Example of modules over PID's From Advanced Modern Algebra (Rotman):

Definition If $M$ is a finitely generated torsion $R$-module, where $R$ is a PID, and $$M= R/(c_1) \oplus R/(c_2) \oplus ... \oplus R/(c_t),$$ where $t \geq 1$ and $c_1 | c_2 | ... | c_t$, then the ideals $(c_1), (c_2), ... , (c_t)$ are called the invariant factors of $M$.



Corollary 8.20 If $M$ is a finitely generated torsion module over a PID $R$, then $$(c_t) = \{r \in R: rM=\{0\}\},$$ where $(c_t)$ is teh last ideal occuring in the decomposition of $M$ [in the definition].
In particular, if $R=k[x]$, where $k$ is a field, then $c_t$ is the monic polynoial of least degree for which $c_tM=\{0\}$.

I understand how these work for abelain groups. Let $G = J(c_1) \oplus ... \oplus J(c_r),$ where $J(c_i)$ is a cyclic group of order $c_i$, and $c_1 | c_2 | ... |c_r$. Then, since $c_iJ(c_i) = 0$, (if we let $c_r = ac_i$ for some $a$) we have $c_rJ(c_i) = a(c_iJ(c_i)) = 0$. It also makes intuitive sense to me.
But when I think about modules, it kind of gets confusing. How can you annhilate a cylic module by multiplying it by a polynomial? In other words, I thought that it would be clearer in my head if I saw an example, so I tried to search the internet but couldn't really find anything useful. So I was wondering if anybody could give me an example (even if its a simple one), just so I can see how multiplying by a polynomial can annihilate a cyclic module.
Thank you in advance
 A: Consider the $k[x]$-module $k[x]/(x^2)$. Then multiplying by $x^2$ will annihilate the module.
I like the following example. If you have a finite dimensional complex vector space $V$ with a linear transformation $f\colon V\to V$, then you can have a structure of $k[x]$-module on $V$, where $x$ acts on $V$ via $f$. Then by the theorem you can decompose $V$ into the sum $V=V/(x-c_1)^{r_1}\oplus\dots\oplus V/(x-c_k)^{r_k}$. This decomposition is exactly the decomposition of $V$ into generalized eigenspaces $V_{c_i}$, where $V_{c_i}=\{v\in V\mid (f-c_i)^Nv=0\mbox{ for some N}\}$. So in this example the theorem gives you the Jordan Normal Form theorem.
A: For a very explcit example $R= \mathbb{Q}[X]$, this is a PID, and let $M = \mathbb{Q}[X]/(X^2)$. You can alternatively think of this as $\{a + bX \colon a, b \in \mathbb{Q}\}$ with component wise addition and the multiplciation by an element of $R$ so an arbitrary polynomial is given by 'multiply as polynomials and keep only constant and linear term'.
This module is cyclic (generated by $1$ or any non-zero constant). And multiplying by $X^2$ always  yields $0$. 
There is nothing special about $X^2$ you can do the same with any polynomial.  
