Suppose $R$ is a von Neumann regular commutative ring with a unit. Prove that every principal ideal $I$ is generated by an idempotent element and for every principal ideal $I$, there exists a principal ideal $J$ such that $R=I+J$ and $I\cap J={0}$.

Would be grateful for your helps and advices.


Let $I$ be a principal left ideal of $R$ .By hypothesis $I=Ra$ for some idempotent $a \in R$. Consider $J=R(1-a)$. Then $R=I+J$ as

Let $r\in R$ .Then $r=ra+r(1-a)$ where $ra\in I;r(1-a)\in J$

Also let $x\in I\cap J$ then $x=r_1a$ and $x=r_2(1-a)$

Then $xa=r_1a $ and $xa=r_2a-r_2{a^2}=r_2a-r_2a=0$

$\implies x=r_1a=0$

$\implies I\cap J=\{0\}$

  • $\begingroup$ By hypothesis I=Ra for some idempotent Actually that seems to be the question, not the hypothesis. $\endgroup$ – rschwieb Dec 2 '14 at 18:36
  • $\begingroup$ I see this as the hypothesis.Maybe its other way round for you $\endgroup$ – Learnmore Dec 2 '14 at 18:38
  • 2
    $\begingroup$ Thanks for your contribution, but that is not the hypothesis but the question itself. $\endgroup$ – Aref Dec 2 '14 at 18:45
  • $\begingroup$ In the English language, "show that..." is used immediately before the question at hand. Hypotheses can be placed before and after the question in various ways. $\endgroup$ – rschwieb Dec 2 '14 at 19:50


Notice that if $a=axa$, then $ax$ is idempotent. Work to show that $axR=aR$.

If $e$ is idempotent, so is $1-e$. Try to solve your problem with $eR$ and $(1-e)R$.


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