Noetherian and Artinian Rings

Introduction

A ring is Noetherian when its ideals satisfy the ascending chain condition, and Artinian when they satisfy the descending chain condition. The two conditions look like mirror images, but they are not: Artinian is the stronger hypothesis, since a theorem of Akizuki and Hopkins states that every Artinian ring is Noetherian. What the two share is that they make the ideal theory of a ring finite in a precise sense, and they are the hypotheses under which the general theorems of this category — primary decomposition, the dimension theory of integral extensions, the finiteness of the class number — become available.

The reason the chain conditions matter is that they replace an appeal to finiteness of the ring by finiteness of its ideal lattice. A field is Noetherian and Artinian; so are the quotient rings $\mathbb{Z}/n\mathbb{Z}$ and $k[x]/(x^n)$; the polynomial ring $k[x_1, \ldots, x_n]$ over a field is Noetherian by the Hilbert basis theorem, though it is not Artinian and is infinite-dimensional as a vector space. The chain conditions are therefore compatible with a ring being very large.

This article defines the two chain conditions, gives the equivalent formulations in terms of finite generation and of maximal or minimal elements, proves the standard closure properties and the Hilbert basis theorem, establishes the Akizuki–Hopkins theorem and the structure theory of Artinian rings as finite products of Artinian local rings, and introduces the length of a ring by way of composition series. Throughout, $R$ is a commutative ring with $1 \neq 0$; ideals and quotient rings are as in Rings, §§6–7, units and zero divisors as in Rings, §§8–9, and localization as in Localization and the Fraction Field. The word module is avoided: every statement below concerns ideals, and the passage to finitely generated modules over a Noetherian ring is not covered here.

---

Chain Conditions

The Ascending Chain Condition

Definition. A commutative ring $R$ is Noetherian if every ascending chain of ideals

$$ \mathrm{A}_1 \subseteq \mathrm{A}_2 \subseteq \mathrm{A}_3 \subseteq \cdots $$

stabilises: there is an index $N$ with $\mathrm{A}_n = \mathrm{A}_N$ for all $n \geq N$. This is the ascending chain condition, abbreviated ACC.

The definition can be recast in two ways that are used constantly.

Theorem. For a commutative ring $R$ the following are equivalent.

(a) $R$ satisfies the ascending chain condition.

(b) Every nonempty set of ideals of $R$ has a maximal element under inclusion.

(c) Every ideal of $R$ is finitely generated.

Proof. (a) $\Rightarrow$ (b): suppose $\Sigma$ is a nonempty set of ideals with no maximal element. Choose $\mathrm{A}_1 \in \Sigma$; since it is not maximal there is $\mathrm{A}_2 \in \Sigma$ with $\mathrm{A}_1 \subsetneq \mathrm{A}_2$; continuing produces a strictly increasing chain, contradicting (a).

(b) $\Rightarrow$ (c): let $I$ be an ideal and let $\Sigma$ be the set of finitely generated ideals contained in $I$. It is nonempty since $(0) \in \Sigma$, so it has a maximal element $\mathrm{B} = (b_1, \ldots, b_n)$. If $x \in I \setminus \mathrm{B}$ then $(b_1, \ldots, b_n, x) \subseteq I$ is finitely generated and strictly larger, a contradiction; hence $I = \mathrm{B}$ is finitely generated.

(c) $\Rightarrow$ (a): let $\mathrm{A}_1 \subseteq \mathrm{A}_2 \subseteq \cdots$ be an ascending chain and put $I = \bigcup_n \mathrm{A}_n$. The union is an ideal: it is closed under addition because two elements lie in some common $\mathrm{A}_n$, and under multiplication by $R$ because each $\mathrm{A}_n$ is an ideal. By hypothesis $I = (x_1, \ldots, x_m)$, and each $x_j$ lies in some $\mathrm{A}_{n_j}$; with $N = \max_j n_j$ all the generators lie in $\mathrm{A}_N$, so $I \subseteq \mathrm{A}_N \subseteq I$ and the chain is constant from $N$ on.

Examples.

(a) A field is Noetherian, its only ideals being $(0)$ and the whole field.

(b) $\mathbb{Z}$ is Noetherian: it is a principal ideal domain, so every ideal is generated by one element.

