Local Rings and the Absence of Nontrivial Idempotents

Q: Is a ring with no nontrivial idempotents automatically local?

No. Every domain — Z, k[x], the Weyl algebra A_1(k) — has only trivial idempotents, and none of these is local. The implication holds only in the presence of a one-sided chain condition, which is exactly the content of (19.19).

Q: Must a local ring be artinian, or noetherian, or commutative?

None of the three. Z_(p) and k[[x]] are local and not artinian; a valuation ring with value group Q is local and not noetherian; the twisted power series ring k[[x;]] for a nontrivial automorphism of a field k is local and not commutative.

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

Completely primary means local *with nilpotent maximal ideal*. Every right artinian local ring is completely primary, and so is kG for a finite p-group in characteristic p. But k[[x]] is local with rad = (x), which is not even nil, so it is not completely primary.

Q: Can a matrix ring over a local ring be local?

Only for n = 1. For n ≥ 2 the matrix unit E_11 is a nontrivial idempotent of M_n(R), so (19.2)(c) rules it out. This shows locality is not preserved by Morita equivalence, even though rad is.

Q: Is there a ring satisfying all of (a), (b) and (c) of (19.2) that is still not local?

Yes. The first Weyl algebra A_1(k) over a field of characteristic 0 is a simple noetherian domain: simplicity gives a unique maximal ideal, being a domain gives Dedekind-finiteness and no nontrivial idempotents, but rad A_1(k) = 0 and A_1(k) is not a division ring.

Executive Summary

Local is the noncommutative generalisation of unique maximal ideal. For a nonzero ring $R$ it means: a unique maximal left ideal, equivalently a unique maximal right ideal, equivalently $R / rad R$ is a division ring, equivalently the non-units $R ∖ U (R)$ form an ideal. The seven conditions of $(19.1)$ are the working definition; which one you use is a matter of convenience, not of substance.

Locality forces three structural consequences: a unique maximal two-sided ideal, Dedekind-finiteness, and — the one that does the most work downstream — **no idempotents other than $0$ and $1$ **. None of the three is sufficient on its own. But under a one-sided chain condition the third becomes sufficient: a nonzero right artinian ring is local precisely when it has no nontrivial idempotents $(19.19)$ . That is the pivot on which idempotent-theoretic arguments in Chapters 7 and 8 turn.

$R ∖ U (R)$ Is an ideal

$0, 1$ Only idempotents

ArtinianMakes the test sufficient

SymmetricLeft versus right

Overview

In commutative algebra local rings are the objects produced by localisation, and localisation is the reason they matter. Noncommutative localisation is far more delicate, so local rings enter this subject by a different door: they are the endomorphism rings of the modules that cannot be broken up. That is what makes them indispensable even to someone who only cares about commutative rings, because endomorphism rings of modules are rarely commutative.

R local ⟺ R \neq 0 and R / rad R is a division ring

(19.1)

The residue-ring form of the definition. It is visibly left-right symmetric, which is why the unique-maximal-left-ideal and unique-maximal-right-ideal conditions agree.

Idempotents are the algebraic shadow of direct-sum decompositions. An idempotent $e \in R$ splits the right regular module as $R_{R} = e R \oplus (1 - e) R$ , and conversely. So no nontrivial idempotents says exactly that $R_{R}$ is an indecomposable module. Locality is a strictly stronger statement — it says $End (R_{R}) ≅ R$ is not merely idempotent-free but has an ideal of non-units — and the gap between the two closes precisely when a chain condition is available, via the Fitting decomposition.

Learning Objectives

State the equivalent conditions of $(19.1)$ and identify which are manifestly side-neutral.
Prove that $R$ local implies $R$ is Dedekind-finite and has only trivial idempotents.
Exhibit rings satisfying each necessary condition separately without being local.
Prove $(19.19)$ : a nonzero right artinian ring is local iff its only idempotents are $0$ and $1$ .
Use the Fitting decomposition to show that an indecomposable module of finite length is strongly indecomposable.
Compute the radical and idempotents of a small finite-dimensional algebra and decide locality.

Definitions

Definition(19.1)Local ring

A ring $R \neq 0$ is local if it has a unique maximal left ideal. That ideal is then $rad R$ , and it is simultaneously the unique maximal right ideal and the unique maximal two-sided ideal. When we want to name the radical we write $(R, 𝔪)$ with $𝔪 = rad R$ .

