Why is there only one way to make $M$ a $\mathbb{Q}$-module, if possible?

Question

Suppose $M$ is a $\mathbb{Q}$-module. Why is the given action of $\mathbb{Q}$ on $M$ (whatever it may be) the only way to make $M$ a $\mathbb{Q}$-module?

I know that an abelian group can only be made into a $\mathbb{Z}$-module in a unique way, since $1\cdot m=m$ for any $m\in M$. So for $p/q\in\mathbb{Q}$, $(p/q)\cdot m=\frac{1}{q}(pm)$. But $p\cdot m$ is necessarily the sum of $m$ a total of $p$ times. Also, $\frac{1}{q}(qm)=(\frac{1}{q}q)\cdot m=1\cdot m=m$. With this, does it somehow follow that there is only a unique way to make an abelian group $M$ a $\mathbb{Q}$-module when it is possible?

Answer 1

Let $M$ be a $\mathbb{Q}$-module. Then the additive group $(M,+)$ is torsion-free, because $nm=0$ for some $n\in\mathbb{N}$ implies $m=\frac{1}{n}nm=0$.

In a torsion-free abelian group $G$ an equation of the form $nx=g$, $n\in\mathbb{N}$, $g\in G$ has at most one solution.

In the $\mathbb{Q}$-module $M$ the element $\frac{1}{n}m$ thus is the unique solution of the equation $nx=m$, $m\in M$. Note that this holds whatever the operation of $\mathbb{Q}$ on $M$ is like, while the equation $nx=m$ does only depend on the operation of $\mathbb{Z}$ on $M$, which is unique as we already know.

Hence in your case the equation $ qx=pm $ has at most one solution $ \frac{1}{q}(pm) $ -- and it has one.

Answer 2

This (old) question already has a nice, explicit answer, but I thought I might offer an alternative perspective for future visitors.

Let $R$ be a commutative ring. To give an $R$-module structure on an abelian group $M$ is equivalent to giving a ring homomorphism $R \to \mathrm{End}_{\mathbb{Z}}(M)$. This gives another explanation why an abelian group $M$ can be given exactly one $\mathbb{Z}$-module structure: $\mathbb{Z}$ is initial in the category of rings, so there is a unique homomorphism $\mathbb{Z} \to \mathrm{End}_{\mathbb{Z}}(M)$.

Now, let $R$ be a ring, and let $S \subset R$ be a multiplicative subset, and let $i \colon R \to S^{-1}R$ be the canonical localization map. Recall that the universal property of localization says that given a ring morphism $\alpha \colon R \to T$ with $\alpha(S) \subset T^{\times}$, there is a unique map $\beta \colon S^{-1}R \to T$ such that $\beta \circ i = \alpha$.

Note that $\mathbb{Q}$ is the localization of $\mathbb{Z}$ at the multiplicative subset $S = \mathbb{Z} \setminus \{0\}$. Again, since $\mathbb{Z}$ is initial in the category of rings, there is exactly one ring morphism $\mathbb{Z} \to T$ for any ring $T$, so by the universal property of localization, there is at most one ring homomorphism $\mathbb{Q} \to T$, which exists if and only if the image of $\mathbb{Z}$ in $T$ lies in $T^{\times}$. Since $\mathbb{Q}$-module structures on an abelian group $M$ are in bijection with ring morphisms $\mathbb{Q} \to \mathrm{End}_{\mathbb{Z}}(M)$, this shows the desired result.

Answer 3

Let $M$ be an abelian group, then

$M$ is a $\mathbb{Q}-$module iff $M$ is torsion-free and divisible.

The $\mathbb{Q}-$module structure is $(p/q).m=y$ where $p.m = q.(y)$. Such a $y$ exists by divisibility requirement and can be checked to be unique by torsion-freeness.