Local Rings: Definition and Equivalent Characterisations

Q: Why is a unique maximal two-sided ideal not the right definition?

Because it is far too weak. Any simple ring — M_n(k), the Weyl algebra A_1(k) in characteristic zero — has exactly one maximal two-sided ideal, namely (0), yet has many maximal left ideals and, in the matrix case, many idempotents. The correct definition uses maximal one-sided ideals; the two-sided uniqueness is then a corollary, (19.2)(a).

Q: Is a local ring necessarily noetherian or artinian?

No. k[[x]] is local and noetherian but not artinian; the ring of germs of holomorphic functions at a point is local and noetherian; a valuation ring of infinite rank in a field is local and neither noetherian nor artinian. Nothing in (19.1) mentions chain conditions.

Q: Is M_n(R) local when R is?

Not for n ≥ 2. The matrix units e_11, , e_nn are nontrivial idempotents, which (19.2)(c) forbids. This is a useful reminder that locality is not Morita invariant, in contrast to the Jacobson radical, which satisfies rad M_n(R) = M_n(rad R).

Q: What is the difference between local and completely primary?

Completely primary means local *and* rad R nilpotent. Endomorphism rings of indecomposable modules of finite length are completely primary by (19.17), and so are group algebras of finite p-groups in characteristic p. But k[[x]] and Z_p are local with non-nilpotent — indeed non-nil — maximal ideal.

Executive Summary

In commutative algebra a local ring is a nonzero ring with one maximal ideal, and localisation manufactures them on demand. Noncommutatively there is no comparably good localisation theory, so local rings have to be found rather than made — and they are found, in bulk, as endomorphism rings of indecomposable modules.

Lam's $(19.1)$ collapses six candidate definitions into one. The load-bearing equivalence is unique maximal left ideal $iff$ * $R / rad R$ is a division ring*: the right-hand condition mentions no side, so the notion is automatically left-right symmetric. The most usable reformulation is the crudest one — the non-units are closed under addition.

6Equivalent conditions in (19.1)

$R ∖ U (R)$ Equals

rad R

SymmetricLeft vs right

$0, 1$ Only idempotents

Overview

Let $R$ be a ring with identity, $rad R$ its Jacobson radical and $U (R)$ its group of units. Call $R \neq 0$ local if it has exactly one maximal left ideal. Because every proper left ideal is contained in a maximal one, that single maximal left ideal must contain every proper left ideal, and it must equal $rad R$ .

R local ⟺ R ∖ U (R) = rad R ⟺ R / rad R is a division ring

(19.1)

Three ways of saying the same thing. Only the first mentions elements; only the last is visibly side-neutral.

The commutative reader should keep the picture but discard the machinery. Localisation at a prime ideal produces local rings in commutative algebra; the noncommutative analogue fails for want of a well-behaved Ore condition in general, so the standard supply of noncommutative local rings comes from three other places: division rings, twisted power series rings, and endomorphism rings — see Examples of Noncommutative Local Rings and Strongly Indecomposable Modules and Local Endomorphism Rings.

Notation: when we want to name the radical we write $(R, 𝔪)$ for a local ring with $𝔪 = rad R$ . The pair notation is borrowed from commutative algebra and is safe here, because $𝔪$ really is the unique maximal left ideal, the unique maximal right ideal and the unique maximal two-sided ideal simultaneously.

Learning Objectives

State the six equivalent conditions of $(19.1)$ and identify which are one-sided as written.
Prove $(1) \Leftrightarrow (3)$ and explain why symmetry is then free.
Show that in a local ring $R ∖ U (R) = rad R$ .
Derive the three necessary conditions of $(19.2)$ : unique maximal ideal, Dedekind-finiteness, no nontrivial idempotents.
Give counterexamples showing each of those three conditions is insufficient on its own.
Apply the sufficient criteria of $(19.3)$ to nilpotent non-units and to valuation rings of a division ring.

Definitions

Definition(19.1)Local ring

A ring $R \neq 0$ is local if it has a unique maximal left ideal. By $(19.1)$ this is equivalent to having a unique maximal right ideal, and we write $(R, 𝔪)$ with $𝔪 = rad R$ when the radical is to be named.

