TITLE: to prove the Cohen theorem
QUESTION [7 upvotes]: Now given a fact that I had proven:
If $I$ is an ideal in a commutative ring $R$ such that $I$ is not finitely generated but every ideal properly containing $I$ is finitely generated, then $I$ is a prime ideal.
My question is how to use this fact to prove the Cohen theorem: If every prime ideal of $R$ is finitely generated, then $R$ is noetherian.
My attempt is: suppose not, then there is an ideal not finitely generated, but didn't know how to derive any contradictions. (I don't think the ideal properly containing it are all finitely generated?)
Thanks for any helps in advance!
Added:
I asked this question for I just want to know if I can reach the conclusion without the axiom of choice, and after thinking these hours and with the comment and answer from you deer, I think that is unavoidable. Thank you all guys!

REPLY [4 votes]: Hint: If there is an ideal not finitely generated then the family $$\{I\subset R; I \text{ is an ideal not finitely generated 
}\}$$ is not empty. Now try using Zorn Lemma over this family.