This is regarding Proposition 3.14 of Atiyah and Macdonald. It says that if $M$ is a finite $R$ module then the formation of localization and annihilators commute. This is well known, and the proof is clear. But I cannot find where the mistake in the proof below is and the proof below does not require $M$ to be finitely generated, so there is an error.
Let $r/s \in Ann(S^{-1}M)$. Thus $r/s.m/t=rm/st=0$, which implies that $urm=0$ for some $u\in S$. So $ru\in Ann(M)$. Hence $r/s=ru/su\in S^{-1}(Ann M)$, which shows that $Ann(S^{-1}M)\subseteq S^{-1}(Ann M)$.
Conversely, let $r/s\in S^{-1}(Ann M)$, where $r\in Ann M$. Consider $m/t\in S^{-1}M$. Note that $r/s.m/t=rm/st=0$. Hence $r/s\in Ann(S^{-1}M)$.
Where is the error? Thanks.
Just to remove this from the unanswered list, here I write down the proof suggested by Darij Grinberg.
Let $M$ be generated by $m_1,\ldots, m_t$. By the first part of the above faulty proof, there is a $u_i\in S$ such that $u_irm_i=0$ for all $1\leq i\leq t$. Thus $ru_i\in Ann(m_i)$ for all $1\leq i\leq t$. Set $u=u_1\cdots u_t$. Note that $ru\in \bigcap_{i=1}^t Ann\; m_i=Ann M$. Note that $r/s=ru/su\in S^{-1}(Ann M)$.