An Example of a Torsion Module 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.
 A: 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 $R = \mathbb{Z}_6$.
A: Here are a couple more interesting examples.

One way to do it (if you are allowing noncommutative rings -- I didn't see any evidence to the contrary) is to pick a proper right ideal of an Artinian simple ring.
Since a semismple ring's elements are either units or zero divisors, that would ensure any generators were torsion. Since the right ideals of a simple Artinian ring are principal, that would guarantee finite generation. And finally, since the ring is simple, the annihilator could only be zero for a nonzero module.
So, for example, let $T$ be the subset of $R=M_2(F)$ consisting of matrices of the form $\begin{bmatrix}a&b\\0&0\end{bmatrix}$ for $a,b\in F$.  It's a torsion, singly generated, simple right $R$ module.

If you really must have a commutative ring, then let me also mention this other example I found out about from Hutchins' Examples of commutative rings (the example is attributed to Kaplansky):
Let $F_2$ be the field of $2$ elements and consider the ring $R=F_2[x,y]/(x^2,xy,y^2)$. Let $a,b,c$ be a basis of $M=F_2^3$ and define the action of the ring on $M$ by $xa=yb=xc=yc=0$ and $xb=ya=c$.
For this ring $R$ and module $M$, you can verify that the annihilator of every element of $M$ is nonzero, but the annihilator of $M$ is zero.
$R$ and $M$ only have $8$ elements.
