I am studying localization of a commutative ring with respect to a multiplicative subset.The concept is motivated by the field of fractions of an integral domain.We introduce inverses of elements of a multiplicative subset.We formalize this things as follows:
Let $R$ be a commutative ring and $S$ be a multiplicative subset i.e. $a,b\in S\implies ab\in S$.Then instead of defining a relation like $(x,s)\sim (y,r) \iff xr=ys$ (which will not be transitive) we define $\sim$ as follows:
$(x,s)\sim (y,r)\iff \exists u\in S$ such that $u(xr-ys)=0$.
We can check by a little calculation that this relation is transitive.But,a concept can have multiple generalizations i.e. generalizations or extensions are not unique.My question is how to predict such a generalization.Can someone give me some motivation to why the relation is modified in this way?Also,I want to know what localization means geometrically in the context of algebraic geometry.
Question: "Can someone give me some motivation to why the relation is modified in this way?Also,I want to know what localization means geometrically in the context of algebraic geometry."
Answer: If $S \subseteq A$ is a multiplicative subset it follows the localized ring $S^{-1}A$ satisfies the following universal property: For any map of commutative rings $f:A \rightarrow B$ where
$$f(s)\text{ is a unit for all }s\in S,$$
$$f(a)=0\text{ it follows there is an element $s\in S$}as=0$$
$$\text{Every element $b\in B$ is of the form $f(a)f(s)^{-1}$}$$
there is a unique isomorphism of rings $F: S^{-1}A \rightarrow B$ such $F \circ p =f$ where $p: A \rightarrow S^{-1}A$ is the canonical map. This property characterize the localization up to isomorphism.
This enables you to localize rings that are not integral domains, and this agrees with your "topological intuition". If $\mathfrak{p} \in U \subseteq Y:=Spec(T)$ with $U$ an open set you get isomorphisms
$$\mathcal{O}_{Y, \mathfrak{p} } \cong \mathcal{O}_{U,\mathfrak{p} }$$
of local rings.
Example: Direct sums. The prime ideals in the direct sum $R:= A\oplus B$ are of the form $I:=\mathfrak{p}⊕B$ (or vice versa), and if you localize $R$ at $I$ you get
$$R_I≅A_{\mathfrak{p}}.$$
The isomorphism $R_I \cong A_{\mathfrak{p}}$ agrees with the intuition that the stalk at a point can be calculated using any open set containing the point: $X_1 \subsetneq X$ is an open subset and you may restrict to $X_1$ to calculate the local ring at $I$: There is an isomorphism
$$R_I \cong \mathcal{O}_{X,I} \cong \mathcal{O}_{X_1, \mathfrak{p}} \cong A_{\mathfrak{p}}.$$
Note: By definition $X:=Spec(A), X_1:=Spec(A),X_2:=Spec(B)$. You may write
$$X\cong X_1 \cup X_2$$
as a disjoint union, and $X_1\subseteq X$ is open and closed. The prime ideal $\mathfrak{p} \in X_1$ is contained in the open set $X_1$.