If every prime ideal in a ring $R$ is finitely generated, then $R$ is Noetherian?

Question

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.

Answer 1

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.