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

Question

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.

Answer 1

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 $(2)$.