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Let $I$ be an ideal of a ring $R$ such that $I$ is not finitely generated but every ideal properly containing $I$ is finitely generated. Then I is prime.

Based on this result it can be proved that if every prime ideal in a ring $R$ is finitely generated, then $R$ is Noetherian.

Assume by contradiction. Let $I$ be a proper ideal of $R$ which is not finitely generated. Then $I$ is contained in a maximal ideal ideal $M$ of $R$ and hence prime. I am having difficulties in showing that $I$ is finitely generated. I think to show that if $J$ is an ideal that properly contains $I$ is finitely generated.

Would you help me please? Thank you in advance.

user404634
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    See https://math.stackexchange.com/questions/1130223/if-r-is-a-commutative-ring-in-which-all-the-prime-ideals-are-finitely-generated. – KCd Aug 22 '17 at 12:57

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If there exists a non-finitely generated ideal $I$, then there exists a maximal-non-finitely generated ideal containing it. (Show Zorn's Lemma applies the poset of non-finitely generated ideals containing $I$.)

By your lemma it is prime.

By your hypothesis, it is finitely generated, a contradiction.

So, no such non-finitely generated ideal exists in that ring.

rschwieb
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  • By which lemma is it prime? – Bernard Aug 22 '17 at 13:37
  • @Bernard The one cited in the first paragraph of the OP: "Let $I$ be an ideal of a ring $R$ such that II is not finitely generated but every ideal properly containing $I$ is finitely generated. Then $I$ is prime." – rschwieb Aug 22 '17 at 14:18
  • That's clear now. Thanks. However, the O.P. did not prove the lemma. – Bernard Aug 22 '17 at 14:32
  • @Bernard I take the "based on this result" to mean "assuming this." They undoubtedly got enough help on it in their previous question – rschwieb Aug 22 '17 at 14:55
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    That's the way I understood your answer with your clarification, after I've taken a look at the duplicate. A slightly different proof is to be found in Bourbaki, Commutative Algebra, ch. II, §1, exercise 6. – Bernard Aug 22 '17 at 15:01