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I am trying to prove that an ideal that is maximal with respect to not being finitely generated must be prime.

What does it mean to be an ideal that is maximal with respect to not being finitely generated?

I am pretty sure that I need to assume that we have a prime ideal $P \subseteq R$ such that $P=$ the intersection of some ideals. Then I can use that for $f \in R$, $(P:(f))$ equals R if $f \in P$ or $P$ if $f \notin P$.

Is this the right approach?

user26857
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math1234567
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    If we denote this 'maximal with respect to not being f.g.' ideal by $P$, it means that if there is some other ideal $I$ that is not f.g. with $P\subseteq I$, then $P=I$. – ah11950 Mar 23 '14 at 11:28
  • Thanks, that makes sense. But how does that help showing that P is prime? – math1234567 Mar 23 '14 at 11:40

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You can actually prove this from elementary principles:

Let $I$ be the ideal maximal with respect to not being finitely generated. Suppose $I$ is not prime. Then $\exists f,g\in R$ such that $f,g\notin I,\ fg\in I$. Then consider $J := I + (f)$. You can show that since $I$ not finitely generated, $J$ is not finitely generated. Moreover, $J \neq R$ (otherwise $g\in I$). Hence, $J$ is a proper ideal strictly greater than $I$ that is not finitely generated, which is a contradiction to our choice of $I$.

chriseur
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  • when you say "you can show that since $I$ is not finitely generated, $J$ is not finitely generated", wouldn't we get that immediately just from how $J$ is defined? – User7238 Dec 01 '21 at 22:12