<s>[INST] Let R be a Boolean ring with unity. Show that $0$ is the only nilpotent element and $1$ is the only unit. A ring is said to be Boolean if $x^2=x$ for all $x \in R$ .
I tried it this way: let $x \neq 0$ ,
$x^n=0$ and 
$x^2=x$ so $x^{2n}=x^{n}=(-x)^n$ How do I show that $x=0$?
For unit part:
Let x be a unit so there exits $y \neq 0$ and and $xy=1$ then $xy=yx$ as we know that the Boolean ring is commutative and then $(xy)^2=xy=1$ and we have $x^2=x$ and $y^2=y$ so $(xy)^2=xy=x^2y^2$ then??
Thanks for any help!
 [/INST] For the first part, $x^2=x$ implies that $x^3=x^2x=x^2=x$ and similarly $x^n=x$ for all $n\geq 1$. Therefore if $x^n=0$ for some $n$ then $x=0$.
For the second part, since $x^2=x$ it follows that $x(1-x)=0$. Therefore if $xy=1$ then multiplying by $1-x$ we get $1-x=0$, so $x=1$.
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