How to prove an element is idempotent In a commutative ring $R$, if $x$ is a non unit element outside Jacobson radical and $Rx +Ry =R$ for all non units $y$ outside Jacobson radical other than x, then how to prove $x$ is idempotent in $R$?
 A: Let $J=\bigcap_{M\in\operatorname{Max}(R)} M$ denote the Jacobson radical of $R$, and suppose $x$ is not idempotent. Then we claim we must have $x^2\in J$; indeed, if $x^2\notin J$, then $x^2$ is a non-unit of $R$ distinct from $x$ and lying outside the Jacobson radical, so we have $R=Rx+Rx^2=Rx$, contradicting that $x$ is not a unit. Thus $x^2\in J$, so $x^2$ lies in every maximal ideal of $R$, so $x$ lies in every maximal ideal of $R$, so $x\in J$, a contradiction. Thus $x$ must indeed be idempotent.

To answer your question below, the hypotheses does indeed force $J=\{0\}$. Indeed, suppose we had $j\in J\setminus\{0\}$. Then certainly $x+j\notin J$, since $x\notin J$, and $x+j\neq x$. Thus, by hypothesis, we have $Rx+R(x+j)=R$, so there are some $a,b\in R$ with $ax+b(x+j)=1$. But this means that $$(a+b)x=1-bj,$$ and we claim $1-bj$ is a unit; indeed, if it were not, then it would be contained in some maximal ideal $M<R$. But we have $j\in J\leqslant M$, so this would mean $1=(1-bj)+bj\in M$, a contradiction. Thus $1-bj$ is a unit, so $x$ is a unit, contrary to the hypothesis, and thus we must have $J=\{0\}$.
(Exercise: Show that, for any commutative ring $R$ and any element $x\in R$, $x$ lies in the Jacobson radical of $R$ if and only if $1-rx$ is a unit for all $r\in R$. One direction of the proof can be argued as above.)

Also, to show that this discussion is not vacuous – ie to show that there actually are rings satisfying the desired conditions – consider the ring $R=\mathbb{Z}\big/6\mathbb{Z}$. The Jacobson radical of $R$ is $\{\overline{0}\}$, and the non-zero non-units of $R$ are $\{\overline{2},\overline{3},\overline{4}\}$, so $\overline{3}$ satisfies the conditions of your problem, since it is coprime with both $\overline{2}$ and $\overline{4}$. For another very tractable example, consider the ring $$R=\mathbb{F}_2[x]\big/\langle x(x+1)\rangle,$$ the non-zero non-units of which are $\overline{x}$ and $\overline{x+1}$, which are of course coprime with each other.
