# Special idempotents in a commutative ring

Let $$R$$ be a commutative ring with $$1$$ and $$e$$ be an idempotent element of $$R$$ with the property that if $$e=x+y$$ (where $$x, y\in R$$), then there exists $$r\in R$$ such that either $$e=rx$$ or $$e=ry$$. Can we deduced that for such an idempotent, the ideal $$\langle e\rangle$$ is a minimal ideal of $$R$$?

• By a minimal ideal of $R$, do you mean a minimal non-zero ideal of $R$? Note that if $R = (\mathbb{Z}/4\mathbb{Z})^2$, then $e = (1, 0)$ satisfies your requirements. – Luc Guyot Jul 2 at 20:16

If you take $$e=1$$, then the condition imposed on $$e$$ translates to: for every $$x \in R,$$ either $$x$$ or $$1-x$$ is a unit. This will be satisfied in any local ring $$R$$, while $$R=(e)$$ will not be minimal unless $$R$$ is a field. This already shows that the conclusion does not hold in general.
(If $$e=1$$ seems too trivial, this example should survive replacing the local ring $$R$$ by any ring of the form $$R\times S$$ and $$e=1$$ by $$(1, 0)$$).
On the other hand, if the Jacobson radical $$J(R)$$ of $$R$$ is $$0$$ and a nonzero idempotent $$e$$ has the imposed property, then $$(e)$$ is indeed minimal:
One needs to check that for $$0 \neq x$$ divisible by $$e$$, i.e. $$ex=x$$, one has $$e$$ divisible by $$x$$. In other words, if $$x=ex$$ is an element not dividing $$e$$, then $$x=0$$. The strategy is to show that $$x$$ lies in this case in the Jacobson radical, hence is $$0$$.
The imposed condition, together with the fact that $$x$$ does not divide $$e$$, implies that $$(e-x)=(e-ex)=e(1-x)$$ divides $$e$$. In particular, $$(1-x)$$ divides $$e$$, and $$e$$ divides $$x$$, so $$(1-x)$$ divides $$x$$. It also divides $$(1-x)$$, so altogether $$1-x$$ divides $$1=(1-x)+x$$. Thus, $$(1-x)$$ is a unit.
To finish the argument, note that $$x$$ in the above argument can be replaced by any multiple $$x'=rx$$. So we obtain that for all $$r,$$ $$1-rx$$ is a unit, i.e. $$x \in J(R)=0$$.