Are the quaternions not uncountably categorical?

Question

Boris Zilber has argued that the field of the complex numbers is "logically perfect". For one thing, the theory of an algebraically closed field of characteristic zero is uncountably categorical: it admits a unique model, up to isomorphism, of some uncountable cardinality - and thus a unique model of each uncountable cardinality, by Morley's theorem!

The field of real numbers is not "logically perfect" in Zilber's sense. More precisely, the theory of a real closed field is not uncountably categorical.

This led Terry Bollinger to ask about the quaternions.

Has anyone studied this? For starters, we need to settle on a 'theory of quaternions'. I want a theory where the only operations are + and ×, and the only constants are 0 and 1. Here's my guess about how to proceed. First we impose the axioms of a division ring (including associativity for multiplication, to rule out the octonions!). We impose a 'characteristic zero' axiom schema that says $1 + 1 + \cdots + 1$ can never equal zero. Then we impose an 'algebraically closure' axiom saying that any polynomial whose coefficients all commute has a root. (Any commuting set of quaternions lies in a subring isomorphic to the complex numbers.) Finally, we impose an axiom saying that multiplication is not commtuative, to rule out the complex numbers.

I haven't thought about this much, but I'm hoping these are enough, with the help of the Frobenius theorem.

Then we can ask if the theory of quaternions is uncountably categorical. My guess is that it's not.

Here's my rough reasoning. The center of the quaternions forms a copy of the real numbers. Moreoever, I'm guessing that we can prove using our 'theory of quaternions" that the center of the quaternions is a real closed field. Conversely, starting from any real closed field, we can build a model of the theory of quaternions. So, I believe we're in trouble, since the theory of a real closed field is not uncountably categorical.

Perhaps someone can say if this sort of argument is valid: if in some structure we can define a subset and this subset, equipped with some operations in our theory, is a model of a second theory that's not uncountably categorical... and any model of that second theory arises in this way... then the original theory could not have been uncountably categorical.

Or, if this doesn't work, maybe someone can straighten things out!

Answer 1

Ignoring axiomatizations, you can just consider all algebras that are first-order equivalent to the quaternions - that is, every first-order statement true in the quaternions is true in them, and vice versa.

Now your argument goes through perfectly. Every first-order statement about the center of an algebra is a first-order statement about that algebra, so the center of an algebra equivalent to the quaternions is a real closed field. Every first-order statement about a finite-dimensional algebra over a field with prescribed generators and relations is a first-order statement about the field, so the quaternion algebra over any real closed field is equivalent to the quaternions.

So by taking two in-equivalent uncountable real closed fields (say, one Archimedean and one not), you get two in-equivalent uncountable quaternion algebras.

Answer 2

I don't think the axioms suffice. Here is some background:

Theorem 16.16 of TY Lam's "First Course in Noncommutative Rings" states: "The centrally finite noncommutative division rings which are right algebraically closed are precisely the division rings over real closed fields. These division rings are also left algebraically closed."

Lam goes on to say that, without the "centrally finite" assumption, not much seems to be known.

In http://eprints.biblio.unitn.it/1626/1/ghiloni-octonions-preprint.pdf, published in J. Algebra Appl., 11, 1250088 (2012), Ghiloni seems to prove the octonionic analogue. Also, Ghiloni attributes the result above to Niven, Baer, and Jacobson, based on I. Niven, Equations in quaternions. Amer. Math. Monthly 48, (1941), 654–661.

For a noncommutative ring $D$, the "right algebraically closed" condition states that for any nonconstant polynomial $a_0 + a_1 X + \cdots + a_n X^n$ with coefficients in $D$, there exists an element $d \in D$ such that $a_0 + a_1 d + \cdots + a_n d^n = 0$. This is stronger than the condition you give, in which the coefficients are required to commute with each other.

The "centrally finite" condition (required in the quaternionic and octonionic results) states that the whole algebra is a finite-dimensional vector space over the center of the algebra.

Lam's remark suggsts that, without the "centrally finite" assumption, there may be right/left algebraically closed division rings which are not quaternion algebras over a real closed field. Since his axioms for algebraic closure are stronger (or at least as strong) as yours, but you omit the central finiteness condition, I think your axioms could allow some exotic examples.

Now, I think an exotic example is given in the paper "Algebraically-closed skew-fields" by L Makar-Limanov, Journal of Algebra Volume 93, Issue 1, March 1985, Pages 117–135. Makar-Limanov gives an even stronger definition of "algebraically closed" (using polynomials in the amalgamated product $A \ast Z[X]$ (amalgamated over $Z$) ). There Makar-Limanov constructs a centrally infinite, algebraically closed (in his strong sense) division algebra (a.k.a. skew-field). So Makar-Limanov's algebra will satisfy the axioms of the question, without being a quaternion algebra over a real closed field.

Add an axiom which implies "centrally finite" and I think it's golden. Unfortunately, I can't think of such an axiom which doesn't explicitly bound the dimension over the center, for example: "For every five elements $a,b,c,d,e$ of the algebra, there exist five elements $u,v,w,x,y$ of the center of the algebra, not all zero, such that $ua + vb + wc + xd + ye = 0$."