symmetric monoidal (∞,1)-category of spectra
(In this section, “ring” means rng, where the presence of a multiplicative identity is not assumed unless we say “unital ring”.)
A (left) noetherian ring is a ring for which every ascending chain of its (left) ideals stabilizes. In other words, it is noetherian if its underlying -module is a noetherian object in the category of left -modules (recall that a left ideal is simply a submodule of ). Similarly for right noetherian rings. Left noetherianness is independent of right noetherianness. A ring is noetherian if it is both left noetherian and right noetherian.
An equivalent condition is that all (left) ideals are finitely generated.
A dual condition is artinian: an artinian ring is a ring satisfying the descending chain condition on ideals. The symmetry is severely broken if one considers unital rings: for example every unital artinian ring is noetherian; artinian rings are intuitively much smaller than generic noetherian rings.
Spectra of noetherian rings are glued together to define locally noetherian schemes.
One of the best-known properties is the Hilbert basis theorem. Let be a (unital) ring.
(Hilbert) If is left Noetherian, then so is the polynomial algebra . (Similarly if “right” is substituted for “left”.)
(We adapt the proof from Wikipedia.) Suppose is a left ideal of that is not finitely generated. Using the axiom of dependent choice, there is a sequence of polynomials such that the left ideals form a strictly increasing chain and is chosen to have degree as small as possible. Putting , we have . Let be the leading coefficient of . The left ideal of is finitely generated; say generates. Thus we may write
The polynomial belongs to , so belongs to . Also has degree or less, and therefore so does . But notice that the coefficient of in is zero, by (1). So in fact has degree less than , contradicting how was chosen.
For a unital ring the following are equivalent:
Direct sums here can be replaced by filtered colimits.
: assume that is Noetherian and are injective modules. In order to verify that is injective it is enough to show that for any ideal any morphism of left modules factors through . Since is Notherian, is finitely generated, so the image of lies in a finite sum . Thus an extension to exists by the injectivity of each .
: if is not left Noetherian then there is a sequence of left ideals . Take . The obvious map factors through , since any element lies in all but finitely many . Now take any injective with . The map cannot extend to the whole , since otherwise its image would be contained in a sum of finitely many . Therefore, is not injective.
: can be computed by taking an injective resolution of . Since direct sums of injective modules are assumed to be injective, we can take a direct sum of injective resolutions of each . It remains to note that Hom out of a finitely generated module commutes with arbitrary direct sums.
: Follows from the fact that is injective iff for any ideal .