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Let $X$ and $Y$ denote $\mathbb{Z}$-modules. Then if $X$ and $Y$ have equal underlying abelian groups, we may deduce that $X=Y$.

Is this still true if we replace $\mathbb{Z}$ with $\mathbb{Q}$?

Equivalently, does there exist an abelian group that can be made into $\mathbb{Q}$-module in more than one way?

goblin GONE
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  • You may also be interested in "epic maps of rings". If $f:R \to S$ is a ring homomorphism, then every $S$-module can be seen as an $R$-module via $r \cdot m := f(r) \cdot m$. If we have two $S$-module structures on $M$ such that $r \cdot_1 m = r \cdot_2 m$ for all $r \in R$, then can we conclude that $s \cdot_1 m = s \cdot_2 m$ for all $s \in S$? The $f$ such that we can do this are called epic ring maps. Fields of fractions and quotient rings are standard examples. – Jack Schmidt Feb 01 '14 at 14:25

1 Answers1

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A $\mathbb Q$-module is a $\mathbb Q$-vector space. Let $V$ be a $\mathbb Q$-vector space. Then for each $n\in\mathbb N$ and $v\in V$, the vector $\frac 1nv$ is the unique(!) vector $w$ with $\underbrace{w+\ldots+w}_n=v$ and then $\frac mnv$ is of course uniquely determined as $\underbrace{w+\ldots+w}_m$ if $m\ge0$ and as the inverse of $\frac{|m|}nv$ if $m<0$.

Hagen von Eitzen
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