In commucative algebra, we usually want to embbed a ring into a field to analysis its property. Remark how we construct $\mathbb Q$ on $\mathbb Z$ (从整数构造有理数域：公理化 - 知乎 (zhihu.com)). We try to find out a method to construct the quotient field of a ring. In this way, mathmatician came up the localization.

Def of Localization

Let $A$ be a ring, S be a multiplicative subset of A. Now, we construct from an ordered pair $(a,s)$ with $a \in A, s \in S$. And define a relation on then, \((a,s) \equiv (b,t) \Leftrightarrow (at - bs) u = 0 \text{ for some } u\in S\) This is a equivalence relation. It implies a equivalence class set $S^{-1}A = {a/s|\text{the equivalence class of }(a,s) a \in A, s \in S}$. Furthermore, we can define addiction and multication, \((a/s) + (b/t) = (at + bs)/st\) \((a/s)(b/t)=ab/st\) Then, $S^{-1}A$ satisfies the axioms of a commutative ring with identity.

Similarly, we can define localization of a A-module M as above, and denote as $S^{-1}M$.

局部化究竟保留了什么

Prop 3.11 iv) The prime ideals of $S^{-1}A$ are in one-to-one correspondence $(\mathfrak{p} \leftrightarrow S^{-1}\mathfrak{p} )$ with the prime ideals of A ideals of $A$ ehich don’t meet $S$.

That is said, quotient ring $A/\mathfrak{a}$ kill the ideals not containing the ideal $\mathfrak{a}$ and keep the ideals containing $\mathfrak{a}$. And, localization kill the prime ideals intersect $S$, and keep the prime ideals not intersect $S$.

局部化的代数特性

Denote A as a ring, S as a multiplicative closed subset of A. M,N is A-modole.

Prop 3.3 The operation $S^{-1}$ is exact.

1. $S^{-1}(N+P) = S^{-1}(N) + S^{-1}(P)$
2. $S^{-1}(N \cap P) = S^{-1}(N) \cap S^{-1}(P)$
3. $S^{-1}(M/N) \simeq (S^{-1}M)/(S^{-1}N)$
4. $S^{-1}M \simeq S^{-1}A \otimes_A M$