On modules which elements are annihilated by a power of the maximal ideal and localization

Question

Let $R$ be a Noetherian ring and let $M$ be an $R$-module in which every element is annihilated by a power of a maximal ideal $m$. Is there a natural way to define a structure of $R_m$-module on $M$?

This question arises from my try proving the functorial isomorphism (in the category of $R$-modules) $H_m^i(\_)\simeq H_{m_m}^i(R_m\otimes_R\_)$ where each functor represents the $i$-th local cohomology. Initially I can prove that $R_m\otimes H_m^i(\_)\simeq H_{m_m}^i(R_m\otimes_R\_)$. I also know that the local cohomology module $H_m^i(M)$ satisfies the hypothesis in my question above, so perhaps it would induce a structure of $R_m$-module on the local cohomology and we are done.

Thank you for any help.

Answer 1

There are lot of different ways to understand your question. Here's one approach which proceeds from elementary principles:

Suppose $\mathfrak{m}$ is a maximal ideal of a Noetherian ring $R$, and $M$ an $R$-module. If, for every $m \in M-\{0\}$, we have that $m$ is annihilated by a power of $\mathfrak{m}$, then $M$ is naturally an $R_{\mathfrak{m}}$-module.

Proof: We claim $M \cong M_{\mathfrak{m}}$ as $R$-modules. First let's note the following: Let $s \in R-\mathfrak{m}$. If $m \in M-\{0\}$, then $\operatorname{Ann}_R(m)$ is $\mathfrak{m}$-primary ideal of $R$, and $mR \cong R/\operatorname{Ann}_R(m)$. But $R/\operatorname{Ann}_R(m)$ is a local ring whose maximal ideal is the image of $\mathfrak{m}$, and so multiplication by any $s$ outside of $\mathfrak{m}$ gives an automorphism on $mR$. In particular, it gives an automorphism on $M$. Let $s_M$ denote the multiplication map defined by $s$ on $M$.

With this idea in mind, the natural map to define is $\phi: M_{\mathfrak{m}} \to M$ by $\dfrac{m}{s} \mapsto s_M^{-1}m$. We easily check this map is $R$-linear. We need to show this map is well-defined and injective; it's clear that it will be surjective, since we can always take $s=1$.

Well-definedness: Suppose $\dfrac{m}{s}=\dfrac{m'}{s'}$. Then there exists a $u \in R-\mathfrak{m}$ such that $u(s'm-sm')=0$. If $s'm-sm' \ne 0$, then, by assumption, we would have $\operatorname{Ann}_R(s'm-sm') \subseteq \mathfrak{m}$, which is not the case. So $s'm=sm'$ which gives $s_M^{-1}m=(s'_M)^{-1}m'$. Thus $\phi$ is well-defined.

Injectivity: If $\dfrac{m}{s} \in \ker \phi$, then $s_M^{-1}m=0$, but $s_M^{-1}$ is injective, so $m=0$.

Thus $\phi$ is an isomorphism.