Ring Theory Conventions and Notation for Noncommutative Rings

Q: If a left ideal is not an ideal, how do I get a quotient ring?

You do not. A left ideal a ⊆ R gives a quotient *left module* R/a, not a quotient ring, and that module is where the theory happens: a is maximal exactly when R/a is a simple module. Quotient rings need two-sided ideals, which is why rad R being two-sided is such a useful theorem.

Q: Is a simple ring the same as a semisimple ring with one factor?

Not in general. A *semisimple* ring is a finite product of matrix rings over division rings, so a semisimple ring that is simple is exactly M_n(D). But there are simple rings that are not semisimple: the Weyl algebra A_1(k) over a field of characteristic 0 is simple, noetherian and infinite-dimensional, with no minimal left ideals at all.

Q: Does an element with two distinct right inverses exist?

Yes, and then it has infinitely many. If ab = 1 and a is not left-invertible, the elements b + (1 - ba)a^n for n ≥ 0 are pairwise distinct right inverses of a. This is Kaplansky's observation, Exercise 1.14 in Lam, and it shows the failure of Dedekind-finiteness is never a near miss.

Q: Why is the zero ring allowed at all?

So that quotients are always defined: R/R has to be a ring. The convention costs nothing because every interesting definition — simple, domain, division ring — carries the clause R ≠ 0 explicitly.

Executive Summary

Noncommutative ring theory is commutative algebra with the safety rail removed. Almost every notion that had one form in the commutative world now has two — a left form and a right form — and the two are genuinely different. This page fixes the conventions that the rest of the collection runs on and isolates the handful of places where sidedness first bites.

Three conventions do most of the work: every ring has an identity, a subring contains that identity, and the unqualified word ideal always means two-sided. Everything one-sided is labelled.

$1 \in R$ Always assumed

2-sidedMeaning of "ideal"

$U (R)$ Unit group notation

$R^{op}$ Side-swapping device

Overview

A ring in this collection is an associative ring with an identity element $1$ , not assumed commutative and not assumed nonzero unless stated. The zero ring, in which $1 = 0$ , is permitted but is excluded from every definition that demands nontriviality — simple rings, domains and division rings are all required to be nonzero.

The reason to insist on an identity is not fastidiousness. Without it, maximal left ideals need not exist, modules need not be unital, and the correspondence between a ring and its category of modules degrades. Radical theory for rings without identity is a real subject, but it is a different one.

Rings are written $R$ , $S$ , $A$ ; the base of an algebra is $k$ , which may itself be noncommutative unless we say otherwise. Modules are covered in Modules over Noncommutative Rings: Left, Right and Bimodules; the symbol inventory is collected in Notation and Symbols Reference.

Learning Objectives

Recall the standing conventions: identity element, subrings containing $1$ , unital modules, unqualified ideal meaning two-sided.
Prove that a nonzero ring $R$ is simple exactly when every $a \neq 0$ admits an equation $\sum b_{i} a c_{i} = 1$ .
Exhibit an element that is a left zero-divisor but not a right zero-divisor.
Distinguish left-invertible, right-invertible and invertible, and state the uniqueness of a two-sided inverse.
Define Dedekind-finiteness and give a ring that fails it.
Verify that a nonzero ring is a division ring as soon as every nonzero element is right-invertible.

Definitions

Definition§1Standing conventions

Ring: associative, with identity $1$ , not necessarily commutative.
Subring: a subring of $R$ contains the identity of $R$ . So $2 ℤ$ is not a subring of $ℤ$ .
Ideal: unqualified, means two-sided. Left ideals and right ideals are always named.
Homomorphism: a ring homomorphism carries $1$ to $1$ .
Module: unital, so $1 \cdot m = m$ .

