Left–Right Symmetry: What Transfers and What Does Not

Q: If left modules over R are right modules over R^op, why is anything asymmetric?

Because that dictionary relates R to a *different* ring. It guarantees that every theorem has a mirror image, not that a given ring satisfies both a property and its mirror. Asymmetry means finding R with P(R) true and P(R^op) false, and such rings exist for the chain conditions, primitivity and perfectness.

Q: Why is the Jacobson radical symmetric when its definition is not?

Because (4.3) replaces the definition with a condition mentioning only units: y ∈ rad R if and only if 1 - xyz is a unit for all x, z. Invertibility is a two-sided notion and the set xyz is unchanged when the product is reversed, so the condition reads identically in R^op.

Q: What about group rings — do the asymmetries ever appear there?

Not for a commutative coefficient ring. The map sending g to g^-1 and extending additively is an anti-automorphism of kG, so kG ≅ (kG)^op and every one-sided property of kG holds on both sides or neither. Any asymmetric behaviour in group rings must come from a noncommutative coefficient ring or a twisted product.

Executive Summary

Every definition in this subject has to pick a side, and the reader must then know whether the choice mattered. Roughly: properties defined through two-sided ideals, units, or the whole module category are symmetric; properties defined through chains of one-sided ideals or through a single faithful module are not.

The technical device is the opposite ring $R^{op}$ . A property $P$ is left-right symmetric precisely when $P (R) \Leftrightarrow P (R^{op})$ for all $R$ . This turns every symmetry question into a computation and every asymmetry claim into a demand for a ring with $R not ≅ R^{op}$ in the relevant respect.

Symmetric

rad R

, semisimplicity

AsymmetricChain conditions, primitivity, perfectness

1965Bergman's primitivity example

$R^{op}$ The transfer device

Overview

The opposite ring $R^{op}$ has the same underlying additive group as $R$ and the multiplication $a \cdot_{op} b = b a$ . It satisfies $(R^{op})^{op} = R$ , and the fundamental dictionary is

{left R -modules} = {right R^{op} -modules}, {left ideals of R} = {right ideals of R^{op}} .

(S.1)

An equality of categories, not merely an equivalence — the module structures are literally the same maps read backwards.

Consequently every theorem proved for left modules is automatically a theorem for right modules, applied to $R^{op}$ . What this does not do is prove that a property of $R$ is equivalent to its mirror image: that requires $R$ and $R^{op}$ to satisfy the property simultaneously, which is a statement about $R$ , not a formality.

Details of the construction are on Opposite Rings and Left–Right Duality; the individual asymmetric cases have their own pages, listed under Comparison.

Learning Objectives

Define $R^{op}$ and state the left-right module dictionary it induces.
Prove that $rad R = rad R^{op}$ as subsets of $R$ .
Prove that a left semisimple ring is right semisimple, using $M_{n} (D)^{op} ≅ M_{n} (D^{op})$ .
Name a ring that is left artinian and not right artinian, and one that is left primitive and not right primitive.
Explain why the existence of an anti-automorphism makes every one-sided distinction vacuous for that ring.
Decide, for a new property, whether to expect symmetry.

Definitions

DefinitionOpposite ring

For a ring $R$ , the opposite ring $R^{op}$ has the same additive group and multiplication $a \cdot_{op} b : = b a$ . It is a ring with the same identity element, and $R$ is commutative if and only if $R = R^{op}$ as rings.

Symmetric property: $P (R) \Leftrightarrow P (R^{op})$ for every ring $R$ . Equivalently, the left and right versions of $P$ agree.
Self-opposite ring: A ring admitting an anti-automorphism, so that $R ≅ R^{op}$ . For such a ring every one-sided property holds on both sides or on neither.
$M_{n} (D)^{op}$: Isomorphic to $M_{n} (D^{op})$ via matrix transposition.
One-sided witness: A ring $R$ possessing a property whose opposite does not; the certificate that the property is asymmetric.

