In the post on threelds, we investigated under what conditions the additive group of one field (the ‘inner field’) could be isomorphic to the multiplicative group of another field (the ‘outer field’). To summarise, this happens in the following cases:

- the outer field is F_3 and the inner field is F_2;
- the outer field is F_(2^p) and the inner field is F_(2^p – 1), where 2^p – 1 is a Mersenne prime;
- the outer field has characteristic 2 and every element has a unique
*n*th root for every positive integer*n*.

This last case, the infinite case, is worth analysing further. We proved that there is at least one example of every infinite cardinality, by appealing to Löwenheim-Skolem, but this was rather non-constructive. In this post, we instead describe an explicit construction of such a field, which is ‘minimal’ in the sense that it embeds into every infinite field with this property.

Firstly, observe that in a field of finite characteristic *p*, every element that is algebraic (i.e. is the solution to some polynomial with coefficients in the ring generated by 1) is an element of some finite subfield, and is therefore a root of unity. As such, in our putative field of characteristic 2 where every element has a unique *n*th root for all *n*, there can be no such elements as then 1 would have too many roots. In other words, every element other than 0 or 1 is transcendental.

If we took such a transcendental element X, then we could construct the field ‘generated’ by X under the field operations together with taking unique *n*th roots. This can be done completely explicitly:

- enumerate all expressions (binary trees where each leaf is labelled with an element of the countable set {1, X_0, X_1, X_2, X_3, …} and each nonleaf node is labelled with an element of {+, −, ×, /});
- define X_0 to be X;
- for each
*n*> 0, define X_*n*to be a*p*th root of the first well-formed* expression (in the above enumeration) that only mentions variables with indices less than*n*and doesn’t already have a*p*th root, where*p*is the (*m*+1)th prime, where*m*is the 2-adic valuation of*n*.

*we include the constraint that no subtree evaluates to 0, thereby avoiding any division-by-zero issues.

Then we obtain a tower of fields F_2(X_0) ⊆ F_2(X_0, X_1) ⊆ F_2(X_0, X_1, X_2) ⊆ … where each field is a finite-degree extension of the previous field. The union of these fields then has the desired property that the multiplicative group forms a vector space over Q.

Each of the extensions that we take is a degree-*p* extension obtained by adjoining a *p*th root of some element. Up to isomorphism, there is only one way to do this, so by induction there is a unique field ‘generated’ by a transcendental element X. This must therefore be a subfield of the outer field of every infinite threeld, as claimed, as every such field has a transcendental element.

Note that this vector space has countably infinite dimension: the irreducible polynomials in F_2[X] form a countably infinite set of linearly independent elements in the multiplicative group (if they didn’t, this would provide a counterexample to unique factorisation in F_2[X]). This answers in the negative Thomas Blok’s question as to whether an infinite threeld can be finite-dimensional.