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Is there any equivalence conditions under which an ideal $I$ in ring of continuous functions be be an idempotent ideal?

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  • $\begingroup$ Not an answer - truly a comment - but perhaps this can add a shred of context: It is easy to find all the idempotent elements in the ring of continuous functions: the two constant functions, $0$ and $1$ (proof: intermediate value theorem). In a Noetherian ring, every idempotent ideal is generated by an idempotent element; unfortunately, this fact is not of direct use here: the ring of continuous functions is not Noetherian. Hence the question at hand may arise. $\endgroup$
    – Benjamin Dickman
    Commented May 25, 2015 at 15:51
  • $\begingroup$ An interesting idempotent ideal is for example the ideal of all functions that vanish at $0$ and that go faster to $0$ than any polynomial. $\endgroup$
    – HenrikRüping
    Commented May 25, 2015 at 17:58
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    $\begingroup$ Continuous functions on what space? If your space is totally disconnected there can be lots of idempotents $\endgroup$
    – Benjamin Steinberg
    Commented May 25, 2015 at 23:00

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