This question was asked in my assignments of commutative algebra and I am not able to make much progress on it.
Question: (a) Show that A is noetherian iff every prime ideal is of finite type.
(b) Let A be a commutative ring and I , J are ideals. Assume that J is of the finite type and A/I , A/J are noetherian. Show that A/IJ is still noetherian.
Attempt:(a) If A is noetherian,then I have shown that every prime ideal is of the finite type. But I am not able to proceed in the opposite direction. Please help me!
(b) assuming that J is of finite type, I have shown that J is noetherian and I have been given that A/I , A/J are noetherian . I showed that $A/IJ \subseteq A+ I$ and as any descending chain in A+I stabilizes then so does in A/IJ and hence A/IJ is noetherian. Is my proof fine?
Thanks!
(a) You have shown the "$\implies$" implication. Let us show the "$\impliedby$" implication.
We are given a ring $A$ such that every prime ideal is of finite type. We are asked to prove $A$ is noetherian, i.e. every ideal is of finite type.
Towards a contradiction, suppose there is an ideal that is not of finite type, i.e., the set of ideals that are not of finite type is not empty. Using Zorn’s lemma we can see that there exists a maximal element $m$ in this set.
Since $m$ is not of finite type, $m$ is not prime. Hence $m$ is properly contained in two ideals $a,b$ such that $ab\subseteq m$. By the maximal property of $m$, we know $a$ and $b$ are finite type and $R/a$ is Noetherian.
Since $b/ab$ as $R/a$-module is finitely generated and $R/a$ is Noetherian, every submodule is finitely generated over $R/a$, hence likewise over $R$.
In particular, $m/ab$ is a finitely generated $R$-module and since $ab$ is finite type, so is $m$. We have thus arrived at a contradiction.
This proof was given in commutative rings with restricted minimum condition by I. S. Cohen, 1950.
(b) "... as any descending chain in $A+I$ stabilizes then so does in $A/IJ$ and hence $A/IJ$ is noetherian." How is "descending chain" relevant to "noetherian"?