For which cardinalities $\kappa$ is $\def\Q{\mathbb Q}\Q^{(\kappa)}$—by which I mean the direct sum of $\kappa$-many copies of $\Q$—isomorphic (as an abelian group) to the multiplicative group of units of some field $F$?

$\Q^{(\kappa)}$ is torsion-free, so for example we can see that the characteristic of $F$ must be 2: since $(-1)^2 = 1$, it is torsion unless $-1 = 1$. $\Q^{(\kappa)}$ is also a divisible group.

From a paper cited in this answer to a previous question of mine, we know that:

  • $\kappa = 0$ works: $\Q^{(0)} \cong (\Bbb Z/2\Bbb Z)^*$,
  • $\kappa = 1$ doesn't work: everything has to be algebraic over $\Bbb Z/2\Bbb Z$ and hence torsion,
  • Some uncountable $\kappa$ works: we can show the following ultraproduct has multiplicative group isomorphic to $\Q^{\aleph_0} \cong \Q^{(\kappa)}$: $$F = \prod_{p\ \text{prime}} \mathrm{GF}(2^p)/U.$$ I think that saying $\kappa = 2^{\aleph_0}$ is assuming CH but whatever.

Are there any other $\kappa$, especially finite, for which we can answer yes or no?

up vote 2 down vote accepted

There is no such field $ F $ for finite $ \kappa $. As you observed, if such a field exists, it must have characteristic $ 2 $. Furthermore, it cannot contain any elements which are algebraic over its prime field $ \mathbb F_2 $ (besides those that are already in $ \mathbb F_2 $), because then $ F^{\times} $ has torsion, which is a contradiction. Since $ F \neq \mathbb F_2 $, it follows that there's an embedding $ \mathbb F_2(T) \to F $, which gives an embedding of multiplicative groups $ \mathbb F_2(T)^{\times} \to F^{\times} \cong \mathbb Q^{\kappa} $. $ \mathbb F_2(T)^{\times} \cong \mathbb Z^{\omega} $, thus it suffices to show that there is no embedding $ \mathbb Z^{\omega} \to \mathbb Q^{\kappa} $ for $ \kappa $ finite. This follows, since the image of $ n $ $ \mathbb Z $-linearly independent elements in $ \mathbb Z^{\omega} $ is linearly independent in $ \mathbb Q^{\kappa} $, thus $ \kappa > n $ for all natural numbers $ n $, i.e $ \kappa $ must be infinite.

  • Excellent! I think that this plus Lowenheim-Skolem (fields with torsion-free divisible multiplicative groups is first-order) tells us that every infinite $\kappa$ works and the only finite one that does is 0. – algorithmshark Feb 6 '17 at 19:02

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