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

Question

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?

Answer 1

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.