noetherian ring



A Noetherian (or often, as below, noetherian) ring (or rng) is one where it is possible to do induction over its ideals, because the ordering of ideals by reverse inclusion is well-founded.


(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 RR is a ring for which every ascending chain of its (left) ideals stabilizes. In other words, it is noetherian if its underlying RR-module RR{}_R R is a noetherian object in the category RModR Mod of left RR-modules (recall that a left ideal is simply a submodule of RR{}_R R). 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.



Every field is a noetherian ring.


Every principal ideal domain is a noetherian ring.


For RR a Noetherian ring (e.g. a field by example 1) then

  1. the polynomial algebra R[X 1,,X n]R[X_1, \cdots, X_n]

  2. the formal power series algebra R[[X 1,,X n]]R[ [ X_1, \cdots, X_n ] ]

over R in a finite number nn of coordinates are Noetherian.



One of the best-known properties is the Hilbert basis theorem. Let RR be a (unital) ring.


(Hilbert) If RR is left Noetherian, then so is the polynomial algebra R[x]R[x]. (Similarly if “right” is substituted for “left”.)


(We adapt the proof from Wikipedia.) Suppose II is a left ideal of R[x]R[x] that is not finitely generated. Using the axiom of dependent choice, there is a sequence of polynomials f nIf_n \in I such that the left ideals I n(f 0,,f n1)I_n \coloneqq (f_0, \ldots, f_{n-1}) form a strictly increasing chain and f nII nf_n \in I \setminus I_n is chosen to have degree as small as possible. Putting d ndeg(f n)d_n \coloneqq \deg(f_n), we have d 0d 1d_0 \leq d_1 \leq \ldots. Let a na_n be the leading coefficient of f nf_n. The left ideal (a 0,a 1,)(a_0, a_1, \ldots) of RR is finitely generated; say (a 0,,a k1)(a_0, \ldots, a_{k-1}) generates. Thus we may write

(1)a k= i=0 k1r ia i a_k = \sum_{i=0}^{k-1} r_i a_i

The polynomial g= i=0 k1r ix d kd if ig = \sum_{i=0}^{k-1} r_i x^{d_k - d_i} f_i belongs to I kI_k, so f kgf_k - g belongs to II kI \setminus I_k. Also gg has degree d kd_k or less, and therefore so does f kgf_k - g. But notice that the coefficient of x d kx^{d_k} in f kgf_k - g is zero, by (1). So in fact f kgf_k - g has degree less than d kd_k, contradicting how f kf_k was chosen.

A homological characterization


For a unital ring RR the following are equivalent:

  1. RR is left Noetherian
  2. Any small direct sum of injective left RR-modules is injective.
  3. Ext R k(A,)\operatorname{Ext}^k_R(A, \cdot) commutes with small direct sums for any finitely generated AA.

Direct sums here can be replaced by filtered colimits.


121 \Rightarrow 2: assume that RR is Noetherian and I αI_\alpha are injective modules. In order to verify that I:= αI αI := \bigoplus_\alpha I_\alpha is injective it is enough to show that for any ideal 𝔧\mathfrak{j} any morphism of left modules f:𝔧If : \mathfrak{j} \to I factors through 𝔧R\mathfrak{j} \to R. Since RR is Notherian, 𝔧\mathfrak{j} is finitely generated, so the image of ff lies in a finite sum I α 1I α nI_{\alpha_1} \oplus \dots \oplus I_{\alpha_n}. Thus an extension to RR exists by the injectivity of each I α kI_{\alpha_k}.

212 \Rightarrow 1: if RR is not left Noetherian then there is a sequence of left ideals 𝔧 1𝔧 2\mathfrak{j}_1 \subsetneq \mathfrak{j}_2 \subsetneq \dots. Take 𝔧:= k𝔧 k\mathfrak{j} := \bigcup_k \mathfrak{j}_k. The obvious map j k(𝔧/𝔧 k)j \to \prod_k (\mathfrak{j} / \mathfrak{j}_k) factors through k(𝔧/𝔧 k)\bigoplus_k (\mathfrak{j} / \mathfrak{j}_k), since any element lies in all but finitely many 𝔧 k\mathfrak{j}_k. Now take any injective I kI_k with 0𝔧/𝔧 kI k0 \to \mathfrak{j} / \mathfrak{j}_k \to I_k. The map 𝔧 kI k\mathfrak{j} \to \bigoplus_k I_k cannot extend to the whole RR, since otherwise its image would be contained in a sum of finitely many I kI_k. Therefore, kI k\bigoplus_k I_k is not injective.

232 \Rightarrow 3: Ext R k(A, αX α)\operatorname{Ext}^k_R(A, \bigoplus_\alpha X_\alpha) can be computed by taking an injective resolution of αX α\bigoplus_\alpha X_\alpha. Since direct sums of injective modules are assumed to be injective, we can take a direct sum of injective resolutions of each X αX_\alpha. It remains to note that Hom out of a finitely generated module commutes with arbitrary direct sums.

323 \Rightarrow 2: Follows from the fact that II is injective iff Ext R 1(R/𝔦,I)=0\operatorname{Ext}^1_R(R / \mathfrak{i}, I) = 0 for any ideal 𝔦\mathfrak{i}.


Revised on September 29, 2017 07:12:53 by Urs Schreiber (