Chain Conditions: Noetherian and Artinian Rings and Modules

Q: If R is left noetherian, is every left R-module noetherian?

No — only the finitely generated ones. Z is a noetherian ring but Q is a non-noetherian Z-module. The correct statement is (1.21), and the finite generation hypothesis cannot be dropped.

Q: Is length additive?

Yes, on short exact sequences: if 0 → N → M → M/N → 0 is exact and M has finite length, then length(M) = length(N) + length(M/N). This follows from Jordan–Hölder by refining a composition series of N together with the preimage of one for M/N, and it makes length the correct generalisation of vector space dimension.

Q: How do chain conditions interact with the Jacobson radical?

Decisively. For a left artinian ring, rad R is nilpotent and R/rad R is semisimple; both statements fail without a chain condition, as k[[x]] shows. The Hopkins–Levitzki proof works by filtering R by powers of the radical and applying (1.20) to the semisimple factors.

Q: Which chain condition does one usually verify in practice?

Noetherian, because of (1.18) and the Hilbert Basis Theorem: most naturally occurring rings are built from noetherian ones by finitely many polynomial, matrix, quotient or localisation steps, all of which preserve the property. Artinian rings are rarer and are usually recognised by being finite-dimensional algebras over a field.

Executive Summary

Wedderburn classified finite-dimensional semisimple algebras. Noether and Artin observed that finite dimensionality is used only through two consequences — that chains of subobjects cannot ascend or descend forever — and replaced it by those consequences. The result is a finiteness hypothesis available for rings that are not algebras over a field at all.

Four facts do almost all the work: chain conditions are equivalent to a maximality property; noetherian is equivalent to every submodule being finitely generated; both conditions pass to submodules, quotients and extensions; and consequently finitely generated modules over a noetherian ring are noetherian. Everything else — Hilbert's Basis Theorem, Hopkins–Levitzki, the Jordan–Hölder theorem — is built on those.

ACCNoetherian

DCCArtinian

BothComposition series exists

(1.18)–(1.21)Lam's numbering

Overview

Let ${C_{i} : i \in I}$ be a family of subsets of a set $C$ . The family satisfies the ascending chain condition if there is no infinite strictly ascending chain $C_{i_{1}} ⊊ C_{i_{2}} ⊊ C_{i_{3}} ⊊ \dots$ inside it. Two reformulations are standard and are used interchangeably:

Every ascending chain $C_{i_{1}} \subseteq C_{i_{2}} \subseteq \dots$ in the family is eventually constant: there is $n$ with $C_{i_{n}} = C_{i_{n + 1}} = \dots$ .
Every nonempty subfamily has a maximal member with respect to inclusion.

The descending chain condition is the same statement with all inclusions reversed, and its second formulation asks for a minimal member. Applied to the family of all submodules of a module $M$ , the two conditions define noetherian and artinian modules; applied to the left ideals of a ring $R$ — that is, to the submodules of $_{R} R$ — they define left noetherian and left artinian rings.

The two conditions are logically independent for modules. For rings they are not: a left artinian ring is always left noetherian, a nontrivial theorem of Hopkins and Levitzki proved a decade after Artin's work and treated separately in Hopkins–Levitzki Theorem. The asymmetry between left and right is a different matter again, and is genuine — see One-Sided Chain Conditions.

Learning Objectives

State the ACC and DCC, and prove the equivalence with the maximal and minimal member formulations.
Prove $(1.18)$ : $M$ is noetherian iff every submodule of $M$ is finitely generated.
Prove $(1.20)$ : for $N \subseteq M$ , the module $M$ is noetherian (artinian) iff $N$ and $M / N$ both are.
Deduce $(1.21)$ for finitely generated modules over a left noetherian or left artinian ring.
State $(1.19)$ linking the two conditions to composition series, and compute one.
Name modules that are noetherian and not artinian, artinian and not noetherian, and neither.

