Can a field be canonically reconstructed from its multiplicative group?

Question

I understand that there are a lot of restrictions on which abelian groups arise as the multiplicative groups of fields. My question is kind of the adjoint:

Suppose I pick a field $k$, but I do not tell you what it is. I give you an abelian group $G$ and tell you that $G\cong k^\times$. But you only receive $G$ as an abstract group; that is, $G$ does not come with the natural action on $(k, +)$ or any other information inherited from $k$.

Given $G$, can you canonically define a field $\hat k$ such that $\hat k \cong k$?

I would not be surprised at all if the answer were "no." But as I'm not convinced either way, I optimistically attempted a natural construction.

I first observed that even though we don't know anything about $k$ as a set, we have a 1-1 correspondence between $G$ and the nonzero elements of $k$. We need to invent a additive neutral element and also somehow produce an addition operation compatible with multiplication.

Let $(\Sigma, +)$ be the free abelian group generated by the elements of $G$; $\sigma: G\to \Sigma$ is the identification of $g \in G$ as $\sigma(g) \in \Sigma$. At this point $\Sigma$ contains the desired $(k, +)$ as a subgroup, although it is way too big. Continuing in the freest possible way, define: $R = G \otimes_\mathbb{Z} \Sigma$. $R$ is naturally distributive, but it is far from having the full additive structure of $k$.

Let $P$ be the ideal generated by all elements $\left\{g\otimes\sigma(g')-gg'\otimes\sigma(1)\mid g, g \in G\right\}$. Set $S=R/P$. By allowing natural distributivity and 0 compatible with $\otimes$, $S$ is a ring with multiplication defined by $(g\otimes 1)(g'\otimes 1) = gg'\otimes 1$ .

And obviously $S$ contains $k$, but it also has so much other contents and structure that I think it's not even a field. Can I get any pointers on what to do next (quotient? localize?)? Or if my objective is impossible, I'd like to see where this attempt breaks down.

Answer 1

Here are two reasons that the answer is no. First, if a construction deserves to be called "canonical", it should be functorial with respect to isomorphisms. In particular, then, every automorphism of the group $G$ should give rise to an automorphism of the associated field $k$. This is obviously not the case: for instance, $x\mapsto x^{-1}$ is an automorphism of any abelian group, but will not extend to a field automorphism for most fields.

Second, even if you consider the group $G$ to be given only "up to isomorphism" so that we cannot ask for an action of the automorphism group, there is still an issue that the group $G$ up to isomorphism does not determine the field $k$ up to isomorphism. For example consider the field $\overline{\mathbb{Q}}$ of algebraic numbers and the smallest subfield $k\subset\overline{\mathbb{Q}}$ which is closed under taking $n$th roots for all $n$. Note that $k$ is a proper subfield of $\overline{\mathbb{Q}}$, because not all polynomials over $\mathbb{Q}$ are solvable by radicals. In particular, $k$ and $\overline{\mathbb{Q}}$ are not isomorphic, since $k$ is not algebraically closed. However, $k^\times$ and $\overline{\mathbb{Q}}^\times$ are isomorphic groups: both are isomorphic to $\mathbb{Q}/\mathbb{Z}\oplus V$ where $V$ is a countably infinite-dimensional vector space over $\mathbb{Q}$. (Here the $\mathbb{Q}/\mathbb{Z}$ summand is the subgroup consisting of the roots of unity. Since $\mathbb{Q}/\mathbb{Z}$ is divisible, it is a direct summand, and its complement is a torsion-free divisible abelian group and hence a $\mathbb{Q}$-vector space, which is easily seen to have countably infinite dimension.)

As for where your attempt breaks down, there isn't much to say. There just isn't any correct "next step" to take after what you have done so far. Note that your construction of $R$ is also just incorrect: by tensoring with $G$ as an abelian group, you are imposing the group structure of $G$ additively instead of multiplicatively. For instance, if $G$ is cyclic of order $2$, then your $R$ will be $2$-torsion, whereas the field (with $3$ elements) which you want to construct should not be $2$-torsion. The correct thing to do is to instead just directly define the multiplication $\Sigma$, using the group operation of $G$. This gives what is known as the group ring $\mathbb{Z}[G]$. It's just a ring, though, not a field, and there is no canonical way to turn it into a field all of whose nonzero elements come from elements of $G$.