No commutativity is assumed and no chain condition is assumed. A commutative local ring in the usual sense is exactly a commutative ring that is local in this sense.

$U (R)$: The group of two-sided invertible elements of $R$ .
$R ∖ U (R)$: The set of non-units. For a general ring this is merely a set; locality is precisely the statement that it is an additive subgroup, and then automatically an ideal.
Dedekind-finite: Every one-sided inverse is a two-sided inverse: if ba = 1 then ab = 1.
Nontrivial idempotent: An element e with e squared equal to e and e different from both 0 and 1.
Valuation ring of a division ring D: A subring R of D such that every nonzero element of D has itself or its inverse in R.
Invariant valuation ring: A valuation ring R of D that is additionally closed under conjugation: the conjugate of R by any nonzero element of D lies in R.
Completely primary: Local with nilpotent maximal ideal.

Rings have an identity; modules are unital. Simple rings and division rings are nonzero by convention.

Core Concepts

Why one maximal left ideal forces a division ring

Every maximal left ideal contains $rad R$ , since the radical is the intersection of all of them. If there is only one such ideal $𝔪$ , then $𝔪 = rad R$ , and $R / 𝔪$ is a nonzero ring whose only left ideals are $0$ and itself. A nonzero ring with no proper nonzero left ideals is a division ring: for $a \neq 0$ the left ideal $R a$ must be everything, giving $b a = 1$ , and the same argument applied to $b$ produces a left inverse of $b$ , which forces $a b = 1$ .

Unique maximal left ideal

𝔪

𝔪 = rad R

R / rad R

has no proper left ideals

R / rad R

is a division ring

Where the side-symmetry comes from

The Jacobson radical is left-right symmetric: the intersection of the maximal left ideals equals the intersection of the maximal right ideals. And being a division ring is a self-opposite property. So condition $(3)$ of $(19.1)$ is invariant under passing to $R^{op}$ , and once $(1) \Leftrightarrow (3)$ is proved, $(2) \Leftrightarrow (3)$ comes for free by applying the same equivalence to $R^{op}$ .

Units detect everything

The bridge from ideals to elements is the standard fact $(4.8)$ that units are detected modulo the radical: $u \in U (R)$ if and only if its image in $R / rad R$ is a unit. If that quotient is a division ring, then every element outside $rad R$ has invertible image and is therefore itself a unit. Hence the non-units are exactly $rad R$ — an ideal, which is why the crude condition $(4)$ is equivalent to the refined ones.

U (R) = {a \in R : \bar{a} \neq 0 in R / rad R} (R local)

(19.1c)

In a local ring, invertibility is a single non-vanishing condition modulo the radical.

Key Results

Theorem(19.1)Characterisations of local rings

Let $R \neq 0$ be a ring. The following are equivalent.

$R$ has a unique maximal left ideal.
$R$ has a unique maximal right ideal.
$R / rad R$ is a division ring.
$R ∖ U (R)$ is an ideal of $R$ .
$R ∖ U (R)$ is closed under addition (a subgroup of $(R, +)$ ).
For every $n \geq 1$ : if $a_{1} + \dots + a_{n} \in U (R)$ then some $a_{i} \in U (R)$ ; equivalently, $a + b \in U (R)$ implies $a \in U (R)$ or $b \in U (R)$ .

A ring satisfying these is called local, written $(R, 𝔪)$ with $𝔪 = rad R$ .

Proof

** $(1) \Rightarrow (3)$ .** Every maximal left ideal contains $rad R$ , so if $𝔪$ is the only one then $rad R = 𝔪$ . Thus $R / rad R$ is nonzero with exactly two left ideals, hence a division ring.

** $(3) \Rightarrow (1)$ .** A division ring has $0$ as its unique maximal left ideal. Maximal left ideals of $R$ correspond bijectively to maximal left ideals of $R / rad R$ , since all of them contain $rad R$ . Hence $rad R$ is the unique maximal left ideal of $R$ .

** $(3) \Rightarrow (4)$ .** By $(4.8)$ an element is a unit exactly when its image modulo $rad R$ is a unit. In a division ring the units are the nonzero elements, so $a \in U (R)$ iff $a \notin rad R$ . Therefore $R ∖ U (R) = rad R$ , an ideal.

