Commutative and anticommutative ring that respects $x^2 = 0$, where $0$ is additive identity

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Can anyone present an example where a ring consists of infinitely many different sets/elements in a universe (ring), is both commutative and anticommutative, and respects $x \cdot x=x^2 =0$ for $\forall x$ in universe? $0$ is additive identity.

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If $G$ is an additive group one can equip it with the multiplication map $a\cdot b=0$ making it a ring, and this will satisfy the desired conditions with flying colors. We can make something less trivial though.

Simply generate a nonunital ring which is commutative, characteristic two, and suffiently nilpotent.

This can be done as follows: let $X$ be an infinite set (whose elements should be thought of as unknowns), and define $A$ to be the collection of all nonempty finite subsets of $X$ (which should be thought of as the collection of nontrivial monomials in our unknowns), form the space $\Bbb F_2A$ freely generated by $A$ (this is the underlying additive group of our ring) and equip it with multiplication extended from

$$U\cdot V:= \begin{cases} U\cup V & U\cap V=\emptyset \\ 0 & U\cap V\ne \emptyset. \end{cases}$$

The above expresses what it would be like to multiply monomials of commuting unknowns all satisfying the relation $x^2=0$: if the monomials share no unknown, you combine them all together, else you get the square $x^2$ of some unknown $x$ if it's in both monomials so we get $0$.

This multiplication is clearly commutative, is distributive by definition, and one easily checks that it's associative, so we do in fact have a ring on our hands.

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Take the field of two elements $F_2$.

Go into the polynomial ring $F_2[x_1,x_2,\ldots,x_n\ldots]$ with countably many variables.

Grab the ideal $I$ generated by $\{x_i^2\mid i\in \Bbb N\}$ and the ideal $J$ generated by $\{x_i\mid i\in \Bbb N\}$.

The quotient rng $J/I$ is a commutative-and-anticommutative rng of characteristic $2$ which satisfies $X^2=0$ for all elements $X$.

The fact that it is characteristic $2$ makes commutativity and anticommutativity the same thing, and it also allows you do use the Freshman's dream theorem to see why every element squares to $0$.