Let $A$ be a commutative ring with identity and let $f_1,...,f_n\in A$ be such that $(f_1,...,f_n)=A$. Also let $I\subsetneq J$ be ideals. In exercise 5.3.H of Ravil Vakil's notes he suggest that we must have $$A_{f_i}I\neq A_{f_i}J$$ for some $i$ (where $A_{f_i}I$ is the ideal in $A_{f_i}$ generated by the image of $I$ under the natural ring homomorphism $A\rightarrow A_{f_i}$). His hint is that $A\rightarrow \Pi A_{f_i}$ is an inclusion, which is easy to see. How does one get $i$ for which $A_{f_i}I\neq A_{f_i}J$ from this though?
(Remark: This seems easy to prove in the case that the ideals are finitely generated, since then the nonzero $A$ module $J/I$ must localize to something nonzero, $(J/I)_{f_i} \neq 0$, for some $i$. And since Vakil's claim is only used to prove that (all $A_{f_i}$ are Noetherian) $\implies$ ($A$ is Noetherian), it seems one can assume the ideals are finitely generated since if there's an infinite ascending chain of ideals in $A$ then there's an infinite ascending chain of finitely generated ideals in $A$. But i'd nonetheless like to know why this more general statement is true.)
Look at the commutative diagram
$\require{AMScd} \begin{CD} A/I @>{i}>> \Pi_i {A_{f_i}/IA_{f_i}} \\ @V{q}VV @V{q_l}VV \\ A/J @>{i}>> \Pi_i {A_{f_i}/JA_{f_i}}; \end{CD} $
The horizontal maps are injective (you have already shown that). If $IA_{f_i}=JA_{f_i} \forall i $ then the map $q_l$ is an isomorphism and hence $q$ must be injective. Thus $I=J$