Let $F$ be the field of **real** algebraic numbers. Is it true that the positive multiplicative group $(F\_{pos}^\*,\cdot,1)$ is isomorphic to the additive group $(F,+,0)$ (as abstract groups, not topological or ordered groups)?


Note that there cannot be any such continuous (or monotone) isomorphism
(i.e $F$ is not [exponentially closed](https://en.wikipedia.org/wiki/Exponentially_closed_field)): any such isomorphism would have to be $E(x)=a^x$ for some $a\in F$, so by the [Gelfond–Schneider theorem](https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem) $E(b)$ won't be algebraic for $b\notin \mathbb{Q}$.



Are the positive multiplicative group and the additive group of the field of real algebraic numbers isomorphic?
To prevent the question from being "unanswered"...


Both groups are abelian torsionfree divisible groups, so they are vector spaces over $\mathbb{Q}$ and completely determined by their dimension. Both are countable, so their dimensions are countable (finite or infinite). To show they are isomorphic as abelian groups it suffices to show that they both have infinite dimension.


For the multiplicative group, the (rational) primes are linearly independent over $\mathbb{Q}$ (by unique factorization); here the action of $\mathbb{Q}$ is as exponents, so if $p\_1,\ldots,p\_r$ are distinct primes, then $p\_1^{q\_1}\cdots p\_r^{q\_r} = 1$ implies $q\_1=\cdots=q\_r=0$. 


For the additive group, you can either take the square roots of the primes, which are linearly independent over $\mathbb{Q}$; or as [Emil Jeřábek](https://mathoverflow.net/users/12705/emil-je%c5%99%c3%a1bek) notes, we know that there are positive algebraic numbers $\alpha$ of degree $n$ for arbitrary $n$, and that if $[\mathbb{Q}(\alpha):\mathbb{Q}]=n$, then $1,\alpha,\ldots,\alpha^{n-1}$ is linearly independent. Hence, $F$ must be infinite dimensional over $\mathbb{Q}$. 


Since both vector spaces have the same dimension over $\mathbb{Q}$ (namely, $\aleph\_0$), they are isomorphic as abelian (divisible) groups. 


