I was working through exercise 3.15 in Aluffi's Chapter 0, which has a step-by-step plan of a proof of the fact that a commutative ring is Noetherian if all the prime ideals are finitely generated. The plan is as follows:
3.15. Recall that a (commutative) ring $R$ is Noetherian if every ideal of $R$ is finitely generated. Assume the seemingly weaker condition that every prime ideal of $R$ is finitely generated. Let $\mathcal F$ be the family of ideals that are not finitely generated in $R$. You will prove $\mathcal F=\emptyset$.
- If $\mathcal F\ne\emptyset$, prove that it has a maximal element $I$.
- Prove that $R/I$ is Noetherian.
- Prove that there are ideals $J_1$, $J_2$ containing $I$, such that $J_1J_2\subseteq I$.
- Give a structure of $R/I$ module to $I/J_1J_2$ and $J_1/J_1J_2$.
- Prove that $$I/J_1J_2$$ is a finitely generated $R/I$-module.
- Prove that $I$ is finitely generated, thereby reaching a contradiction.
However, it seems that half of the steps are actually redundant. This makes me unsure whether my proof is correct, so if anybody could help find the error (if there is one), that would be great.
The union of every chain of $\mathcal F$ is not finitely generated since otherwise some element in the chain would be finitely generated. So every chain has an upper bound and we have a maximal element $I$.
$R/I$ is Noetherian since the preimage of any non-finitely generated ideal would be a non-finitely generated ideal in $R$ larger than $I$.
$I$ is not prime since otherwise it would be finitely generated by the statement of the theorem. Hence we have $a,b\notin I$ s.t. $ab\in I$.Then $I$ is a proper subset of ideals $I+(a)$ and $I+(b)$. This is where my proof differs from the plan: but these 2 larger ideals must be finitely generated, or that would contradict the maximality of $I$. Now note that $(I+(a))(I+(b))=I+(a)I+(b)I+(ab)=I+(ab)=I$, and $I$ is finitely generated as a product of finitely generated ideals. We arrived at a contradiction, so the assumption $\mathcal F\neq \emptyset$ is false.