# 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:

1. $(2,X)$ is the ideal that contain all polynomial with even constant
2. $(2,X)$ is not principle ideal, so $\mathbb{Z}[X]$ is not a PID

I am not sure will these facts help to answer to question.

• How many generators does the ideal $(2,X)$ in $\mathbb{Z}[X]$ have (as a $\mathbb{Z}[X]$-module)? Are there any relations amongst these generators?
– Kyle
Commented Jun 17, 2014 at 22:51
• 2 generators? If I want to show $(2,X)$ is not free, am I suppose to show it has no linearly independent basis ?
– SamC
Commented Jun 17, 2014 at 22:53
• Right. As a related question, you may want to think about how many different free $R$-modules there are (up to isomorphism) with a linearly independent basis of fixed cardinality.
– Kyle
Commented Jun 17, 2014 at 22:59
• Here's another good exercise: Show that an ideal $I$ in an integral domain $R$ is a free $R$-module if and only if it is a principal ideal.
– Kyle
Commented Jun 17, 2014 at 23:19
• Thanks, i will think about it. One question through, by fixed cardinality, you mean the underlying set for the $R$-module is finite?
– SamC
Commented Jun 17, 2014 at 23:31

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

• Thanks, in the $\Rightarrow$ direction, $R$ is an integral domain maybe over kill, what I really need is $R$ is commutative. Since this is a exam question, and the proof for $(2,X)$ is not principal in $\mathbb{Z}[X]$ already went over in lecture, so I just take it as a fact.
– SamC
Commented Jun 18, 2014 at 11:42
• Does there exist such an example only over non-PIDs? Commented Nov 25, 2020 at 23:32
• Is there a torsion free $Z$-module ($Z$ is a PID) which is not free ? Commented Dec 14, 2021 at 2:38

$(2,X)$ 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.