Form of basic open set of affine scheme: The intersection of two basic open sets.

Let $$R$$ be a commutative ring and $$\text{Spec} R$$ be the affine scheme associated to it. If we define the basic open sets as $$D_f=\{\text{prime ideals of R not containing f for f\in R}\}$$, then it is well-known that $$D_f=\text{Spec} R_f$$. Now suppose I have two basic open sets $$D_f$$ and $$D_g$$ and consider the intersection $$D_f\cap D_g$$, is there a way to describe this intersection as the Spec of some local ring?

• We have $D_f\cap D_g = D_{fg}$. Commented Apr 2, 2021 at 13:23

Question: "Now suppose I have two basic open sets Df and Dg and consider the intersection Df∩Dg, is there a way to describe this intersection as the Spec of some local ring?"

Answer: If $$X:=Spec(R)$$ and $$f\in R$$ is an element you define $$D(f):=Spec(R_f)$$. There is an isomorphism of rings

$$R_f \cong R[t]/(tf-1).$$

Let $$D(g):=Spec(R_g)$$ and let $$i: D(f) \rightarrow X, j: D(g) \rightarrow X$$ be the inclusion maps. You may define the intersection using the fiber product of the maps $$i,j$$: $$D(f)\cap D(g):=Spec(R_f\otimes_R R_g).$$

There is a "canonical" isomorphism

$$R_f\otimes_R R_g\cong R[t,u]/(tf-1,ug-1) \cong R[t]/(Tfg-1)\cong R_{fg}$$

defined by

$$\phi(t):=gT, \phi(u):=fT$$ and $$\psi(T):=tu$$. You may verify that $$\phi \circ \psi=\psi \circ \phi=Id$$ equals the identity, hence $$R_f\otimes_R R_g \cong R_{fg}$$. It follows

$$D(f) \cap D(g):=D(f)\times_X D(g) \cong D(fg)$$

is an isomorphism of schemes.