Definitions

Definition—Chain conditions on a module

Let $R$ be a ring and $M$ a left or right $R$ -module. $M$ is noetherian if the family of all submodules of $M$ satisfies the ACC, and artinian if that family satisfies the DCC. A ring $R$ is left noetherian (resp. left artinian) if $_{R} R$ is noetherian (resp. artinian) as a left module over itself, that is, if the family of left ideals satisfies ACC (resp. DCC). Noetherian without qualification means both left and right noetherian.

Simple module: A nonzero module whose only submodules are $0$ and itself. Simple modules are both noetherian and artinian, trivially.
Composition series: A chain $0 = M_{0} ⊊ M_{1} ⊊ \dots ⊊ M_{n} = M$ with every $M_{i} / M_{i - 1}$ simple. Its length is $n$ .
Length: The common length of all composition series of $M$ , well defined by the Jordan–Hölder theorem and written $length (M)$ or $ℓ (M)$ .
Finitely generated: $M = R x_{1} + \dots + R x_{n}$ for finitely many $x_{i} \in M$ ; equivalently $M$ is a quotient of a free module $R^{n}$ .
Semilocal ring: A ring with $R / rad R$ semisimple; every left artinian ring is semilocal, but not conversely.

The equivalence of the chain formulation with the maximal element formulation uses the axiom of dependent choice in one direction. This is standard and universally assumed; it is worth knowing that the assumption is there.

Core Concepts

Why ACC and maximality are the same condition

Suppose every nonempty subfamily has a maximal member, and let $C_{1} \subseteq C_{2} \subseteq \dots$ be a chain. The set ${C_{i}}$ has a maximal member $C_{n}$ , and since the family is a chain, $C_{m} \subseteq C_{n}$ for all $m$ ; combined with $C_{n} \subseteq C_{m}$ for $m \geq n$ this gives $C_{m} = C_{n}$ for $m \geq n$ . Conversely, if some nonempty subfamily had no maximal member, then for each of its members one could choose a strictly larger one, generating an infinite strictly ascending chain.

Noetherian is a statement about generators

The reason noetherian is easier to work with is $(1.18)$ : it converts a condition about infinitely many submodules into a condition about each one separately. Artinian admits no such reformulation. There is no useful finitely cogenerated criterion at the elementary level, and this is precisely why artinian rings needed the extra Hopkins–Levitzki insight before they could be handled.

ACC on submodulesevery submodule finitely generatedevery nonempty family has a maximal member

Both conditions are extension-closed

The single most-used fact is $(1.20)$ : for a submodule $N \subseteq M$ , the module $M$ satisfies a chain condition iff both $N$ and $M / N$ do. Neither half alone suffices — $ℤ \subseteq ℚ$ has $ℤ$ noetherian while $ℚ$ is not — and the proof of the hard direction is a small but genuinely non-obvious argument, given in full below.

Where the two conditions meet

A module has a composition series precisely when it satisfies both chain conditions, $(1.19)$ . That is the correct generalisation of finite dimensional: a finite-dimensional vector space has a composition series of length equal to its dimension, and the Jordan–Hölder theorem supplies a well-defined length in the general case.

Key Results

Proposition(1.18)The finite generation criterion

Let $R$ be a ring and $M$ a left $R$ -module. Then $M$ is noetherian if and only if every submodule of $M$ is finitely generated.

Proof

**( $\Rightarrow$ ).** Let $N \subseteq M$ and consider the family $ℱ$ of finitely generated submodules of $N$ . It is nonempty ( $0 \in ℱ$ ), so by ACC it has a maximal member $N_{0}$ . If $N_{0} \neq N$ , pick $x \in N ∖ N_{0}$ ; then $N_{0} + R x$ is a finitely generated submodule of $N$ strictly containing $N_{0}$ , contradicting maximality. Hence $N = N_{0}$ is finitely generated.

