Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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.
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 $$\mathrm{Ann}(M_2)=(y)\not\subset (xy,y^2)=(y)(x,y)=\mathrm{Ann}(M_1)\mathrm{Ann}(M_3).$$
