Unital ring $R$ such that $a^2=a$ for every $a\in R$

Question

Question: Let $R$ be a unital ring such that $a^2=a$ for every $a\in R$. Prove that $\mathrm{char}(R)=2$ and $R$ is commutative. Give infinitely many examples of such rings.

It's very easy to show that $\mathrm{char}(R)=2$ and $R$ is commutative, but I'm a little stuck as to how to give infinitely many examples.

Answer 1

Such rings are called Boolean rings.

$$a+a=(a+a)^2= a^2+a^2+a^2+a^2= a+a+a+a$$ $$\implies a+a=0$$

In particular $1+1=0$ and $\operatorname{char}(R)=2$.

Commutativity I leave as an exercise. You can do something analogously. Let me know if you need more hints on this.

For the examples, here is a general construction: Let $X$ be a set with $|X|\geq2$ (to avoid the trivial ring) and consider the power set $2^X$. Make this into a ring by defining

$$A+B= A\triangle B:= (A\setminus B)\cup(B\setminus A)$$ $$A\cdot B= A \cap B$$

Check that this is a ring with identity $X$. Clearly we have $A^2=A$ for every subset of $X$, so this ring satisfies your property.

Now, you can make an infinite collection of non-isomorphic rings by considering this ring and making the power set of this ring into a ring in the same way. By Cantor's theorem, these two rings have different cardinalities hence are not isomorphic. Continue this construction to get an infinite countable family of such rings.

---

Addendum: We have to exclude the trivial ring $R=\{0\}$ since this ring has characteristic $0$ (or $1$, depending on your definition) and satisfies your condition.

Answer 2

Elaborating on Mindlack's answer, from $(x+y)^2 = x^2 + y^2$, we obtain $$a^2 + a^2 = (a+a)^2 = (a+a)(a+a)= a^2 + a^2 + a^2 + a^2,$$ whence $a+a = a^2 + a^2 = 0$, and therefore the characteristic is $2$.

Concerning commutativity, from $x+x=0$, we get $x=-x$, whence $$a+b = (a+b)^2 = a^2 + b^2 + ab + ba = a + b + ab + ba$$ implies that $$ab = -ba = ba.$$

To give infinitely many such examples, consider Boolean rings (indeed, these rings satisfying $x^2=x$ are precisely the Boolean rings).
These are related to Boolean algebras, so you only have to think about infinitely many Boolean algebras: for example, the powerset algebras of finite sets.