**( $\Leftarrow$ ).** Let $M_{1} \subseteq M_{2} \subseteq \dots$ be an ascending chain of submodules. Their union $N = ⋃_{i} M_{i}$ is a submodule, because the chain is directed. By hypothesis $N = R x_{1} + \dots + R x_{k}$ for finitely many $x_{j}$ , and each $x_{j}$ lies in some $M_{i_{j}}$ . Taking $n = \max_{j} i_{j}$ gives all $x_{j} \in M_{n}$ , hence $N \subseteq M_{n} \subseteq N$ and the chain is constant from $n$ on. $□$

Proposition(1.20)Chain conditions in short exact sequences

Let $R$ be a ring, $M$ a left $R$ -module and $N \subseteq M$ a submodule. Then $M$ is noetherian if and only if both $N$ and $M / N$ are noetherian. The same statement holds with noetherian replaced by artinian throughout. In particular the direct sum of two noetherian (resp. artinian) modules is noetherian (resp. artinian), and so by induction is any finite direct sum.

Proof

**( $\Rightarrow$ ).** Submodules of $N$ are submodules of $M$ , so $N$ inherits the chain condition. Submodules of $M / N$ correspond bijectively and inclusion-preservingly to submodules of $M$ containing $N$ , so $M / N$ inherits it too.

**( $\Leftarrow$ ), noetherian case.** Let $M_{1} \subseteq M_{2} \subseteq \dots$ be an ascending chain in $M$ . The chains $M_{i} \cap N$ in $N$ and $(M_{i} + N) / N$ in $M / N$ both stabilise; choose $n$ beyond both stabilisation points. We claim $M_{n} = M_{n + j}$ for all $j \geq 0$ .

Let $x \in M_{n + j}$ . Its class in $M / N$ lies in $(M_{n + j} + N) / N = (M_{n} + N) / N$ , so $x = y + z$ with $y \in M_{n}$ and $z \in N$ . Then $z = x - y \in M_{n + j}$ since $M_{n} \subseteq M_{n + j}$ , so $z \in M_{n + j} \cap N = M_{n} \cap N \subseteq M_{n}$ . Hence $x = y + z \in M_{n}$ , proving $M_{n + j} \subseteq M_{n}$ and therefore equality.

Artinian case. Identical, with all inclusions reversed: for a descending chain, choose $n$ beyond the stabilisation of $M_{i} \cap N$ and $(M_{i} + N) / N$ , and run the same computation to get $M_{n} \subseteq M_{n + j}$ .

For the direct sum statement, apply the result to $N = M' \subseteq M' \oplus M''$ with quotient $M''$ . $□$

Proposition(1.21)Finitely generated modules over a noetherian ring

Let $R$ be a left noetherian ring and $M$ a finitely generated left $R$ -module. Then $M$ is a noetherian module. The same holds with noetherian replaced by artinian on both sides: a finitely generated left module over a left artinian ring is artinian.

Proof

Suppose $M$ is generated by $n$ elements. Then there is a surjection $R^{n} ↠ M$ of left $R$ -modules. The module $_{R} R$ is noetherian by hypothesis, so $R^{n}$ is noetherian by the direct sum case of $(1.20)$ , and $M$ , being a quotient of $R^{n}$ , is noetherian by $(1.20)$ again. The artinian argument is word for word the same. $□$

Proposition(1.19)Composition series

A left $R$ -module $M$ has a finite composition series if and only if $M$ is both noetherian and artinian. When such a series exists, any two composition series of $M$ have the same length and, after reordering, isomorphic factors — the Jordan–Hölder theorem.

Proof

**( $\Leftarrow$ ).** Assume both chain conditions and $M \neq 0$ . Set $M_{0} = 0$ . Given $M_{i} ⊊ M$ , the family of submodules strictly containing $M_{i}$ is nonempty, so by DCC it has a minimal member $M_{i + 1}$ ; minimality says exactly that $M_{i + 1} / M_{i}$ is simple. This produces a strictly ascending chain $M_{0} ⊊ M_{1} ⊊ \dots$ , which by ACC must terminate — and it can terminate only by reaching $M$ .

