Is k[x;] ever right noetherian when is not surjective?

No. Part (3) of (1.25) produces an infinite direct sum of nonzero right ideals from any b ∈ k (k), and such a b exists precisely when fails to be surjective. Conversely, if is an automorphism then the mirror-image argument makes k[x;] a principal right ideal domain as well.

One-Sided Chain Conditions and the Dieudonné Example

Q: Why does the left Euclidean algorithm work in k[x;] but not the right one?

Dividing on the left by a monic f requires subtracting cx^n-df, whose leading coefficient is c^n-d(1) = c — no condition on c. Dividing on the right requires subtracting f x^n-dc', whose leading coefficient is ^n-d(c'), so one must solve ^n-d(c') = c. That is possible for all c exactly when is surjective.

Q: Does an example exist that is left noetherian, right noetherian, and left artinian but not right artinian?

No, and for a structural reason: by Hopkins–Levitzki a left artinian ring is left noetherian, but that gives nothing on the right. A ring that is left artinian and right noetherian is in fact right artinian — this is a theorem, since a right noetherian ring with nilpotent radical and semisimple quotient satisfies the right DCC. The genuinely available asymmetry is left artinian with *neither* right condition, as in the triangular example.

Q: Why is the direct sum in (1.25) built from x^i(bx) rather than x^i b?

Because the contradiction comes from comparing leading coefficients of b,x f and x g, both of which have all coefficients modified by . With the generators x^i b one obtains b c_r = (a_r-1) instead, which only says b ∈ (k)c_r^-1 and yields no contradiction. Concretely, over k = Q(t) with (t) = t^2 and b = t, one has b(tx) = t^2 x = x b, so bR and xbR already intersect nontrivially.

Q: What replaces symmetry when a property is one-sided?

The opposite ring. Every left statement about R is a right statement about R^op, so results proved on one side transfer automatically to the other side of a different ring. What does not transfer is a left result to a right result about the *same* ring; that requires a genuine proof, and for chain conditions no such proof exists.

Q: Are there natural rings, not built as counterexamples, that are noetherian on one side only?

They are rare but real. Rings of differential operators on non-smooth varieties, certain skew group rings over non-noetherian coefficient rings, and some enveloping algebras of infinite-dimensional Lie algebras exhibit one-sided behaviour. Nearly all rings arising in commutative algebra, algebraic geometry and finite group representation theory are two-sided noetherian, which is why the asymmetry is easy to forget.

Executive Summary

Chain conditions are stated on one side because they are one-sided. There exist rings satisfying the ascending chain condition on left ideals and failing it on right ideals, and rings satisfying the descending chain condition on the left while failing both chain conditions on the right. This page gives the two constructions Lam uses to prove it, both self-contained and both instructive about why the asymmetry is available.

The mechanism is the same in both cases: an operation that is injective but not surjective. In $k [x; σ]$ the culprit is an endomorphism $σ$ of a division ring with $σ (k) ⊊ k$ ; in Dieudonné's ring it is the relation $y x = 0$ , which annihilates one side of an ideal while leaving the other side free. In Triangular Rings the same phenomenon appears a third time, driven by a bimodule that is finitely generated over one corner and not the other.

3Independent constructions

$σ (k) ⊊ k$ The mechanism in (1.25)

$y^{2} = y x = 0$ Dieudonné's relations

(1.25)–(1.26)Lam's numbering

Overview

A ring $R$ is left noetherian if $_{R} R$ satisfies the ACC on submodules, that is, if the left ideals of $R$ satisfy the ACC; right noetherian is the same condition for right ideals; and noetherian without qualification means both. The definitions and basic closure properties are collected in Chain Conditions. What that page deliberately leaves open is whether the two halves are independent.

They are. The obstruction to a general proof is easy to locate: every argument that transfers a property from left to right factors through the opposite ring $R^{op}$ , and $R^{op}$ is left noetherian precisely when $R$ is right noetherian. So a proof of symmetry would have to exhibit a canonical isomorphism-invariant reason why $R$ and $R^{op}$ share the property — as happens for the Jacobson radical, which is characterised by a condition mentioning only units.

Three constructions cover essentially every one-sided example encountered in practice. Two are on this page; the third, the triangular ring, is on its own page and is the cheapest of the three.

Learning Objectives

Prove that $k [x; σ]$ is a principal left ideal domain when $k$ is a division ring, with no hypothesis on $σ$ beyond being an endomorphism.
Construct an infinite direct sum of nonzero right ideals in $k [x; σ]$ when $σ$ is not surjective.
Determine the additive structure of $ℤ ⟨ x, y ⟩ / (y^{2}, y x)$ and prove it is left noetherian.
Show that the ideal $ℤ [x] y$ is not a finitely generated right ideal.
Prove that a left noetherian ring is Dedekind-finite.
Classify standard ring properties as left-right symmetric or not.

Definitions

Definition—Principal left ideal domain

A ring $R$ is a principal left ideal domain if $R$ is a domain — nonzero, with no zero-divisors — and every left ideal of $R$ has the form $R f$ for a single $f \in R$ . Such a ring is left noetherian by the finite generation criterion, but nothing follows about its right ideals.

$k [x; σ]$: Left polynomials $\sum a_{i} x^{i}$ over a ring $k$ , with $x b = σ (b) x$ for a ring endomorphism $σ$ of $k$ . Constructed in Skew Polynomial Rings and Hilbert's Twist.
$σ (k)$: The image of $σ$ , a subring of $k$ . For $k$ a division ring and $σ \neq 0$ , it is a division subring, and it is proper exactly when $σ$ is not surjective.
$R^{op}$: The opposite ring. $R$ is right noetherian iff $R^{op}$ is left noetherian, so a one-sided example and its opposite realise both asymmetries.
Left Goldie: ACC on left annihilators plus no infinite direct sum of nonzero left ideals. Left noetherian implies left Goldie; the converse fails.
Left T-nilpotent: For every sequence $a_{1}, a_{2}, \dots$ in the ideal, some product $a_{n} \dots a_{1}$ vanishes. This is the condition in Bass's left perfect rings, another asymmetric notion.