$U (R)$: The group of units of $R$ , that is the two-sided invertible elements. Also written $R^{\times}$ .
$Z (R)$: The centre, ${z \in R : z r = r z for all r \in R}$ ; a commutative subring.
Left-invertible: $a$ with $b a = 1$ for some $b$ . Right-invertible means $a b = 1$ for some $b$ .
Left zero-divisor: A nonzero $a$ with $a b = 0$ for some nonzero $b$ ; right zero-divisor is the mirror notion.
Domain: A nonzero ring in which $a b = 0$ forces $a = 0$ or $b = 0$ . Not assumed commutative.
Reduced: $a^{2} = 0 \Rightarrow a = 0$ . Every domain is reduced; a direct product of domains is reduced but is not a domain unless one factor is.
Simple ring: $R \neq 0$ whose only ideals are $0$ and $R$ . Simple does not mean semisimple and does not imply any chain condition.
Division ring: $R \neq 0$ with $U (R) = R ∖ {0}$ . Also called a skew field.

A commutative simple ring is a field; a commutative domain is an integral domain. The noncommutative classes are strictly larger and much harder to describe.

Core Concepts

Quotients and the universal property

Only two-sided ideals can be quotiented by. If $I ⊴ R$ then $\bar{R} = R / I$ carries a well-defined multiplication and the projection $π : R \to \bar{R}$ is a surjective ring homomorphism with kernel $I$ . Its universal property is the reason ideals matter at all: any homomorphism $ϕ : R \to R'$ with $ϕ (I) = 0$ factors uniquely as $ϕ = \bar{ϕ} \circ π$ .

ϕ : R ⟶ R', ϕ (I) = 0 ⟹ \exists! \bar{ϕ} : R / I ⟶ R' with \bar{ϕ} \circ π = ϕ .

(1.0)

A left ideal has no quotient ring; it only has a quotient module.

Three layers of invertibility

In a commutative ring there is one notion of inverse. Here there are three, and they separate. If $a$ has a right inverse $b$ and a left inverse $b'$ then the two coincide:

b' = b' (a b) = (b' a) b = b,

(1.0a)

so a two-sided inverse, when it exists, is unique — which is what licenses the notation $a^{- 1}$ .

The gap between right-invertible and invertible is exactly the failure of Dedekind-finiteness, and it is a real gap: infinite-dimensional endomorphism rings supply counterexamples. See the shift operators in the worked example below.

Zero-divisors split by side

An element can annihilate something on its right without annihilating anything on its left. The standard witness is a triangular ring built from an abelian group with torsion, and it is worth carrying around: many "obvious" symmetric statements about zero-divisors die on it.

invertibleleft-invertiblenot a right zero-divisor

The second implication is Exercise 1.4(b) in Lam, read backwards: an element that is left-invertible and not a right zero-divisor is in fact a unit. Neither arrow reverses in general.

Key Results

Proposition§1Element criterion for simplicity

Let $R$ be a nonzero ring. Then $R$ is simple if and only if for every $a \neq 0$ in $R$ there are finitely many elements $b_{1}, \dots, b_{n}$ and $c_{1}, \dots, c_{n}$ of $R$ with $\sum_{i = 1}^{n} b_{i} a c_{i} = 1$ .

Proof

The ideal generated by $a$ is precisely the set $R a R$ of finite sums $\sum b_{i} a c_{i}$ : that set is an additive subgroup, is closed under left and right multiplication, and contains $a = 1 \cdot a \cdot 1$ . If $R$ is simple and $a \neq 0$ then $R a R$ is a nonzero ideal, hence equals $R$ , hence contains $1$ , which is the stated equation.

Conversely, suppose every nonzero $a$ admits such an equation and let $I \neq 0$ be an ideal. Choose $a \in I$ with $a \neq 0$ . Then $1 = \sum b_{i} a c_{i} \in I$ because $I$ absorbs multiplication on both sides, so $I = R$ .

Corollary§1The commutative case collapses

A commutative ring $R \neq 0$ is simple if and only if it is a field. Indeed the criterion reduces to: for every $a \neq 0$ there is $b$ with $b a = 1$ .