**( $\Rightarrow$ ).** Induct on the length $n$ of a composition series. For $n = 1$ , $M$ is simple, hence both noetherian and artinian. For $n > 1$ , apply $(1.20)$ to $M_{n - 1} \subseteq M$ : the submodule has a composition series of length $n - 1$ and the quotient is simple, so both satisfy both chain conditions, and therefore so does $M$ .

The Jordan–Hölder statement follows from the Schreier refinement theorem, itself a consequence of the Zassenhaus butterfly lemma; the argument is purely lattice-theoretic and is not reproduced here. $□$

Proposition—Surjective endomorphisms of noetherian modules

Let $M$ be a noetherian left $R$ -module and $f : M \to M$ a surjective $R$ -module endomorphism. Then $f$ is injective, hence an automorphism.

Proof

The kernels form an ascending chain $\ker f \subseteq \ker f^{2} \subseteq \dots$ , which stabilises: $\ker f^{n} = \ker f^{n + 1}$ for some $n$ . Let $x \in \ker f$ . Since $f$ is surjective so is $f^{n}$ , so $x = f^{n} (y)$ for some $y \in M$ . Then $f^{n + 1} (y) = f (x) = 0$ , so $y \in \ker f^{n + 1} = \ker f^{n}$ , whence $x = f^{n} (y) = 0$ . $□$

Remark—The one-sided theorem of Hopkins and Levitzki

A left artinian ring is left noetherian. This is emphatically false for modules: the Prüfer group is an artinian $ℤ$ -module that is not noetherian. The ring statement requires the Jacobson radical and is proved in Hopkins–Levitzki Theorem. Note that it is one-sided in the strong sense: left artinian implies left noetherian and says nothing about the right side.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Maximal member instead of a chain

Converting ACC into every nonempty family has a maximal member removes the need to construct chains. The proof of $(1.18)$ is three lines in that formulation and awkward in the other.

Move 2

Test against $N$ and $M / N$

Given a chain in $M$ , intersect it with $N$ and push it to $M / N$ . Both derived chains stabilise; the little computation with $x = y + z$ then forces the original to stabilise.

Move 3

Reduce to $R^{n}$

A finitely generated module is a quotient of a free module of finite rank. Any property closed under finite direct sums and quotients therefore passes from $_{R} R$ to all finitely generated modules.

Move 2 is worth internalising in its exact-sequence form. If $0 \to N \to M \to M / N \to 0$ is exact then $M$ has a chain condition iff the two outer terms do. Properties with this behaviour are called Serre classes, and both chain conditions define one.

A word of warning about Move 3: it fails for infinite direct sums. $⨁_{i = 1}^{\infty} ℤ$ is not a noetherian $ℤ$ -module, even though each summand is. Finiteness of the index set is doing real work.

Worked Example

A composition series and Jordan–Hölder in $ℤ / 12 ℤ$

Regard $M = ℤ / 12 ℤ$ as a $ℤ$ -module. It is finite, hence both noetherian and artinian, so $(1.19)$ guarantees a composition series. Two of them:

0 ⊊ (4) ⊊ (2) ⊊ ℤ / 12 ℤ, 0 ⊊ (6) ⊊ (3) ⊊ ℤ / 12 ℤ .

(E.1)

Here $(4) = {0, 4, 8}$ has order $3$ ; $(2)$ has order $6$ ; $(6) = {0, 6}$ has order $2$ ; $(3)$ has order $4$ . The successive quotients have orders $3, 2, 2$ in the first series and $2, 2, 3$ in the second, so the factor multisets are ${ℤ / 3, ℤ / 2, ℤ / 2}$ in both cases and both series have length $3$ — exactly as Jordan–Hölder predicts. Note $length (ℤ / 12 ℤ) = 3 =$ the number of prime factors of $12$ counted with multiplicity.

Noetherian but not artinian

$ℤ$ as a module over itself: every ideal is $(n)$ , hence cyclic, so $ℤ$ is noetherian by $(1.18)$ . It is not artinian: $(2) ⊋ (4) ⊋ (8) ⊋ \dots$ never stabilises. The same chain shows the ring $ℤ$ is not left artinian.

