for every $x$ in ring $R$, we have $x^2 = x$. Show any finitely generated ideal is cyclic.

Question

With the given condition, $R$ should be commutative and for every $x$ in $R$, $2x = 0$. I'm trying to first solve this problem for the case where ideal $I$ is generated with only two elements, like $I = < x, y>$. In this case, I found a hint that if we consider $z = x + y + xy$, we can prove $I = < z >$
I know $xz = x$ and $yz = y$. How can I use this information to show $I = < z>$?

Answer 1

That shows $x,y\in (z)$, therefore $(x,y) \subseteq (z)$.

On the other hand, $z\in (x,y)$ obviously. So $(z)\subseteq (x,y)$.

Therefore you have equality.