Localization of a finitely generated module is nonzero iff the anihilator is contained in the prime ideal? 
I'm a bit stuck on both directions.
I think I need to use the fact that M is finitely generated and maybe find a way to show each of the generators is killed by everything and that would imply $M_p$ is infact 0?
https://dornsife.usc.edu/assets/sites/363/docs/F17_510ab.pdf
 A: First let us recall the definition of $M_P$. Given an $R$-module $M$, and a prime ideal $P \subset R$, we define the multiplicative set (contains $1$ and closed under multiplication) $S = R-P$. The localization of $M$ at the prime ideal $P$ denoted as $M_P$ is a set of equivalent classes of the form $\frac{m}{s}$ where $m\in M$ and $s\in S$, and 
$$\frac{m}{s} \sim \frac{n}{t}$$
if there exists a $c\in S$ such that 
$$tcm = scn \quad \text{ in } M.$$
$(\Rightarrow)$ By contrapositive, if $Ann(M)$ is not contained in $P$, so there exists some $Ann(M) \ni r\in R-P = :S$.  Now given any $\frac{m}{s}$ where $m\in M$ and $s\in S$, we will show 
$$\frac{m}{s} \sim \frac{0}{1} \quad \text{ this is the zero element in $M_P$ }.$$
To show this, we will show there exists a $c\in S$ such that $cm = s0= 0$ in $M$, just take $c = r\in Ann(M)$.
$(\Leftarrow)$ Again by contrapositive, suppose $M_P = 0$, this means for each generator of $M$, call it $m_i$, we must have 
$$\frac{m_i}{1} \sim \frac{0}{1}$$
this means there exists a $c_i\in S$ such that $c_im_i = 0$ in $M$. Since $M$ is finitely generated, then $Ann(M) \ni \prod_i c_i \in S$, so $Ann(M)$ is not contained in $P$.
(So the above proof is basically showed "$M_P = 0$ if and only if $Ann(M)\not\subset P$.)
