If a module is nonzero, then a localization module is nonzero Let $R$ be a commutative ring, when $\mathfrak p$ is a prime ideal, there is the localization $M_{\mathfrak p}:=S^{-1}M$, where $S=R\setminus\mathfrak p$. Show: 

If $M$ is a nonzero $R$-module, then there exists a prime ideal $\mathfrak p$ such that $M_{\mathfrak p}\neq 0$.

I don't understand this exercise, in which case is $M_{\mathfrak p}$ zero ? $M$ is nonzero, this is good. So does it remain to prove that the set $R\setminus\mathfrak p$ is nonempty ?
$\textbf{EDIT:}$ I think I got it: 
If $M$ is nonzero $\implies\exists m\in M,m\neq 0$
let $I=\{a\in R:aM=0\}$
since $1\in R\Rightarrow1\cdot m=m\neq 0\implies I\neq R$
and $I$ is an ideal, so $\exists$ maximal ideal $\mathfrak p$, s.t. $I\subseteq\mathfrak p$
but every maximal ideal is prime, so suppose $M_{\mathfrak p}=0$ then ;
$m/1=0$ in $M_{\mathfrak p}\implies m/1=0/1$
by definition $\exists s\in R\setminus\mathfrak p$ such that $0=s(1\cdot m-1\cdot 0)=sm$ 
Hence $s\in I\subseteq\mathfrak p$ but $s\in R\setminus\mathfrak p\implies s\notin\mathfrak p$ which is absurd.
 A: Since the question now has an answer, I'll just add an illustrative example.    
The $\mathbb Z$-module $M=\mathbb Z/2\mathbb Z$ has non-zero  localization $M_{2\mathbb Z}\neq 0$ at the prime $2\mathbb Z=(2)\subset \mathbb Z$ .
But all its other localizations are zero: $M_{\mathfrak p}=0$ for all   primes $\mathfrak p\neq (2)$ namely for $\mathfrak p=(0),(3),(5),(7),(11),\cdots,(4999),\cdots\subset \mathbb Z$ .  
A: $M_p=0$ if and only if $x/s=0/1$ for all $x\in M$ and $s\in R-p$ if and only if for all $x\in M$ there is $t_p\in R-p$ such that $t_px=0$.

If all localizations $M_p$ are $0$, then $M=0$. 

Pick an element $x\in M$. Since $x/s=0/1$ in $M_p$ there is $t_p\in R-p$ such that $t_px=0$. Now notice that the ideal generated by all $t_p$, when $p$ runs through the set of prime ideals of $R$, equals the whole ring. Write $1$ as a (finite) linear combination of $t_p$ and multiplying by $x$ get $x=0$.
Edit. An alternative proof (using annihilators): 
If $M\ne 0$ pick a non-zero element $x\in M$ and consider its annihilator $\operatorname{Ann}(x)=\{a\in R\mid ax=0\}$. This is an ideal of $R$ and clearly $\operatorname{Ann}(x)\ne R$. There is a prime ideal $p$ of $R$ such that $\operatorname{Ann}(x)\subseteq p$. Then $x/1\ne0/1$ in $M_p$, otherwise there is $t_p\in R-p$ such that $t_px=0$, a contradiction. This proves that $M_p\ne0$.