** $(4) \Rightarrow (5) \Rightarrow (6)$ ** are immediate: an ideal is closed under addition, and closure under addition applied contrapositively is exactly the statement that a sum of non-units is a non-unit.

** $(6) \Rightarrow (3)$ .** Let $a \notin rad R$ and choose a maximal left ideal $𝔪$ with $a \notin 𝔪$ . Maximality gives $𝔪 + R a = R$ , so $1 = m + b a$ with $m \in 𝔪$ , $b \in R$ . Now $m$ is a non-unit, since a unit inside a proper left ideal would collapse it; and $1 \in U (R)$ , so $(6)$ forces $b a \in U (R)$ . In particular $a$ has a left inverse in $R$ , so the image $\bar{a}$ has a left inverse in $\bar{R} = R / rad R$ . Thus every nonzero element of $\bar{R}$ is left-invertible. Given $\bar{a} \neq 0$ with $\bar{b} \bar{a} = 1$ , note $\bar{b} \neq 0$ , so $\bar{c} \bar{b} = 1$ for some $\bar{c}$ ; then $\bar{c} = \bar{c} (\bar{b} \bar{a}) = (\bar{c} \bar{b}) \bar{a} = \bar{a}$ , whence $\bar{a} \bar{b} = 1$ and $\bar{a} \in U (\bar{R})$ . So $\bar{R}$ is a division ring.

** $(2) \Leftrightarrow (3)$ .** Condition $(3)$ is unchanged on replacing $R$ by $R^{op}$ , because $rad (R^{op}) = rad R$ and the opposite of a division ring is a division ring. Applying the proved equivalence $(1) \Leftrightarrow (3)$ to $R^{op}$ gives $(2) \Leftrightarrow (3)$ .

Proposition(19.2)Necessary conditions

Let $R$ be a local ring. Then:

$R$ has a unique maximal two-sided ideal, namely $rad R$ ;
$R$ is Dedekind-finite: if $a$ has a left inverse in $R$ then $a \in U (R)$ ;
$R$ has no nontrivial idempotents: $e = e^{2}$ implies $e = 0$ or $e = 1$ .

Proof

(a) A maximal two-sided ideal $M$ is proper, so it contains no unit; hence $M \subseteq R ∖ U (R) = rad R$ . Since $rad R$ is itself a proper two-sided ideal and $M$ is maximal, $M = rad R$ .

(b) Suppose $b a = 1$ . Then $\bar{b} \bar{a} = 1$ in the division ring $R / rad R$ , so $\bar{a} \neq 0$ and $\bar{a}$ is a unit there; by $(4.8)$ , $a \in U (R)$ .

(c) Let $e = e^{2}$ and put $f = 1 - e$ . Then $e + f = 1 \in U (R)$ , so by $(19.1) (6)$ one of $e, f$ is a unit. Since $e f = e (1 - e) = e - e^{2} = 0$ and a unit annihilating an element kills it, $e$ a unit gives $f = 0$ , i.e. $e = 1$ ; and $f$ a unit gives $e = 0$ .

Remark(19.2′)None of the three is sufficient

(a) alone: every simple ring has a unique maximal ideal, namely $(0)$ . But $M_{2} (k)$ is simple and not local — it has two distinct maximal left ideals and plenty of idempotents.
(b) alone: every commutative ring is Dedekind-finite, and $ℤ$ is not local.
(c) alone: every domain lacks nontrivial idempotents, and $ℤ$ is again not local.
All three together: the first Weyl algebra $A_{1} (k)$ over a field $k$ of characteristic $0$ is a simple domain, hence satisfies (a), (b) and (c); but $rad A_{1} (k) = 0$ and $A_{1} (k)$ is not a division ring, so it is not local.

Proposition(19.3)Two sufficient criteria

Let $R \neq 0$ be a ring in which every non-unit is nilpotent. Then $R$ is local.
Let $D$ be a division ring and $R \subseteq D$ a subring such that for every $d \in D^{*}$ either $d \in R$ or $d^{- 1} \in R$ . Then $R$ is local.

