Threelds

Complex Projective 4-Space 2022-05-26

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 $\mathbb{F}_3$ and the inner field is isomorphic to $\mathbb{F}_2$ .

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 $\mathbb{F}_{2^p}$ and inner field $\mathbb{F}_{2^p-1}$ where $2^p - 1$ 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 pth 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:

$N = (2 - 1)(3 - 1)(5 - 1)(7 - 1)(11 - 1) \cdots (p - 1) + 1$

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 pth 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 pth 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 $\mathbb{Q}$ .

Now, because every vector space over $\mathbb{Q}$ 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.