Can we have an associative form of "octonions" and hypercomplexes, if we eliminate division?

Question

I'm interested in hypercomplexes, or number systems with many square roots of $-1$. Now, I know that quaternions are non-commutative, but associative. I'm wondering if it's possible to have a number system with more "elements" (or square roots of $-1$) that remains associative.

My understanding is that hypercomplex systems lose many properties in order to keep division. So I'm wondering, if we discard division, can we keep the associative property and add more square roots of $-1$?

In other words, Can we have a number system with $2^n$ square roots of $-1$ that is associative, for arbitrary $n$?

Answer 1

Depends on what you mean by "number system". You can certainly construct the ring

$$ \mathbb R[X_1,X_2,\ldots, X_n]/\langle X_1^2-1, X_2^2-1, \ldots, X_n^2 - 1\rangle $$

in which $\prod_{i\in A} X_i$ is a square root of $-1$ for each of the $2^{n-1}$ odd-sized subsets $A\subseteq \{1,\ldots,n\}$. It will be associative and commutative.

But it will be horribly division-less.

In fact, for any commutative ring, if there are two different elements with the same square that are not each other's negative, it will contain zero divisors, because $$ (a+b)(a-b) = a^2 - b^2 = 0 $$ if $a^2=b^2$, and unless $a$ and $b$ are the same or the negative of each other, both of $a+b$ and $a-b$ will be nonzero.

Answer 2

Tessarines, also known as bicomplex numbers, are commutative and associative. You can build a tessarine-line system for any dimension $2^n$.

You also can build any dimensional analogs of split-complex numbers.

You also can build any dimensional analogs of dual numbers.

You can combine them. They will remain commutative and associative.

You can build 3 and 6-dimensional numbers in 3 ways that also would be commutative and associative.

All these systems have zero divisors, that is non-zero elements by which we cannot divide. So, yes, by sacrificing universally applicable division you can build a commutative and associative algebra of higher dimensions.

This does not mean that you cannot divide in these systems. Generally, you can, you only have to check the denominator to be not a zero divisor.

For instance, here is a division formula in a 3-dimensional system isomorphic to $\mathbb{R}\times\mathbb{C}$:

$\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\frac{a_2^2 \left(a+b j_2+c j_1\right)-a_2 \left(b_2 \left(a j_2+b j_1+c\right)+c_2 \left(a j_1+b+c j_2\right)\right)+c_2^2 \left(a j_2+b j_1+c\right)-b_2 c_2 \left(a+b j_2+c j_1\right)+b_2^2 \left(a j_1+b+c j_2\right)}{a_2^3+b_2^3+c_2^3-3 a_2 b_2 c_2}$

When doing the division you have to check that $a_2^3+b_2^3+c_2^3-3 a_2 b_2 c_2$ is not zero, which is the criterion for the denominator $a_2+b_2 j+c_2 k$ not to be a zero divisor.