Proof

(a) It suffices to show $R ∖ U (R) \subseteq rad R$ , since the reverse inclusion always holds and $(19.1) (4)$ then applies. Let $a \notin U (R)$ and let $k \geq 1$ be minimal with $a^{k} = 0$ . First, $R a$ contains no unit: if $r a \in U (R)$ then from $(r a) a^{k - 1} = r a^{k} = 0$ we get $a^{k - 1} = 0$ , contradicting minimality of $k$ . So every element of $R a$ is a non-unit, hence nilpotent by hypothesis; thus $R a$ is a nil left ideal and $R a \subseteq rad R$ by $(4.11)$ . In particular $a \in rad R$ .

(b) We verify $(19.1) (6)$ . Suppose $a + b \in U (R)$ ; multiplying on the right by $(a + b)^{- 1}$ we may assume $a + b = 1$ , and we may assume $a, b \neq 0$ (otherwise the other summand is $1$ ). Both $a$ and $b$ are invertible **in $D$ **, and the question is whether an inverse lies in $R$ . Set $c = a^{- 1} b \in D$ ; by hypothesis $c \in R$ or $c^{- 1} \in R$ .

If $c \in R$ , then $a^{- 1} = a^{- 1} (a + b) = 1 + a^{- 1} b = 1 + c \in R$ , so $a \in U (R)$ . If instead $c^{- 1} = b^{- 1} a \in R$ , then $b^{- 1} = b^{- 1} (a + b) = b^{- 1} a + 1 = c^{- 1} + 1 \in R$ , so $b \in U (R)$ . Either way $(19.1) (6)$ holds and $R$ is local.

Remark(19.3′)Valuation rings

When $D$ is a field, a subring $R$ satisfying the hypothesis of $(19.3) (b)$ is precisely a valuation ring of $D$ ; these arise from Krull valuations and are always local. For a genuine division ring $D$ the valuation rings that come from valuations $v : D^{*} \to G$ into an ordered group carry the extra invariance property $d^{- 1} R d \subseteq R$ for all $d \in D^{*}$ , and are then called invariant valuation rings. Locality, however, needs only the weaker one-or-the-other hypothesis.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Push everything through $R / rad R$

Unit-detection $(4.8)$ makes the quotient by the radical a faithful test for invertibility. Any statement about units in a local ring becomes a statement about nonzero elements of a division ring.

Move 2

Maximality gives $1 = m + b a$

If $a$ escapes a maximal left ideal $𝔪$ , then $𝔪 + R a = R$ . Writing $1$ as a sum of a non-unit and something is exactly the shape condition $(6)$ is designed to attack.

Move 3

Left inverses bootstrap

In a nonzero ring where every nonzero element has a left inverse, associativity alone upgrades left inverses to two-sided ones. This is what converts $(6)$ into the division-ring condition.

Move 4

Nil ideals sink into the radical

$(4.11)$ : any nil one-sided ideal lies in $rad R$ . To prove locality from a nilpotence hypothesis, build a nil left ideal around each non-unit.

Move 5

Opposite-ring symmetry

Prove one side, then observe that the criterion you landed on is self-opposite. This is cheaper than repeating the argument and is the standard way $(2)$ is obtained.

Move 6

Idempotents versus $e + (1 - e) = 1$

Splitting $1$ as $e + (1 - e)$ and feeding it to $(6)$ kills nontrivial idempotents in one line. The same trick recurs whenever a decomposition of $1$ is available.

Move 4 is the one worth internalising because it is what makes $(19.3) (a)$ work at all. The hypothesis every non-unit is nilpotent says nothing about ideals; the proof manufactures an ideal by observing that $R a$ cannot contain a unit when $a$ is nilpotent, and only then invokes the radical.

Worked Example

A four-dimensional noncommutative local algebra

Let $σ : ℂ \to ℂ$ be complex conjugation and put

R = {(\begin{matrix} a & b \\ 0 & \bar{a} \end{matrix}) : a, b \in ℂ} \subseteq M_{2} (ℂ) .

(E.1)

This is closed under multiplication because $\bar{a a'} = \bar{a} \bar{a}'$ : the product of two such matrices has $(1, 1)$ entry $a a'$ and $(2, 2)$ entry $\bar{a} \bar{a}' = \bar{a a'}$ . As a real algebra $\dim_{ℝ} R = 4$ .

