=A⊕B$$ by definition. Hence you do not need functors: The disjoint union $$X∪Y$$ can be constructed directly.