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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).

user26857
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jgeilberg
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    Some related questions: http://math.stackexchange.com/questions/79538/annihilator-of-quotient-module-m-im and http://math.stackexchange.com/questions/811875/annihilator-of-a-quotient-module. However I did not find the answer there. – jgeilberg Sep 14 '15 at 19:40
  • This answer contains a counterexample. – user26857 Sep 16 '15 at 16:02

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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 $Ann(M/IM)=(x,y,z)$.

Mohan
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