Question

Are non Noetherian rings useful?

Answer 1

By far the most important ring in class field theory is the ring of adeles of Q (there are also adelic rings attached to other 'global fields'), which is non-Noetherian. This and the rings the other commenter mentioned are very important, but in algebraic geometry, they matter not so much for scheme theory.

Noetherian assumptions are OK in algebraic geometry because a lot of the time algebraic geometry is reasoning about varieties, which are Noetherian schemes since (by Hilbert's basis theorem) polynomial rings over a field in finitely many variables are Noetherian.

Where non-Noetherian schemes really arise is in arithmetic geometry. Perfectoid spaces, which you may have heard a big buzz about, are not usually Noetherian! Additionally, even if you are considering only varieties and curves, oftentimes moduli spaces of natural objects are non-Noetherian, even if you were working in a Noetherian context to begin with.

Answer 2

Yes, and i'll give you an example of geometric nature.

There are geometric situations where you want to prove a certain statement about all schemes, not just noetherian ones. A typical situation is something like "a certain functor from Sch/S to Set is representable". I don't know how much AG you have seen, but trust me that some of the most important theorems in AG are of this form. Since you need to work with all S-schemes to prove statements like the one above, you need to be able to juggle with non-noetherian objects

The other answers give you other reasons, arithmetic geometry being another big one topic where non noetherianness comes into play

Answer 3

Other people have mentioned natural examples from number theory. There are also tons of examples in geometric representation theory. We study various moduli spaces related to algebraic groups - a basic example is an algebraic loop group, which parameterizes maps from a punctured formal disk into an algebraic group. They are analogous to p-adic groups in arithmetic, and relate to the geometric Langlands program, QFT, etc. These kinds of spaces are infinite-dimensional, and accordingly rings of (algebraic) functions on them are non-Noetherian.

TL;DR: lots of moduli spaces that occur in nature are infinite-dimensional, so algebraic functions on them form non-Noetherian rings.

Answer 4

A more analytic example: The ring of holomorphic functions is not Noetherian

Answer 5

Lazard ring is not Noetherian and very important. Rings of smooth functions in manifolds are not noetherian.

Answer 6

Any valuation ring associated to a non-discrete valuation is non-Noetherian. Valuations and valuation rings are pretty important: they come up in number theory, for example the algebraic closure of the p-adics carries a non-discrete valuation. Valuations are used in non-archimedean analysis, non-archimedean geometry (rigid analytic geometry, Berkovich geometry, adic spaces e.g. perfectoid spaces) and tropical geometry. Even in algebraic geometry, you have e.g. the valuative criterion for properness, which requires you to consider the spectrum of general valuation rings, even if you just want to show a morphism between Noetherian schemes is proper.

Answer 7

Oh, also, the ring of stable homotopy groups of spheres is non-Noetherian.  That's...a pretty important one.

Answer 8

Even if you only like to work with smooth complex projective varieties, it's still possible to run into non-noetherian spaces.

Let X be a smooth projective variety over C, where C is the complex numbers. Then the tangent bundle TX is a natural vector bundle on X, representing first-order tangent information. Viewing it as a scheme, it admits a map TX -> X. Another perspective on TX is that it represents the functor

Y -> Hom(Y x\_C C\[t\]/t^(2), X)

(Check: the C-points of this are just tangent vectors on X.)

Now it's natural to consider also higher order tangent information. This gives rise to the notions of bundles of m-jets, which similarly correspond to m-order tangent information. If X\_m is the bundle of m-jets over X, then it represents the functor

Y -> Hom(Y x\_C C\[t\]/t^(m+1), X).

Now there are natural maps X\_m -> X\_n for m > n, and as a result there's a natural way to take the limit of the objects lim\_{m -> infty} X\_m. The resulting object turns out to be a scheme called the arc scheme, and you can think of the points of the arc scheme as being infinitesimal arcs on X, with all orders of tangent information. The arc scheme also represents a functor: it represents

Y -> Hom(Y x\_C C\[\[t\]\], X),

where C\[\[t\]\] is the ring of formal power series over C. However, the arc scheme is basically never noetherian! You can check that the dimensions of the bundles of m-jets increase as m goes to infinity, and since the arc scheme admits maps to each X\_m, you might see why it won't be noetherian in general.

Answer 9

Yes.

The easiest example is a ring of polynomials in infinitely many variables.  The chain (x\_0) < (x\_0, x\_1) < (x\_0, x\_1, x\_2) < ... does not terminate.

The ring of continuous real functions is also non-Noetherian, as is the ring of algebraic integers.  It should also be noted that Noetherian rings can have non-Noetherian subrings (if I remember correctly).

EDIT: I guess I didn't quite address utility in any way.  Hopefully it is clear that those three rings are useful at times.  I simply wanted to point out that there are useful non-Noetherian rings and that restricting yourself to Noetherian rings does exclude some naturally occurring rings.

Answer 10

According to my advisor, most people these days (at least in spectral AG) prefer to work without any Noetherian assumptions.