Let $\phi:N\to M$ be an injective $O-$module homomoprhism where $O$ is assumed to be dedekind, $N,M$ are f.g. One knows that $\hat{O}\otimes_O M\cong\varprojlim_r\frac{M}{p^rM}$ where $\hat{O}$ is completion against $0\neq p\in Spec(O)$. Let $\tilde{N}=\hat{O}\otimes N$ and similarly for $\tilde{M}$.
Then $\tilde{\phi}:\tilde{N}\to\tilde{M}$ is injection. There is a part of proof is not obvious to me. Let $\pi_r:\tilde{N}\to\frac{N}{p^rN}$ be the $r-th$ component.
"Suppose $a\in Ker(\tilde{\phi})$. Then $\pi_{n+k}(a) \in Ker(\frac{N}{Np^{n+k}}\to\frac{M}{Mp^{n+k}})$. So $a(n)\in\frac{N\cap Mp^{n+k}+Np^n}{Np^n}$. Then allowing $k$ varies to see $a(n)\in\frac{(\cap_{k\geq 0}Mp^{n+k})\cap N+Np^n}{Np^n}$. Then if $M$ is finitely generated module over dedekind domain $O$ and let $tM$ be the torsion part of $M$, one has $\cap_{r}p^rM=(tM^{(p)})$ where $p^sI=Ann(tM),p\not\vert I$ and $tM^{(p)}$ is the component of $tM$ annhilated by $I$. Thus $a(n)\in\frac{(tM^{(p)}\cap N)+Np^n}{Np^n}=\frac{tM^{(p)}\cap N}{tM^{(p)}\cap Np^n}$ by 2nd isomorphism theorem. '$\textbf{Then the module}$ $\frac{tM^{(p)}\cap N}{tM^{(p)}\cap Np^n}=0$'."
$\textbf{Q:}$ Why is the module $\frac{tM^{(p)}\cap N}{tM^{(p)}\cap Np^n}=0$ is obvious here? I used the fact that $ptM^{(p)}=tM^{(p)}$ and $I,p$ are coprime. This is related to taylor frohlich Algebraic Number Theory pg 92 bottom of 4.7's proof for flatness of completion ring.