# localization of a module

Let $R$ be a commutative ring and $M$ an $R$-module. Let $S\subset R$ be a non-empty 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 $M\times S$ is an equivalence relation.

###### Proof.

Clearly $(m,s)\sim(m,s)$ as $t(sm-sm)=0$ for any $t\in S$, where $S\neq\varnothing$. 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 $us_{3}$ and the second by $ts_{1}$, we get $tus_{2}(s_{3}m_{1}-s_{1}m_{3})=0$. Since $tus_{2}\in S$, $(m_{1},s_{1})\sim(m_{3},s_{3})$ 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

 $[(m,s)]\mbox{ or more commonly }\frac{m}{s}$

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 $+$ and $\cdot$ defined above is a unital module over $R_{S}$.

###### Proof.

That $+$ and $\cdot$ are well-defined is based on the following: if $(m_{1},s_{1})\sim(m_{2},s_{2})$, then

 $\frac{m}{s}+\frac{m_{1}}{s_{1}}=\frac{m}{s}+\frac{m_{2}}{s_{2}},\qquad\frac{m_% {1}}{s_{1}}+\frac{m}{s}=\frac{m_{2}}{s_{2}}+\frac{m}{s},\quad\mbox{and}\quad% \frac{r}{s}\cdot\frac{m_{1}}{s_{1}}=\frac{r}{s}\cdot\frac{m_{2}}{s_{2}},$

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, $\displaystyle{\frac{0}{s}}$ is the additive identity in $M_{S}$ and $\displaystyle{\frac{-m}{s}}\in M_{S}$ is the additive inverse of $\displaystyle{\frac{m}{s}}$. So $M_{S}$ is a module over $R_{S}$. Finally, since $(mt,st)\sim(m,s)$ for any $t\in S$, $\displaystyle{\frac{t}{t}\cdot\frac{m}{s}=\frac{m}{s}}$ 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 $S=R-\mathfrak{p}$, where $\mathfrak{p}$ is a prime ideal in $R$, then $M_{S}$ is usually written $M_{\mathfrak{p}}$.

Title localization of a module LocalizationOfAModule 2013-03-22 17:26:59 2013-03-22 17:26:59 CWoo (3771) CWoo (3771) 7 CWoo (3771) Definition msc 13B30