localization of a module
Let $R$ be a commutative ring and $M$ an $R$module. Let $S\subset R$ be a nonempty multiplicative set. Form the Cartesian product^{} $M\times S$, and define a binary relation^{} $\sim $ on $M\times S$ as follows:
$({m}_{1},{s}_{1})\sim ({m}_{2},{s}_{2})$ if and only if there is some $t\in S$ such that $t({s}_{2}{m}_{1}{s}_{1}{m}_{2})=0$
Proposition 1.
$\sim $ on $M\mathrm{\times}S$ is an equivalence relation^{}.
Proof.
Clearly $(m,s)\sim (m,s)$ as $t(smsm)=0$ for any $t\in S$, where $S\ne \mathrm{\varnothing}$. Also, $({m}_{1},{s}_{1})\sim ({m}_{2},{s}_{2})$ implies that $({m}_{2},{s}_{2})\sim ({m}_{1},{s}_{1})$, since $t({s}_{2}{m}_{1}{s}_{1}{m}_{2})=0$ implies that $t({s}_{1}{m}_{2}{s}_{2}{m}_{1})=0$. Finally, given $({m}_{1},{s}_{1})\sim ({m}_{2},{s}_{2})$ and $({m}_{2},{s}_{2})\sim ({m}_{3},{s}_{3})$, we are led to two equations $t({s}_{2}{m}_{1}{s}_{1}{m}_{2})=0$ and $u({s}_{3}{m}_{2}{s}_{2}{m}_{3})=0$ for some $t,u\in S$. Expanding and rearranging these, then multiplying the first equation by $u{s}_{3}$ and the second by $t{s}_{1}$, we get $tu{s}_{2}({s}_{3}{m}_{1}{s}_{1}{m}_{3})=0$. Since $tu{s}_{2}\in S$, $({m}_{1},{s}_{1})\sim ({m}_{3},{s}_{3})$ as required. ∎
Let ${M}_{S}$ be the set of equivalence classes^{} in $M\times S$ under $\sim $. For each $(m,s)\in M\times S$, write
$$[(m,s)]\text{or more commonly}\frac{m}{s}$$ 
the equivalence class in ${M}_{S}$ containing $(m,s)$. Next,

•
define a binary operation^{} $+$ on ${M}_{S}$ as follows:
$$\frac{{m}_{1}}{{s}_{1}}+\frac{{m}_{2}}{{s}_{2}}:=\frac{{s}_{2}{m}_{1}+{s}_{1}{m}_{2}}{{s}_{1}{s}_{2}}.$$ 
•
define a function $\cdot :{R}_{S}\times {M}_{S}\to {M}_{S}$ as follows:
$$\frac{r}{s}\cdot \frac{m}{t}:=\frac{rm}{st}$$ where ${R}_{S}$ is the localization^{} of $R$ over $S$.
Proposition 2.
${M}_{S}$ together with $\mathrm{+}$ and $\mathrm{\cdot}$ defined above is a unital module over ${R}_{S}$.
Proof.
That $+$ and $\cdot $ are welldefined is based on the following: if $({m}_{1},{s}_{1})\sim ({m}_{2},{s}_{2})$, then
$$\frac{m}{s}+\frac{{m}_{1}}{{s}_{1}}=\frac{m}{s}+\frac{{m}_{2}}{{s}_{2}},\frac{{m}_{1}}{{s}_{1}}+\frac{m}{s}=\frac{{m}_{2}}{{s}_{2}}+\frac{m}{s},\text{and}\mathit{\hspace{1em}}\frac{r}{s}\cdot \frac{{m}_{1}}{{s}_{1}}=\frac{r}{s}\cdot \frac{{m}_{2}}{{s}_{2}},$$ 
which are clear by Proposition^{} $1$. Furthermore $+$ is commutative^{} and associative and that $\cdot $ distributes over $+$ on both sides, which are all properties inherited from $M$. Next, $\frac{0}{s}$ is the additive identity in ${M}_{S}$ and $\frac{m}{s}}\in {M}_{S$ is the additive inverse of $\frac{m}{s}$. So ${M}_{S}$ is a module over ${R}_{S}$. Finally, since $(mt,st)\sim (m,s)$ for any $t\in S$, $\frac{t}{t}}\cdot {\displaystyle \frac{m}{s}}={\displaystyle \frac{m}{s}$ so that ${M}_{S}$ is unital. ∎
Definition. ${M}_{S}$, as an ${R}_{S}$module, is called the localization of $M$ at $S$. ${M}_{S}$ is also written ${S}^{1}M$.
Remarks.

•
The notion of the localization of a module generalizes that of a ring in the sense that ${R}_{S}$ is the localization of $R$ at $S$ as an ${R}_{S}$module.

•
If $S=R\U0001d52d$, where $\U0001d52d$ is a prime ideal^{} in $R$, then ${M}_{S}$ is usually written ${M}_{\U0001d52d}$.
Title  localization of a module 

Canonical name  LocalizationOfAModule 
Date of creation  20130322 17:26:59 
Last modified on  20130322 17:26:59 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  7 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 13B30 