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

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.

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.
Here's a key insight that helps a lot. Let $$a=(1,0)$$ and $$b=(0,1)$$. Then $$a$$ and $$b$$ generate $$R$$ as an abelian group, and so the distributive law implies that the multiplication is determined entirely by products of these two elements. For instance, we can compute $$(1,1)\cdot (0,1)=(a+b)\cdot b=a\cdot b+b\cdot b$$ from knowing what $$a\cdot b$$ and $$b\cdot b$$ are.
So, a multiplication satisfying the distributive law is determined by just four products: $$a\cdot a$$, $$a\cdot b$$, $$b\cdot a$$, and $$b\cdot b$$. (Less obviously, every possible collection of values you could give to these four products does extend to a multiplication satisfying the distributive law--this is related to the fact that $$a$$ and $$b$$ are not just generators but a basis for $$R$$ as a vector space over $$\mathbb{Z}_2$$.) Now you can experiment with different values you can give these to violate commutativity and associativity. (For instance, the only way to violate commutativity will be to make $$a\cdot b$$ different from $$b\cdot a$$.) It should not take much trial and error to find an example that works.