If $R$ is a ring such that $x^2=x$ $\forall x \in R$ and $I$ is a prime ideal. Show that $R/I$ has two elements.
$R/I = \{ r+I:r\in R \}$
Let $a \in R$
if $a\in I$ then $a+I = 0+I=I$.
if $a\notin I$. We know that $a^2=a$ then $a^2-a=a(a-1)=0 \in I$ since $I$ is an ideal, it is a subgroup under addition. It must contain $0$. As $a\notin I$ then $a-1 \in I$ because $I$ is a prime ideal
this implies that $a+I = 1+I$ so $$R/I = \{ I,1+I \}$$
what do you think, is it correct?...If it's not. How would you do it?