Let $R$ be a commutative ring with the multiplicative identity. Let $x$ be a non unit element of $R$ such that $xR+aR=R$ for all $a\in R\smallsetminus (J(R)\cup U(R)\cup\{x\})$ where $U(R)$ is the set of units in $R$ and $ J(R) $ is the Jacobson radical of $R$.
My question is how to show $x$ is a idempotent element in $R$.
Any hints/ideas are much appreciated. Thanks in advance for any replies.
My definitions...
An ideal $ M $ in a ring $ R $ is said to be maximal if $ M\neq R $ and for every ideal $ N $ of $ R $ such that $ M\subset N\subset R $ either $ N=M $ or $ N=R $.
Let $ R $ be a commutative ring. Then the Jacobson radical $ J(R) $ of $ R $ is the intersection of all maximal ideals of $ R $.
- Let $ R $ be a ring. An element $ a $ of $ R $ is said to be idempotent if $ a^{2}=a $.
Let's suppose for a moment that you could prove x idempotent. Then x could not be outside of $J(R)$, because then you could set $x=a$ and get $aR=R$, and $a$ is a unit (contradiction.)
But if x is in $J(R)$ and is idempotent, it would have to be zero, and then $aR=R$ for all nonunits outside of the Jacobson radical. The only way this is true is if no such $a$ exists, and so $R$ is local.
This is incompatible with the second two things you want to prove. In both cases, we would be talking about a field. But that contradicts the existence of x. The question is simply inconsistent as written.
Anyhow, here is an example to consider. Localize $\Bbb Z$ at the multiplicative set given by the intersection of the complements of $(2)$ and $(3)$. Then with $x=2$, the possibilities for $a$ are the odd multiples of 3, which are cop rime to $x$, so $xR+aR=R$. But clearly x isn't idempotent, and the radical isn't zero.