$U (R)$: The group of units of $R$ : elements with a two-sided inverse.
Idempotent: $e \in R$ with $e^{2} = e$ . It is trivial if $e = 0$ or $e = 1$ , and nontrivial otherwise.
Dedekind-finite: $a b = 1$ implies $b a = 1$ . Equivalently, one-sided inverses are two-sided.
Indecomposable module: $M \neq 0$ that is not the direct sum of two nonzero submodules. Equivalently, $End (M)$ has no nontrivial idempotent.
Strongly indecomposable: $M \neq 0$ with $End (M)$ local. Strictly stronger than indecomposable in general.
Completely primary: Local with nilpotent maximal ideal. Endomorphism rings of finite-length indecomposables are of this type.

All rings have an identity, all modules are unital, and artinian always carries a side. Where a result is genuinely one-sided this page says so.

Core Concepts

Idempotents are decompositions

The dictionary between idempotents and direct sums is exact and is used in both directions throughout this chapter. Given $e^{2} = e$ in $R$ , the right regular module splits as $R_{R} = e R \oplus (1 - e) R$ ; given a splitting $R_{R} = A \oplus B$ , write $1 = e + f$ with $e \in A$ , $f \in B$ and check that $e$ is idempotent with $A = e R$ .

e^{2} = e

R_{R} = e R \oplus (1 - e) R

R_{R}

decomposable

R

not local

Only the last arrow needs proof, and only the last arrow is not reversible without further hypotheses. Reversing it is what $(19.19)$ does, at the cost of assuming a chain condition.

Why the non-units must be closed under addition

Condition $(5'')$ of $(19.1)$ — if $a + b \in U (R)$ then $a \in U (R)$ or $b \in U (R)$ — looks like an odd way to define a class of rings, but it is the condition that kills idempotents in one line: $e + (1 - e) = 1$ is a unit, so one of $e$ , $1 - e$ is a unit; since $e (1 - e) = 0$ , a unit among them forces the other to be $0$ .

Local, completely primary, and the artinian world

Local rings need not be artinian: $ℤ_{(p)}$ , $k [[x]]$ , the $p$ -adic integers $ℤ_{p}$ and rank-one valuation rings with value group $ℚ$ are all local with non-nilpotent — indeed non-nil — radical. Local rings arising as $End (M)$ for $M$ of finite length always have nilpotent radical, and are called completely primary; group algebras of finite $p$ -groups in characteristic $p$ are the other standard supply.

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.
For every $n$ , if $a_{1} + \dots + a_{n} \in U (R)$ then some $a_{i} \in U (R)$ .
If $a + b \in U (R)$ then $a \in U (R)$ or $b \in U (R)$ .

Proof

It suffices to prove (1) $\Leftrightarrow$ (3); the equivalence (2) $\Leftrightarrow$ (3) then follows because condition (3) is unchanged on replacing $R$ by $R^{op}$ .

**(3) $\Rightarrow$ (1).** Every maximal left ideal contains $rad R$ . If $R / rad R$ is a division ring it has no left ideals besides $0$ and itself, so the only maximal left ideal of $R$ is $rad R$ .

**(1) $\Rightarrow$ (3).** If $𝔪$ is the unique maximal left ideal then $rad R = 𝔪$ , so $R / rad R$ has exactly two left ideals and is therefore a division ring.

**(3) $\Rightarrow$ (4).** Since units lift modulo the radical, any $a \notin rad R$ has invertible image in the division ring $R / rad R$ , hence $b a = 1 - r$ and $a c = 1 - r'$ with $r, r' \in rad R$ ; both $1 - r$ and $1 - r'$ are units, so $a$ has a left and a right inverse and thus $a \in U (R)$ . Conversely no element of $rad R$ is a unit. Hence $R ∖ U (R) = rad R$ , an ideal.

**(4) $\Rightarrow$ (5) $\Rightarrow$ (5 $'$ ) $\Rightarrow$ (5 $''$ )** are immediate.

**(5 $''$ ) $\Rightarrow$ (3).** Let $a \notin rad R$ and choose a maximal left ideal $𝔪$ with $a \notin 𝔪$ . Then $𝔪 + R a = R$ , so $1 = m + b a$ for some $m \in 𝔪$ , $b \in R$ . As $m$ lies in a proper left ideal it is not a unit, so $(5'')$ forces $b a \in U (R)$ . Passing to $\bar{R} = R / rad R$ , every nonzero element has a left inverse, so $\bar{R} ∖ {0}$ is a group under multiplication and $\bar{R}$ is a division ring.