Proposition§1One-sided test for a division ring

Let $R \neq 0$ . The following are equivalent: (1) $R$ is a division ring; (2) every $a \neq 0$ in $R$ is right-invertible; (3) the only right ideals of $R$ are ${0}$ and $R$ . The same holds with right replaced by left throughout.

Proof

(1) $\Rightarrow$ (2) is trivial. For (2) $\Rightarrow$ (1), the hypothesis says $R ∖ {0}$ is a set closed under multiplication — if $a b = 0$ with $a, b \neq 0$ , pick $c$ with $b c = 1$ , then $0 = (a b) c = a$ , a contradiction — in which every element has a right inverse and which contains $1$ . A semigroup with identity in which every element is right-invertible is a group: given $a \neq 0$ choose $b$ with $a b = 1$ , then choose $c$ with $b c = 1$ ; now $a = a (b c) = (a b) c = c$ , so $b a = b c = 1$ as well. Hence $a \in U (R)$ .

(1) $\Rightarrow$ (3): a nonzero right ideal contains some $a \neq 0$ , hence contains $a a^{- 1} = 1$ , hence is $R$ . (3) $\Rightarrow$ (2): for $a \neq 0$ the right ideal $a R$ is nonzero, so $a R = R$ and $1 = a b$ for some $b$ .

PropositionEx. 1.6Jacobson's lemma on

1 - a b

and

1 - b a

Let $a, b \in R$ . If $1 - b a$ is left-invertible then so is $1 - a b$ ; if $1 - b a$ is invertible then so is $1 - a b$ , and an explicit inverse is

(1 - a b)^{- 1} = 1 + a (1 - b a)^{- 1} b .

(1.0b)

Proof

Suppose $u (1 - b a) = 1$ . Then $b = u (1 - b a) b = u b (1 - a b)$ , so $a b = a u b (1 - a b)$ and therefore

1 = (1 - a b) + a b = (1 - a b) + a u b (1 - a b) = (1 + a u b) (1 - a b),

which exhibits a left inverse for $1 - a b$ . If $u = (1 - b a)^{- 1}$ is a genuine two-sided inverse, the symmetric computation with the roles of the factors exchanged gives $(1 - a b) (1 + a u b) = 1$ as well, so $1 + a (1 - b a)^{- 1} b$ is the inverse. Nothing here uses commutativity, and the identity is the algebraic shadow of the fact that $a b$ and $b a$ have the same nonzero spectrum.

Remark§1Dedekind-finiteness is a hypothesis, not a fact

A ring is Dedekind-finite if $a b = 1$ implies $b a = 1$ . Domains are Dedekind-finite: from $a b = 1$ we get $a (b a - 1) = (a b) a - a = 0$ and $a \neq 0$ , so $b a = 1$ . Left noetherian rings are Dedekind-finite (Lam, Exercise 1.12), as are algebraic algebras over a field. But the property genuinely fails in general — see the worked example.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Generate an ideal, then find $1$ in it

To prove a ring is simple, or that an ideal is everything, produce an equation $\sum b_{i} a c_{i} = 1$ . To prove an ideal is proper, find an invariant it cannot reach.

Move 2

Upgrade one-sided to two-sided

A right inverse of a right inverse is the original element. This two-step trick converts every element is right-invertible into every element is a unit, and it is the whole content of the division-ring test.

Move 3

Specialise to a concrete ring

To show that two words in a free construction are different, map the free ring into a small explicit ring where they visibly differ. Used constantly in Free Rings and Rings Defined by Generators and Relations.

Move 2 deserves emphasis because it is the reason so many one-sided-looking definitions turn out to be side-neutral. It works because the identity is available; it is exactly what fails for rings without $1$ .

