# Annihilators and exact sequences

Let $R$ be a commutative ring. Let $M_1$, $M_2$ and $M_3$ be $R$-modules. Let the following sequence be exact:

$$0\longrightarrow M_1 \overset{f}{\longrightarrow}M_2\overset{g}{\longrightarrow}M_3\longrightarrow 0.$$

I have proven that

$$\mbox{Ann}(M_1)\mbox{Ann}(M_3) \subset \mbox{Ann}(M_2).$$

And I know that the following does not hold in general

$$\mbox{Ann}(M_2) \subset\mbox{Ann}(M_1)\mbox{Ann}(M_3).$$

I am looking for a counterexample.

• I'm guessing (?) that you're talking of modules over some ring $\;R\;$ here? A commutative ring? With unit? Something...? Feb 28, 2014 at 14:52

Let $R=k[x,y]$, and let $$M_1=(x,y)/(y),\qquad M_2=k[x,y]/(y),\qquad M_3=k[x,y]/(x,y).$$ Then $M_1$ is the kernel of the obvious quotient map $M_2\to M_3$ (third isomorphism theorem), so we do have an exact sequence $$0\longrightarrow M_1 \xrightarrow{\text{inclusion}}M_2\xrightarrow{\text{quotient}}M_3\longrightarrow 0$$ but $$\mathrm{Ann}(M_1)=(y),\qquad \mathrm{Ann}(M_2)=(y),\qquad \mathrm{Ann}(M_3)=(x,y)$$ so that $$\mathrm{Ann}(M_2)=(y)\not\subset (xy,y^2)=(y)(x,y)=\mathrm{Ann}(M_1)\mathrm{Ann}(M_3).$$