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This problem comes from this commutative algebra problem

Let $R$ be a commutative ring with identity, $I$ is a finite generated ideal of $R$ such that $I^2=I$, then exists $e\in R$ such that $I=Re$.

This problem can be proved using the Nakayama lemma, but the condition for finite generated is necessary? If I isn't finitely generated, does the conclusion still hold? follow

Let $R$ be a commutative ring with identity, $I$ is an ideal of $R$ such that $I^2=I$, then exists $e\in R$ such that $I=Re$.

How to prove this? Or there is a counter-example to disprove.

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    $\begingroup$ What about the ideal $K[\mathbf{Q}_{>0}]$ in $K[\mathbf{Q}_{\ge 0}]$? $\endgroup$
    – YCor
    Commented Jul 19, 2023 at 11:48
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    $\begingroup$ Of course the question is strangely posed. If the ideal is not f.g., obviously it cannot be generated by an idempotent. The question is rather whether "finitely generated" can be removed from the assumption (and the answer is no, i.e. there exists an infinitely generated example). $\endgroup$
    – YCor
    Commented Jul 19, 2023 at 11:50
  • $\begingroup$ GuoJi: no, this is certainly not what I wrote. $\endgroup$
    – YCor
    Commented Jul 19, 2023 at 12:04
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    $\begingroup$ @HenrikRüping Does anybody use “infinitely generated” in that way? Are there really people who would say that $\mathbb{Z}$ is an infinitely generated abelian group? $\endgroup$
    – Jeremy Rickard
    Commented Jul 19, 2023 at 13:36
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    $\begingroup$ @GuoJi: The title speaks about an idempotent, but the body of the question does not say anything about the element $e$. Do you want $e^2 = e$? $\endgroup$
    – Alex M.
    Commented Jul 19, 2023 at 14:55

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I have a counterexample from analysis. Let $R= C[0,1]$, the ring of real-valued continuous functions on the unit interval, and let $$I = \{ f \in C[0,1] : f(0) = 0 \}.$$ Then $I$ cannot be generated by any idempotent $p$, for if it were $p$ would have to satisfy $p(0) =0$ and $p(x) = 1 \ (x \neq 0)$, contradicting continuity. However you can check that $I^2 = I$: if $f\in I$ and $f(x) \geq 0 \ (x \in [0,1])$, then $f = (\sqrt{f})^2$ shows that $f \in I^2$; for general $f \in I$, decompose the function into its positive and negative parts.

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  • $\begingroup$ Edited as I realised there was a much simpler answer than my original response. $\endgroup$
    – Jared White
    Commented Jul 19, 2023 at 14:37
  • $\begingroup$ Your quantification is not clear to me, but I think "$p(x) = 1$ ($x \ne 0$)" means "$p(x) = 1$ for some $x \ne 0$" (rather than, as I originally read it, "$p(x) = 1$ for all $x \ne 0$"). \\ Incidentally, while a simple counterexample is good, I don't think there's anything wrong with giving an additional counterexample, too. $\endgroup$
    – LSpice
    Commented Jul 19, 2023 at 15:01
  • $\begingroup$ It means $p(x) = 1$ for all $x \neq 0$. If $p$ generates $I$ then it cannot vanish at some $x \neq 0$, since this would force $f(x) = 0$ for every $f \in I$. $\endgroup$
    – Jared White
    Commented Jul 19, 2023 at 15:56
  • $\begingroup$ Re, ah, good point. Even the existential quantification would be enough, since a function that is an idempotent in $C([0, 1])$ can only be $\{0, 1\}$-valued, hence cannot assume both $0$ and $1$ as values; but, as you point out, a more explicit approach is available. $\endgroup$
    – LSpice
    Commented Jul 19, 2023 at 16:08

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