Non-associative ring whose non-zero elements form non-commutative quasigroup (with regard to multiplication) without identity?

Question

An example of a non-associative ring R whose non-zero elements form commutative quasigroup (with regard to multiplication in R) without identity is easy to find.

I'm looking for examples of a non-associative ring S whose non-zero elements form a non-commutative quasigroup without unity.

Put another way: A non-associative non-commutative division ring without unity. 'division ring' in the spirit of 'ring R (associative or not) whose non-zero elements form non-empty quasigroup by means of multiplication in R'.

Can you help me ?

Answer 1

The ring $\Bbb O$ of Cayley octaves is indeed the example you want, because it contains Hamilton quaternions $\Bbb H$ as a subring, whose multiplicative group (w/o $0$) is already noncommutative. $\Bbb O$ is actually constructed from $\Bbb H$ via the Cayley-Dickson process.

Answer 2

Let $(R,+,\cdot)$ be any non-associative division ring (including any field) - then $(R \setminus \{0\},+,/)$ is, in the general case, a non-associative non-commutative division ring without unity.

Answer 3

What the OP is looking is a noncommutative presemifield, and then a semifield is the case where there is a multiplicative identity element. Here is one of the two smallest examples, the other being given by the transpose of the table below.

On $\mathbb{Z}_2\times \mathbb{Z}_2$ with its usual addition, define $(0,0)\cdot (x,y) = (0,0) = (x,y)\cdot (0,0)$ for all $x,y\in \mathbb{Z}_2$ and define all other products according to the following table: $$ \begin{array}{c|cccc} \cdot & (0,1) & (1,0) & (1,1) \\ \hline (0,1) & (1,0) & (0,1) & (1,1) \\ (1,0) & (1,1) & (1,0) & (0,1) \\ (1,1) & (0,1) & (1,1) & (1,0) \end{array} $$ The nonzero elements evidently form a noncommutative quasigroup without an identity element. Nonassociativity is also easy to see: $$ ((0,1)\cdot (0,1))\cdot (0,1) = (1,0)\cdot (0,1) = (1,1) \neq (0,1) = (0,1)\cdot (1,0) = (0,1)\cdot ((0,1)\cdot (0,1))\,. $$ It's not obvious by inspection that both distributive laws hold, but this can indeed be checked.