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.
 A: 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.
A: 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.
