Let $(R,m,k)$ be a commutative Noetherian local ring and let $M$ be an $R$-module in which every element is annihilated by some power o $m$. It induces a structure of $\widehat{R}$-module on $M$ where $\widehat{R}$ is the $m$-adic completion of $R$: given $r\in\widehat{R}$ and $x\in M$ then $rx=r_nx$ where $n$ is the smallest integer such that $m^nx=0$ and $r\mapsto r_n+m^n$ through the canonical map $\widehat{R}\rightarrow R/m^n$.
Is it true that $M$ is isomorphic (as $R$-module as $\widehat{R}$-module) to $\widehat{R}\otimes_R M$?
I've tried to prove it considering the composition $M\rightarrow\widehat{R}\otimes_R M\rightarrow\widehat{R}\otimes_{\widehat{R}}M$ given by $$x\mapsto1\otimes x\mapsto1\otimes x$$ since it's an isomorphism and thus $M\rightarrow\widehat{R}\otimes_R M$ is injective. In order to prove that it is also surjective, any element in $\widehat{R}\otimes_R M$ of the form $r\otimes x$ must be reached by some element in $M$. The natural candidate is $rx=r_nx\in M$, but $rx\mapsto1\otimes(r_nx)=(1\cdot r_n)\otimes x=r_n\otimes x$. Is it clear that $r_n\otimes x=r\otimes x?$ How could I proceed?
Indeed it is true and to prove this is enough to check that $r_n\otimes x=r\otimes x$. As $rx=r_nx$ we have $r-r_n\in\widehat{m}^n=m^n\widehat{R}$, i.e., $r-r_n=\sum_ir_is_i$ with $r_i\in m^n$ and $s_i\in\widehat{R}$. Then
$$x\otimes(r-r_n)=x\otimes(\sum_ir_is_i)=\sum_i(r_ix\otimes s_i)=0.$$