Worked Example

A left zero-divisor that is not a right zero-divisor

Let $M = ℤ / 2 ℤ$ , regarded as a $(ℤ, ℤ)$ -bimodule, and form the triangular ring

R = (\begin{matrix} ℤ & ℤ / 2 ℤ \\ 0 & ℤ \end{matrix}) = {(\begin{matrix} x & y \\ 0 & z \end{matrix}) : x, z \in ℤ, y \in ℤ / 2 ℤ},

(E.1)

with formal matrix multiplication; the $(1, 2)$ entry is computed modulo $2$ .

Set

a = (\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}), b = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) .

(E.2)

Then $a b = (\begin{matrix} 0 & 2 \cdot 1 \\ 0 & 0 \end{matrix}) = 0$ because $2 \cdot 1 = 0$ in $ℤ / 2 ℤ$ , so $a$ is a left zero-divisor. But $a$ is not a right zero-divisor: if $c = (\begin{matrix} x & y \\ 0 & z \end{matrix})$ satisfies

c a = (\begin{matrix} x & y \\ 0 & z \end{matrix}) (\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}) = (\begin{matrix} 2 x & y \\ 0 & z \end{matrix}) = 0,

(E.3)

then $2 x = 0$ in $ℤ$ forces $x = 0$ , and $y = 0$ , $z = 0$ , so $c = 0$ . Note by contrast that $b^{2} = 0$ , so $b$ is a zero-divisor on both sides. The asymmetry is entirely caused by $2$ acting injectively on $ℤ$ from one side and as zero on $ℤ / 2 ℤ$ from the other.

A ring that is not Dedekind-finite

Let $k$ be a field, $V = ⨁_{i \geq 1} k e_{i}$ a countably infinite-dimensional $k$ -vector space, and $R = {End}_{k} (V)$ with composition as multiplication. Define $a, b \in R$ on the basis by

b (e_{i}) = e_{i + 1} (i \geq 1), a (e_{1}) = 0, a (e_{i}) = e_{i - 1} (i \geq 2) .

(E.4)

Then $(a b) (e_{i}) = a (e_{i + 1}) = e_{i}$ for every $i$ , so $a b = 1$ . But $(b a) (e_{1}) = b (0) = 0 \neq e_{1}$ , so $b a \neq 1$ . Hence $a$ is right-invertible without being invertible, and $R$ is not Dedekind-finite.

Comparison and Classification

One notion in the commutative world, two here
Commutative notion	Left version	Right version	Do they agree?
Ideal	Left ideal	Right ideal	No — differ in almost every noncommutative ring
Inverse	$b a = 1$	$a b = 1$	Not in general; agree iff $R$ is Dedekind-finite
Zero-divisor	$a b = 0$ , $b \neq 0$	$c a = 0$ , $c \neq 0$	No — see the worked example
Noetherian	ACC on left ideals	ACC on right ideals	No — independent conditions
Artinian	DCC on left ideals	DCC on right ideals	No — independent conditions
Simple	no proper nonzero ideal	same	Yes — the definition is two-sided already
Division ring	every nonzero element left-invertible	every nonzero element right-invertible	Yes — either one suffices

Which classes of rings satisfy which conditions
	Domain	Reduced	Dedekind-finite	Simple
Division ring $D$	yes	yes	yes	yes
$M_{n} (D)$ , $n \geq 2$	no	no	yes	yes
$ℤ$	yes	yes	yes	no
$ℤ / 6 ℤ$	no	yes	yes	no
$ℤ / 4 ℤ$	no	no	yes	no
Upper triangular $T_{2} (k)$	no	no	yes	no
${End}_{k} (V)$ , $\dim_{k} V = \infty$	no	no	no	no
Weyl algebra $A_{1} (k)$ , $char k = 0$	yes	yes	yes	yes

Which classes of rings satisfy which conditions

Relationship Map

The classes defined here nest as follows. Every containment is strict, and each is witnessed by a ring in the table above.

