Can an abelian group be turned into two nonisomorphic vector spaces by different actions of the same field? I was wondering if you could find an abelian group and define two different actions of a field on it such that the resulting vector spaces are of different dimensions over the field.
I tried various extensions of $\mathbb Q$. Now, since the set $V$ is already an abelian group, the action of $\mathbb Z$, and therefore $\mathbb Q$, are already determined. So I tried $\mathbb Q(\alpha)$, but I got stuck here too: if $\alpha$ is algebraic, then its action is restricted to satisfy its polynomial, so the vector spaces would still be isomorphic. If $\alpha$ is transcendental, I tried having it act trivially, but that doesn't work since it's impossible to extend this to rational functions in $\alpha$, as is required.
Is what I'm trying to do even possible? If so, could I have some hints on an example?
 A: Let $F=k(T)$ and $E=k(T^2$), where $k$ is any field. Then $F$ is a one-dimensional vector space over itself and, being an algebraic field extension $F=E(T)\cong E[X]/(X^2-T^2)$,  a twodimensional vector space over $E$. But $F\cong E$ via $T\mapsto T^2$, obviously, so we have endowed the abelian group $F$ with two distinct and nonisomorphic $F$-vector space structures.
A: Consider $\mathbb{R}$ and $\mathbb{R}^2$. As $\mathbb{Q}$-vector spaces they have the same dimension, so they are isomorphic as abelian groups a fortiori. Of course, they are not isomorphic as $\mathbb{R}$-vector spaces.
A: Let $\sigma:F\to F$ be a field endomorphism such that $[F:\sigma(F)]=\infty$.
The "twisted polynomial ring" $F[x;\sigma]$ is the set of polynomials with coefficients written on the left of powers of $x$, and the coefficients are not assumed to commute with $x$. Instead, $xa:=\sigma(a)x$ for all $a\in F$.
Take a look at the ideal generated by $x$ in the ring $F[x;\sigma]/(x^2)$. For all intents and purposes, it is the set $Fx+(x^2)$.
On the left
the action of scalars looks like this:
$(\alpha+ (x^2))\cdot(\gamma x+(x^2))=\alpha\gamma x+(x^2)$
Consequently, $Fx+(x^2)$ is $1$ dimensional as a left $F$ vector space.
On the right
the action of scalars looks like this:
$(\gamma x+(x^2))\cdot(\alpha+ (x^2))=\gamma\sigma(\alpha) x+(x^2)$
This is very different from the left side. By hypothesis, a basis of $F$ over $\sigma(F)$ has infinitely many elements. Let's fix such a basis $\mathcal{B}=\{\gamma_i \mid i\in I\}$. Now consider the set of elements $\{\gamma_i x+(x^2)\mid i\in I\}$. A right $F$ linear combination equal to zero looks like this:
$0+(x^2)=\sum_{j=1}^n (\gamma_ix+(x^2))\cdot (\alpha_i+(x^2))=\sum_{j=1}^n \gamma_i\sigma(\alpha_i)x+(x^2)=(\sum_{j=1}^n \gamma_i\sigma(\alpha_i))x+(x^2)$
Comparing the coefficients of both sides, this says $\sum_{j=1}^n \gamma_i\sigma(\alpha_i)=0$, and by linear independence of the $\gamma_i$ over $\sigma(F)$, $\alpha_i=0$ for all $i\in I$. So what we have shown is that $\{\gamma_i x+(x^2)\mid i\in I\}$ form an $F$-linearly independent set when $F$ acts on the right. On the right, $Fx+(x^2)$ is an infinite dimensional $F$ vector space.
(Additional note: you could let $[F:\sigma(F)]=n<\infty$ also, but you wouldn't want $n=1$ of course.)
