Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 7 / 1 / b / Solution

The forward implication is immediate: the localization of a module that is zero remains zero. Conversely, suppose $M\ne 0$ and choose $0\ne x\in M$. Its annihilatoris 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, provingIn the suggested cyclic description, $Ax\cong A/J$ and part (a) gives $(Ax{)}_{\mathfrak{m}}\ne 0$; 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.