3
$\begingroup$

A commutative ring $R$ is called Noetherian if every ascending chain of ideals in $R$ stabilizes, that is, $$ I_1\subseteq I_2\subseteq I_3\subseteq\cdots $$ implies the existence of $n\in\mathbb{N}$ such that $I_n=I_{n+1}=I_{n+2}=\cdots$.

My question is the following:

Suppose $R$ is commutative ring such that for every ascending chain of ideals $$ I_1\subseteq I_2\subseteq I_3\subseteq\cdots $$ there exist infinitely many $n\in\mathbb{N}$ such that $I_{n}=I_{n+1}$. Does it follow that $R$ is Noetherian?

Motivation. Let's call this condition on ascending chain "stabilizing temporarily" infinitely many times. If a counter-example to my question is found, we would obtain a nice family of non-Noetherian rings. The canonical example of non-Noetherian ring I was taught is given by $k[x_1, x_2, x_3,...]$, and the ascending chain $$(x_1)\subsetneq (x_1, x_2)\subsetneq (x_1, x_2, x_3)\subsetneq\cdots $$ which doesn't stabilize. Note that it doesn't "stabilize temporarily" (not even once).

$\endgroup$
2
$\begingroup$

The answer is yes, but for boring reasons.

Suppose $R$ is non-Noetherian and every ascending chain has an infinite number of successive equalities. Our hypothesis implies there is an ascending chain $I_1\subseteq I_2\subseteq\cdots$ that never stabilizes.

Then define the ascending chain $J_1\subset J_2\subset\cdots$ obtained by deleting all repeats from the original ascending chain. Since the original chain never fully stabilized, it must have an infinite number of distinct ideals, so this new chain never stabilizes and it does not have any successive equalites, let alone an infinite number of them guaranteed by our second hypothesis on $R$, a contradiction.

$\endgroup$
  • $\begingroup$ Thanks for your answer! I am just worried about one thing. When you say "deleting all repeats" is something like that allowed? I have always been warned to be cautious when dealing with the "infinite". $\endgroup$ – Prism Jul 9 '13 at 2:28
  • $\begingroup$ @Prism Sure. It can even be defined rigorously. Let $n(1)=1$ and for $i>1$ let $n(i)$ be the smallest $n$ such that $I_{n(i-1)}\ne I_n$. Then we can set $J_i=I_{n(i)}$. $\endgroup$ – anon Jul 9 '13 at 2:31
  • $\begingroup$ Thanks! (+1) so we just iterate one at a time :) $\endgroup$ – Prism Jul 9 '13 at 2:35
  • 1
    $\begingroup$ @Prism There is any easy way to visualize "deleting duplicates." Just imagine the set $S=\bigcup_i\{J_i\}$. Since sets don't carry duplicates, the elements of $S$ form a strictly increasing subchain of the original chain. (If you demand chains to be indexed by $\Bbb N$ then I would have to restate somehow, because this set could be finite. If you allow "chains" to have finitely many elements then we're OK.) Thinking of the $J_i$ as links in the chain, the ACC is equivalent to this condition: "Given any chain, the set of its links is a finite set." $\endgroup$ – rschwieb Jul 9 '13 at 14:01
  • $\begingroup$ @rschwieb: Thanks a lot for the insight! I really like the idea of collecting the ideals into one set which would automatically delete the duplicates. $\endgroup$ – Prism Jul 9 '13 at 15:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.