Assume $R$ is a commutative local ring, $I$ is a proper ideal in $R$ and $M$ is a finitely generated $R$-module. Is it true that $\operatorname{Ann}(M/IM)=I+\operatorname{Ann}(M)$?
(Note that $\supseteq$ is clear.)
If it's true I would like (a hint for) an elementary proof (preferably avoiding any homological algebra).
Typically only a power of $Ann(M/IM)$ will be contained in $I+Ann (M)$. For a simple example, take the `universal' example. Let $R$ be the power series over a field in three variables $x,y,z$. Consider the module $M$ given as the quotient of the free module $Re_1\oplus Re_2$ by $xe_1+ye_2, ze_1+xe_2$. You can easily check that $Ann(M)$ is the ideal generated by $x^2-yz$, and if $I=(y,z)$, then $Ann(M/IM)=(x,y,z)$.