# Examples of Non-Associative Algebras with (Non-Associative) Division and with Unity without Classical Division

Let $$\mathbb{K}$$ be a field. By an $$\mathbb{K}$$-algebra I mean a $$\mathbb{K}$$-module $$A$$ equipped with a binary operation $$∗:A\times A\rightarrow A$$ that is $$\mathbb{K}$$-bilinear and denoted by $$(x,y)\mapsto xy$$.

Let $$A$$ be an algebra.

1. $$A$$ is said to have (non-associative) division if $$A\neq0$$ and for any $$a\in A$$ with $$a\neq0$$ the functions $$L_a:x\mapsto ax$$ and $$R_a:x\mapsto xa$$ are both bijective.

2. $$A$$ is said to classical division if $$A$$ has unity and for every $$a\in A$$ with $$a\neq0$$ there is an element $$b\in A$$ such that $$ab=ba=1$$.

I know that, if $$A$$ is alternative, that is, satisfies $$(aa)b=a(ab)$$ and $$(ab)b=a(bb)$$ for any $$a,b\in A$$, then (1) and (2) are equivalent.

What would be examples of non-associative algebras that:

a) Has (non-associative) division and unity but does not have classical division?

b) Has classical division but does not have (non-associative) division?

I already saw the question here, but it does not have a satisfactory answer yet.

• What is $\mathbb{K}$? A field? A commutative ring? Commented Aug 22, 2022 at 16:19
• For a) you can use Baez example on modified Quaternions and for b) you can take (imho) the Sedenions, since nonzero elements are invertible ($a^ {-1}=a^\star/||a||$)and it has zero divisors so $R_a$ and $L_a$ can''t be bijections. All of the above are unital algebras. Commented Aug 22, 2022 at 17:00

Consider the real algebra $$A$$ given by the usual Quaternion multiplication $$\mathbb{H}$$ except that $$i^2=-1+\frac{j}{2}$$. I will denote the product of $$a,b\in A$$ by "$$a.b$$" and the product of $$a,b\in \mathbb{H}$$ by "$$ab$$" and write $$v=v_1+v_2i+v_3j+v_4k$$ for a generic vetor since the underlying vector space structure on $$A$$ and $$\mathbb{H}$$ coincide. Also, consider the Sedenions $$\mathbb{S}$$ with its usual structure.
Af 1.: The sedenions $$\mathbb{S}$$ satisfy $$(2)$$ but not $$(1)$$.
Given $$0\neq a\in \mathbb{S}$$ we have $$a(a^\star/||a||)=1$$ where $$a^\star$$ is the conjugation of $$a$$ (it is an easy exercise to show that $$aa^\star=a^\star a$$). Notice that taking $$e_0=1,\ldots,e_{15}$$ as the usual basis for $$\mathbb{S}$$ we have $$(e_3+e_{10})(e_{6}-e_{15})=0$$ therefore $$L_{e_3+e_{10}}$$ is not bijective.
Af 2.: $$A$$ satisfies $$(1)$$ but not $$(2)$$.
Take $$a,b\in A$$ such that $$a.b=0$$ but $$a,b\neq 0$$.
Notice that $$a.b=ab+\frac{a_2b_2j}{2}$$. Clearly $$a_2=0$$ or $$b_2=0$$ implies $$||ab||=||a||\cdot||b||=0$$ hence $$a=0$$ or $$b=0$$. WLOG $$a,b$$ may satisfy $$a_2=1=b_2$$. If $$0=a.b=ab+\frac{j}{2}$$ then $$||ab||=||a||\cdot ||b||=1/2$$ but this is a contradiction for $$||a||=(a_1^2+1+a_3^2+a_4^2)^{1/2}\geq 1$$ and the same applies to $$b$$. This shows that $$L_x,R_x$$ are indeed bijective in $$A$$. Clearly '$$i$$' has a right inverse named $$(-i-k/2)$$ and a left inverse $$(-i+k/2)$$, also by injectivity of $$R_i,L_i$$ they are unique, hence $$A$$ can't satisfy $$(2)$$.