Proposition(19.2)Necessary conditions

Let $R$ be a local ring. Then: (a) $R$ has a unique maximal (two-sided) ideal, namely $rad R$ ; (b) $R$ is Dedekind-finite; (c) the only idempotents of $R$ are $0$ and $1$ .

Proof

(a) A maximal ideal $M$ contains no unit, so $M \subseteq R ∖ U (R) = rad R$ ; maximality gives $M = rad R$ .

(b) Suppose $b a = 1$ . If $a \notin U (R)$ then $a \in rad R$ , whence $1 = b a \in rad R$ , contradicting $R \neq 0$ . So $a \in U (R)$ and $b = a^{- 1}$ .

(c) Let $e^{2} = e$ and put $f = 1 - e$ , so $e + f = 1 \in U (R)$ . By $(19.1) (5'')$ one of $e, f$ is a unit. If $e \in U (R)$ then $e f = 0$ gives $f = 0$ , i.e. $e = 1$ ; if $f \in U (R)$ then the same equation gives $e = 0$ .

Remark(19.2′)None of (a), (b), (c) is sufficient

Condition (a) holds for every simple ring, e.g. $M_{2} (k)$ , which is not local. Condition (b) holds for every commutative ring, e.g. $ℤ$ . Condition (c) holds for every domain, e.g. $k [x]$ . Even all three together are insufficient: the first Weyl algebra $A_{1} (k)$ over a field of characteristic $0$ is a simple noetherian domain, so it satisfies (a), (b) and (c), yet $rad A_{1} (k) = 0$ and $A_{1} (k)$ is not a division ring, so it is not local.

Proposition(19.3)Nilpotence criterion

Let $R \neq 0$ be a ring in which every non-unit is nilpotent. Then $R$ is local, and $rad R$ is a nil ideal.

Proof

Let $a \notin U (R)$ and let $k \geq 1$ be least with $a^{k} = 0$ . Every element of $R a$ is a non-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. Hence $R a$ consists of non-units, so of nilpotent elements, so $R a$ is a nil left ideal and therefore $R a \subseteq rad R$ by $(4.11)$ . Thus $R ∖ U (R) \subseteq rad R$ ; the reverse inclusion is automatic, so $R ∖ U (R) = rad R$ is an ideal and $(19.1) (4)$ applies.

Theorem(19.16)Fitting decomposition

Let $R$ be a ring and $M_{R}$ a right $R$ -module of finite composition length. For every $f \in End (M_{R})$ there is an integer $n$ with $M = \ker (f^{n}) \oplus im (f^{n})$ ; any $n$ at which both chains $im (f) \supseteq im (f^{2}) \supseteq \dots$ and $\ker (f) \subseteq \ker (f^{2}) \subseteq \dots$ have stabilised will do.

Theorem(19.17)Endomorphism rings of finite-length indecomposables

Let $M_{R}$ be an indecomposable right $R$ -module of finite composition length $n$ . Then $E = End (M_{R})$ is a local ring and $(rad E)^{n} = 0$ . In particular $M$ is strongly indecomposable and $E$ is completely primary.

Proof

We prove the part needed below: every $f \in E ∖ U (E)$ is nilpotent, whence $E$ is local by $(19.3)$ . Choose $n$ as in $(19.16)$ , so $M = \ker (f^{n}) \oplus im (f^{n})$ . Since $M$ is indecomposable one summand is $0$ .

If $\ker (f^{n}) = 0$ then $im (f^{n}) = M$ , so $f^{n}$ is bijective; then $f$ is injective (because $f^{n}$ is) and surjective (because $im (f) \supseteq im (f^{n}) = M$ ), so $f \in U (E)$ , contrary to hypothesis. Therefore $\ker (f^{n}) \neq 0$ , so $im (f^{n}) = 0$ , i.e. $f^{n} = 0$ .

The bound $(rad E)^{n} = 0$ follows by viewing $M$ as a left $E$ -module and applying Nakayama's Lemma repeatedly to the chain $M \supseteq 𝔪 M \supseteq 𝔪^{2} M \supseteq \dots$ with $𝔪 = rad E$ : each inclusion is strict until the term is $0$ , and $M$ has only $n$ steps of room.

