Infinite rings with lots of zero divisors Today I was trying to find an infinite ring $R$ whose all nonzero and nonidentity elements were zero divisors and actually found one: $\mathcal R =\text{Fun}(\mathbb N, \mathbb Z/2\mathbb Z)$. Given a function $f \in \mathcal R$ we can define another function $g \in \mathcal R$ by $g(x) = 1 - f(x)$. As everyone can easily check, $$(fg)(x) = f(x) - f(x)^2 = 0.$$

One can of course replace $\mathbb N$ with any infinite set. Now I'm curious if there are more examples of such strange behaving ring, that aren't similar to $\mathcal R$. Is there a notion of such rings in abstract algebra?
A very obvious example is any Boolean ring. Since $x^2=x$ for every element in the ring, $x(1-x)=0$. Thus if $x$ is neither $0$ nor $1$, $x$ must be a zero divisor.

These rings are very easy to come by. You can take any subring of a direct product of any number of copies of the field of two elements.
In [Malik, Mordeson, Sen, "Fundamentals of Abstract Algebra"] and some other books, they give following definition:


  Let $(R,+)$ be a commutative group. Define multiplication on $R$ by $ab=0$ for all $a,b \in R$, where $0$ denotes the identity element of the group $(R,+)$. Then $(R,+,.)$ is a ring called the "zero ring". If $R$ contains more than one element, then $R$ is a commutative ring without $1$ and every nonzero element of $R$ is a zero divisor.


In this sense, $(\mathbb Z(p^{\infty}),+,.) $ seems to satisfy this property by definition.