A field F consists of two compatible Abelian groups — an additive group on F and a multiplicative group on F \ {0} — such that multiplication distributes over addition.

In certain cases, though, this multiplicative group can be the additive group of a field structure on F \ {0}, giving a third Abelian group structure on F \ {0,1}. Let us call such an algebraic structure a **threeld**, because it is a generalisation of a field with three compatible Abelian group operations.

We’ll refer to the field structure on F as the *outer field*, and the field structure on F \ {0} as the *inner field*; the multiplicative group of the outer field coincides with the additive group of the inner field by definition.

If the outer field does not have characteristic 2, then it contains distinct elements {−1, +1}; these form an order-2 subgroup of the multiplicative group of the outer field, or equivalently an order-2 subgroup of the additive group of the inner field. It follows, therefore, that the inner field must have characteristic 2. However, there can only be at most two elements (±1) in the outer field which square to 1, so the inner field contains exactly two elements: the outer field is isomorphic to and the inner field is isomorphic to .

In all other threelds, the outer field has characteristic 2, so we shall henceforth concentrate on this case.

We can fully characterise the remaining finite threelds. Because the multiplicative group of a finite field is cyclic, it must have prime order to support an inner field, so the finite threelds have outer field and inner field where is a Mersenne prime.

### Infinite threelds

What about infinite threelds? If there are infinitely many Mersenne primes, then there are arbitrarily large finite threelds and the upward Löwenheim-Skolem theorem implies the existence of an infinite threeld (and indeed threelds of arbitrarily large infinite cardinalities).

Note that in an infinite threeld, the inner field must have characteristic 0; if it were of characteristic *p*, then every element in the outer field would be a *p*th root of unity, but there can only be at most *p* such roots.

Can we prove the existence of an infinite threeld *without* assuming the existence of infinitely many Mersenne primes?

Consider the first-order theory of fields of characteristic 2, augmented with the following additional infinite schema of axioms:

- every element has a unique square root;
- every element has a unique cube root;
- every element has a unique 5th root;
- every element has a unique 7th root;
- every element has a unique 11th root;
- […]
- every element has a unique pth root;
- […]

Every finite initial segment of these axioms has a model — firstly define:

and then note that 2^N − 1 ≡ 1 (mod *p*) for each of these primes by Fermat’s little theorem. It follows that the (cyclic!) multiplicative group of the finite field on 2^N elements has order not divisible by *p*, so we can find a unique *p*th root of any element, and therefore this finite field is a model of that initial segment of axioms.

So, by compactness of first-order predicate logic, there exists a model satisfying all of these axioms. It must be a field of characteristic 2, and these axioms ensure that its multiplicative group contains *p*th roots of all elements. As an abelian group, it is torsion-free and divisible, and is therefore the additive group of a vector space over .

Now, because every vector space over can be made into the additive group of a field (a number field if the dimension is finite, or a transcendental field otherwise), the infinite field that we obtained by the compactness theorem can indeed be upgraded into a threeld. The Löwenheim-Skolem theorem then gives threelds of all infinite cardinalities.

Is there a way to actually construct such an infinite threeld? Or to prove that the vector space over Q of the middle operation (inner additive/outer multiplicative) group must have infinite dimension?

I think I can prove that it can’t have dimension 1: Assume it did, and take an element f (not equal to one in the outer field). Then f+1 (in the outer field) must also be in the inner field. Since the vector space over Q of the middle operation has dimension 1, it must be the case that (f+1)^m = f^n for some m,n. As the outer field has characteristic 2, this means that the element f has finite middle operation order, contradicting the fact that the middle operation is a vector space over Q.

I’d be very interested to know if such a proof of nonexistence can be extended to any finite dimension, or if there exists some spectacular construction of finite dimension.

Even if the vector space must have infinite dimension, can it not still be an algebraic field? For example, the field of all real algebraic numbers has infinite dimension over Q, but is certainly not transcendental.

Another question I have: Is it possible to construct a field such that the middle operation forms not just a vector space over Q, but also a vector space over R (which I believe would force it to be isomorphic to R or C)?

I can answer the last of your questions: using the same first-order theory (of fields of characteristic 2 where every element has a unique pth root for each p), we can use Löwenheim-Skolem to specify the cardinality of the resulting infinite field. By choosing a model with cardinality equal to that of the continuum, the middle operation forms a vector space over Q which is isomorphic

as an abelian groupto the additive group operation on R and C, so you can indeed get the inner field to be R (or C).If I’m not mistaken, the multiplicative group of R{0} is isomorphic to the additive group of F_2 * R via log, so this gives an explicit threeld structure on R.

Unfortunately, F_2*R does not form the additive group of a field, as it doesn’t have a defined characteristic. So you’d need to construct a different field such that the multiplicative group structure is isomorphic to the additive group structure on R.

You say “it is a generalisation of a field”, but since the threelds constitute a subset of the fields, should that not be specialization instead?