Torsion-free but not free The question is asking me to give an example of a finitely generated $R$-module that is torsion-free but not free. 
I remember in lecture, lecturer say something about the ideal $(2,X)$ in $\mathbb{Z}[X]$considered as a $\mathbb{Z}[X]$-module is torsion-free but not free. But I don't know why, what I know about these guys are:


*

*$(2,X)$ is the ideal that contain all polynomial with even constant

*$(2,X)$ is not principle ideal, so $\mathbb{Z}[X]$ is not a PID


I am not sure will these facts help to answer to question.
 A: I'll take tomasz' suggestion and write my comments here, as an answer.
Hint: (Exercise) Let $R$ be an integral domain. Then an ideal $I\subset R$ is a free $R$-module if and only if it is principal.
Your proof of this fact, @SamC, is essentially solid. However, you should actually be using the fact that $R$ is an integral domain in the $(\Leftarrow)$ direction, not $(\Rightarrow)$; take a quick second look at the definition of linear independence.  
Also, before applying the hint, one must also show that $(2,X)$ is not a principal ideal in $\mathbb{Z}[X]$. Sounds silly, I know. But it's worthwhile to rigorously show that you can't write $(2,X)=(f)$ for some $f \in \mathbb{Z}[X]$.
A: $(2,X)$ in $\mathbb{Z}[X]$ as a $\mathbb{Z}[X]$-module is torsion free but not free. First, it is quite clear that $(2,X)$ is torsion free as $0$ is the only torsion element. What about free? To show this I whish to prove that:

An ideal $I$ in an integral domain $R$ is a free $R$-module if and only if it is a principal ideal. - user153841

proof:
$(\Rightarrow)$ Let $I$ be an ideal in an integral domain $R$, and it is a free $R$-module. Let $B=\{e_i \mid i \in \mathbb{N}\}$ be a linearly independent basis for $I$. Since $I$ is an ideal, $e_2e_1$ and $e_1e_2 \in I$ also $e_2e_1-e_1e_2 \in I$
$$e_2e_1-e_1e_2=0$$
Which is impossible since $B$ is a linearly independent basis and $R$ is an integral domain (so commutative). Hence $B=\{e\}$ only has one element. So $I=(e)$ which is a principal ideal.
$(\Leftarrow)$ Let $I$ be an principal ideal. Suppose $I=(e)$, which mean $B=\{e\}$ is a basis for $I$ and clearly it is linearly independent, since $R$ is an integral domain meaning that there are no zero-divisors. Therefore $I$ is a free $R$-module.

So, since $(2,X)$ is not a principal ideal in $\mathbb{Z}[X]$, it can not be free.
