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Let $R$ be a commutative ring. We consider the set of ideals which are not finitely generated: $$ \Theta :=\{I ~:~ I\textrm{ is an ideal of } R\textrm{ which is not finitely generated} \}. $$ We assume that $\Theta \neq \varnothing$. By Zorn's lemma, let $I$ be a maximal element of $\Theta$.

Show that $I$ is a maximal ideal of $R$.

user26857
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Sya Ba
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    Hint: the only thing to prove is that if $I\subset J$ are two ideals, and if $I$ is not finitely generated, then $J$ is not finitely generated either. Now I really think that you should try to work this out by yourself and edit your question to indicate where you are stuck. This question is likely to get closed for lack of context otherwise. – Arnaud Mortier Nov 19 '19 at 15:25
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    @ArnaudMortier That's obviously wrong. $I \subseteq R$ and $R$ is always finitely generated. –  Nov 19 '19 at 15:27
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    @PaulK with $J\neq R$ of course, as in the definition of "maximal ideal". By the way how is $R$ finitely generated if it doesn't have a unit? – Arnaud Mortier Nov 19 '19 at 15:28
  • I already tried to show that " if $ I\subset J$ are two ideals, and if $I$ is not finitely generated, then $J$ is not finitely generated " but I can not – Sya Ba Nov 19 '19 at 15:33
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    @ArnaudMortier That still doesn't generally hold. Consider $R = \mathscr{O}(\mathbb{C})$, $J = { f : f(0) = 0}$, and $I = { f : f(0) = 0 \land (\exists k \in \mathbb{N})(\forall n \in \mathbb{Z})(\lvert n\rvert > k \implies f(n) = 0)}$. Then $I \subset J \subsetneqq R$, $I$ isn't finitely generated, but $J = \langle z\rangle$. The assumption that $I$ is maximal in $\Theta$ is essential. – Daniel Fischer Nov 19 '19 at 15:35
  • @DanielFischer Thanks for this example. My bad here. – Arnaud Mortier Nov 19 '19 at 15:37

1 Answers1

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There exist non-Noetherian rings whose maximal ideals are all finitely generated.

In each such ring, the poset of non-finitely generated ideals is nonempty, but indeed even if you find a maximal element, it cannot be a maximal ideal.

So the problem is evidently incorrect.

user26857
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rschwieb
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