# Homomorphism from module to localization

Let $M$ be an $R$-module of finite length and let $\mbox{Ann}(M)\subset P_1,...,P_k$ be maximal ideals. If $n\in\mathbb{Z}_+$ is such that $P_1^n\cdots P_k^n\subset\mbox{Ann}(M)$, then the sequences $0\rightarrow P_i^nM \rightarrow M \rightarrow M_{P_i} \rightarrow 0$ are exact and $M/P_iM \cong M_{P_i}$.

I wonder what the maps are in these sequences. Especially what is the homomorphism $M \rightarrow M_{P_i}$? I especially wonder because I can not find such a surjective homomorphism.

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Welcome to math.stackexchange! To get better answers, I would suggest editing your question to make it clearer what is given and what you understand and don't understand. For example, you use $P_i$ and $p_i$, is there any difference? Are you talking about localization? Then use $M_{p_i}$ Are you asking what are the maps in the sequence? etc. – levap Dec 3 '12 at 20:03
because I can not log in to my account from the computer in my school for some reason! I am very sorry about that! And also, I wanted to email you, but didn't find any link on your page for doing that. I would be very thankful for your help though! – lena Dec 3 '12 at 20:26

For a prime $P$ of $R$ containing $\mbox{Ann}(M)$ one denotes by $\overline{P}$ its image in $\overline{R}=R/\mbox{Ann}(M)$. Since $M$ has finite length, $\overline{R}$ is artinian and therefore there is a positive integer $n$ such that $\overline{P_1}^n\cdots\overline{P_k}^n=(0)$. Then $\overline{R}\simeq\overline{R}/\overline{P_1}^n\times\cdots\times\overline{R}/\overline{P_k}^n$. This is a ring isomorphism, but it is also an $R$-module isomorphism. Tensor with $M$ and obtain $M\simeq M/P_1^nM\times\cdots\times M/P_k^nM$. Localizing at $P_i$ one gets $M_{P_i}\simeq(M/P_i^nM)_{P_i}$. The last step is to show that the canonical map $M/P_i^nM\to(M/P_i^nM)_{P_i}$ is an isomorphism. In order to prove this try to simplify as follows: take $N$ an $S$-module and $\mathfrak{m}$ a maximal ideal of $S$ such that $\mathfrak{m}^nN=0$. Then the canonical map $N\to N_\mathfrak{m}$ is an isomorphism. (Hint: For an element $u\in S-\mathfrak{m}$ we have $(u)+\mathfrak{m}^n=S$.)