Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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

1 Answer 1

up vote 5 down vote accepted

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 $y\notin I$.

share|improve this answer
Dear @QiL, I confess that I am baffled by the equivalence you evoke in your parenthesis of the second line. Dare I ask you for a few words of explanation ? –  Georges Elencwajg Nov 6 '11 at 17:31
Dear Georges, thank you for pointing out my mistake. I edited my answer accordingly at two places. –  user18119 Nov 6 '11 at 22:51
Thanks, @QiL. My finger was itching to click on a certain upward directed arrow... –  Georges Elencwajg Nov 6 '11 at 23:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.