Corollary(19.19)Artinian rings: locality equals idempotent-freeness

Let $R \neq 0$ be a right artinian ring. Then $R$ is local if and only if $R$ has no idempotents other than $0$ and $1$ .

Proof

Necessity is $(19.2) (c)$ and needs no chain condition.

Sufficiency. Take $M = R_{R}$ . By the Hopkins–Levitzki Theorem $(4.15)$ a right artinian ring is right noetherian, so $M$ has finite composition length. Idempotents of $End (M_{R}) ≅ R$ correspond to direct-sum decompositions of $M$ , so the hypothesis says exactly that $M$ is indecomposable. By $(19.17)$ , $End (M_{R}) ≅ R$ is a local ring.

Corollary(19.19′)The left-handed version

The same statement holds with left artinian in place of right artinian: apply $(19.19)$ to $R^{op}$ , which is right artinian exactly when $R$ is left artinian, has the same idempotents as $R$ , and is local exactly when $R$ is by the symmetry in $(19.1)$ .

Remark—Completely primary

A right artinian local ring has nilpotent radical, so it is completely primary. The converse of *local $\Rightarrow$ completely primary* fails in the other direction too: $ℤ_{(p)}$ is local but its radical $p ℤ_{(p)}$ is not nil, so it is not completely primary and in particular not artinian.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Test membership by units

To show a set equals $rad R$ , show it is exactly the complement of $U (R)$ . Every equivalence in $(19.1)$ is proved by moving between is a unit and lies outside the radical.

Move 2

Split off with an idempotent

To contradict locality, produce a nontrivial idempotent. To produce one, find a direct-sum decomposition of some module and take the projection; $End$ turns the decomposition into the idempotent.

Move 3

Fitting: kernel or image dies

On an indecomposable module of finite length, one of $\ker (f^{n})$ , $im (f^{n})$ must vanish. That dichotomy converts no idempotents into every non-unit is nilpotent, which is the hypothesis of $(19.3)$ .

Move 3 is the load-bearing one and it is worth noticing what it consumes. It needs both chain conditions on $M$ — ascending to stabilise kernels, descending to stabilise images. Assuming only one of them is not enough, and $(19.18)$ records the counterexamples: $ℤ$ as a module over itself satisfies the ACC and is indecomposable, yet $End (ℤ) ≅ ℤ$ is not local.

Worked Example

A finite-dimensional algebra that is local

Let $k$ be a field and let $A \subseteq M_{3} (k)$ be the algebra of upper triangular matrices with constant diagonal:

A = {(\begin{matrix} a & b & c \\ 0 & a & d \\ 0 & 0 & a \end{matrix}) : a, b, c, d \in k} = k \cdot 1 \oplus J, J = {(\begin{matrix} 0 & b & c \\ 0 & 0 & d \\ 0 & 0 & 0 \end{matrix})} .

(E.1)

$\dim_{k} A = 4$ . $A$ is noncommutative: the $(1, 3)$ entries of the two products differ by $b d' - b' d$ .

$J$ is an ideal of $A$ with $J^{2} = k E_{13}$ and $J^{3} = 0$ , so $J$ is nilpotent and therefore $J \subseteq rad A$ . Since $A / J ≅ k$ is a field, $J$ is maximal and hence $rad A = J$ . By $(19.1) (3)$ , $A$ is local; being finite-dimensional it is artinian, so it is completely primary with $(rad A)^{3} = 0$ .

Now check $(19.19)$ from the other side. Let $e = a \cdot 1 + ν \in A$ be idempotent, $ν \in J$ . Reducing modulo $J$ gives $a^{2} = a$ in $k$ , so $a \in {0, 1}$ . If $a = 0$ then $e = ν$ is both nilpotent and idempotent, so $e = 0$ . If $a = 1$ then $1 - e \in J$ is nilpotent and idempotent, so $e = 1$ . The only idempotents are trivial — exactly as $(19.19)$ predicts for an artinian local ring.

Two contrasts that pin down the hypotheses

Artinian, but with idempotents. $M_{2} (k)$ is artinian and $E_{11}$ is a nontrivial idempotent, so $M_{2} (k)$ is not local — consistent with $(19.19)$ , and with the fact that $M_{2} (k)$ has two maximal left ideals (indeed infinitely many).
Idempotent-free, but not artinian. $ℤ$ has no nontrivial idempotents and is not local: it has one maximal ideal for each prime. Dropping artinian from $(19.19)$ destroys the sufficiency direction immediately.
Local, but not artinian. $ℤ_{(p)}$ has unique maximal ideal $p ℤ_{(p)}$ and no nontrivial idempotents, but the chain $p ℤ_{(p)} ⊋ p^{2} ℤ_{(p)} ⊋ \dots$ never stabilises.

Process and Workflow

Is my ring $R \neq 0$ local?

It is right or left artinianSearch for idempotents. If the only ones are

0

and

1

R

is local by

(19.19)

; a single nontrivial idempotent settles it the other way.

Every non-unit is nilpotent

R

is local by

(19.3) (a)

, and

rad R

is nil.

$R$ sits inside a division ring $D$ with $d \in R$ or $d^{- 1} \in R$ for all $d \in D^{\times}$

R

is local by

(19.3) (b)

— the valuation-ring criterion.

None of the aboveCompute

rad R

and test whether

R / rad R

is a division ring. There is no shortcut in general.

Locate the radicalFor a finite-dimensional algebra this is a linear-algebra computation; for a valuation-type ring read it off from the value group.

Form the residue ringCompute

R / rad R

and ask whether it is a division ring. If it splits as a product,

R

is not local and the splitting produces idempotents.

Cross-check with unitsVerify

R ∖ U (R) = rad R

on a few elements. This catches an incorrectly computed radical fast.

Record the extra structureIf

rad R

is nilpotent, note that

R

is completely primary — several later theorems ask for exactly that.

Comparison and Classification

Locality against the three necessary conditions
Ring	Unique max. ideal	Dedekind-finite	No nontrivial $e$	Local?
Division ring $D$	yes	yes	yes	yes
$k [[x]]$ , $k [[x; σ]]$	yes	yes	yes	yes
$ℤ_{(p)}$ , $ℤ_{p}$	yes	yes	yes	yes
$M_{2} (k)$	yes (simple)	yes	no	no
$A_{1} (k)$ , $char k = 0$	yes (simple)	yes	yes	no
$ℤ$	no	yes	yes	no
$k \times k$	no	yes	no	no
Upper triangular $T_{2} (k)$	no	yes	no	no

Which criterion is usable in which setting
	Finite-dim. algebra	Commutative noetherian	Endomorphism ring	General ring
Unique maximal left ideal	partial	yes	no	yes
$R / rad R$ a division ring	yes	yes	partial	yes
Non-units closed under addition	yes	yes	yes	yes
Every non-unit nilpotent	yes	no	partial	no
No nontrivial idempotent	yes	no	yes	no

Which criterion is usable in which setting

yes means the criterion is both valid and practical there; part means valid but awkward to verify; no means it is not sufficient in that setting.

Relationship Map

The implications below are unconditional. Only the reverse of the last one needs a hypothesis, and that hypothesis is a chain condition.

R

localno nontrivial idempotent

R_{R}

indecomposable

$R$ is local — $R \neq 0$ , $R ∖ U (R) = rad R$
- always implies
  - unique maximal two-sided ideal (19.2)(a)
  - Dedekind-finite (19.2)(b)
  - only trivial idempotents (19.2)(c)
  - $R^{op}$ is local
  - $Z (R)$ is a local ring
- is implied by
  - every non-unit nilpotent, $R \neq 0$ (19.3)(a)
  - $R$ right artinian with no nontrivial idempotent (19.19)
  - $R = End (M)$ , $M$ indecomposable of finite length (19.17)
  - $R = k G$ , $char k = p$ , $G$ a finite $p$ -group (19.10)
- never implies
  - artinian — see $ℤ_{(p)}$
  - noetherian — rank-two valuation rings
  - commutative — see $k [[x; σ]]$

Rings with no nontrivial idempotentincludes all domains

Local rings

R / rad R

a division ring

Completely primary

rad R

nilpotent

Artinian locale.g.

k [t] / (t^{n})

k G

for a

p

-group

Division rings

rad R = 0

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

The engine of Krull–Schmidt

Azumaya's uniqueness theorem needs the summands to have local endomorphism rings. $(19.17)$ supplies that hypothesis for every module of finite length, which is why Krull–Schmidt works over artinian rings at all.