Note the asymmetry of the asymmetry: proving a property symmetric is a theorem about all rings, while proving it asymmetric requires only one ring. The second is often much harder, because the required ring must be constructed.

Core Concepts

Three sources of symmetry

Unit-theoretic descriptions. Invertibility is a two-sided notion, so any property expressible through $U (R)$ transfers. This is why $rad R$ is symmetric: $(4.3)$ recharacterises it as ${y : 1 - x y z \in U (R) for all x, z}$ .
Two-sided ideal descriptions. Prime, semiprime, simple, and all four radicals are defined by conditions on two-sided ideals, which are the same objects in $R$ and $R^{op}$ .
Classification. Once a class is classified by a list of data closed under $op$ — as semisimple rings are, by $\prod M_{n_{i}} (D_{i})$ — symmetry follows from the classification rather than from the definition.

Two sources of asymmetry

Chains of one-sided ideals. The left ideals of $R$ are the right ideals of $R^{op}$ , so ACC on left ideals of $R$ is ACC on right ideals of $R^{op}$ — a different condition. Triangular rings realise every combination.
A single distinguished module. Left primitivity asks for one faithful simple left module. Nothing relates the simple left modules of $R$ to its simple right modules, and Bergman's 1965 construction shows the two demands are genuinely different.

Definition mentions one sideRecharacterise via units or two-sided idealsSymmetry provedor: build

R

with

R^{op}

failing it

Key Results

Theorem(4.3), (4.4)The Jacobson radical is symmetric

For any ring $R$ , the intersection of the maximal left ideals of $R$ equals the intersection of the maximal right ideals of $R$ . Equivalently, $rad (R^{op}) = rad (R)$ as subsets of the common additive group.

Proof

By $(4.3)$ , for $y \in R$ the condition $y \in rad R$ is equivalent to

1 - x y z \in U (R) for all x, z \in R .

Now read this condition in $R^{op}$ . The set of units is the same subset in both rings, since $a b = b a = 1$ is a condition unchanged by reversing the product. And $x \cdot_{op} y \cdot_{op} z = z y x$ , so as $x$ and $z$ range over $R$ the element $x \cdot_{op} y \cdot_{op} z$ ranges over exactly the same set as $x y z$ does. Therefore $y$ satisfies the displayed condition in $R$ if and only if it satisfies it in $R^{op}$ , and $rad R = rad R^{op}$ . Since $rad R^{op}$ is the intersection of the maximal left ideals of $R^{op}$ , which are the maximal right ideals of $R$ , the theorem follows.

Theorem(3.5)Semisimplicity is symmetric

A ring $R$ is left semisimple if and only if it is right semisimple. Accordingly semisimple may be used without qualification.

Proof

Suppose $R$ is left semisimple. By the Wedderburn–Artin Theorem $(3.5)$ there are division rings $D_{1}, \dots, D_{r}$ and integers $n_{i} \geq 1$ with $R ≅ M_{n_{1}} (D_{1}) \times \dots \times M_{n_{r}} (D_{r})$ . Passing to opposites commutes with finite products, so

R^{op} ≅ M_{n_{1}} (D_{1})^{op} \times \dots \times M_{n_{r}} (D_{r})^{op} .

Transposition $A \mapsto A^{T}$ is an isomorphism $M_{n} (D)^{op} \to M_{n} (D^{op})$ : the $(i, j)$ entry of the product of $A^{T}$ and $B^{T}$ computed in $M_{n} (D^{op})$ is $\sum_{k} B_{j k} A_{k i} = (B A)_{j i}$ , which is the $(i, j)$ entry of $(B A)^{T}$ , and $B A$ is precisely the product of $A$ and $B$ in $M_{n} (D)^{op}$ . Each $D_{i}^{op}$ is again a division ring, since invertibility is unchanged. Hence $R^{op}$ is a finite product of matrix rings over division rings and is therefore left semisimple; that is, $R$ is right semisimple. The converse follows by applying the same argument to $R^{op}$ .