Step 1 — identify the units

A matrix in $R$ is invertible in $M_{2} (ℂ)$ iff $a \bar{a} \neq 0$ , i.e. $a \neq 0$ , and then the inverse

{(\begin{matrix} a & b \\ 0 & \bar{a} \end{matrix})}^{- 1} = (\begin{matrix} a^{- 1} & - a^{- 1} b {\bar{a}}^{- 1} \\ 0 & {\bar{a}}^{- 1} \end{matrix})

(E.2)

The $(2, 2)$ entry is ${\bar{a}}^{- 1} = \bar{a^{- 1}}$ , so the inverse lies back in $R$ .

So $U (R) = {matrices with a \neq 0}$ and the non-units are exactly the matrices with $a = 0$ .

Step 2 — check the criteria

The non-units form the set $𝔪 = {(\begin{matrix} 0 & b \\ 0 & 0 \end{matrix})}$ , visibly closed under addition and under multiplication by $R$ on both sides, so $(19.1) (4)$ holds and $R$ is local with $rad R = 𝔪$ . Independently, every non-unit squares to $0$ , so $(19.3) (a)$ applies as well. And $R / 𝔪 ≅ ℂ$ is a division ring, which is $(19.1) (3)$ .

Step 3 — confirm it is genuinely noncommutative

(\begin{matrix} i & 0 \\ 0 & - i \end{matrix}) (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) = (\begin{matrix} 0 & i \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) (\begin{matrix} i & 0 \\ 0 & - i \end{matrix}) = (\begin{matrix} 0 & - i \\ 0 & 0 \end{matrix}) .

(E.3)

The two products differ, so $R$ is a noncommutative local ring. Since $𝔪^{2} = 0$ it is completely primary, and $R$ is finite-dimensional over $ℝ$ , hence artinian.

The same construction with $ℂ / ℝ$ replaced by any field automorphism $σ$ of any field $k$ gives a local ring $k ⋉_{1} k_{σ}$ , noncommutative exactly when $σ \neq id$ .

Process and Workflow

How do I test whether a given ring $R \neq 0$ is local?

You can compute $rad R$ Form

R / rad R

and ask whether it is a division ring. For finite-dimensional algebras this is the fastest route — the Wedderburn decomposition of the semisimple quotient must be a single

1 \times 1

block.

You can describe the unitsCheck that the non-units are closed under addition. This is

(19.1) (5)

and is usually the shortest hand argument for concrete matrix or power-series rings.

Every non-unit looks nilpotentApply

(19.3) (a)

directly. Typical for finite-dimensional algebras with a single simple module.

$R$ sits inside a division ringCheck the valuation condition of

(19.3) (b)

: for each

d \neq 0

, at least one of

d

d^{- 1}

is in

R

$R$ is an endomorphism ringIf

R = End (M_{R})

with

M

indecomposable of finite length,

(19.17)

gives locality at once — no computation needed.

Rule out the cheap obstructionsLook for a nontrivial idempotent, or two elements summing to a unit with neither a unit. Either kills locality immediately.

Locate the candidate maximal idealUsually the obvious two-sided ideal — augmentation ideal, strictly-upper-triangular part, series with zero constant term.

Show the quotient is a division ringIdentify

R / 𝔪

explicitly. If it is a division ring, then

𝔪 \supseteq rad R

forces

𝔪 = rad R

Confirm $𝔪 \subseteq rad R$ Typically by nilpotence or nility, using

(4.11)

; then equality holds and

R

is local.

