the action of $\mathbb{Q}$ on the abelian group should be extended uniquely from $\mathbb Z$

Question

The action of $\mathbb{Q}$ on the abelian group $(M,+)$ should be extended uniquely from $\mathbb Z$ by the axioms of a module. But I do not know how precisely? I.e. how is $$\cdot:{\mathbb Q}\times M\to M $$ defined? I understand that $1\cdot x=x$ and $(r\cdot s)\cdot x=r\cdot (s\cdot x)$

Answer 1

I see, let's suppose that $M$ is secretly already a $\mathbb Q$-vector space and show that the vector space structure is uniquely determined by $M$ as a $\mathbb{Z}$-module (i.e. as an abelian group). What that means is we would like to show that the $\mathbb{Q}$-action on $M$ is uniquely determined by the addition on $M$.

So, fix $m \in M$. Clearly $1\cdot m = m$, and for a positive integer $k$, we have the following. $$k\cdot m = \underbrace{m+\dots+m}_{k \text{ times}}$$ This is simply because multiplication is forced to be compatible with addition. Similarly, $-k\cdot m = -(k\cdot m)$. Finally, for $q \neq 0$ an integer, because $M$ is assumed to be a $\mathbb{Q}$-vector space under some other multiplication, there is $m' \in M$ with $q\cdot m' = m$. This forces $\left(\frac 1q\right)\cdot m = m'$. Now you can convince yourself that this forces the $\mathbb{Q}$-vector space structure to be uniquely determined by the addition.

Answer 2

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).