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Let $I$ be a finitely generated ideal in $R$, such that $I^2 = I$. Using the fact there exists $x\in R$ such that $e = 1 - x\in I$ and $xI = 0$, use Nakayama's Lemma to show that $I$ is principal, generated by an idempotent.

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marked as duplicate by YACP, Amzoti, Micah, Davide Giraudo, Joe May 12 '13 at 20:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

could you please include some information regarding what you have tried so far, or whether your trouble lies in understanding NAK, how to apply it, etc... – John Martin Feb 8 '13 at 0:31
@DoctorBatmanGod Huh? The linked question says nothing about avoiding Nakayama's lemma, and one of the proofs mentioned in the answer does use the lemma. – Math Gems Feb 8 '13 at 3:51
@MathGems Sorry, I had someting else open in a new tab and I thought you linked to it where the question was asking to avoid the use of Nakayama's lemma. But I would agree, this can be considered a duplicate. – Sam DeHority Feb 8 '13 at 4:16
up vote 3 down vote accepted

Well, you know that by Nakayama's that there is an $x$ such that $1-x \in I, xI = 0$. For any element $y$ in $I$ we have that $$(1-x)y = y-xy = y$$ Which means that any element of $I$ can be written of the form $$(1-x)z ~~~z\in R$$ which gives the inclusion $I \subseteq (1-x)$ and the inverse inclusion comes from the fact that $1-x \in I$. We also have idempotence because $$(1-x)(1-x) = 1-x$$ By our first claim. Therefore $I = (1-x)$ and satisfies all listed properties.

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