Comparison and Classification

Locality across familiar rings
Ring	$rad$	Local?	Why
Division ring $D$	$0$	yes	$D / 0 = D$ is a division ring
$ℤ_{(p)}$	$p ℤ_{(p)}$	yes	valuation ring of $ℚ$
$ℤ / p^{n} ℤ$	$(p)$	yes	every non-unit is nilpotent, $(19.3) (a)$
$ℤ / 6 ℤ$	$0$	no	two maximal ideals; $2 + 3 = 5$ is a unit but neither is
$ℤ$	$0$	no	infinitely many maximal ideals
$M_{n} (D)$ , $n \geq 2$	$0$	no	nontrivial idempotents, $(19.2) (c)$
Upper triangular $T_{n} (k)$ , $n \geq 2$	strictly upper triangular	no	diagonal idempotents
$T_{n} (k)$ with constant diagonal	strictly upper triangular	yes	quotient is $k$
$k [[x]]$	$(x)$	yes	non-units are the series with zero constant term
$k [[x; σ]]$ , $σ \neq id$	$(x)$	yes	same, and noncommutative
$A_{1} (k)$ , $char k = 0$	$0$	no	simple domain but not a division ring

Which of the necessary conditions of (19.2) each ring satisfies
	Unique max. ideal	Dedekind-finite	No nontrivial $e = e^{2}$	Local
$M_{2} (k)$	yes	yes	no	no
$ℤ$	no	yes	yes	no
$A_{1} (k)$ , char $0$	yes	yes	yes	no
$k [[x]]$	yes	yes	yes	yes
${End}_{k} (V)$ , $\dim V = \infty$	no	no	no	no

Which of the necessary conditions of (19.2) each ring satisfies

The third row is the point: all three necessary conditions can hold at once in a ring that is not local. Locality is strictly stronger than their conjunction.

Relationship Map

Local rings sit at the bottom of the semilocal hierarchy: they are exactly the semilocal rings whose semisimple quotient is a division ring.

All rings

rad R

defined and two-sided

Semilocal

R / rad R

semisimple

Local

R / rad R

a division ring

Completely primary…and

rad R

nilpotent

Division ring

rad R = 0

$R$ local — the implications, one way only
- implies
  - unique maximal two-sided ideal $(19.2) (a)$
  - Dedekind-finite $(19.2) (b)$
  - no nontrivial idempotents $(19.2) (c)$
  - every finitely generated projective module is free $(19.29)$
  - $Z (R)$ is a commutative local ring (Exercise 19.6)
- is implied by
  - $R \neq 0$ and every non-unit nilpotent $(19.3) (a)$
  - $R$ a valuation ring of a division ring $(19.3) (b)$
  - $R = End (M_{R})$ for $M$ indecomposable of finite length $(19.17)$
  - $R$ right artinian with no nontrivial idempotents $(19.19)$
- is preserved by
  - passing to $R^{op}$ (Exercise 19.1)
  - power series $R [[x]]$ and twisted power series $R [[x; σ]]$
  - quotients $R / I$ for $I$ a proper ideal

The last item deserves a check: if $R$ is local and $I ⊊ R$ is a proper ideal, then $I \subseteq rad R$ (a proper ideal contains no unit), so $R / I$ is nonzero with $(R / I) / (rad R / I) ≅ R / rad R$ a division ring — hence $R / I$ is local. This is the mechanism behind Exercise 19.5.

Applications and Industry Use

Applications here means where this structure is used — inside mathematics and in the engineering and computing disciplines that consume it.

Module theory

Unique decomposition

The Krull–Schmidt–Azumaya theorem needs its summands to have local endomorphism rings; that hypothesis is what makes the exchange argument run. Locality is the precise condition under which unique decomposition survives.

Coding theory

Codes over chain rings

Linear codes over finite chain rings such as $ℤ / p^{n}$ and Galois rings are analysed via the radical filtration of a local ring; the residue division ring carries the associated code over a field.

Commutative algebra and geometry

Stalks and germs

Local rings of a variety at a point, and rings of germs of holomorphic functions, are the geometric prototypes. Their maximal ideal records vanishing at the point.

Symbolic computation

Algebra recognition

Deciding locality of a finite-dimensional algebra reduces to computing the radical and checking that the semisimple quotient has a single simple module — a standard primitive in GAP, Magma and Sage.

The honest reading is that local rings are infrastructure rather than an end in themselves: they are the coefficient rings over which decomposition arguments behave, and the residue-field mechanism that lets you argue modulo the radical and lift back.

Design Considerations

Design considerations here means the choices made when modelling a problem with these algebraic structures.

