Are Clifford Fractions (defined via left and right reductions) well defined? Are they algebraically closed?

Given a Clifford algebra, it seems to me that we can define 2 different algebras of fractions via either left reduction and then right reduction, or vise versa (right reduction and then left reduction). Assuming the reductions are well defined, the result is similar to a field in that all elements have inverses, however, addition is non-commutative, so it is not a field. I conjecture that the Clifford Fractions are algebraically closed.

For Clifford integers (Clifford numbers with integral coefficients), n, d, m, b such that d and b are non-zero

Addition (now non-commutative):

$ \frac{n}{d} + \frac{m}{b} = \frac{nb + dm}{db}$


$ \frac{n}{d} \frac{m}{b} = \frac{nm}{db}$


$ \frac{n}{d} \equiv \frac{m}{b} \ \text{mod left}$ iff $ \exists k \neq 0 : kn=m, \ kd=b$


$ \frac{n}{d} \equiv \frac{m}{b} \ \text{mod right}$ iff $ \exists k \neq 0 : nk=m, \ dk=b$

Left-then-Right-Reduction and Right-then-Left-Reduction:

I’m not sure this is well defined, but it seems that we can define two different “lowest terms reductions” for each element (first left, then right, or first right, then left), thus giving us two different algebras that resemble fields of fractions, except that both addition and multiplication are non-commutative.

Addition and Multiplication are not Closed:

Note, since, in the definitions above, $ db$ can be 0. Therefore, addition and multiplication are not closed.

Algebraic Closure:

The non-standard analysis proof of the Fundamental Theorem of Algebra (Algebraic Closure of the Complex Numbers) is simply the Continuous Newton’s Method:

To find a root of a polynomial f, choose an x. If f(x) is non-zero, calculate

$ dx = (\frac{-f}{f^{(n)}})^\frac{1}{n} dt$ where n is the smallest natural number such that the nth derivative of f(x), $ f^{(n)}(x)$ is non-zero at x

Update x as x+dx, then repeat. Since f is a polynomial and we’re using non-standard analysis, this eventually terminates at a finite value of x.

Over the polynomials of Clifford Fractions, it seems that the correct algorithm is to choose dx as

$ dx = (\frac{-f}{\nabla^n f})^\frac{1}{n} dt$ where n is the smallest natural number such that the nth vector-derivative of f(x), $ \nabla^n f(x)$ is non-zero at x

dx seems well defined in the Clifford Fractions (assuming the reductions that define the Clifford Fractions are well defined). However, addition is not closed, and I don’t have a proof that x + dx always exists within the algebra.