# Why $\mathbb{Q}$-vector space has no $\mathbb{Q}$-torsion?

Why $$\mathbb{Q}$$-vector space has no $$\mathbb{Q}$$-torsion ?

What does the notion of $$\mathbb{Q}$$-torsion technically mean ?

• what is your definition of $\mathbb{Q}$-torsion? – Guido A. Aug 31 at 17:19
• @GuidoA. It is actually taken from an answer to another post which is copied below: "You certainly have extra assumptions on $M$, because it is not possible in general to put a strucutre of $\mathbb{Q}$-module extending the action of $\mathbb{Z}$, aka $\mathbb{Q}$-vector space, because a $\mathbb{Q}$-vector space has no $\mathbb{Q}$-torsion, and thereofre no $\mathbb{Z}$-torsion. In particular, it is impossible if $M$ is a finite abelian group (which makes already a lot of counterexamples)." – user3357120 Aug 31 at 17:22

In general, if $$M$$ is a module over a ring $$R$$, we say that an element $$m \in M \setminus \{0\}$$ is of torsion if there exists $$r \in R$$ such that $$r \cdot m = 0$$. For example, if we consider $$\mathbb{Z}/2\mathbb{Z}$$ as a $$\mathbb{Z}$$-module, then
$$2 \cdot \bar{1} = \bar{2} = \bar{0}.$$
If $$M$$ has some torsion element, we say that $$M$$ has $$R$$-torsion. This can never happen when $$R$$ is a field: suppose that we have a $$\Bbbk$$-vector space $$V$$ and $$v \in V, \lambda \in \Bbbk$$ such that $$\lambda \cdot v = 0$$. Multipying by $$\lambda^{-1}$$ we obtain
$$v = 1 \cdot v = (\lambda^{-1}\lambda) \cdot v = \lambda^{-1}(\lambda \cdot v) = \lambda^{-1} \cdot 0 = 0.$$