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We know that any $mathbbZ$ module structure on an abelian group is unique, and furthermore the same is true for $mathbbQ$.

Any complex vector space structure is not unique, we can just compose with an automorphism of $mathbbC$ (for example the conjugation map).

However, $mathbbR$ has trivial automorphism group, see:

Is an automorphism of the field of real numbers the identity map?

So my question is, given an abelian group made into an $mathbbR$ vector space, is this the only way we can do this?

We would get another structure if we can embed $mathbbR$ into itself, but I'm not sure if this is possible.

If not, then of all the different $mathbbR$ structures, must they all have the same dimension?

I know that both of these are not true for general fields.

Thanks in advance.

(Related: number of differents vector space structures over the same field $mathbbF$ on an abelian group)

Sure. Note that instead of using an automorphism of $mathbbR$, you can use an automorphism of the abelian group. Given $mathbbR$-vector spaces $V$ and $W$ any abelian group isomorphism $f:Vto W$, you can define a new $mathbbR$-vector space structure on $V$ with scalar multiplication $mathbbRtimes Vto V$ given by $(r,v)mapsto f^-1(rf(v))$. This will not be the same as the original scalar multiplication on $V$ unless $f$ is $mathbbR$-linear.

In particular, as an abelian group, an $mathbbR$-vector space $V$ of dimension $n$ is just a $mathbbQ$-vector space of dimension $ncdot 2^aleph_0$. If $n>0$, this gives loads of automorphisms of $V$ that are not $mathbbR$-linear (for instance, for any nonzero $vin V$ and any irrational $alphainmathbbR$, $v$ and $alpha v$ are linearly independent over $mathbbQ$ so there is an automorphism that exchanges $v$ and $alpha v$, and this cannot be $mathbbR$-linear), and thus loads of isomorphic but distinct $mathbbR$-vector space structures on $V$. Or, if $0<n,m leq 2^aleph_0$, then $ncdot 2^aleph_0=mcdot 2^aleph_0=2^aleph_0$, so $V$ is actually isomorphic as an abelian group to any $mathbbR$-vector space $W$ of dimension $m$, and so $V$ can also be given an $mathbbR$-vector space structure of dimension $m$.

$BbbR$ and $BbbR^2$ are isomorphic as abelian groups but not as real vector spaces.

You can non-trivially embed the real field into itself. This is because every field extension has a transcendence basis.

So let $S ⊆ ℝ$ be a transcendence basis for $ℝ/ℚ$. Let $F = ℚ(S)$. Then $ℝ/F$ is an algebraic extension. Now every map $S → S$ naturally extends to a field morphism $F → F$ and then to a field morphism $ℝ → ℝ$.

Since $S$ is uncountable, this yields uncountably many field morphisms $ℝ → ℝ$, all of which give rise to different $ℝ$-linear structures on the abelian group $ℝ$ itself.