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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): any such isomorphism would have to be $E(x)=a^x$ for some $a\in F$, so by the Gelfond–Schneider theorem $E(b)$ won't be algebraic for $b\notin \mathbb{Q}$.

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    $\begingroup$ Yes, since the theory of divisible torsion-free abelian groups (aka $\mathbb Q$-linear spaces) is uncountably categorical. $\endgroup$
    – Emil Jeřábek
    Commented Mar 24, 2016 at 16:04
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    $\begingroup$ They are both torsionfree abelian divisible groups, so they are both isomorphic to a direct sum of copies of $\mathbb{Q}$; since they have the same cardinality, they have to be isomorphic. $\endgroup$
    – Arturo Magidin
    Commented Mar 24, 2016 at 16:04
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    $\begingroup$ @S.Carnahan: Yes, you're right. The additive group is of course infinite dimensional: for example, $\{\sqrt{p}\mid p\text{ is prime}\}$ is (additively) independent over $\mathbb{Q}$. As for the multiplicative group, the primes themselves are independent. $\endgroup$
    – Arturo Magidin
    Commented Mar 24, 2016 at 16:17
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    $\begingroup$ For the additive group, an argument easier to check is that for any $d\ge1$, there are real algebraic numbers $\alpha$ of degree $d$ (such as $2^{1/d}$), in which case $\{1,\alpha,\dots,\alpha^{d-1}\}$ is linearly independent. $\endgroup$
    – Emil Jeřábek
    Commented Mar 24, 2016 at 16:23
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    $\begingroup$ So... should this be written up as a formal answer? $\endgroup$
    – Arturo Magidin
    Commented Mar 24, 2016 at 23:37

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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 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.

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  • $\begingroup$ Unique factorization in $F$??? $\endgroup$
    – abx
    Commented Apr 11, 2016 at 20:03
  • $\begingroup$ @abx: In $\mathbb{Z}$; raise to a power to clear all denominators in the $q_i$, and that gives you an expression for $1$ as a product of (integral) powers of primes. $\endgroup$
    – Arturo Magidin
    Commented Apr 11, 2016 at 20:54
  • $\begingroup$ Sorry I am confused, but what is a prime in $F$? $\endgroup$
    – abx
    Commented Apr 11, 2016 at 21:55
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    $\begingroup$ @abx I said "(rational) primes". In algebraic number theory, this is the way one refers to the primes in ℤ (to distinguish them from prime elements or prime ideals in other number fields). So it's not "prime in F". It's the usual prime integers. $\endgroup$
    – Arturo Magidin
    Commented Apr 11, 2016 at 22:19
  • $\begingroup$ Oh, OK, sorry -- I misunderstood what you were saying. $\endgroup$
    – abx
    Commented Apr 11, 2016 at 22:45

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