I'm working on a problem (in a graduate Linear Algebra text) as follows.
The author ask us to find a module $M$ which is finitely generated by torsion elements but $\text{Ann}(M)=\{0\}$.
I can't seem to find such an example. I did find module with finitely generated by its torsion elements but $\text{Ann}(M)\neq \{0\}$, also examples of module with $\text{Ann}(M) = \{0\}$ but not finitely generated.
What i had in mind is that the ring $R$ should not be an integral domain, otherwise $\text{ann}(M)\neq \{0\}$. I quite new to this and i think this is the point where i need some help. Any help will be appreciated. Thanks.
Best Answer
First, you correctly note that no examples will exist if $R$ is an integral domain: if $M$ is finitely generated with $M = Rx_1 + \dots + Rx_n$ and $x_1, \dots, x_n \in t(M)$, then there exist $a_i \in R$ such that $a_ix_i = 0$ for each $i$. Since $R$ is a domain, we have that $a_1 \dots a_n \neq 0$ and for any $x = \sum_i b_ix_i$ we get
$$ (a_1 \cdots a_n) \sum_ib_ix_i = \sum_i b_i(a_1 \cdots \hat{a_i} \cdots a_n)a_ix_i = 0 $$
which proves that $Ann(M) \ni a_1 \cdots a_n \neq 0$.
However, take $M = \mathbb{Z}_2 \oplus \mathbb{Z}_3$ as a $\mathbb{Z}_6$ module (you can think of it as a $\mathbb{Z}$-module and the note that the representation factors through the quotient $\mathbb{Z} \to \mathbb{Z}/6\mathbb{Z}$ since $2,3 | 6$). Then $M$ is generated by $(1,0)$ and $(0,1)$ which are of torsion, annihilated by $2$ and $3$ (mod $6$) respectively. But any element $ x \in Ann(M)$ has to annihilate this two together, and thus we have that $2 | x$ and $3 | x$. Consequently, $x$ is divided by $6$ and so it is zero in $R = \mathbb{Z}_6$.