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A threeld is a generalization of a field, with three operations, such that the $F$ is a field with respect to the first (outer) and second (middle) operations (call it the outer field), and $F\setminus\{0\}$ is a field with respect to the second and third (inner) operations (the inner field). The following definitions and results are originally from https://cp4space.hatsya.com/2022/05/25/threelds/

If the outer field does not have characteristic $2$, then the element $-1$ is the only element of order two with respect to the middle operation, and therefore with the exception of the field $\mathbb{F}_3$ this cannot form a threeld (the characteristics would not agree). Therefore, all other threelds have outer characteristic $2$. For finite threelds, this thus restricts threelds to $\mathbb{F}_{2^p}$, where $2^p-1$ is a Mersenne prime.

Infinite threelds of arbitrary cardinality exist (see the article for proof via the Löwenheim–Skolem theorem). Countable examples must have the middle operation isomorphic to a vector space over $\mathbb{Q}$. My question is, is it possible for that vector space to be finite dimensional? If so, is there an explicit construction of this?

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The notion of threeld is of little relevance to this question (and to anything at all, maybe). The question is just equivalent to:

does there exist a field $K$ (necessarily of char. 2) whose group of units is isomorphic to $\mathbf{Q}^d$ for some integer $d>0$?

The answer is no. Because such a field has some element $x\notin\{0,1\}$. Then $x$ having infinite order in $K^*$, it is transcendental over $\mathbf{F}_2$. So $K^*$ contains $\mathbf{F}_2(x)^*$ as a subgroup. The latter has a infinite $\mathbf{Z}$-free family, namely given by the family of irreducible polynomials in $\mathbf{F}_2[x]$.

It is easy to prove along the same lines:

Let $K$ be a field that is not algebraic over a finite field. Then $K^*$ has infinite $\mathbf{Q}$-rank (i.e., has an infinite $\mathbf{Z}$-free family).

(Conversely if $K$ is algebraic over a finite field, $K^*$ is torsion, i.e., has $\mathbf{Q}$-rank zero.)

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