Which side? Locality is symmetric, so you may verify it on whichever side is convenient. Do not carry that licence over to artinian, noetherian, primitive or perfect, all of which are genuinely one-sided.
Which criterion to adopt as the definition? For hand computation take the non-unit condition $(19.1) (5)$ ; for structural arguments take $R / rad R$ a division ring; for module-theoretic work take unique maximal left ideal, which pairs directly with the unique simple module.
How much nilpotence to demand. Requiring $rad R$ nilpotent gives completely primary, which is the right class for endomorphism rings of finite-length modules. Demanding it globally would exclude $k [[x]]$ and the $p$ -adic integers.
Where to place the module. Endomorphism rings act on the opposite side to the module structure; fix the convention once. Lam works with right modules and lets $End (M_{R})$ act on the left, which is why $M$ becomes a left $E$ -module without extra work.
When to insist on an identity. Without $1$ , the unit-based criteria are meaningless and only the quasi-regularity formulation survives. This collection assumes an identity throughout.

Standards and Notation

Standards here covers notation, symbol and markup standards, and reference implementations, rather than material or design codes.

Preferred notation

(R, 𝔪)

with

𝔪 = rad R

(Lam)

Common variant

(R, J)

with

J = J (R)

(Anderson–Fuller, homological sources)

Residue object

R / 𝔪

is a division ring, not a residue field, unless

R

is commutative

Terminology hazardSome commutative-algebra texts reserve local for noetherian rings and say quasi-local otherwise

MarkupPresentation MathML per ISO/IEC 40314; symbol conventions per ISO 80000-2

GAP / SageRadicalOfAlgebra, A.radical(); locality by checking the semisimple quotient

Failure Modes and Common Mistakes

Do not assume a local ring is noetherian, artinian or commutative; none of these is implied by $(19.1)$ .
Do not conclude from $R$ local that $M_{n} (R)$ is local: for $n \geq 2$ it has nontrivial idempotents. Locality is not a Morita invariant, unlike the radical itself.
Do not confuse local with semilocal. Semilocal only asks that $R / rad R$ be semisimple; local demands a single simple module.
Do not read $(19.1) (6)$ as a sum of units is a unit. It is the contrapositive statement: a sum can only be a unit if a summand already is. In $ℤ / p^{n}$ , $1 + (p^{n} - 1) = 0$ is a sum of units that is not a unit.

Historical Notes and Lessons Learned

1930sKrull's local ringsKrull formalises local rings in commutative algebra as the objects attached to a prime ideal by localisation, in support of dimension theory and valuation theory.
1937–45Valuations on division ringsValuation theory is extended beyond fields; the resulting valuation rings inside division rings supply the first systematic family of noncommutative local rings, with the invariance condition appearing as an extra requirement.
1945Jacobson's radicalWith the radical available for arbitrary rings, the condition * $R / rad R$ is a division ring* becomes a workable definition and severs locality from any chain condition.
1950AzumayaAzumaya proves the general Krull–Schmidt uniqueness theorem under the hypothesis that the summands have local endomorphism rings, giving noncommutative local rings a decisive structural role.
1958Kaplansky on projectivesKaplansky shows every projective module over a local ring is free, without a finite generation hypothesis, generalising the finitely generated case that follows from Nakayama's Lemma.

The methodological lesson: the commutative definition (one maximal ideal) and the noncommutative one (one maximal left ideal) look like the same sentence but are not, and the two-sided version is strictly weaker. Choosing the one-sided formulation is what preserves the theorems.

Quick Reference

Definition

R \neq 0

with a unique maximal left ideal

Radical form

R / rad R

is a division ring

Element test

a + b \in U (R) \Rightarrow a \in U (R)

b \in U (R)

Non-units

R ∖ U (R) = rad R = 𝔪

SymmetryLeft and right versions agree;

R^{op}

is local too

IdempotentsOnly

0

and

1

ProjectivesFinitely generated projective

\Rightarrow

free

Not impliedNoetherian, artinian, commutative,

𝔪

nilpotent

The conditions of (19.1) at a glance
Label	Condition	Sided?
(1)	unique maximal left ideal	left
(2)	unique maximal right ideal	right
(3)	$R / rad R$ a division ring	neither
(4)	$R ∖ U (R)$ is an ideal	neither
(5)	$R ∖ U (R)$ closed under addition	neither
(6)	$a + b \in U (R) \Rightarrow a$ or $b \in U (R)$	neither

Frequently Asked Questions

Why is a unique maximal two-sided ideal not the right definition?

