# Why are noetherian rings such natural objects in algebraic geometry?

I assume it is partially because they are good generalizations of polynomial rings, but what makes this generalization better than graded algebras or other generalizations of polynomial rings?

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Have you seen mathoverflow.net/questions/63715/… ? – Qiaochu Yuan Jul 2 '11 at 2:19
@Qiaochu - Thanks for the link, I didn't know it. But it doesn't quite answer what I had in mind. I was actually thinking of saying in the question "graded algebra (with perhaps some finiteness condition)", perhaps I should have. – teil Jul 2 '11 at 2:56
I don't claim my comment gives a general explanation, it only indicates an example in which the Noetherian condition seems to be essential: Noetherian rings are characterized by the fact that arbitrary direct sums of injective modules is injective too. How is related this fact to some generalizations of Grothendieck duality, it may be seen from papers like H. Krause, "The stable derived category of a noetherian scheme", Compos. Math. 141 (2005), 1128-1162, or A. Neeman, "The homotopy category of flat modules, and Grothendieck duality," Invent. Math., 174 (2008), pp. 255-308. – George C. Modoi Jul 2 '11 at 7:31

The best answer I've ever been able to come up with is that the class of noetherian rings contains the classical number rings $\mathbf{Z}$ and $\mathbf{R}$ and is closed under the formation of polynomial rings, localization, completion, and quotients. So it contains many of the rings you will come across in ordinary situations (whatever that means). It also has the advantage that the definition is tractable enough that if someone hands you an explicit ring, it's not out of the question to try to work out from scratch whether it's noetherian. If you're the kind of person who likes abstract fields, then they're also included.

On the other hand, I don't think of it as a truly fundamental concept, like say finite presentation. But there is no denying its convenience. If you need to avoid some infinitary phenomena but you still want a broad class of rings, it's often hard to beat noetherianness. It's also quite good in situations where you're too lazy to work out exactly what finiteness conditions you care about.

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I am not entirely sure what you have in mind when you contrast being noetherian to being graded. These belong to different aspects of being a ring. It's kind of like saying "He came in a hurry and [in] a winter coat".

Also, graded algebras are actually natural objects in algebraic geometry. That's where projective schemes/varieties come from.

Anyway, let me say something possibly useful, too: I believe that the truly important notion is being finitely presented. See the original definition of coherence. If you accept that, then one can say that the importance of being noetherian is that being finitely presented follows from being finitely generated and it is inherited by both quotients and subobjects.

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+1 for the zeugma (and the good answer) :) – Zev Chonoles Jul 2 '11 at 5:27
Zev, thanks in turn for the thanks! I remembered that some such word existed for a long time, but could not for the life of me remember it (or exactly what it meant, which made it hard to search for it). – L Spice Jul 3 '11 at 3:46

I'd just like to briefly add that Noetherian rings can be surprisingly non-geometric. In particular, they can fail to be excellent. Thus

1. The regular (non-singular) locus can fail to be open.
2. Notions of dimension need not be reasonable (two maximal chains of primes with the same top and bottom members can be the different lengths).
3. Normalization need not be a module-finite extension.

Of course, excellent is just a hodge-podge of conditions that avoid these particular pathologies (and avoid these after some standard operations).

The usual rings (finite type over a field or $\mathbb{Z}$) are excellent, as are complete local rings. However, it can be hard to prove that an arbitrary ring is excellent.

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That's probably because not every ring es excellent :) – Mariano Suárez-Alvarez Jul 2 '11 at 13:11
Very very true. – Karl Schwede Jul 2 '11 at 17:14

Roughly, Noetherian schemes are (locally) given by finitely many equations, and this makes it possible to make inductions on the number of equations. Also, it fits very well to the classical intuition for algebraic varieties.

To be more precise, a more fundamental concept is that of a morphism of finite type $X \to S$. For a fixed $S$, these make up a category, but it is not so well behaved and there are quite some subcategories such as the category of abelian schemes which behave only well if you assume that $S$ is, say, locally noetherian. On the other hand, you can also consider the category of morphisms of finite presentation $X \to S$ and it turns out that quite often this is more natural since now you can drop any finiteness conditions on $S$. The reason is, roughly, that here $X$ is described by finitely many equations, and that these equations also satisfy only a finite number of relations, the latter being important for reduction and induction arguments in the spirit of the Five Lemma. If $S$ is noetherian, we get the latter for free and that is the reason why you often just assume it.

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Oh, now I realize that Sándor basically has written the same answer. Perhaps I will delete this one. – Martin Brandenburg Jul 2 '11 at 9:30
Don't erase -- it is carefully written and will be clarifying to some. – Zoran Skoda Jul 2 '11 at 14:27

I study mainly modules and abelian groups, but even I find the Noetherian property useful. For me, probably the main feature lurking in the background is that submodules of a finitely generated $R$-module, $R$ Noetherian, are finitely generated. This fact also allows us to conclude that a finitely generated module over a Noetherian ring $R$ is finitely presented. For a finitely generated module over an arbitrary ring, we always have the exact sequence $$0 \rightarrow K \rightarrow R^n \rightarrow M \rightarrow 0$$ where $K$ is the kernel of the map $R^n \rightarrow M$ that sends the standard basis elements of $R^n$ to the generators of $M$. We say $M$ is finitely presented if $K$ is finitely generated. If $R$ is Noetherian, this is automatic.

It is also nice that when $R$ is Noetherian, any finitely generated $R$-module has a primary decomposition.

Facts like this are standard fare in commutative algebra. Now it has been a good while since I've done much algebraic geometry, but if memory serves, when decomposing an affine variety into irreducible subvarieties, one uses a primary decomposition of the ideal$I$ of the variety (in the polynomial ring in $n$ variables over the field underlying the affine space; this field is usually assumed algebraically closed). But the ideal $I$ turns out to be equal to its own radical, so the primary decomposition involves only prime ideals, and each of these corresponds to an irreducible (affine) variety, each being a component of the original variety.

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