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Let $R$ be a commutative ring with $1 \ne 0$.

I'm trying to prove that if $R$ contains an ideal $I$ that is not finitely generated, then $R$ contains a proper ideal $J$ such that $R/J$ is Noetherian.

If $J$ is a maximal ideal, then $R/J$ would be a field, which would be Noetherian. Locating a maximal ideal is proving difficult for me, however. I know that since $I$ is not finitely generated, $R$ is not Noetherian. Thus I can't guarantee myself a maximal element under inclusion for a given nonempty set of ideals. I would appreciate some help finding a suitable maximal ideal, or a nudge in another direction if you know one.

Thank you.

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    $\begingroup$ The existence of a maximal ideal in any commutative ring with $1$ is given by Zorn's lemma. Any reason why you cannot use that? $\endgroup$ – Ayman Hourieh Mar 19 '15 at 22:44
  • $\begingroup$ The result as I know it says that in a ring with 1 every proper ideal is contained in a maximal ideal. Can I just use the ideal $\{0\}$ (since I don't know that $I$ is proper), to guarantee the maximal $J$? If so, it seems like I don't need the existence of $I$ at all (except to keep $R$ from being Noetherian and making the problem trivial by letting $J=\{0\}$. $\endgroup$ – jamisans Mar 19 '15 at 22:50
  • $\begingroup$ Exactly. Use $\{0\}$. I think $I$ is mentioned just to tell you that $R$ is not Noetherian. $\endgroup$ – Ayman Hourieh Mar 19 '15 at 22:52
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Take any maximal ideal $\mathfrak m$ of $R$. Then $R/\mathfrak m$ is a field, which is both noetherian and artinian.

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