# Wrong proof Every Boolean ring has 2 elements

I need help to understand where my argument is wrong, because I proved that every boolean ring has only $2$ elements which is wrong.

My proof goes like:

Let $R$ be a boolean ring. Then by definition $x^2=x$ which implies $x^n=x$ for all $n\ge1$. Therefore the only nilpotent element is $0$ hence the nilradical is just $0$. But the nilradical is a prime ideal therefore $R/0=R$ is an integral domain. Now consider an arbitrary element $x\in R$. We have $x^2-x=0$. So $x(x-1)=0$ therefore since $R$ is an integral domain $x$ must either be zero or $1$.

In fact the nilradical is the intersection of all minimal primes, i.e. it is prime if and only if there is precisely one minimal prime. Geometrically, this means that $Spec(R)$ is irreducible.