= Solution
If $M=0$, all its <localizations> are zero. Conversely, take $0\ne m\in M$. Its <annihilator> $I=\operatorname{Ann}_R(m)$ is a proper <ideal>, so it is contained in a maximal <ideal> $P$, which is prime. The zero criterion for <module localization> says that $m/1=0$ in $M_P$ only if some $s\notin P$ satisfies $sm=0$. This would put $s\in I\subseteq P$, a contradiction. Therefore $M_P\ne0$, proving
$$
\boxed{M=0\iff M_P=0\text{ for every prime ideal }P.}
$$
In fact maximal <ideals> suffice, and neither finite generation nor <Noetherianity> is needed for this implication.
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