Counterexample(1.24)Chain conditions are asymmetric

Let $S \subseteq R$ be fields with $\dim_{S} R$ infinite — for instance $ℚ \subseteq ℝ$ — and let $A = (\begin{matrix} R & R \\ 0 & S \end{matrix})$ . Then $A$ is left noetherian and left artinian, and neither right noetherian nor right artinian.

Counterexample(1.25)Skew polynomials

Let $σ$ be an endomorphism of a division ring $k$ that is not surjective. Then $R = k [x; σ]$ , with $x a = σ (a) x$ , is a principal left ideal domain — in particular left noetherian — and is not right noetherian.

Remark(11.2)Primitivity and perfectness

Semiprimitivity is symmetric because $rad R$ is, but primitivity is not: Bergman constructed a left primitive ring that is not right primitive in 1965, and further examples are due to Jategaonkar. Perfectness is likewise asymmetric: Lam's example $(23.22)$ — the ring $k \cdot 1 + J$ where $J$ consists of the infinite matrices over $k$ with finitely many nonzero entries, all strictly above the diagonal — is right perfect and not left perfect, because $J$ is right T-nilpotent and not left T-nilpotent.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Recharacterise through units

Any statement mentioning only $U (R)$ transfers verbatim, because $U (R^{op}) = U (R)$ as a subset. This is the entire proof for $rad R$ .

Move 2

Transfer through the classification

When the class is classified, check that the classifying data is closed under $op$ . Semisimplicity is symmetric only because $M_{n} (D)^{op} ≅ M_{n} (D^{op})$ .

Move 3

Build asymmetry into a triangular ring

In $(\begin{matrix} R & M \\ 0 & S \end{matrix})$ the bimodule $M$ is seen from the left over $R$ and from the right over $S$ . Choosing $M$ finite-dimensional on one side and infinite on the other manufactures any desired asymmetry.

Move 4

Use a non-surjective endomorphism

$k [x; σ]$ inherits a one-sided division algorithm only on the side where $σ$ can be undone. Non-surjectivity of $σ$ is exactly the asymmetry.

Move 5

Look for an involution first

Before hunting for an asymmetric example, check that the candidate ring has no anti-automorphism. Group rings and matrix rings over commutative rings are immediately disqualified.

Move 6

Impose a chain condition to collapse

Under the DCC most asymmetries vanish: for left artinian rings, prime, left primitive, right primitive and simple all coincide $(11.7)$ .

Worked Example

Computing the opposite of a triangular ring

Let $A = (\begin{matrix} ℝ & ℝ \\ 0 & ℚ \end{matrix})$ , the ring of $(1.24)$ with $S = ℚ$ , $R = ℝ$ and the bimodule $M = ℝ$ , an $(ℝ, ℚ)$ -bimodule. Reversing multiplication and relabelling the corners gives

A^{op} ≅ (\begin{matrix} ℚ & ℝ \\ 0 & ℝ \end{matrix}),

(E.1)

The same bimodule $ℝ$ , now viewed as a $(ℚ, ℝ)$ -bimodule: the sides have swapped.

Now apply the triangular criterion $(1.22)$ to $A^{op}$ on the left. It requires $ℚ$ and $ℝ$ to be left artinian — true, both are fields — and requires $M = ℝ$ to be artinian as a **left $ℚ$ -module**. But $ℝ$ is an infinite-dimensional $ℚ$ -vector space, so it has no DCC on subspaces. Hence $A^{op}$ is not left artinian, which is exactly the statement that $A$ is not right artinian.

The radical, computed twice

Let $N = (\begin{matrix} 0 & ℝ \\ 0 & 0 \end{matrix})$ , the strictly upper triangular part of $A$ . Then $N^{2} = 0$ and $A / N ≅ ℝ \times ℚ$ , a product of fields, which is semisimple with zero radical. So $N \subseteq rad A$ because $N$ is nilpotent, and $rad A \subseteq N$ because the image of $rad A$ in $A / N$ lies in $rad (A / N) = 0$ . Hence

rad A = N = rad A^{op},

(E.2)

Confirming $(4.4)$ on a ring whose chain conditions are as asymmetric as possible.

