an idempotent ideal is generated by an idempotent?

Question

How can i prove in a commutative ring that if an idempotent ideal I=Ra and a is not contained in jacobson radical of R ,also a is not unitary,I is generated by an idempotent?

Answer 1

Let R be a commutative ring, and suppose $I = (a)$ is an idempotent ideal, i.e., $I^2=I$.

The goal is to show $I$ can be generated by an idempotent element.

Is that the correct problem?

If so, the proof is easy . . . \begin{align*} &I^2 =I\\[4pt] \implies\;&(a)^2 =(a)\\[4pt] \implies\;&(a^2) = (a)\\[4pt] \implies\;&a = ra^2,\;\text{for some $r \in R$}\\[8pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{Claims:}\\[4pt] &(1)\;\;(a) = (ra)\text{.}\\[4pt] &(2)\;\;ra\;\text{is idempotent.}\\[8pt] &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \text{Proof of $(1)$:}\\[8pt] &ra \in (a)\\[4pt] \implies\;&(ra) \subseteq (a)\\[8pt] &a = ra^2\\[4pt] \implies\;&a = a(ra)\\[4pt] \implies\;&a \in (ra)\\[4pt] \implies\;&(a) \subseteq (ra)\\[8pt] &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \text{Proof of $(2)$:}\\[8pt] &(ra)^2 =r(ra^2) = r(a) = ra\\[4pt] \end{align*} Therefore the ideal $(a)$ has an idempotent generator, as was to be shown.

Note: I never used the part of the hypothesis relating to the Jacobson radical.