Modular representation theory

Blocks and $p$ -groups

$k G$ for a finite $p$ -group in characteristic $p$ is local; the block decomposition of a general $k G$ is precisely the decomposition of $1$ into primitive central idempotents, so finding idempotents is the practical form of failing to be local.

Computer algebra

Algebra recognition

GAP, Magma and Sage decide indecomposability of a finite-dimensional algebra by computing the radical and inspecting the semisimple quotient; a local algebra is the base case where the recursion stops.

Commutative geometry

Stalks and germs

Local rings of a variety at a point, and rings of germs of holomorphic functions, are the motivating commutative examples; the noncommutative theory keeps their idempotent-freeness while abandoning localisation.

The honest summary is that locality is a hypothesis-supplying property. Very few theorems are about local rings for their own sake; a great many are about arbitrary rings and become tractable because some auxiliary endomorphism ring turns out to be local.

Computational Notes

Computational notes cover algorithms, cost and library behaviour rather than manufacturing process.

For a finite-dimensional algebra $A$ over a field given by structure constants with $\dim_{k} A = n$ , deciding locality costs one radical computation ( $O (n^{3})$ field operations in characteristic $0$ via the trace form; Friedl–Rónyai in characteristic $p$ ) plus a check that $A / rad A$ is a division ring.
Searching for a nontrivial idempotent directly is the wrong algorithm: the naive approach solves a quadratic system in $n$ unknowns. Reduce modulo the radical first, split the semisimple quotient, then lift idempotents — lifting is always possible modulo a nil ideal.
Over a finite field, deciding whether $A / rad A$ is a division ring is effective because finite division rings are fields (Wedderburn's little theorem), so the test reduces to commutativity plus absence of zero divisors.
For infinitely generated or finitely presented rings, locality is not decidable in general; the word problem for finitely presented rings already is not.

Failure Modes and Common Mistakes

Do not assume $rad R$ is nilpotent for a general local ring; $k [[x]]$ and $ℤ_{(p)}$ have radicals with no nonzero nilpotent element at all. Completely primary is the term for the stronger condition.
Do not conflate indecomposable with strongly indecomposable: $ℤ$ as a $ℤ$ -module is indecomposable but $End (ℤ) ≅ ℤ$ is not local.
Do not apply the Fitting argument under only one chain condition; $(19.18)$ shows both ACC-only and DCC-only fail to give the full conclusion of $(19.17)$ .
Do not expect a local ring to be commutative, noetherian, or a domain — $k [[x; σ]]$ , rank-two valuation rings and $k [t] / (t^{2})$ respectively refute each.

Best Practices

Say which characterisation of local you are invoking; proofs that silently move between the ideal-theoretic and unit-theoretic forms are hard to audit.
When you use $(19.19)$ , record the side of the artinian hypothesis, even though the conclusion is symmetric.
Verify a claimed radical by checking $R ∖ U (R) = rad R$ on a handful of elements before building on it.
If your local ring came from $End (M)$ with $M$ of finite length, note that it is completely primary — that extra nilpotence is often the hypothesis a later step needs.
State nontrivial idempotent rather than idempotent whenever $0$ and $1$ are in scope; the ambiguity causes real errors in write-ups.

Quick Reference

Definition

R \neq 0

with a unique maximal left ideal

Residue form

R / rad R

is a division ring

Unit form

R ∖ U (R) = rad R

is an ideal

Sum test

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

b \in U (R)

IdempotentsOnly

0

and

1

; necessary, not sufficient

Artinian test

R

artinian: local

\Leftrightarrow

no nontrivial idempotent

Symmetry

R

local

\Leftrightarrow

R^{op}

local

Completely primaryLocal with

rad R

nilpotent

Numbered results used on this page
Reference	Statement	Hypotheses
(19.1)	Seven equivalent forms of local	$R \neq 0$
(19.2)	Unique max ideal, Dedekind-finite, no nontrivial $e$	$R$ local
(19.3)(a)	Every non-unit nilpotent $\Rightarrow$ local	$R \neq 0$
(19.16)	$M = \ker (f^{n}) \oplus im (f^{n})$	$M$ of finite length
(19.17)	$End (M)$ local, $(rad End M)^{n} = 0$	$M$ indecomposable, length $n$
(19.19)	Local $\Leftrightarrow$ no nontrivial idempotent	$R \neq 0$ right artinian

Frequently Asked Questions

Is a ring with no nontrivial idempotents automatically local?