Artinian but not noetherian

Fix a prime $p$ and let $M = ℤ (p^{\infty}) = ℤ [1 / p] / ℤ$ , the Prüfer $p$ -group, as a $ℤ$ -module. Its proper submodules are exactly the finite cyclic groups

0 ⊊ ⟨ 1 / p ⟩ ⊊ ⟨ 1 / p^{2} ⟩ ⊊ \dots ⊊ ℤ (p^{\infty}), ⟨ 1 / p^{n} ⟩ ≅ ℤ / p^{n} ℤ .

(E.2)

The submodule lattice is a single chain of order type $ω + 1$ .

A strictly descending chain of proper submodules corresponds to a strictly decreasing sequence of non-negative integers $n$ , which must terminate; so $M$ is artinian. The chain displayed in $(E . 2)$ is strictly ascending and infinite, so $M$ is not noetherian, and by $(1.18)$ it is not finitely generated. This is the standard demonstration that Hopkins–Levitzki has no module analogue.

Neither

$ℚ$ as a $ℤ$ -module is not finitely generated, hence not noetherian; and $ℤ ⊋ 2 ℤ ⊋ 4 ℤ ⊋ \dots$ is an infinite strictly descending chain of submodules, so it is not artinian either.

Process and Workflow

Identify the ambient ringChain conditions are relative:

ℝ

is noetherian over

ℝ

and not over

ℚ

in any triangular ring built from it.

Check the ring firstIf

R

is left noetherian,

(1.21)

settles every finitely generated left module at once.

Otherwise filterFind a submodule

N

with

N

and

M / N

both under control, then apply

(1.20)

. Iterating gives a finite filtration with manageable factors.

Falling back on generatorsFor noetherian, verify instead that every submodule is finitely generated. There is no corresponding shortcut for artinian.

Which chain condition should you try to establish?

You need finite generation of submodulesNoetherian. Use

(1.18)

; over a noetherian ring,

(1.21)

does the work for all finitely generated modules.

You need a minimal submodule to existArtinian. The DCC hands you minimal members of nonempty families — this is how socles and minimal left ideals are produced.

You need a length functionBoth.

(1.19)

gives a composition series, Jordan–Hölder makes the length well defined, and length is additive on short exact sequences.

You are working with a ring, not a moduleLeft artinian is the stronger hypothesis: it implies left noetherian by Hopkins–Levitzki. The converse fails —

ℤ

is noetherian and not artinian.

Comparison and Classification

Standard modules and rings against the two conditions
	Noetherian	Artinian	Finitely generated	Composition series
$ℤ$ over $ℤ$	yes	no	yes	no
$ℤ / 12 ℤ$ over $ℤ$	yes	yes	yes	yes
$ℚ$ over $ℤ$	no	no	no	no
$ℤ (p^{\infty})$ over $ℤ$	no	yes	no	no
$k [x]$ over $k [x]$	yes	no	yes	no
$k ⟨ x, y ⟩$ over itself	no	no	yes	no

Standard modules and rings against the two conditions

How the two conditions differ in practice
Feature	Noetherian	Artinian
Elementwise criterion	every submodule finitely generated $(1.18)$	none at this level
Guarantees	maximal submodules exist in any nonempty family	minimal submodules exist; socle is nonzero
Polynomial extension	$R$ left noetherian $\Rightarrow R [x]$ left noetherian	$R [x]$ is never artinian for $R \neq 0$
For rings	does not imply artinian ( $ℤ$ )	implies noetherian (Hopkins–Levitzki)
For modules	does not imply artinian ( $ℤ$ )	does not imply noetherian ( $ℤ (p^{\infty})$ )
Left–right	genuinely independent	genuinely independent

Relationship Map

The implications below hold for rings; each arrow needs the hypothesis written on it, and none reverses.

