Making idempotent element by a relation Let $R$ be a commutative ring with identity and let $a, b \in R$ such that $a=ab$. How can we make a non zero idempotent element of $R$ by this relation?
 A: The existence of $a,b∈R$ with $a=ab$ does not imply the existence of an idempotent other than 0 and 1. For example, consider the ring $\mathbb Z[x,y]/(xy)$. It has two such elements, for example $a=x$ and $b=1−y$, but it has no non-trivial idempotents. Any element $p$ in this ring can be written uniquely in the form of a polynomial in $x$ and $y$ with no cross-terms (since $xy=0$), say $a+b_1x+⋯+b_mx^m+c_1y+⋯+c_ny^n$, with coefficients in $\mathbb Z$. If $m>0$, then $p^2$ will have an $x^{2m}$ term, which $p$ doesn't, so $p$ can't be idempotent. Similarly, an idempotent can't have $n>0$. So the only possible idempotents are those of $\mathbb Z$, namely 0 and 1.
(I tried to post this as a comment, but for some reason the TeX was made unreadable there.)
A: Given the relation $a = ab$, which implies $a - ab = a(1 - b) = 0$, the only zero divisors you are guaranteed are multiples of $a$ and $1 - b$.
For example if $R$ is a domain, and $a,b$ are the images of $x,y$ in $R[x,y]/x(1-y)$, then the only zero divisors will be multiples of $a$ and $b$.
But if we do have two zero divisors $a$, $1-b$, then we can obtain idempotents whenever $(a, 1-b) = (1)$, by the Chinese Remainder Theorem. Find $r,s$ so that $ra + s(1-b) = 1$. Then $ra$ and $s(1-b)$ are your idempotents. 
