= Solution
The forward implication is immediate: the <localization of a module> that is zero remains zero. Conversely, suppose $M\ne0$ and choose $0\ne x\in M$. Its <annihilator>
$$
J=\operatorname{Ann}_A(x)=\{a\in A:ax=0\}
$$
is a proper <ideal>, since $1\notin J$. Choose a <maximal ideal> $\mathfrak m$ containing $J$. Then $x/1$ cannot be zero in $M_{\mathfrak m}$: otherwise the <vanishing criterion in a module localization> would give $sx=0$ for some $s\notin\mathfrak m$, contradicting $s\in J\subseteq\mathfrak m$. Thus some maximal <localization> is nonzero, proving
$$
\boxed{M=0\iff M_{\mathfrak m}=0\text{ for every maximal ideal }\mathfrak m.}
$$
In the suggested cyclic description, $Ax\cong A/J$ and part (a) gives $(Ax)_{\mathfrak m}\ne0$; the same fraction argument embeds it in $M_{\mathfrak m}$. This is <localization detects zero elements>, applied to an arbitrary <module>; finite generation is unnecessary.
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