Past exam of the commutative algebra course of the University of Oxford / 2023 / 3 / e / Solution

Set $L=M/\mathfrak{a}M$, a quotient module. The exactness of localization gives, for every maximal ideal $\mathfrak{m}$,If $\mathfrak{m}\supseteq \mathfrak{a}$, the hypothesis $M_{\mathfrak{m}}=0$ makes this quotient module zero. If $\mathfrak{m}⊉\mathfrak{a}$, choose $s\in \mathfrak{a}\setminus \mathfrak{m}$. Its image in $R_{\mathfrak{m}}$ is a unit, so $\mathfrak{a}{R}_{\mathfrak{m}}=R_{\mathfrak{m}}$ and the displayed quotient module is again zero. ThusFor 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 $\ell /1\ne 0$ in $L_{\mathfrak{m}}$: any element annihilating $\ell$ lies inside $\mathfrak{m}$. This contradicts the vanishing just proved. Hence $L=0$, andThe use of localization detects zero elements is valid for every module; neither finite generation nor a Noetherian hypothesis is needed.