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

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

  • $\begingroup$ what is your definition of $\mathbb{Q}$-torsion? $\endgroup$ – Guido A. Aug 31 at 17:19
  • $\begingroup$ @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)." $\endgroup$ – 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

$$ v = 1 \cdot v = (\lambda^{-1}\lambda) \cdot v = \lambda^{-1}(\lambda \cdot v) = \lambda^{-1} \cdot 0 = 0. $$


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