= Solution
Set $L=M/\mathfrak aM$, a <quotient module>. The <exactness of localization> gives, for every <maximal ideal> $\mathfrak m$,
$$
L_{\mathfrak m}\cong M_{\mathfrak m}/\mathfrak aR_{\mathfrak m}M_{\mathfrak m}.
$$
If $\mathfrak m\supseteq\mathfrak a$, the hypothesis $M_{\mathfrak m}=0$ makes this <quotient module> zero. If $\mathfrak m\not\supseteq\mathfrak a$, choose $s\in\mathfrak a\setminus\mathfrak m$. Its image in $R_{\mathfrak m}$ is a <unit>, so $\mathfrak aR_{\mathfrak m}=R_{\mathfrak m}$ and the displayed <quotient module> is again zero. Thus
$$
L_{\mathfrak m}=0\qquad\text{for every maximal ideal }\mathfrak m.
$$
For completeness, if $0\ne\ell\in L$, its <annihilator> is a proper <ideal>, and hence lies in a <maximal ideal> $\mathfrak m$. The <vanishing criterion in a module localization> shows $\ell/1\ne0$ in $L_{\mathfrak m}$: any element annihilating $\ell$ lies inside $\mathfrak m$. This contradicts the vanishing just proved. Hence $L=0$, and
$$
\boxed{M/\mathfrak aM=0,\qquad\mathfrak aM=M.}
$$
The use of <localization detects zero elements> is valid for every <module>; neither finite generation nor a <Noetherian> hypothesis is needed.
Back to article page
