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Prove that if $R$ is a commutative ring in which all the prime ideals are finitely generated, then $R$ is Noetherian.

Here is what I been told to do: Suppose that $R$ is not Noetherian, and use Zorn’s lemma to obtain a maximal element $I$ in the collection of all ideals of $R$ that are not finitely generated. Then use the following proposition: Let $I$ be an ideal of a commutative ring $R$, and let $r ∈ R$. If the ideals $I +rR$ and {$s ∈ R : sr ∈ I$} are finitely generated, then $I$ is a finitely generated ideal.

Can anyone help please. thanks a lot.

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  • $\begingroup$ does complete proof help you? $\endgroup$
    – user 1
    Feb 2, 2015 at 12:48
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    $\begingroup$ yes, that would be help a lot. thank you. $\endgroup$
    – user138017
    Feb 2, 2015 at 12:48
  • $\begingroup$ This is sometimes called Cohen's theorem, depending on the context. $\endgroup$
    – rschwieb
    Feb 2, 2015 at 13:24

1 Answer 1

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Book: Steps in Commutative Algebra (by Sharp)

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  • $\begingroup$ What is $K=(P:a)$? $\endgroup$
    – user482152
    Sep 12, 2018 at 16:47
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    $\begingroup$ all elements like $r$ that $ra\in P$ $\endgroup$
    – user 1
    Sep 13, 2018 at 6:32
  • $\begingroup$ Did you really show p is finitely generated though? P=Rp1+Rp2 +...........+Rpn + aK not Ra? $\endgroup$ Feb 16, 2020 at 13:03

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