Verify the unit criterion by hand: $1 + (\begin{matrix} 0 & b \\ 0 & 0 \end{matrix}) = (\begin{matrix} 1 & b \\ 0 & 1 \end{matrix})$ has inverse $(\begin{matrix} 1 & - b \\ 0 & 1 \end{matrix})$ in $A$ , and the same two matrices are mutually inverse in $A^{op}$ since the pair of equations is symmetric. The radical does not notice the side; the chain conditions do.

Process and Workflow

You have a one-sided hypothesis. Does the side matter?

The hypothesis mentions only units or two-sided idealsIt is symmetric. Prove it by reading the condition in

R^{op}

and observing that nothing changes.

The ring has an anti-automorphismThe side does not matter for this ring, whatever the property. Check for a transpose, an involution, or the inversion map on a group ring.

The ring is left artinianMost distinctions collapse: primitivity, primeness and simplicity coincide

(11.7)

, and both chain conditions hold on the left.

The hypothesis is a chain conditionThe side matters. Verify it on the side you need and do not transport it.

The hypothesis is the existence of a faithful simple moduleThe side matters. Bergman's example shows the two versions are inequivalent.

Write the property as a formulaMake explicit every quantifier over left ideals, right modules or elements.

Substitute $a \cdot_{op} b = b a$ Rewrite the formula in

R^{op}

. If it is literally unchanged, the property is symmetric.

If it changes, look for a collapse hypothesisA chain condition, an anti-automorphism, or commutativity may make the two versions agree in your setting.

Otherwise, demand a witnessAssume the two versions differ only when a published example says so; asymmetry claims without a witness are usually false.

Comparison and Classification

The ledger
Property	Symmetric?	Reason or witness
$rad R$	Yes	Unit criterion $(4.3)$ is invariant under $op$
Semiprimitive	Yes	$rad R = 0$ is a symmetric condition
Semisimple	Yes	Wedderburn data closed under $op$ $(3.5)$
Simple	Yes	Defined by two-sided ideals
Prime, semiprime	Yes	Defined by two-sided ideals; ${Nil}_{*} R$ symmetric
${Nil}^{*} R$ , Levitzki radical	Yes	Sums of nil, respectively locally nilpotent, two-sided ideals
Local, semilocal, semiperfect	Yes	Conditions on $rad R$ , on $R / rad R$ and on lifting
Semiprimary	Yes	Nilpotence of an ideal is symmetric
Von Neumann regular	Yes	$a = a x a$ is unchanged by reversing the product
Dedekind finite	Yes	$a b = 1 \Rightarrow b a = 1$ is self-mirroring
Noetherian	No	$k [x; σ]$ with $σ$ non-surjective $(1.25)$ ; Dieudonné's ring $(1.26)$
Artinian	No	$(\begin{matrix} ℝ & ℝ \\ 0 & ℚ \end{matrix})$ $(1.24)$
Primitive	No	Bergman's 1965 ring; further examples by Jategaonkar
Perfect	No	Lam's infinite triangular ring $(23.22)$
Principal one-sided ideal domain	No	$k [x; σ]$ is a principal left ideal domain only
Goldie	No	The ring of $(1.24)$ is left Goldie and not right Goldie

When does the asymmetry disappear?
	Chain conditions	Primitivity	Perfectness
$R$ commutative	yes	yes	yes
$R$ has an anti-automorphism	yes	yes	yes
$R$ left artinian	yes	yes	yes
$R$ semisimple	yes	yes	yes
$R$ semiprimary	no	no	yes
$R$ general	no	no	no

When does the asymmetry disappear?

