Question

Question about abstract algebra:

Consider Q (set of rational numbers) as a Z-module (Z being the set of integers). Show that Q is torsion free.

Is there something screwy about this question?

Answer 1

Since \(\mathbb{Q}\) is a \(\mathbb{Z}\)-module and \(\mathbb{Z}\) is an integer domain, we can difine the "torsion submodule" of \(\mathbb{Q}\) as
\[ \large t\mathbb{Q}=\{q\in\mathbb{Q}:(\exists m\in\mathbb{Z}\setminus\{0\})
(mq=0)\}. \]
So in order for \(\mathbb{Q}\) to be torsion-free we have to prove that \(t\mathbb{Q}=\{0\}\).

Answer 2

It is trivial that \(0\in t\mathbb{Q}\). So let \(q\in\mathbb{Q}\) (\(q\neq0\)) be in \(t\mathbb{Q}\). This means that \(q=a/b\) where both \(a\) and \(b\) are non-zero. Now for some integer \(m\neq0\) we have that
\[ \large 0=mq=m\frac{a}{b}=\frac{ma}{b} \]
but this implies that \(ma=0\) from which it follows that \(m=0\), a contradiction.

Therefore \(t\mathbb{Q}=\{0\}\), and \(\mathbb{Q}\) is torsion free.

Answer 3

It is trivial that \(0\in t\mathbb{Q}\). So let \(q\in\mathbb{Q}\) (\(q\neq0\)) be in \(t\mathbb{Q}\). This means that \(q=a/b\) where both \(a\) and \(b\) are non-zero. Now for some integer \(m\neq0\) we have that
\[ \large 0=mq=m\frac{a}{b}=\frac{ma}{b} \]
but this implies that \(ma=0\) from which it follows that \(m=0\), a contradiction.

Therefore \(t\mathbb{Q}=\{0\}\), and \(\mathbb{Q}\) is torsion free.

Answer 4

Thanks for your response. My brain was malfunctioning after exam revision. Your explanation is completely right I think!

I think that, for some reason, I was thinking of the torsion elements as any rational such that there exists an integer that sends the rational to another rational. Like the identity of the group of cosets Z / Q, in which case obviously all rational numbers are of that type. As I said, my brain was not working properly!