Because it is far too weak. Any simple ring — $M_{n} (k)$ , the Weyl algebra $A_{1} (k)$ in characteristic zero — has exactly one maximal two-sided ideal, namely $(0)$ , yet has many maximal left ideals and, in the matrix case, many idempotents. The correct definition uses maximal one-sided ideals; the two-sided uniqueness is then a corollary, $(19.2) (a)$ .

Is a local ring necessarily noetherian or artinian?

No. $k [[x]]$ is local and noetherian but not artinian; the ring of germs of holomorphic functions at a point is local and noetherian; a valuation ring of infinite rank in a field is local and neither noetherian nor artinian. Nothing in $(19.1)$ mentions chain conditions.

M_{n} (R)

local when

R

is?

Not for $n \geq 2$ . The matrix units $e_{11}, \dots, e_{n n}$ are nontrivial idempotents, which $(19.2) (c)$ forbids. This is a useful reminder that locality is not Morita invariant, in contrast to the Jacobson radical, which satisfies $rad M_{n} (R) = M_{n} (rad R)$ .

What is the difference between local and completely primary?

Completely primary means local and $rad R$ nilpotent. Endomorphism rings of indecomposable modules of finite length are completely primary by $(19.17)$ , and so are group algebras of finite $p$ -groups in characteristic $p$ . But $k [[x]]$ and $ℤ_{p}$ are local with non-nilpotent — indeed non-nil — maximal ideal.

Does every commutative ring embed in a local ring in a useful way?

Commutatively, yes: localise at a prime. That is exactly the tool that has no good general noncommutative analogue, since the multiplicative sets one wants rarely satisfy an Ore condition. This is the historical reason local rings play a smaller role noncommutatively, and why the noncommutative examples come from endomorphism rings and completions rather than from localisation.

R

is local, what are its simple modules?

There is exactly one simple left module up to isomorphism, namely $R / rad R$ viewed as a left $R$ -module — a one-dimensional space over the residue division ring. The same statement holds on the right. This is the module-theoretic content of locality and the reason local rings are the building blocks in decomposition theory.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §19 (pp. 293–310).
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §15 and §27.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.
O. F. G. Schilling, The Theory of Valuations, Mathematical Surveys 4, American Mathematical Society, 1950.
I. Kaplansky, “Projective modules”, Annals of Mathematics 68 (1958), 372–377.

AI Suggested Questions

Construct a ring with a unique maximal ideal that is Dedekind-finite and has no nontrivial idempotents, yet is not local, other than the Weyl algebra.
Show directly that the opposite ring of a local ring is local, without invoking the symmetry of the Jacobson radical.
Which local rings are von Neumann regular, and why does that force the ring to be a division ring?
Give an example of a local ring whose maximal ideal is nil but not nilpotent.
How does the centre of a local ring behave, and why is it again local?
Work out the structure of local rings that are also right discrete valuation rings, and construct a noncommutative one using twisted power series.
Explain why locality fails to be a Morita invariant while the Jacobson radical is one.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why one maximal left ideal forces a division ring

Where the side-symmetry comes from

Units detect everything

06Key Results

07Proof Techniques and Method

Push everything through R/radR

Maximality gives 1=m+ba

Left inverses bootstrap

Nil ideals sink into the radical

Opposite-ring symmetry

Idempotents versus e+(1−e)=1

08Worked Example

A four-dimensional noncommutative local algebra

Step 1 — identify the units

Step 2 — check the criteria

Step 3 — confirm it is genuinely noncommutative

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Unique decomposition

Codes over chain rings

Stalks and germs

Algebra recognition

13Design Considerations

14Standards and Notation

15Failure Modes and Common Mistakes

16Historical Notes and Lessons Learned

17Quick Reference

18Frequently Asked Questions

19Related KEVOS Topics

20References

21AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Push everything through $R / rad R$

Maximality gives $1 = m + b a$

Idempotents versus $e + (1 - e) = 1$

Worked Example

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Design Considerations

Standards and Notation

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions