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.
Best Answer
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 $f \in \mathbb{Z}[X]$.