I'm somewhat confused over the definition of Artinian and Noetherian rings.

A Noetherian ring is a ring in which there are no infinite chains of nested ideals. That is, if $I_i$ are some ideals in a ring $R$, which satisfy the condition $I_1\subset I_2 \subset ... \subset I_n \subset...$, then $\exists$ integer $N$ such that $I_N = I_m$ for all $m\ge N$.

An Artinian ring is a ring in which there are no infinite chains of nested ideals, but... if $I_i$ are some ideals in a ring $R$, which satisfy the condition $I_1\supset I_2 \supset ... \supset I_n \supset...$, then $\exists$ integer $N$ such that $I_N = I_m$ for all $m\ge N$.

But what is the difference between the Noetherian and the Artinian conditions? Can't we rewrite the ascending chain in the Noetherian case as a descending chain, and vice versa?

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    $\begingroup$ It suffices to give an example of one that is not the other. $\mathbb{Z}$ is Noetherian but not Artinian. There is no infinite ascending chain, but $\ldots \subset(8)\subset(4)\subset(2)$ is a descending chain. Ponder these two statements for a while. $\endgroup$ – Bob Jones Feb 26 '17 at 23:30
  • $\begingroup$ @BobJones Do you mean there is no infinite descending chain? I thought an ascending chain is a chain which begins from a smallest set and ascends to a bigger one. $\endgroup$ – sequence Feb 27 '17 at 1:02
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    $\begingroup$ You are correct in your definition. But do you believe $(2)\supset (4)\supset (8)\supset\ldots$ is ascending or descending? $\endgroup$ – Bob Jones Feb 27 '17 at 1:24
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    $\begingroup$ Exactly. So $\mathbb{Z}$ HAS infinitely descending ideal chains, namely the one I just gave. So it is not Artinian. It is a PID, though, so it is Noetherian. $\endgroup$ – Bob Jones Feb 27 '17 at 1:31
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    $\begingroup$ It turns out that artinian rings are noetherian, but it is far from obvious. $\endgroup$ – rschwieb Feb 27 '17 at 2:46

No, you can't just rewrite an infinite ascending chain as an infinite descending chain or vice versa. For instance, suppose you have an ascending chain $$I_1\subset I_2 \subset \dots \subset I_n \subset\dots$$ and want to get a descending chain $$J_1\supset J_2 \supset \dots \supset J_n \supset\dots$$ What are you going to define $J_1$ to be? You can't make it $I_1$, since that's the smallest ideal in the chain, and $J_1$ needs to be the largest instead. If you made $J_1$ be $I_2$, you could then make $J_2$ be $I_1$, but then you wouldn't be able to define $J_3$. If you made $J_1$ be $I_3$ you could define $J_2=I_2$ and $J_3=I_1$, but then you would have no way to define $J_4$. And so on.

For a similar phenomenon that should be familiar, there is an infinite ascending chain of natural numbers, namely $0\leq 1\leq 2\leq 3\leq\dots$. But there is no infinite descending chain of natural numbers, since if you start such a chain at $n$, you can only ascend $n$ times before you run out of natural numbers!

  • $\begingroup$ My understanding was that we could rewrite $I_1\subset I_2 \subset \dots \subset I_n \subset\dots$ as $...\supset I_n\supset \dots\supset I_2 \supset I_1$, hence my confusion. Also, if $R$ is Noetherian then $I_1\subset I_2 \subset I_n$ can be rewritten as $I_n\supset \dots\supset I_2 \supset I_1$, so it is also Artinian. $\endgroup$ – sequence Feb 27 '17 at 1:08
  • $\begingroup$ You can write that, but that's not an "infinite descending chain" for the purposes of the definition of an Artinian ring. $\endgroup$ – Eric Wofsey Feb 27 '17 at 1:11
  • $\begingroup$ So the idea is that every infinite ascending or descending chain must have a beginning element? Then this greatly clarifies the definition. Unfortunately, this important note was not given in my book or elsewhere I searched. $\endgroup$ – sequence Feb 27 '17 at 1:13
  • $\begingroup$ Well, the definition you quoted does not actually use the term "infinite descending chain". It just writes $I_1\supset I_2 \supset ... \supset I_n \supset...$. Notice that this is simply not the same as $...\supset I_n\supset \dots\supset I_2 \supset I_1$. $\endgroup$ – Eric Wofsey Feb 27 '17 at 1:14
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    $\begingroup$ No? As I've said, you can't interchange infinite ascending chains and infinite descending chains to deduce Artinian from Noetherian or vice versa. $\endgroup$ – Eric Wofsey Feb 27 '17 at 1:29

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