Proposition

Let $S$ be a multiplicatively closed subset of a ring $A$, and let $M$ be a finitely generated $A$-module. Prove that $S^{-1}M = 0$ if and only if there exists $s \in S$ such that $sM = 0$.

Solution

Suppose that there exists $t \in S$ such that $tM = 0$. Let $(m, s) \in S^{-1}M$ be given. Then $t(m - 0) = 0$, so $(m, s) \equiv (0, 1)$.

On the other hand, suppose $S^{-1}M = 0$. Since $M$ is finitely generated, there exist $m_1, \cdots, m_n \in M$ such that $M = \{ a_1m_1 + \cdots + a_nm_n \mid a_1, \cdots, a_n \in A \}$. For each $m_i$, $(m_i, 1) \equiv (0, 1)$, so there must exist $t_i \in S$ such that $t_im_i = 0$. Let $t = t_1 \cdots t_n$. Then $\forall a_1m_1 + \cdots + a_nm_n \in M, t(a_1m_1 + \cdots + a_nm_n) = a_1(tm_1) + \cdots + a_n(tm_n) = 0$.