# Hopkins-Levitzki: an uncanny asymmetry?

Not every left Noetherian ring is left Artinian. Take $\mathbb{Z}$ as a quick example.

But:

Hopkins-Levitzki theorem: a left Artinian ring is left Noetherian.

I find this quite amazing. I find this asymmetry shocking. It just seems plain unreasonable that there are rings where every ascending chain of ideals stabilizes but not every descending chain stabilizes, and at the same time every ring with stabilizing descending chains has stabilizing ascending chains.

I know asymmetries abound in ring/module theory, but this one strikes me as more elementary and uncanny.

My question is:

Why does this happen?

Of course, this question is at an informal level; I'm not asking for a proof of the theorem. I just want to understand why one chain condition implies the other, but not the other way around. At first glance, it just seems so symmetrical, that I would have expected the conditions to be equivalent, or to have neither condition implying the other.

My very naive first observation is that for noetherian rings, we have the characterization "every ideal is finitely generated", but for artinian rings there is not (that I know of) a simple analog, which is perhaps the first spark of an asymmetry...

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The principal ideal theorem implies that a (commutative) Noetherian ring does satisfy the descending chain condition for prime ideals. – Keenan Kidwell Jan 17 '12 at 1:40
Note that there is an obvious way to get a descending chain (say, $(x)\supset (2x)\supset \cdots$), but not an ascending one. – Alex Becker Jan 17 '12 at 1:59
Dear @Keenan: +1 for your extremely judicious comment on the descending chain condition for primes in a noetherian ring. – Georges Elencwajg Jan 17 '12 at 12:23
Neither artinian nor noetherian implies the other for modules, so maybe it makes more sense to ask what is different about the regular module versus other modules. I tend to think of it as the "1" being the main difference (non-unital rings need not satisfy hopkins-levitzki), but one has to be careful that a cyclic artinian module need not be noetherian (in the noncomm case), so it is something special about the generator of the regular module, not just that it is cyclic. – Jack Schmidt Jan 17 '12 at 18:34
I would like to add an evil asymmetry that will further blow your mind. A ring which satisfies the DCC on left principal ideals satisfies the ACC on right principal ideals! – rschwieb May 4 '12 at 19:06

I hope you agree that it suffices to explain the asymmetry in the commutative case : if the concepts artinian and noetherian are asymmetric there, symmetry cannot be restored by considering non-commutative rings.
And in the commutative case I think algebraic geometry comes to the rescue.

In the geometric translation from the ring $R$ to the affine scheme $Spec(R)$ noetherian means that every decreasing sequence of closed subschemes is stationary, which is quite reasonable, and artinian means that every increasing sequence of closed subschemes is stationary which is very unreasonable or at least very, very special.

In order to see that intuitively, it suffices to consider an algebraic variety $V$ over a field and look at subvarieties $W\subset V$.
It is pretty clear that you cannot visualize an infinite decreasing sequence $V\supsetneq W_1 \supsetneq W_2\supsetneq$ of closed varieties. This incapacity is noetherianness at work.
On the other hand you can easily visualize an increasing sequence of subvarieties: just take finite sets: $\lbrace P_1\rbrace \subsetneq \lbrace P_1,P_2\rbrace \subsetneq \lbrace P_1,P_2, P_3\rbrace, ...$ . This capacity is non-artinianness at work.

So you see that noetherian and artinian are completely asymetric concepts .

Finally the very, very special artinian case alluded to above is when $V$ is just a finite set to begin with.
In that case obviously you cannot have infinite strictly increasing sequences of subvarieties but "even less" infinite decreasing sequences ("even less" is vaguely supposed to illustrate that artinian implies noetherian.)

Edit As Keenan very pertinently comments, although in a noetherian ring you may very well have an infinite strictly decreasing sequence of ideals , these ideals cannot all be prime.
Geometrically you can have a strictly increasing sequence of subvarieties of $V$, but they cannot all be irreducible.
The proof of this (maybe underrated) result can be found in Eisenbud's book, Corollary 10.3 here [Google allows you to read everything around this result for free]

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Thank you for your answer! Even if my knowledge of algebraic geometry is (still) null, what you describe is geometrically reasonable to me. I guess that "to be included in" is geometrically easier than "to include"! – Bruno Stonek Jan 17 '12 at 13:14
Dear @Bruno, I am afraid I don't quite see what your remark on "to be included in" and to "include" refers to. As for algebraic geometry, many mathematicians (albeit not all of them) find pure commutative algebra a bit dry and are more comfortable when they see a geometric translation of its results into algebraic geometry. But I won't deny that learning (even basic) algebraic geometry is a non trivial investment in terms of time and energy. As a first step you might like to browse the exercises in chapter 1 of Atiyah-Macdonald's Commutative Algebra, starting from Exercise 15. – Georges Elencwajg Jan 17 '12 at 13:31
Dear Georges: I was surely too rash to comment on something I don't really understand; just ignore that comment. As for commutative algebra & algebraic geometry, I haven't studied them yet because I have been studying other things, but oh, don't worry, the interest is surely there, and I will get to it sooner than later! – Bruno Stonek Jan 17 '12 at 13:38