I know they are important in abstract algebra, but why do people study them? Why are they so important to study? Do they make certain things easier to understand?

1You might find something here: http://mathoverflow.net/questions/69307/whyarenoetherianringssuchnaturalobjectsinalgebraicgeometry – Ian Coley Dec 08 '13 at 21:58

2Well many many many nontrivial theorems in commutative algebra only hold for noetherian rings (open any book, for example Eisenbud's or Matsumura's). A nice and basic example is $\dim(R[t])=\dim(R)+1$ (for general $R$ we only have $\dim(R)+1 \leq \dim(R[t]) \leq 2 \dim(R)+1$ and every value occurs). – Martin Brandenburg Dec 08 '13 at 22:22

1See also http://en.wikipedia.org/wiki/Noetherian_induction for a proof technique which is not available for arbitrary rings. – Martin Brandenburg Dec 09 '13 at 01:15
4 Answers
The answer to this question (which I have also asked myself at various times) might also be a "No". Maybe Noetherian rings aren't that important after all, and their perceived ubiquity is a meme in commutativealgebraic literature?
Let me quote a paragraph from the Foreword of the (so far unfinished) translation of Commutative algebra  Constructive methods by Henri Lombardi and Claude Quitté (the French version, Algèbre Commutative, Méthodes constructives, is available from Calvage & Mounet, and some additions can be found on http://hlombardi.free.fr/publis/LivresBrochures.html ). I hope the authors don't mind this.
Finally, let us mention two striking traits of this work compared to classical works on commutative algebra.
The first is to have left Noetherianity on the backburner. Experience shows that indeed Noetherianity is often too strong an assumption, which hides the true algorithmic nature of things. For example, such a theorem usually stated for Noetherian rings and finitely generated modules, when its proof is examined to extract an algorithm, turns out to be a theorem on coherent rings and finitely presented modules. The usual theorem is but a corollary of the right theorem, but with two nonconstructive arguments allowing to deduce coherence and finite presentation from Noetherianity and finite generation in classical mathematics. A proof in the more satisfying framework of coherence and finitely presented modules is often already published in research articles, although rarely in an entirely constructive form, but “the right statement” is generally missing in the reference works.
I certainly endorse such a position as a constructivist, as a noncommutative algebraist and as a combinatorialist. (Most advantages of the concept of Noetherianity crumble into dust when one crosses the boundaries of commutative algebra into either direction.)
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Interesting! I don't think I know anything about the constructive picture of algebra. If Noetherianity is too strong, do constructivists have a alternative condition? Or is it just a casebycase basis when constructive proofs are possible? – rschwieb Dec 09 '13 at 13:37

1As far as I understand, Lombardi and Quitte are mostly proposing coherence and finite presentation conditions finetuned to the actual statements being made. (Often, of course, Noetherianity can be disposed of completely.) – darij grinberg Dec 09 '13 at 17:25


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The ring of integers, Dedekind domains, and a host of other rings from number theory and ideal theory that are Noetherian all evidence that they simply show up "in nature" a lot.
Algebraic geometers care a great deal about polynomial rings $\Bbb F[x_1,\dots x_n]$ over fields, which are Noetherian. In a nice case like $\Bbb F=\Bbb C$, the resulting connection between the Noetherianness of the ring and the Noetherianness of topologies on algebraic varieties allows decompositions of the varieties. You might also be familiar with primary decomposition, which is possible in Noetherian rings.
Noetherianness is also nicely stable over a lot of ring constructions. For a commutative Noetherian ring $R$:
 $R[X]$ and $R[[X]]$ are Noetherian
 the localization $R_S$ at a multiplicative set $S$ is Noetherian
 the $I$adic completion of $R$ at an ideal $I$ is Noetherian
Neither of the first two work very well for Artinian rings. $F[x]$ is not Artinian at all even though $F$ is, and localization doesn't work because in Artinian rings you only have units and zero divisors: zero divisors are not good for multiplicative sets, and units are already invertible.
Noetherianness also can be applied well to domains. Many commutative Noetherian domains are of interest in number theory and ideal theory, but people who study Artinian domains are called "field theorists" for obvious reasons.
Another thing is that commutative Noetherian rings have the DCC on prime ideals. Commutative artinian rings are much more restrictive: the prime ideals are all maximal.
Noncommutative Noetherian rings are interesting too, but I don't think I can do justice to them here. I would recommend Goodearl and Warfield's An introduction to noncommutative Noetherian rings for an awesome account of those.
One final distant connection: I remember being struck when someone first mentioned "the axiom of regularity implies that set membership is a Noetherian relation." At that point, I started thinking that the Noetherian condition is really something special.
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Well noetherian rings have a special property called the a.c.c. I'm sure you already know that, but once this property has been established you can assume a lot of things about those rings. Also they make things easier to understand and to even answer when you use noetherian rings. One big reason why they are important is that if R is noetherian the R[X] is also noetherian which then helps us see that any infinite set of polynomial equations may be associated to a finite set of polynomial equations with precisely the same solution set (the solution set of a collection of polynomials in n variables is generally a geometric object (such as a curve or a surface) in nspace).
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I'm adding this as an answer because StackExchange won't let me comment.
@Martin Brandenburg, $dim(R[t]) = dim(R) + 1$ is a bad example of a property that requires Noetherianness. The class of rings which satisfy this are called Jaffard rings and include (not necessarily Noetherian) Prüfer domains such as the ring of (all) algebraic integers.
As pointed out in @darij grinberg's answer, while there are many theorems for which Noetherianness is a sufficient condition, there aren't many theorems for which Noetherianness is a necessary condition.
I would say rather that the class of Noetherian rings has a nice balance of (1) being easy to define, (2) containing many of the rings that come up in practice cf. @rschwieb's answer, and (3) satisfying many important theorems that are used often in practice.
We are in the situation where large swathes of the literature are stated for Noetherian rings even when its not necessary (for example, this is a common criticism of Hartshorne's book Algebraic Geometry). So in practice it's often much easier to just make the blanket assumption that all rings are Noetherian instead of having potentially complicated lists of more general conditions at the beginning of each statement, and trying to chase down (or add) nonNoetherian references to the literature.
So basically, what Jim said here.
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