When is the localization of a module trivial? Let $R$ be a commutative ring and $M$ a finitely generated $R$-module. Let $S^{-1}M$ be localization of $M$, where $S$ is a multiplicatively closed subset of $R$. 

How to show that $S^{−1}M =0$ if and only if there exists $s \in S$ such that $sM =0$?

Context
The idea of localization is to make elements of $s$ invertible. To this end, $S^{-1}M$ is defined as the set of equivalence classes of pairs $(m, s)\in M\times S$, where two  pairs $(m, s)$ and $(n, t)$ are  equivalent if $u(sn-tm) = 0$ for some $u\in S$. 
The localization of a ring is trivial if and only if $S$ contains $0$. However, here the assumption is a bit different: it may be that $0\notin S$. 

By definition
$$\frac{a_1}{s_2}=\frac{a_2}{s_2}\;\;,\;\;\frac{a_i}{s_i}\in S^{-1}M\,\Longleftrightarrow\,\exists\,\,r\in S\,\,\,s.t.\,\,\,r(s_2a_1-s_1a_2)=0$$
Say thus that $\,M=\langle\,m_1,\ldots,m_n\,\rangle_R\,$ , and suppose there exists $$\,t\in S\subset R\;\; s.t.\;\; tM=0\Longleftrightarrow tm_i=0\,\,,\,\,\forall\,i=1,\ldots,n$$
then:
$$\forall \;m=\sum_{k=1}^nr_im_i\in M\;\;,\;\;\frac{m}{s}=\frac{0}{s}=0\Longleftrightarrow$$
$$\,\exists\,\,s'\in S\,\,\,s.t.\,\,\,s'(sm)=s'sm=\sum_{k=1}^ns'sr_im_i=0$$
Now just choose $\,s'=t\,$ above and polish...
The other direction:
$$S^{-1}M=0\Longleftrightarrow \sum_{k=1}^n\frac{r_im_i}{s_i}=0\,\,,\,\,\forall\,i=1,\ldots,n$$
In particular
$$\forall\,i=1,\ldots,n\;\exists\,x_i\in S\,\,s.t.\,\,\,\,x_ir_im_i=0$$
Now take a look at $\,\displaystyle{t:=\prod_{k=1}^nx_i}\,\ldots$