A “yes” means the left and right versions of that property agree for every ring in the row. Semiprimary rings are perfect on both sides by (23.19) yet can still fail the chain conditions on one side only.

Relationship Map

Why a property is symmetric — Classify the reason, and the verdict for a new property becomes predictable
- Stated with units
  - $rad R$
  - Local rings
  - Dedekind finiteness
- Stated with two-sided ideals
  - Simple, prime, semiprime
  - All four radicals
  - Indecomposability as a ring
- Symmetric via a classification
  - Semisimple, through Wedderburn–Artin
  - Simple artinian, through $M_{n} (D)$
- Asymmetric: one-sided chains
  - Noetherian, artinian, Goldie
  - T-nilpotence and hence perfectness
- Asymmetric: one distinguished module
  - Primitivity
  - One-sided self-injectivity

Left artinianprime

=

primitive

=

simpleall one-sided distinctions collapse

Design Considerations

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

Fix the side once, globally. Mixing left modules with right ideals in the same argument is the single most common source of error in this subject. Lam works with left modules throughout, and so does this collection.
Prefer symmetric hypotheses when you have the choice. Semiprimary is symmetric and does the work of artinian in most radical arguments; use it and state the side only when it is genuinely needed.
Record the side in the name. Right perfect and left artinian should never be abbreviated to perfect and artinian in a written proof, even when the ring at hand happens to be symmetric.
Exploit anti-automorphisms deliberately. If your ring is a group algebra or a matrix ring over a commutative ring, say so once and then drop all side qualifications; the reader will thank you.
When modelling, ask which side the data lives on. In a skew polynomial model of a time-varying system, the direction of $σ$ decides which side is noetherian, and that decides which computations terminate.

Standards and Notation

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

Opposite ring

R^{op}

; older sources use

R^{*}

R'

Module sides

_{R} M

left,

M_{R}

right,

_{R} M_{S}

bimodule

Dictionary

_{R} M

is the same data as

M_{R^{op}}

Matrices

M_{n} (R)^{op} ≅ M_{n} (R^{op})

by transposition

Endomorphisms

End (_{R} R) ≅ R^{op}

when maps are written on the left

Convention hazardWriting homomorphisms on the right removes the

op

in the previous line; sources differ

ImplementationsGAP and Magma expose an opposite-algebra constructor; side conventions for module categories differ between systems

Failure Modes and Common Mistakes

Do not assume a subring or a quotient inherits an anti-automorphism; the collapse argument applies to the ring you actually have.
Do not infer from left artinian implies left noetherian that any implication crosses sides. It never does.
Do not quote a mid-century source's primitive without checking the side convention; several early papers do not state one.

Quick Reference

Definition

a \cdot_{op} b = b a

;

(R^{op})^{op} = R

DictionaryLeft

R

-modules

=

right

R^{op}

-modules

Always symmetric

rad R

, all four radicals, semisimple, simple, prime, local, semilocal, semiperfect, von Neumann regular

Never guaranteedNoetherian, artinian, Goldie, primitive, perfect, principal one-sided ideal domain

Collapse 1An anti-automorphism makes everything symmetric for that ring

Collapse 2Left artinian: prime

=

primitive

=

simple

(11.7)

Matrix rule

M_{n} (D)^{op} ≅ M_{n} (D^{op})

TrapRight perfect

\Leftrightarrow

DCC on principal left ideals

Witnesses at a glance
Asymmetry	Witness	Reference
Artinian	$(\begin{matrix} ℝ & ℝ \\ 0 & ℚ \end{matrix})$	$(1.24)$
Noetherian	$k [x; σ]$ , $σ$ non-surjective	$(1.25)$
Noetherian, finitely presented	$ℤ ⟨ x, y ⟩ / (y^{2}, y x)$	$(1.26)$
Primitivity	Bergman's ring	§11, remark after $(11.2)$
Perfectness	$k \cdot 1 + J$ , $J$ infinite strictly upper triangular	$(23.22)$

Frequently Asked Questions

If left modules over

R

are right modules over

R^{op}

, why is anything asymmetric?

