A relation between an element and an annihilator

Question

Let $M$ be a faithful and finitely generated $R$-module, where $R$ is a commutative ring with $1$, and let $\text{ann}_R (m)=\{r\in R\,|\, r \cdot m=0\}$ for $m\in M$. It is easily seen that $r\in \text{ann}_R (\text{ann}_R (r))$ for $r\in R$, therefore $r=0$ if $\text{ann}_R (\text{ann}_R (r))=0$. Is this statement true for $m\in M$, that is, if $\text{ann}_R (\text{ann}_R (m))=0$, can we deduced that $m=0$?

Where $\text{ann}_R (\text{ann}_R(m))=\{r\in R\,|\, r \cdot \text{ann}_R (m)=0\}$.

Answer 1

$ann$ with respect to... ?
$ann_{\color {red}R} (m)=\{r\in {\color {red}R}| rm=0\}.$

• $ann_{\color {red}M} (ann_{\color {red}R} (m))=0$: Let $x\in ann_R (m).$ Then $xm=0$. So $m\in ann_M (ann_R (m))=0.$ In this case $m=0.$

• $ann_{\color {red}R} (ann_{\color {red}R} (m))=0$: It is possible that $x$ be regular element and $x\in ann_R (m).$ ($m$ is called atorsion element) So in this case $m$ can be non_zero.

Answer 2

Let $R:=\mathbf{Z}$ and $M:=\mathbf{Z}×\mathbf{Z}_2$. Clearly $M$ is finitely generated and faithful. Set $m=(0,1)$. Then $m\not=(0,0)$, but $ann_R(ann_R(m))=0$.