Consider $\{D_i:i\in I\}$ as a collection of multiplicative closed subsets of a commutative ring $R$. Show that the following statements are equivalent
(1) If $D^{-1}_iM=0$ for every $i\in I$ and for all modules $M$, then $M=0$;
(2) For any $(d_i)_{i\in I}\in \prod_{i\in I}D_i,$ the set $\langle\{d_i:i\in I\}\rangle=R,$ where $\langle\{d_i:i\in I\}\rangle$ is the ideal generated by $\{d_i:i\in I\}$.
(1)$\rightarrow$(2): Let $(d_i)_i$ be an element of $\prod_i D_i$ and let $I$ be the ideal generated by the $d_i$. Then $R/I$ is an $R$-module and for all $i$ we have $D_i^{-1}R/I = 0$ (because $0$ mod $I$ is an element of $D_i$ for all $i$). Thus by our assumption $R/I = 0$ and therefore $I=R$.
(2)$\rightarrow$ (1): Since $D_i^{-1}M = 0$ for all $i$ it follows that for all $i$ there is a $d_i\in D_i$, such that $d_iM = 0$. Now consider the ideal $I$ generated by the $d_i$'s. Then by (2) $I=R$, thus there is a linear combination $\sum_i a_id_i =1$. So $M=1M=\sum_i a_id_i M=0$.