(c) Every principal ideal domain is Noetherian, by the same reason; this covers $F[x]$ for a field $F$ by Unique Factorisation Domains.

(d) A quotient $R/I$ of a Noetherian ring is Noetherian: the ideals of $R/I$ correspond to the ideals of $R$ containing $I$, and a chain in $R/I$ lifts to a chain in $R$.

(e) The polynomial ring $k[x_1, x_2, x_3, \ldots]$ in infinitely many variables over a field $k$ is not Noetherian: the chain $(x_1) \subsetneq (x_1, x_2) \subsetneq (x_1, x_2, x_3) \subsetneq \cdots$ never stabilises, and the ideal $(x_1, x_2, x_3, \ldots)$ is not finitely generated.

(f) The ring of all algebraic integers is not Noetherian; it contains the chain of principal ideals generated by $2^{1/2^n}$ with $n \geq 1$, which is strictly increasing because $2^{1/2^n} = (2^{1/2^{n+1}})^2$.

The Descending Chain Condition

Definition. A commutative ring $R$ is Artinian if every descending chain of ideals

$$ \mathrm{A}_1 \supseteq \mathrm{A}_2 \supseteq \mathrm{A}_3 \supseteq \cdots $$

stabilises. Equivalently, by the same argument as above with inclusions reversed, every nonempty set of ideals has a minimal element under inclusion. This is the descending chain condition, abbreviated DCC.

The descending chain condition has no description by finite generation: $\mathbb{Z}$ is Noetherian but not Artinian, since the chain $(2) \supsetneq (4) \supsetneq (8) \supsetneq \cdots$ of principal ideals never stabilises. The following elementwise consequence is used repeatedly.

Proposition. Let $R$ be an Artinian commutative ring and let $x \in R$. Then the chain of principal ideals

$$ (x) \supseteq (x^2) \supseteq (x^3) \supseteq \cdots $$

stabilises; that is, there is $n \geq 1$ with $x^n = x^{n+1} y$ for some $y \in R$.

Proof. The chain is descending, so it stabilises by the DCC; equality $(x^n) = (x^{n+1})$ then exhibits $y$ with $x^n = x^{n+1}y$.

Examples.

(a) A field is Artinian and Noetherian.

(b) $\mathbb{Z}/n\mathbb{Z}$ is Artinian for every $n \geq 1$: it is finite, and a finite ring has only finitely many ideals, so every chain of ideals is necessarily eventually constant.

(c) Every finite ring is Artinian.

(d) The ring $k[x]/(x^n)$ is Artinian: its ideals are the principal ideals $(x^i)$ for $0 \leq i \leq n$, a finite chain.

(e) $\mathbb{Z}$ is not Artinian, and neither is $k[x]$, by the chains above.

(f) A ring that is a vector space over a field is Artinian if and only if it is finite-dimensional over that field, since its ideals are in particular subspaces and a descending chain of subspaces of a finite-dimensional space must stabilise.

Length and Composition Series

The two chain conditions are the finiteness conditions for a generalisation of dimension, which for vector spaces is the dimension and for abelian groups is the number of cyclic factors.

Definition. A composition series of an ideal $I$ is a finite strictly decreasing chain

$$ I = I_0 \supsetneq I_1 \supsetneq \cdots \supsetneq I_\ell = (0) $$

of ideals of $R$ such that no ideal lies strictly between $I_{i}$ and $I_{i+1}$ for any $i$. The integer $\ell$ is the length of the series, and $R$ has finite length if it has a composition series; the length $\ell(R)$ is then defined and is independent of the series.

That the length is well defined is the content of the Jordan–Hölder theorem, whose proof for the ideals of a ring is the same as the proof of the corresponding theorem for the subnormal series of a group in Groups.

Theorem (Jordan–Hölder). If $R$ has a composition series, then any two composition series of $R$ have the same length, and every strictly decreasing chain of ideals can be refined to a composition series.

Proof sketch. The argument of Zassenhaus and Schreier is transcribed from groups to ideals: for two factor chains one forms the "butterfly" intersections $I_i \cap J_j$ and shows the factor quotients of a common refinement are isomorphic up to order, using the isomorphism $(A \cap B)/(A \cap C) \cong (C + (A \cap B))/(C + (A \cap C))$ for $C \subseteq B \subseteq A$.

Theorem. Let $R$ be a commutative ring. Then $R$ has finite length if and only if it is both Noetherian and Artinian.

Proof sketch. If $R$ is Noetherian and Artinian, build a chain $R = I_0 \supsetneq I_1 \supsetneq \cdots$ by choosing $I_{i+1}$ maximal among the proper ideals of $I_i$; this is possible by the ascending chain condition, and the descending chain condition forces the process to terminate at $(0)$ after finitely many steps, producing a composition series. Conversely, a composition series of length $\ell$ has at most $\ell + 1$ terms, and every chain of ideals can be refined to one of length at most $\ell$; between two ideals of such a chain only finitely many distinct intermediate ideals can occur, so both chain conditions hold.

Thus for rings of finite length the two chain conditions coincide, and the length is a numerical invariant. A field has length $1$, the ring $\mathbb{Z}/p^n\mathbb{Z}$ has length $n$, and the polynomial ring $k[x]$ has infinite length.

---

The Noetherian Condition

Closure Properties

Theorem. Let $R$ be a Noetherian commutative ring.

(a) Every quotient $R/I$ is Noetherian.

(b) Every localization $S^{-1}R$ is Noetherian.

(c) A finite direct product $R_1 \times \cdots \times R_m$ is Noetherian if and only if each factor is.

Proof. (a) The ideals of $R/I$ correspond bijectively to the ideals of $R$ containing $I$, and an ascending chain of the latter is an ascending chain of ideals of $R$, hence stabilises.

(b) Let $\mathrm{B}_1 \subseteq \mathrm{B}_2 \subseteq \cdots$ be an ascending chain of ideals of $S^{-1}R$, and put $\mathrm{A}_n = \iota^{-1}(\mathrm{B}_n)$, where $\iota : R \to S^{-1}R$ is the structure map. These are ideals of $R$ forming an ascending chain, so $\mathrm{A}_n = \mathrm{A}_N$ for $n \geq N$. By the ideal correspondence of Localization and the Fraction Field, $S^{-1}\mathrm{A}_n = \mathrm{B}_n$, so $\mathrm{B}_n = S^{-1}\mathrm{A}_n = S^{-1}\mathrm{A}_N = \mathrm{B}_N$ for $n \geq N$.

(c) The ideals of a product $A \times B$ are of the form $I \times J$ with $I$ an ideal of $A$ and $J$ an ideal of $B$, and a chain in the product projects to chains in the factors. If both factors are Noetherian, a chain in $A \times B$ stabilises since both projections do, using that an ideal of a product is determined by its two projections. Conversely if $A \times B$ is Noetherian then $A$ and $B$ are quotients of it, hence Noetherian by (a).

Theorem (finitely generated extensions). If $R$ is Noetherian and $A$ is a finitely generated commutative $R$-algebra, then $A$ is Noetherian.

Proof. By definition $A \cong R[x_1, \ldots, x_n]/I$ for some $n$ and some ideal $I$. The polynomial ring $R[x_1, \ldots, x_n]$ is Noetherian by the Hilbert basis theorem iterated $n$ times, and a quotient of a Noetherian ring is Noetherian by (a).

The Hilbert Basis Theorem

Theorem (Hilbert basis theorem). If $R$ is a Noetherian commutative ring, then the polynomial ring $R[x]$ is Noetherian.

Proof. Let $I \subseteq R[x]$ be a nonzero ideal. We show $I$ is finitely generated. For each $n \geq 0$ let $L_n \subseteq R$ be the set of leading coefficients of the elements of $I$ of degree $n$, together with $0$; this is an ideal of $R$, the ideal of leading coefficients in degree $n$. Since the product of an element of degree $n$ with $x$ has degree $n+1$ and the same leading coefficient, $L_n \subseteq L_{n+1}$; the ideals $L_n$ therefore form an ascending chain and, as $R$ is Noetherian, there is $N$ with $L_n = L_N$ for all $n \geq N$.

