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?
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The proof is involved in showing if $I$ is an ideal of a commutative ring with unity and $I^2=I,$ then the ideal $I$ is principal. See here. – user623904 Oct 08 '22 at 00:57
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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.
quasi
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Then my proof should be OK as is. There's no need to mention units or Jacobson radical. – quasi Aug 13 '17 at 08:25
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@quasi 1) Some brackets used when work with elements bring confusion. 2) Some hints would have been in order. – user26857 Aug 13 '17 at 11:16
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@Arrow: No, just definitions. The element $a$ is an element of the ideal $(a)$, hence, since $(a^2)=(a)$, it follows that $a$ is a multiple of $a^2$. – quasi Nov 26 '17 at 16:50
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@quasi ah, you're right of course. I was thinking of this question where $I$ is not apriori principal. – Arrow Nov 26 '17 at 18:23