No. Every domain — $ℤ$ , $k [x]$ , the Weyl algebra $A_{1} (k)$ — has only trivial idempotents, and none of these is local. The implication holds only in the presence of a one-sided chain condition, which is exactly the content of $(19.19)$ .

Why is the definition stated with left ideals if locality is symmetric?

A definition has to pick a side to be stated at all. Symmetry is a theorem: conditions (3) to (5 $''$ ) of $(19.1)$ mention only units and the radical, both of which are unchanged under passing to $R^{op}$ . Contrast this with primitivity or perfectness, which really are one-sided.

Must a local ring be artinian, or noetherian, or commutative?

None of the three. $ℤ_{(p)}$ and $k [[x]]$ are local and not artinian; a valuation ring with value group $ℚ$ is local and not noetherian; the twisted power series ring $k [[x; σ]]$ for a nontrivial automorphism $σ$ of a field $k$ is local and not commutative.

What is the difference between local and completely primary?

Completely primary means local with nilpotent maximal ideal. Every right artinian local ring is completely primary, and so is $k G$ for a finite $p$ -group in characteristic $p$ . But $k [[x]]$ is local with $rad = (x)$ , which is not even nil, so it is not completely primary.

Can a matrix ring over a local ring be local?

Only for $n = 1$ . For $n \geq 2$ the matrix unit $E_{11}$ is a nontrivial idempotent of $M_{n} (R)$ , so $(19.2) (c)$ rules it out. This shows locality is not preserved by Morita equivalence, even though $rad$ is.

Is there a ring satisfying all of (a), (b) and (c) of

(19.2)

that is still not local?

Yes. The first Weyl algebra $A_{1} (k)$ over a field of characteristic $0$ is a simple noetherian domain: simplicity gives a unique maximal ideal, being a domain gives Dedekind-finiteness and no nontrivial idempotents, but $rad A_{1} (k) = 0$ and $A_{1} (k)$ is not a division ring.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §19 (pp. 294–310), especially (19.1), (19.2), (19.17) and (19.19).
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992 (local rings, Fitting's lemma and semiperfect rings).
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988 (idempotents, Peirce decompositions and local rings).
A. Facchini, Module Theory: Endomorphism Rings and Direct Sum Decompositions in Some Classes of Modules, Progress in Mathematics 167, Birkhäuser, 1998.

AI Suggested Questions

Give a full proof that idempotents lift modulo a nil ideal, and explain why this makes $(19.19)$ a statement about the semisimple quotient.
Which local rings arise as endomorphism rings of indecomposable modules over a fixed artinian ring?
Construct a local ring whose maximal ideal is nil but not nilpotent.
How does $(19.19)$ interact with the block decomposition of a group algebra in characteristic $p$ ?
Why is the centre of a ring with a unique maximal two-sided ideal always local, and where does the argument use commutativity of the centre?
Compare local rings with semiperfect rings: what exactly does semiperfect add once idempotents are allowed to be nontrivial?
Is there an algorithm that, given structure constants for a finite-dimensional algebra over $ℚ$ , decides locality in polynomial time?

Local Rings and Idempotents

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Idempotents are decompositions

Why the non-units must be closed under addition

Local, completely primary, and the artinian world

Key Results

Proof Techniques and Method

Test membership by units

Split off with an idempotent

Fitting: kernel or image dies

Worked Example

A finite-dimensional algebra that is local

Two contrasts that pin down the hypotheses

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

The engine of Krull–Schmidt

Blocks and $p$ -groups

Algebra recognition

Stalks and germs

Computational Notes

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Idempotents are decompositions

Why the non-units must be closed under addition

Local, completely primary, and the artinian world

06Key Results

07Proof Techniques and Method

Test membership by units

Split off with an idempotent

Fitting: kernel or image dies

08Worked Example

A finite-dimensional algebra that is local

Two contrasts that pin down the hypotheses

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

The engine of Krull–Schmidt

Blocks and p-groups

Algebra recognition

Stalks and germs

13Computational Notes

14Failure Modes and Common Mistakes

15Best Practices

16Quick Reference

17Frequently Asked Questions

18Related KEVOS Topics

19References

20AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Blocks and $p$ -groups

Computational Notes

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions