# Annihilator of quotient module M/IM

Let $A$ be a commutative ring, $I$ an ideal of $A$ and $M$ an module over $A$. Is it true that $\operatorname{Ann}(M/IM) = \operatorname{Ann}(M) + I$?

One inclusion is certainly true, but I don't know about the other one. If the above statement does not hold in general, does it perhaps for finitely generated modules?

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Counter-example: $M=k[x]$ is a k-module. Let $I=(x)$. Then $M/IM=k$ which has $(0)$ as annihilator. But $Ann(M)=(0)$ also, so the right hand side is $(x) \neq (0)$. M is not finitely generated here however. –  Fredrik Meyer Nov 6 '11 at 15:54
@Fredrik Is $A = k$ here? How does $I = (x)$ make sense, then? –  Dylan Moreland Nov 6 '11 at 16:06
@Dylan: Thanks. I should've been more careful. It seems like a bit of modifying will work: Let $A=M=\mathbb{Z}$ and $I=(x)$. Then the same reasoning should hold. –  Fredrik Meyer Nov 6 '11 at 19:59

Replace $A$ with $A/\operatorname{Ann}(M)$. Then $M$ is an $A$-module with trivial annihilator ideal. Let $a\in \operatorname{Ann}(M/IM)$ (equivalent to (EDIT) $(aM+IM)/IM=0$). You would like to conclude that $a\in I$, or equivalently, that $(aA+I)/I=0$.
This is true by definition if $M$ is faithfully flat over $A$ (e.g. if $M$ is free of positive rank over $A$) EDIT because then $(aA+I)/I\otimes_A M$ is isomorphic to $(aM+IM)/IM=0$ by flatness, hence $(aA+I)/I=0$ by faithful flatness. Otherwise it is false even when $M$ is finitely generated over $A$.
Example: let $A=k[x,y]\subset k[t]$ where $k$ is a field, $x=t^2, y=t^3$. Let $M=k[t]$. It is finitely generated over $A$ (a system of generators is $1, t$). Let $I=xA$. Then $IM=t^2k[t]$ and $yM=t^3k[t]\subseteq IM$. So $y\in \operatorname{Ann}(M/IM)$. But $y\notin I$.