Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
The short answer is "yes." –  Rasmus Oct 24 '11 at 7:28
add comment

2 Answers 2

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 $qm=pm$ has at most one solution $m$ -- and it has one.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.