Because that dictionary relates $R$ to a different ring. It guarantees that every theorem has a mirror image, not that a given ring satisfies both a property and its mirror. Asymmetry means finding $R$ with $P (R)$ true and $P (R^{op})$ false, and such rings exist for the chain conditions, primitivity and perfectness.

Why is the Jacobson radical symmetric when its definition is not?

Because $(4.3)$ replaces the definition with a condition mentioning only units: $y \in rad R$ if and only if $1 - x y z$ is a unit for all $x, z$ . Invertibility is a two-sided notion and the set ${x y z}$ is unchanged when the product is reversed, so the condition reads identically in $R^{op}$ .

Are there natural rings with no anti-automorphism?

Yes, and they are exactly the ones the asymmetric examples exploit. A triangular ring $(\begin{matrix} R & M \\ 0 & S \end{matrix})$ with $R not ≅ S$ cannot have one, because an anti-automorphism would have to exchange the two diagonal quotients. This is the structural reason triangular rings are the standard source of one-sided examples.

Does semisimplicity really need Wedderburn–Artin to be proved symmetric?

Not strictly — there are direct arguments — but the classification proof is the shortest honest route and shows exactly why the symmetry holds: the class of finite products of matrix rings over division rings is closed under passing to opposites, because transposition identifies $M_{n} (D)^{op}$ with $M_{n} (D^{op})$ and $D^{op}$ is again a division ring.

What about group rings — do the asymmetries ever appear there?

Not for a commutative coefficient ring. The map sending $g$ to $g^{- 1}$ and extending additively is an anti-automorphism of $k G$ , so $k G ≅ (k G)^{op}$ and every one-sided property of $k G$ holds on both sides or neither. Any asymmetric behaviour in group rings must come from a noncommutative coefficient ring or a twisted product.

Is one-sided self-injectivity really asymmetric?

Yes; rings that are left self-injective and not right self-injective are known, and are treated in Lam's companion volume on modules and rings rather than in the present text. It is a further instance of the general pattern: the condition singles out a distinguished module on one side, so nothing forces the mirror statement.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §1 (pp. 3–5, 18–24), §4 (pp. 50–55), §11 (pp. 183–202) and §23 (pp. 356–357).
T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics 189, Springer-Verlag, 1999.
G. M. Bergman, “A ring primitive on the right but not on the left”, Proceedings of the American Mathematical Society, 1964, with a subsequent correction — the first example of a ring primitive on one side only.
A. V. Jategaonkar, Left Principal Ideal Rings, Lecture Notes in Mathematics 123, Springer-Verlag, 1970 — the source of further one-sided examples.
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.

AI Suggested Questions

Describe Bergman's 1965 construction of a left primitive ring that is not right primitive.
Which homological invariants — global dimension, weak dimension, finitistic dimension — are left-right symmetric?
Characterise the rings admitting an anti-automorphism among finite-dimensional algebras over a field.
Give a finitely presented ring that is left artinian and not right artinian, or explain why none can exist.
How does one-sided behaviour interact with Morita equivalence, given that $M_{n} (R)^{op} ≅ M_{n} (R^{op})$ ?
For skew polynomial rings $k [x; σ, δ]$ , exactly which conditions on $σ$ and $δ$ produce one-sided noetherianness?
Is there a general criterion predicting whether a newly defined ring-theoretic property will be symmetric?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Three sources of symmetry

Two sources of asymmetry

06Key Results

07Proof Techniques and Method

Recharacterise through units

Transfer through the classification

Build asymmetry into a triangular ring

Use a non-surjective endomorphism

Look for an involution first

Impose a chain condition to collapse

08Worked Example

Computing the opposite of a triangular ring

The radical, computed twice

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Design Considerations

13Standards and Notation

14Failure Modes and Common Mistakes

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

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