semisimpleleft artinianleft noetherianleft ideals finitely generated

Chain conditions — where they lead
- left artinian
  - left noetherian — Hopkins–Levitzki Theorem
  - $rad R$ is nilpotent
  - $R / rad R$ is semisimple, so Wedderburn–Artin applies
- left noetherian
  - every finitely generated left module is noetherian $(1.21)$
  - $R$ is Dedekind-finite — see One-Sided Chain Conditions
  - $R [x]$ is left noetherian (Hilbert Basis Theorem)
- both, on modules
  - a composition series exists $(1.19)$
  - length is well defined and additive on short exact sequences

In the other direction, chain conditions are what make the radical usable: $rad R$ is nilpotent for left artinian $R$ , and the whole Wedderburn–Artin classification rests on that.

Failure Modes and Common Mistakes

Infinite direct sums destroy both conditions: $⨁_{i \geq 1} ℤ$ is not noetherian even though each summand is. Only finite direct sums are covered by $(1.20)$ .
A noetherian module need not have a composition series — $ℤ$ does not. Both conditions are needed for $(1.19)$ .
Finitely generated is strictly weaker than noetherian: the free algebra $k ⟨ x, y ⟩$ is generated by two elements as a ring but is neither left nor right noetherian.
The equivalence of ACC with the maximal element condition uses dependent choice; in a choice-free setting the two are not interchangeable.

Historical Notes and Lessons Learned

1890Hilbert's Basis TheoremHilbert proves that ideals of a polynomial ring in finitely many variables over a field are finitely generated — the first substantial theorem whose content is an ascending chain condition.
1921Noether's IdealtheorieEmmy Noether isolates the ACC as an axiom in its own right and develops the ideal theory of commutative rings from it, replacing ad hoc finiteness arguments.
1927Artin's minimum conditionArtin replaces finite dimensionality by the DCC on one-sided ideals and proves Wedderburn's structure theorem in that generality — the birth of the Wedderburn–Artin theory.
1928–1934Schreier and ZassenhausThe refinement theorem and the butterfly lemma make the Jordan–Hölder theorem a formal consequence of lattice properties, and give length its unambiguous meaning.
1939Hopkins and LevitzkiIndependently, both prove that a ring with the DCC on left ideals has the ACC on left ideals — a fact unknown to Noether and Artin themselves.
1960Bass's perfect ringsBass shows that the DCC on principal one-sided ideals already gives much of the theory, and identifies left perfect rings as the correct home for projective covers.

The lesson is one of axiomatic economy. Wedderburn's proofs used finite dimensionality in exactly two ways, and each use was an instance of a chain condition. Isolating those two uses turned a theorem about algebras over fields into a theorem about rings — and, unexpectedly, revealed that for rings the two conditions are not independent after all.

Quick Reference

ACCno infinite

C_{1} ⊊ C_{2} ⊊ \dots

; every nonempty family has a maximal member

DCCno infinite

C_{1} ⊋ C_{2} ⊋ \dots

; every nonempty family has a minimal member

Noetherian test

(1.18)

: every submodule finitely generated

Extensions

(1.20)

M

has it

\Leftrightarrow

N

and

M / N

have it

Finitely generated

(1.21)

: f.g. over left noetherian (artinian)

R

\Rightarrow

noetherian (artinian)

Composition series

(1.19)

: exists

\Leftrightarrow

noetherian and artinian

Rings onlyleft artinian

\Rightarrow

left noetherian (Hopkins–Levitzki)

Neverartinian

\Rightarrow

noetherian for modules; left

\Rightarrow

right for either

Closure properties
Operation	Noetherian	Artinian
Submodule	inherited	inherited
Quotient module	inherited	inherited
Extension	inherited	inherited
Finite direct sum	inherited	inherited
Infinite direct sum	fails	fails
Quotient ring	inherited	inherited
Subring	fails	fails
Polynomial ring $R [x]$	inherited	fails always

Frequently Asked Questions

Why is there no artinian analogue of the finite generation criterion?

