How do I show that given a prime ideal $P\subset R$, the map of $M\to M_P, m\mapsto m/1$ is injective if and only if $R\setminus P$ contains no elements killing a nonzero element of $M$?
One direction looks easy. Suppose the map is not injective. Then $m/1=m '/1$ for some $m\ne m'$. So there is $v\in R\setminus P$ with $v(m-m')=0$ in $M$. So $v$ kills the nonzero element $m-m'$ of $M$.
The other direction isn't that easy. Suppose $v\in R\setminus P$ kills an element $m\ne 0$. How to prove that the map isn't injective?
That is because $\dfrac m1=0$ in $M_P$ by definition since $vm=0$ in $M$, hence the kernel of the canonical map (which is $R$-linear) is not reduced to $0$.