If we let $R$ be local and suppose $x ∈ R$ satisfies $x^2 = x$, then I have to show that $x = 0$ or $x = 1$.
A commutative ring $R$ with $1$ is called local if $R − R^×$ is an ideal of $R$.
If we let $x, y ∈ R$, then I was thinking of the expansion of the equation $(x + y)^2$ = $x + y$. But I am not sure how to proceed from here and if this is even the right way to tackle this problem.
Any help would be grateful.
Rearranging, we have $$0= x^2 - x = x(x-1)$$
If $x\neq 0,1$, that means that both $x$ and $x-1$ are both nonzero zero divisors, and in particular non-units. Since $R$ is local, they must both be in the unique maximal ideal. But that means $$x - (x-1) = 1$$ is in that maximal ideal, so the maximal ideal isn't proper, a contradiction.
So it is only possible that $x=0$ or $x=1$, which of course both satisfy that relation.