Let $A$ be a discrete valuation ring, $p\in A$ be irreducible, $M$ be an $A$-module and $L=p^nA$ for some $n\geq 0$. There is a unique $A$-linear map $\psi:L\otimes M\to M$ satisfying $\psi(a\otimes m)=am$ for all $a\in L$ and $m\in M$.
I have read that we can 'identify' $\psi$ with the map $m\mapsto p^nm:M\to M$, so that $\psi$ is injective if and only if $m\mapsto p^nm$ is. The map $m\mapsto p^n\otimes m:M\to L\otimes M$ is clearly $A$-linear and surjective. But to show that this is injective, I need to prove that $p^n\otimes m=0$ implies $m=0$. If $p^n\otimes m=0$, then applying $\psi$ shows that $p^nm=0$. But how do we know that $m=0$ from this? Am I missing something?
For a bit of context, I am trying to prove that $M$ is flat iff $\{m\in M:pm=0\}=0$.
Since $A$ is integral, there is an $A$-module isomorphism $L\simeq A$ sending $p^n$ to $1$ ($L$ is a free $A$-module of rank $1$, with basis element $p^n$). Then the fact that, in $L\otimes_A M$, $p^n\otimes m=0$ implies $m=0$ is exactly the same as the fact that, in $A\otimes_A M$, $1\otimes m=0$ implies $m=0$.