Finite generation is a statement about the smallest submodule containing a given finite set, which is exactly what an ascending chain probes. The dual notion is finite cogeneration — every family of submodules with zero intersection has a finite subfamily with zero intersection — and a module is artinian iff every quotient is finitely cogenerated. That criterion exists but is far less usable, because cogeneration is not visible from elements.

Does a noetherian ring have to be commutative or an algebra?

No. The definition refers only to left ideals. Noncommutative examples include all left artinian rings, all skew polynomial rings $k [x; σ]$ over a division ring, group rings of polycyclic-by-finite groups over noetherian coefficient rings, and the Weyl algebras $A_{n} (k)$ over a field of characteristic zero.

R

is left noetherian, is every left

R

-module noetherian?

No — only the finitely generated ones. $ℤ$ is a noetherian ring but $ℚ$ is a non-noetherian $ℤ$ -module. The correct statement is $(1.21)$ , and the finite generation hypothesis cannot be dropped.

Is length additive?

Yes, on short exact sequences: if $0 \to N \to M \to M / N \to 0$ is exact and $M$ has finite length, then $length (M) = length (N) + length (M / N)$ . This follows from Jordan–Hölder by refining a composition series of $N$ together with the preimage of one for $M / N$ , and it makes length the correct generalisation of vector space dimension.

How do chain conditions interact with the Jacobson radical?

Decisively. For a left artinian ring, $rad R$ is nilpotent and $R / rad R$ is semisimple; both statements fail without a chain condition, as $k [[x]]$ shows. The Hopkins–Levitzki proof works by filtering $R$ by powers of the radical and applying $(1.20)$ to the semisimple factors.

Which chain condition does one usually verify in practice?

Noetherian, because of $(1.18)$ and the Hilbert Basis Theorem: most naturally occurring rings are built from noetherian ones by finitely many polynomial, matrix, quotient or localisation steps, all of which preserve the property. Artinian rings are rarer and are usually recognised by being finite-dimensional algebras over a field.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §1, results (1.18)–(1.22) (pp. 20–22).
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §10 and §11.
J. C. McConnell and J. C. Robson, Noncommutative Noetherian Rings, revised edition, Graduate Studies in Mathematics 30, American Mathematical Society, 2001, Chapters 1–2.
N. Jacobson, Basic Algebra II, 2nd edition, W. H. Freeman, 1989, Chapter 3.
E. Noether, “Idealtheorie in Ringbereichen”, Mathematische Annalen 83 (1921), 24–66.
E. Artin, “Zur Theorie der hyperkomplexen Zahlen”, Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg 5 (1927), 251–260.

AI Suggested Questions

Prove the Hilbert Basis Theorem in the noncommutative setting: $R$ left noetherian implies $R [x]$ left noetherian.
Show that a module is artinian iff every quotient of it is finitely cogenerated.
Give a ring that is left noetherian, right noetherian, and neither left nor right artinian, other than $ℤ$ .
Which group rings $k G$ are noetherian? State the polycyclic-by-finite criterion and what is known about its converse.
Prove that a commutative ring is artinian iff it is noetherian of Krull dimension zero.
How does the Krull dimension of a noncommutative noetherian ring differ from the commutative notion?
Work through the Zassenhaus lemma and derive the Jordan–Hölder theorem from it.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why ACC and maximality are the same condition

Noetherian is a statement about generators

Both conditions are extension-closed

Where the two conditions meet

06Key Results

07Proof Techniques and Method

Maximal member instead of a chain

Test against N and M/N

Reduce to Rn

08Worked Example

A composition series and Jordan–Hölder in ℤ/12ℤ

Noetherian but not artinian

Artinian but not noetherian

Neither

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Failure Modes and Common Mistakes

13Historical Notes and Lessons Learned

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Test against $N$ and $M / N$

Reduce to $R^{n}$

Worked Example

A composition series and Jordan–Hölder in $ℤ / 12 ℤ$

Process and Workflow

Comparison and Classification

Relationship Map

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions