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I search for sufficient conditions for a ring $R$ so that any idempotent ideal constitutes only of idempotent elements of $R$. Of course, in the commutative case, any finitely generated idempotent ideal is generated by an idempotent element.

Any internet searches in the relevant literature (those in my hand) have come to naught. I would be appreciated, in advance, to any help or suggested information.

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  • $\begingroup$ "Of course, in the commutative case, any finitely generated idempotent ideal is generated by an idempotent element." But what does that have to do with your question? There are finitely generated idempotent ideals in commutative rings which contain non-idempotent elements. Take any field other than $F_2$, for example. Or, if you prefer a nonfield, any finite product of fields other than $F_2$... $\endgroup$ – rschwieb Jan 5 '16 at 16:50
  • $\begingroup$ "Constitutes only of" literally means that all elements are idempotent. Did you mean "is generated by idempotent elements"? That would fit better with the comment about commutative rings and f.g. idempotent ideals... $\endgroup$ – rschwieb Jan 5 '16 at 16:52
  • $\begingroup$ Since $R$ itself is an ideal of $R$, your condition would mean that every element of $R$ is idempotent... $\endgroup$ – J.-E. Pin Jan 5 '16 at 16:55
  • $\begingroup$ Yes, I mean that all elements of the ideal are idempotent. @rshwieb Yes, but I want to just state the mere result, no more. Any way, thanks for your examples. The idea of J.-E. Pin is a good one. $\endgroup$ – karparvar Jan 5 '16 at 18:11
  • $\begingroup$ @karparvar Unfortunately the notification was typoed, so I didn't see this until now. If you are sticking with what's written, you're stuck with the class of Boolean rings in which all elements are idempotent. $\endgroup$ – rschwieb Jan 6 '16 at 16:38
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Let me expand my comment into a slightly more complete answer.

Since $R$ itself is an ideal of $R$, the condition implies that every element of $R$ is idempotent and thus $R$ is a Boolean ring. Now every Boolean ring is commutative and defines a Boolean algebra where $x \wedge y = xy$ and $x \vee y = x + y + xy$ and $\bar x = 1 + x$. Conversely, every Boolean algebra $(B, \vee, \wedge, \overline{})$ defines a Boolean ring by setting $x + y = (x ∨ y) ∧ \overline{(x ∧ y)}$ and $xy = x \wedge y$. Finally, Stone's representation theorem for Boolean algebras gives a complete characterisation of Boolean rings (or Boolean algebras) as a fields of sets. A fields of sets is a Boolean subalgebra of the Boolean algebra $\mathcal{P}(E)$ of all subsets of a given set $E$, equipped with the operations $X \vee Y = X \cup Y$, $X \wedge Y = X \cap Y$ and $\overline{X} = E - X$, the complement of $X$).

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