Rings with identitythe ambient class of this collection

Dedekind-finite

a b = 1 \Rightarrow b a = 1

; includes all left noetherian rings

Reducedno nonzero nilpotents

Domainno zero-divisors on either side

Division ring

U (R) = R ∖ {0}

Fieldcommutative division ring

Simple rings — only ideals are $0$ and $R$
- include
  - every division ring
  - $M_{n} (D)$ for $D$ a division ring
  - the Weyl algebra $A_{1} (k)$ in characteristic $0$
- do not include
  - $ℤ$ and every commutative ring that is not a field
  - upper triangular matrix rings of size $\geq 2$
- are not required to be
  - artinian or noetherian
  - semisimple
  - finite-dimensional over anything

Standards and Notation

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

Unit group

U (R)

in this collection;

R^{\times}

and

R^{*}

are common variants

Centre

Z (R)

; occasionally

Cent (R)

Matrix ring

M_{n} (R)

; the blackboard-bold variant is avoided here

Opposite ring

R^{op}

; some authors write

R^{\circ}

Ideal relation

I ⊴ R

for a two-sided ideal;

I \subseteq_{ℓ} R

appears for left ideals in some texts

SymbolsISO 80000-2 fixes

ℤ

ℚ

ℝ

ℂ

; it says nothing about ring-theoretic notation

SystemsGAP, Magma, Sage and Macaulay2 all default to ideal meaning two-sided, and require an explicit side for one-sided ideals

Failure Modes and Common Mistakes

Do not assume a left ideal is an ideal. In $M_{2} (k)$ the columns are minimal left ideals and there are no proper nonzero two-sided ideals at all.
Do not transport a commutative-algebra theorem by analogy. Every prime ideal is contained in a maximal ideal survives; the intersection of the minimal primes is the nilradical needs care about which nilradical is meant.
Do not confuse reduced with domain. $ℤ \times ℤ$ is reduced and full of zero-divisors.
Do not read $\sum b_{i} a c_{i} = 1$ as $b a c = 1$ . Simplicity generally needs a genuine sum; a single term suffices only in special cases.

Best Practices

Label every one-sided hypothesis and every one-sided conclusion explicitly, even when you expect symmetry.
When a result is proved on one side, record whether the proof dualises verbatim; if it does, invoke $R^{op}$ once rather than rewriting the argument.
Keep a small stock of standing counterexamples: a triangular ring for asymmetry, ${End}_{k} (V)$ with $\dim V = \infty$ for one-sided inverses, $M_{n} (D)$ for simple-but-not-a-domain.
State whether $k$ in $k$ -algebra is assumed commutative; several constructions in §1 allow a noncommutative base, and the word algebra is then a mild abuse.
Before claiming a ring is a division ring, check only one side — the one-sided test is a genuine saving.

Quick Reference

Ringassociative, with

1

, not necessarily commutative

Subringcontains the ambient

1

Idealtwo-sided unless labelled left or right

Simple

R \neq 0

and for all

a \neq 0

\sum b_{i} a c_{i} = 1

is solvable

Domain

R \neq 0

and

a b = 0 \Rightarrow a = 0

b = 0

Division ring

R \neq 0

and

U (R) = R ∖ {0}

Dedekind-finite

a b = 1 \Rightarrow b a = 1

Unit identity

(1 - a b)^{- 1} = 1 + a (1 - b a)^{- 1} b

Standing counterexamples worth memorising
Phenomenon	Witness	Why it works
Left but not right zero-divisor	$(\begin{matrix} ℤ & ℤ / 2 ℤ \\ 0 & ℤ \end{matrix})$ , $a = diag (2, 1)$	$2$ kills $ℤ / 2 ℤ$ on one side only
Right-invertible, not invertible	shift operators in ${End}_{k} (V)$ , $\dim_{k} V = \infty$	one-sided shift has a kernel but no cokernel
Simple but not artinian	Weyl algebra $A_{1} (k)$ , $char k = 0$	a simple noetherian domain of infinite dimension
Reduced but not a domain	$ℤ \times ℤ$	idempotents give zero-divisors without nilpotents
Left noetherian, not right	$(\begin{matrix} ℚ & ℚ \\ 0 & ℤ \end{matrix})$	$ℚ$ is simple as a left $ℚ$ -module, huge as a right $ℤ$ -module

Frequently Asked Questions

Why insist that every ring has an identity?