Each $L_n$ for $n \leq N$ is finitely generated, say $L_n = (a_{n,1}, \ldots, a_{n,k_n})$. Choose $f_{n,j} \in I$ of degree $n$ with leading coefficient $a_{n,j}$, and put $J = (f_{n,j} : 0 \leq n \leq N,\ 1 \leq j \leq k_n) \subseteq I$. We claim $J = I$.

Let $f \in I$ be nonzero, of degree $d$, with leading coefficient $a$. If $d \leq N$, then $a \in L_d$, so $a = \sum_j c_j a_{d,j}$ with $c_j \in R$, and $f - \sum_j c_j f_{d,j}$ has degree strictly less than $d$ and still lies in $I$. If $d > N$, then $a \in L_d = L_N$, so $a = \sum_j c_j a_{N,j}$ and $f - \sum_j c_j x^{d-N} f_{N,j}$ lies in $I$ and has degree strictly less than $d$. Repeating this reduction finitely many times expresses $f$ as an element of $J$. Hence $I = J$ is finitely generated, and $R[x]$ is Noetherian.

Corollary. If $R$ is Noetherian, then $R[x_1, \ldots, x_n]$ is Noetherian; more generally every finitely generated commutative $R$-algebra is Noetherian.

Corollary. If $k$ is a field, then $k[x_1, \ldots, x_n]$ is Noetherian, hence every ideal in it is finitely generated. In particular every algebraic set in $k^n$, defined as the common zero set of a family of polynomials, is already the common zero set of finitely many of them: one takes the family to generate an ideal, which is finitely generated.

The last corollary is the geometric content of the theorem, and it is what makes the ideal theory of polynomial rings algorithmic; the algorithmic side lies outside this article.

Graded and Filtered Rings

The Hilbert basis theorem is the first instance of a general principle: finite generation over a Noetherian base is inherited by the structures built from it. Two of its refinements are used in other articles of this category.

Theorem. Let $R = \bigoplus_{n \geq 0} R_n$ be a graded commutative ring with $R_0$ Noetherian and $R$ generated as an $R_0$-algebra by finitely many homogeneous elements of positive degree. Then $R$ is Noetherian.

Proof. A finitely generated graded algebra over $R_0$ is a quotient of a polynomial ring $R_0[x_1, \ldots, x_n]$ with the standard grading, which is Noetherian by the Hilbert basis theorem; a quotient of a Noetherian ring is Noetherian.

Corollary. If $k$ is a field, the ring $k[x_1, \ldots, x_n]$ with its standard grading is Noetherian, and so is any quotient of it by a homogeneous ideal. Consequently a graded $k$-algebra generated by finitely many elements of positive degree has the ascending chain condition on homogeneous ideals.

That the ascending chain condition descends from a graded ring to its degree-zero part is false, and the obstruction is exactly the failure of $R_0$ to be a quotient of $R$ in the graded sense; this is why the hypothesis above places the finite generation on the positive-degree generators.

---

The Artinian Condition

The Akizuki–Hopkins Theorem

An Artinian ring is not merely a ring satisfying a dual condition; it is automatically Noetherian, and the proof runs through the nilradical.

Definition. The nilradical of $R$ is the ideal

$$ \operatorname{nil}(R) = \{x \in R : x^n = 0 \text{ for some } n \geq 1\}. $$

It is an ideal because if $x^m = 0$ and $y^n = 0$ then $(x + y)^{m+n} = 0$ by the binomial expansion, and because $(rx)^m = r^m x^m = 0$. The nilradical is also the intersection of all prime ideals of $R$, a standard result of the ideal theory of Rings.

Definition. The Jacobson radical of $R$ is the intersection of all maximal ideals,

$$ \operatorname{Jac}(R) = \bigcap_{\mathrm{M} \text{ maximal}} \mathrm{M}. $$

Theorem (Akizuki–Hopkins; the Hopkins–Levitzki theorem). Every Artinian commutative ring is Noetherian.

Proof sketch. Let $R$ be Artinian, and put $\mathrm{N} = \operatorname{nil}(R)$.

The first step is that every prime ideal of $R$ is maximal. If $\mathrm{P}$ is prime and $R/\mathrm{P}$ is not a field, then $R/\mathrm{P}$ is an Artinian domain that is not a field, which is impossible: in an Artinian domain every nonzero element $a$ generates a descending chain $(a) \supseteq (a^2) \supseteq \cdots$, which stabilises, so $a^n = a^{n+1} b$ for some $n$ and some $b$, and cancellation of $a^n \neq 0$ in the domain gives $1 = ab$.

The second step is that there are only finitely many maximal ideals. If $\mathrm{M}_1, \mathrm{M}_2, \ldots$ were infinitely many distinct maximal ideals, the products $\mathrm{M}_1 \supseteq \mathrm{M}_1\mathrm{M}_2 \supseteq \mathrm{M}_1\mathrm{M}_2\mathrm{M}_3 \supseteq \cdots$ form a descending chain; if it stabilised at stage $n$, then $\mathrm{M}_1 \cdots \mathrm{M}_n = \mathrm{M}_1 \cdots \mathrm{M}_{n+1}$ and, since the $\mathrm{M}_i$ are distinct maximal ideals, they are pairwise comaximal, so $\mathrm{M}_1 \cdots \mathrm{M}_n + \mathrm{M}_{n+1} = R$, and multiplying the stabilised equality by an element of $R$ shows $\mathrm{M}_1 \cdots \mathrm{M}_n \subseteq \mathrm{M}_{n+1}$, forcing $\mathrm{M}_1 \cdots \mathrm{M}_n = \mathrm{M}_1\cdots\mathrm{M}_n \cap \mathrm{M}_{n+1}$ and hence $\mathrm{M}_1 \cdots \mathrm{M}_n \subseteq \mathrm{M}_{n+1}$, which contradicts comaximality. So there are finitely many, say $\mathrm{M}_1, \ldots, \mathrm{M}_r$.

The third step is that $\mathrm{N}$ is nilpotent. The descending chain $\mathrm{N} \supseteq \mathrm{N}^2 \supseteq \mathrm{N}^3 \supseteq \cdots$ stabilises, say $\mathrm{N}^n = \mathrm{N}^{n+1} = \cdots = \mathrm{I}$. If $\mathrm{I} \neq (0)$, consider the nonempty set of ideals $\mathrm{B}$ with $\mathrm{I}\mathrm{B} \neq (0)$ and choose one minimal by the descending chain condition. Since $\mathrm{I}\mathrm{B} \neq (0)$ there is $x \in \mathrm{B}$ with $x\mathrm{I} \neq (0)$; then $x\mathrm{I}$ is an ideal contained in $\mathrm{B}$ with $x\mathrm{I} \neq (0)$, so minimality gives $\mathrm{B} = (x)$. Also $x\mathrm{I}$ is an ideal with $(x\mathrm{I})\mathrm{I} = x\mathrm{I}^2 = x\mathrm{I} \neq (0)$, so by minimality $(x) \subseteq x\mathrm{I}$; hence $x = xa$ for some $a \in \mathrm{I}$, that is, $x(1 - a) = 0$. Now $\mathrm{I} = \mathrm{N}^n \subseteq \mathrm{N}$, so $a$ is nilpotent, and $1 - a$ is then a unit, because $(1-a)(1 + a + \cdots + a^{m-1}) = 1 - a^m = 1$ for $m$ large. Hence $x = 0$, contradicting $x\mathrm{I} \neq (0)$. So $\mathrm{I} = (0)$, and $\mathrm{N}^n = (0)$.

The fourth step concludes. The chain $R \supseteq \mathrm{N} \supseteq \mathrm{N}^2 \supseteq \cdots \supseteq \mathrm{N}^n = (0)$ has factors $\mathrm{N}^i/\mathrm{N}^{i+1}$, each annihilated by $\mathrm{N}$ and hence an ideal of the reduced Artinian ring $R/\mathrm{N}$. A reduced Artinian ring is a finite product of fields — its finitely many minimal primes are maximal, pairwise comaximal and have zero intersection — and an ideal of a finite product of fields that satisfies the descending chain condition is a finite direct sum of copies of those fields, hence has a composition series. Concatenating composition series along the chain exhibits a composition series of $R$, so $R$ has finite length and is Noetherian by the theorem on finite length above.

Corollary. A commutative ring is Artinian if and only if it is Noetherian of Krull dimension zero. Consequently an Artinian ring has finitely many prime ideals, all of them maximal, and its nilradical is nilpotent.

Here the Krull dimension of a ring is the supremum of the lengths of chains of prime ideals, developed; the corollary is stated here for its content and proved there.

The Structure of Artinian Rings

The Akizuki–Hopkins theorem reduces the study of Artinian rings to that of Noetherian rings of dimension zero, and these have a complete structure theory.

Theorem. Let $R$ be a commutative Artinian ring. Then $R$ has finitely many maximal ideals $\mathrm{M}_1, \ldots, \mathrm{M}_r$, the ideals $\mathrm{M}_1, \ldots, \mathrm{M}_r$ are pairwise comaximal, and the natural map

$$ R \longrightarrow R/\mathrm{M}_1^{n_1} \times \cdots \times R/\mathrm{M}_r^{n_r} $$

is an isomorphism for every $n_i \geq 1$ with $\mathrm{N}^{n_i} \subseteq \mathrm{M}_i^{n_i}$; in particular

$$ R \;\cong\; R_{\mathrm{M}_1} \times \cdots \times R_{\mathrm{M}_r}, $$

a finite product of Artinian local rings, where $R_{\mathrm{M}_i}$ is the localization of $R$ at $\mathrm{M}_i$.

Proof sketch. By the Akizuki–Hopkins proof there are finitely many maximal ideals and $\mathrm{N}^{n} = (0)$ for some $n$. Distinct maximal ideals are comaximal, and the Chinese remainder theorem gives $R/\mathrm{N}^n \cong \prod_i R/\mathrm{M}_i^{n}$ up to the multiplicities that $\mathrm{N}^n = (0)$ forces; each factor is a local Artinian ring with maximal ideal the image of $\mathrm{M}_i$. Finally, since every element outside $\mathrm{M}_i$ is invertible in $R_{\mathrm{M}_i}$ and since only the powers of $\mathrm{M}_i$ survive in the corresponding factor, $R_{\mathrm{M}_i} \cong R/\mathrm{M}_i^{N}$ for $N$ large enough.

Corollary. A commutative Artinian ring is local if and only if its nilradical is its unique maximal ideal. A reduced Artinian ring — one with $\operatorname{nil}(R) = (0)$ — is a finite product of fields.

Examples.

(a) $\mathbb{Z}/12\mathbb{Z} \cong \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$, a product of the local Artinian rings $\mathbb{Z}/4\mathbb{Z}$ and the field $\mathbb{Z}/3\mathbb{Z}$. The maximal ideals are $(2)$ and $(3)$.

(b) The ring $k[x]/(x^3)$ is local Artinian with maximal ideal $(x)$ and nilradical $(x)$; its elements are the classes of $a_0 + a_1 x + a_2 x^2$.

(c) The dual numbers $\mathbb{D}'_F = F[x]/(x^2)$ of Dual Numbers Algebra are local Artinian with maximal ideal $(\varepsilon)$ of square zero. This is the smallest local Artinian ring that is not a field, and it is the model for the infinitesimal thickening used in deformation theory.

(d) A finite product of fields is reduced Artinian; over a field $k$, $k \times k$ is Artinian with two maximal ideals.

Zero-Dimensional Rings and the Principal Ideal Theorem

The Artinian rings are the rings of Krull dimension zero, and for Noetherian rings the two notions almost coincide.

Theorem. A Noetherian commutative ring of Krull dimension zero is Artinian.

Proof. Let $R$ be Noetherian of dimension zero. Every prime ideal is maximal, and there are finitely many of them: if there were infinitely many, choose $\mathrm{P}_1, \mathrm{P}_2, \ldots$ distinct, and consider the chain of radical-type ideals $\mathrm{P}_1 \supseteq \mathrm{P}_1 \cap \mathrm{P}_2 \supseteq \cdots$; since $R$ is Noetherian, the ideal $I_n = \mathrm{P}_1 \cap \cdots \cap \mathrm{P}_n$ stabilises, giving $I_n \subseteq \mathrm{P}_{n+1}$; but then $\mathrm{P}_{n+1}$ contains some $\mathrm{P}_i$ with $i \leq n$, and since both are maximal, $\mathrm{P}_{n+1} = \mathrm{P}_i$, a contradiction. So there are finitely many maximal ideals $\mathrm{M}_1, \ldots, \mathrm{M}_r$. For each $i$ the localization $R_{\mathrm{M}_i}$ is a Noetherian local ring of dimension zero; its maximal ideal is nilpotent, since a noetherian local ring of dimension zero has nilpotent maximal ideal by the principal ideal theorem. Hence $\mathrm{M}_i^{n_i} \subseteq \mathrm{M}_i R_{\mathrm{M}_i}$ is zero in $R_{\mathrm{M}_i}$, and the kernel of $R \to \prod_i R_{\mathrm{M}_i}$ is contained in every maximal ideal, hence in $\operatorname{Jac}(R)$; but that kernel is $\operatorname{nil}(R)$, which is the intersection of the maximal ideals, and it is nilpotent, so $R \cong \prod_i R_{\mathrm{M}_i}$ is a finite product of Artinian local rings, hence Artinian.

Corollary. For a Noetherian commutative ring, Artinian is equivalent to Krull dimension zero, equivalently to every prime ideal being maximal.

The principal ideal theorem of Krull used above states that in a Noetherian ring a minimal prime ideal over a principal ideal $(a)$ has height at most $1$. Its proof belongs with the dimension theory, and it is cited here as standard. The theorem is the reason the two chain conditions interact so closely: the ascending chain condition bounds the height of primes, while the descending chain condition bounds the ring to dimension zero.

---

Summary

A commutative ring is Noetherian when its ideals satisfy the ascending chain condition, equivalently when every ideal is finitely generated, equivalently when every nonempty set of ideals has a maximal element. It is Artinian when its ideals satisfy the descending chain condition, equivalently when every nonempty set of ideals has a minimal element. The Noetherian condition passes to quotients, localizations, finite products and finitely generated algebra extensions; the Hilbert basis theorem is the statement that if $R$ is Noetherian then so is $R[x]$, and its proof is the leading-coefficient reduction by degree.

Every Artinian ring is Noetherian, by the Akizuki–Hopkins theorem: in an Artinian ring every prime is maximal, there are finitely many maximal ideals, the nilradical is nilpotent, and the filtered chain by powers of the nilradical has vector-space factors. A Noetherian ring is Artinian exactly when its Krull dimension is zero, and an Artinian ring decomposes as a finite product of Artinian local rings, corresponding to its finitely many maximal ideals; a reduced Artinian ring is a finite product of fields.

The intermediate notion is finite length: a ring has a composition series exactly when it is both Noetherian and Artinian, and the length is then well defined by the Jordan–Hölder theorem. A field has length $1$ and $\mathbb{Z}/p^n\mathbb{Z}$ has length $n$.

Summary of Notation

Further Reading

• Emmy Noether, "Idealtheorie in Ringbereichen", Mathematische Annalen 83 (1921), 24–66, for the ascending chain condition and its role in ideal theory.
• Wolfgang Krull, "Zur Theorie der zweiseitigen Ideale in nichtkommutativen Bereichen", Mathematische Zeitschrift 28 (1928), 481–503, for the descending chain condition and the principal ideal theorem.
• Yasuo Akizuki, "Teilerkettensatz und Vielfachenkettensatz", Proceedings of the Physico-Mathematical Society of Japan 17 (1935), 337–345, for the theorem that Artinian rings are Noetherian.
• Charles Hopkins, "Rings with minimal condition for admissible left ideals", Duke Mathematical Journal 4 (1938), 664–667, for the independent proof of the same theorem.
• David Hilbert, "Über die Theorie der algebraischen Formen", Mathematische Annalen 36 (1890), 473–534, for the basis theorem and the founding of the ideal-theoretic approach.
• Nicolas Bourbaki, Commutative Algebra, Chapters 1–7 (Springer, 1998), for chain conditions, primary decomposition and the structure of Artinian rings.
• Hideyuki Matsumura, Commutative Ring Theory (Cambridge University Press, 1989), for the Hilbert basis theorem, composition series and the Akizuki–Hopkins theorem in the standard modern form.
• Irving Kaplansky, Commutative Rings (University of Chicago Press, rev. ed. 1974), for the equivalent formulations of the Noetherian condition and the structure of Artinian rings.