Dtyjlui

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 $1cdot m=m$ for any $min M$. So for $p/qinmathbb{Q}$, $(p/q)cdot m=frac{1}{q}(pm)$. But $pcdot m$ is necessarily the sum of $m$ a total of $p$ times. Also, $frac{1}{q}(qm)=(frac{1}{q}q)cdot m=1cdot m=m$. 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?

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

In a torsion-free abelian group $G$ an equation of the form $nx=g$, $ninmathbb{N}$, $gin G$ 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 $nx=m$, $min M$. 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.

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 divisbility requirement and can be checked to be well defined by torsion freeness.

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.