Because almost everything downstream depends on it. Maximal left ideals exist by Zorn's Lemma only when $1$ is available; modules can be assumed unital; the free constructions have clean universal properties; and the correspondence between $R$ and its module category behaves. Rings without identity are studied, and radical theory for them is well developed, but the theorems are different and the proofs are longer.

If a left ideal is not an ideal, how do I get a quotient ring?

You do not. A left ideal $𝔞 \subseteq R$ gives a quotient left module $R / 𝔞$ , not a quotient ring, and that module is where the theory happens: $𝔞$ is maximal exactly when $R / 𝔞$ is a simple module. Quotient rings need two-sided ideals, which is why $rad R$ being two-sided is such a useful theorem.

Is a simple ring the same as a semisimple ring with one factor?

Not in general. A semisimple ring is a finite product of matrix rings over division rings, so a semisimple ring that is simple is exactly $M_{n} (D)$ . But there are simple rings that are not semisimple: the Weyl algebra $A_{1} (k)$ over a field of characteristic $0$ is simple, noetherian and infinite-dimensional, with no minimal left ideals at all.

How do I tell quickly whether a ring is Dedekind-finite?

Four sufficient conditions cover most cases: the ring is a domain; the ring is left (or right) noetherian; the ring is an algebraic algebra over a field; the ring is finite. Failure requires something genuinely infinite in a one-directional way, and the canonical failure is ${End}_{k} (V)$ with $\dim_{k} V$ infinite.

Does an element with two distinct right inverses exist?

Yes, and then it has infinitely many. If $a b = 1$ and $a$ is not left-invertible, the elements $b + (1 - b a) a^{n}$ for $n \geq 0$ are pairwise distinct right inverses of $a$ . This is Kaplansky's observation, Exercise 1.14 in Lam, and it shows the failure of Dedekind-finiteness is never a near miss.

Why is the zero ring allowed at all?

So that quotients are always defined: $R / R$ has to be a ring. The convention costs nothing because every interesting definition — simple, domain, division ring — carries the clause $R \neq 0$ explicitly.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §1 (pp. 1–25).
T. Y. Lam, Exercises in Classical Ring Theory, 2nd edition, Problem Books in Mathematics, Springer-Verlag, 2003, Chapter 1.
N. Jacobson, Basic Algebra II, 2nd edition, W. H. Freeman, 1989, Chapter 2.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §1–§2.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 1.

AI Suggested Questions

Give an example of a ring in which a left ideal is a two-sided ideal but its square is not.
Prove Kaplansky's result that an element with more than one right inverse has infinitely many.
Which finiteness conditions on a ring imply Dedekind-finiteness, and which do not?
Construct a simple ring that is neither left nor right noetherian.
How does the definition of the Jacobson radical change for rings without an identity?
Show that a left artinian domain must be a division ring.
What is the smallest noncommutative ring with identity, and why is every ring of order $p^{2}$ commutative?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Quotients and the universal property

Three layers of invertibility

Zero-divisors split by side

06Key Results

07Proof Techniques and Method

Generate an ideal, then find 1 in it

Upgrade one-sided to two-sided

Specialise to a concrete ring

08Worked Example

A left zero-divisor that is not a right zero-divisor

A ring that is not Dedekind-finite

09Comparison and Classification

10Relationship Map

11Standards and Notation

12Failure Modes and Common Mistakes

13Best Practices

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

Generate an ideal, then find $1$ in it

Worked Example

Comparison and Classification

Relationship Map

Standards and Notation

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions