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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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    I found this "symmetric" property interesting : if $R$ is Artinian, then it has finitely many maximal (= prime) ideals ; if $R$ is Noetherian, then it has [finitely many minimal prime ideals](http://math.stackexchange.com/q/525862).2017-01-24

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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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    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!2012-01-17