TITLE: How many ways are there to consider $\Bbb Q$ as an $\Bbb R$-module?
QUESTION [11 upvotes]: How many ways are there to consider $\Bbb Q$ as an $\Bbb R$-module?
I guess there is only one way, and that is the  trivial case. i.e. 
$$\forall r\in \Bbb R,\, \forall a,b\in \Bbb Q \qquad r\cdot \frac{a}{b}=0$$
With an idea I can proof this:
$$\exists r\not=s\in \Bbb R \quad\text{s.t.}\quad r\cdot \frac{a}{b}=s\cdot \frac {a}{b}$$
but I think this idea is wrong.

REPLY [5 votes]: Here is another way to think about things.
First note that $\text{End}_\mathbf{Ab}(\mathbb{Q})$ (the abelian endomorphisms of $\mathbb{Q}$) is countable. Indeed if $g\in\text{End}_\mathbf{Ab}(\mathbb{Q})$ then for any $p\in\mathbb{Z}$ we have that $g(p)=pg(1)$ and thus if $\displaystyle qg\left(\frac{p}{q}\right)=g(p)=pg(1)$--in other words, $\displaystyle g\left(\frac{p}{q}\right)=\frac{p}{q}g(1)$. Thus, $\#\text{End}_\mathbf{Ab}(\mathbb{Q})=\#\mathbb{Q}$.
Now, an $\mathbb{R}$-module structure amounts to a ring map $\mathbb{R}\to\text{End}_\mathbf{Ab}(\mathbb{Q})$. This ring map clearly can't be unital, else it is an injection because $\mathbb{R}$ is a field which is impossible by cardinality issues. That said, note that if $1\mapsto g\in\text{End}_\mathbf{Ab}(\mathbb{Q})$ under our ring map, then $g^2=g$, and so $g(g-1)=0$. Thus, $g(g(1)-1)=g(g(1))-g(1)=0$, but as we have already observed $g(g(1))=g(1)g(1)=g(1)^2$, thus $g(1)^2=g(1)$. If $g(1)=1$ then $g=\text{id}$ and so our ring map was unital--bad. So, $g(1)=0$, and from the above we see that $g=0$, and so $1\mapsto 0$, and so the ring map is the zero map--this corresponds to the trivial module structure Zev alluded to.