descending chain of non-finitely generated ideals

Question

Let $R$ be a commutative ring (not necessarily with unity) such that $ab\ne 0,\forall a,b\in R\setminus\{0\}$. If every descending chain of non-finitely generated ideals of $R$ terminate, then is it true that every ascending chain of ideals in $R$ terminates ?

Answer 1

Yes, this is true. As a hint, if $I\subseteq R$ is any non-finitely generated ideal, consider the descending chain $$I\supseteq aI\supseteq a^2I\supseteq a^3I\supseteq\dots$$ for an appropriately chosen $a\in R$.

A full proof is hidden below.

Let $a\in I$ be any nonzero element. Then $I\neq aI$, since otherwise $a$ would generate $I$ as an ideal. Thus $I\supset aI$, and it follows that all the other inclusions in the chain above are also strict, since they are obtained from $I\supset aI$ by multiplication by $a$ and multiplication by $a$ is injective. Moreover, since multiplication by $a$ gives isomorphisms of $R$-modules $I\cong aI\cong a^2I\cong\dots$, all of these ideas are non-finitely generated.

So, this proves that if there exists a non-finitely generated ideal of $R$, then there exists a non-terminating infinite descending chain of such ideals. This is exactly the contrapositive of your statement.

Answer 2

At first I wondered if you didn’t mean the ACC, but then I saw it doesn’t seem to make a difference. Even considering this for rings with unity, the statement becomes silly.

If $R$ has even one non-finitely-generated ideal, it isn’t Noetherian.

If it is a ring with a nonempty chain of such ideals, then it has such an ideal.

So, this proposition could only hold in a ring with no non-finitely-generated ideals at all, whence it is Noetherian. But it is silly to characterize a ring property tautologically in terms of empty collections.

Answer 3

Eric Wofsey has already answered the question affirmatively for integral domains. user521337 then modified the original question (in the comments to Wofsey's answer) to ask what happens when the ring is allowed to have nontrivial zero divisors. In this situation the answer is negative. One can derive this from this answer. This link will take you to a description of a unital, commutative, local ring with exactly one nonfinitely generated ideal. Such a ring has no infinite descending chain of nonfinitely generated ideals, but must have an infinite ascending chain of ideals.