Non associative, non commutative “ring” without unit whose additive group is $\mathbb{Z}_2 \times \mathbb{Z}_2$

Question

My ring theory teacher has a very broad definition of rings: he doesn't require them to be associative. As such, he told us to work out a definition of a product operation $\cdot$ on $R = \mathbb{Z}_2 \times \mathbb{Z}_2$ that is distributive over the additive operation $+$ (the operation that makes $R$ an abelian group) and that makes the resulting ring
$(R, +, \cdot)$ non associative, non commutative and without a unit element. Im having some trouble doing this and was wondering if someone could shed some light on the way to doing it – preferably without having to check for lots of cases or trial and error.

Answer 1

The distribution law tells us that $$ a(0,0)=a(0,0)+a(0,0)=0\\ (0,0)a=(0,0)a+(0,0)a=0\\ a(1,1)=a(1,0)+a(0,1)\\ (1,1)a=(1,0)a+(0,1)a $$ Hence it is enough to choose $(1,0)(0,1)$, $(1,0)(1,0)$, $(0,1)(1,0)$, $(0,1)(0,1)$. Here in order for the multiplication not to be commutative, $(1,0)(0,1)$ and $(0,1)(1,0)$ must be different.

One choice is as following. $$ (0,1)(0,1)=(1,0)\\ (0,1)(1,0)=(1,0)\\ (1,0)(0,1)=(0,1)\\ (1,0)(1,0)=(0,1) $$ You can check that this multiplication do satisfy the distribution law, is non-associative and non-commutative, and has no identity.