Throughout, a domain is a nonzero ring in which $a b = 0$ implies $a = 0$ or $b = 0$ ; commutativity is not assumed.

Core Concepts

Where the asymmetry comes from

In $R = k [x; σ]$ the variable acts differently on the two sides. Left multiplication by $x$ sends $\sum a_{i} x^{i}$ to $\sum σ (a_{i}) x^{i + 1}$ , so $x R$ consists of polynomials with zero constant term *and all coefficients in $σ (k)$ *. Right multiplication by $x$ sends $\sum a_{i} x^{i}$ to $\sum a_{i} x^{i + 1}$ , so $R x$ is simply the polynomials with zero constant term. The image $σ (k)$ appears on one side and not the other; if it is a proper subring, the sides genuinely differ.

σ

not surjective

x R ⊊ R x

in coefficient termsinfinite direct sum of right idealsnot right noetherian

The Euclidean algorithm survives on one side only

Division with remainder works in $k [x; σ]$ for a monic divisor with the quotient written on the left: given $g$ of degree $n$ with leading coefficient $c$ and $f$ monic of degree $d \leq n$ , the polynomial $g - c x^{n - d} f$ has degree $< n$ , because $c x^{n - d} f$ has leading term $c σ^{n - d} (1) x^{n} = c x^{n}$ . Iterating gives $g = q f + r$ with $\deg r < d$ . Nothing here needs $σ$ to be surjective.

The mirror-image algorithm, writing the quotient on the right, would need to solve $σ^{n - d} (c') = c$ for $c'$ — that is, it would need $c \in σ^{n - d} (k)$ . When $σ$ is not surjective this fails, and with it the right ideal theory.

Dieudonné's mechanism: a one-sided annihilator

In $R = ℤ ⟨ x, y ⟩ / (y^{2}, y x)$ the relations kill every occurrence of $y$ except at the end of a word. The surviving monomials are $x^{i}$ and $x^{i} y$ , so $R = ℤ [x] \oplus ℤ [x] y$ . The ideal $I = ℤ [x] y$ is a free $ℤ [x]$ -module of rank one on the left, and on the right it is annihilated by both generators — so as a right module it is just an abelian group of infinite rank.

Key Results

Theorem(1.25)Skew polynomials over a division ring

Let $k$ be a division ring and $σ$ a ring endomorphism of $k$ . Set $R = k [x; σ]$ . Then:

$R$ is a domain and $\deg (f g) = \deg f + \deg g$ for nonzero $f, g$ ;
every left ideal of $R$ is principal, so $R$ is a principal left ideal domain and in particular left noetherian;
if $σ$ is not surjective then $R$ is not right noetherian;
$R$ is neither left nor right artinian.

Proof

(1). A ring homomorphism out of a division ring is injective, so $σ$ is injective. If $f$ has leading term $a_{m} x^{m}$ and $g$ has leading term $b_{n} x^{n}$ , then the coefficient of $x^{m + n}$ in $f g$ is $a_{m} σ^{m} (b_{n})$ , which is nonzero because $k$ is a division ring and $σ^{m}$ is injective. Hence degrees add and $R$ has no zero-divisors.

(2). Let $I \neq 0$ be a left ideal and choose $f \in I ∖ {0}$ of least degree $d$ . Multiplying on the left by the inverse of its leading coefficient — legitimate since $k$ is a division ring — we may take $f$ monic. Given $g \in I$ nonzero of degree $n$ : if $n \geq d$ with leading coefficient $c$ , then $g - c x^{n - d} f \in I$ has degree $< n$ , since $c x^{n - d} f$ has leading term $c σ^{n - d} (1) x^{n} = c x^{n}$ . Iterating produces $q$ with $g - q f \in I$ of degree $< d$ , hence zero by minimality. So $I = R f$ .

(3). Choose $b \in k ∖ σ (k)$ ; note $b \neq 0$ since $0 = σ (0) \in σ (k)$ . We claim the right ideals $x^{i} (b x) R$ , $i \geq 0$ , have direct sum. Suppose not, and take a relation

x^{n} (b x) f_{n} + x^{n + 1} (b x) f_{n + 1} + \dots + x^{n + m} (b x) f_{n + m} = 0, f_{n} \neq 0,

(1.25a)

with $n$ the least index carrying a nonzero term.

Factor $x^{n}$ out on the left and cancel it, which is legitimate because $R$ is a domain. What remains is $b x f_{n} + x h = 0$ for some $h \in R$ , so $b x f_{n} = x g$ with $g = - h$ .

Let $c_{r}$ be the leading coefficient of $f_{n}$ , of degree $r$ . Since $x f_{n} = \sum_{s} σ (c_{s}) x^{s + 1}$ , the left side $b x f_{n}$ has degree $r + 1$ and leading coefficient $b σ (c_{r}) \neq 0$ . The right side $x g = \sum_{s} σ (a_{s}) x^{s + 1}$ has degree $\deg g + 1$ and leading coefficient $σ (a_{\deg g})$ . Comparing gives $\deg g = r$ and

b σ (c_{r}) = σ (a_{r}) ⟹ b = σ (a_{r}) σ (c_{r})^{- 1} = σ (a_{r} c_{r}^{- 1}) \in σ (k),

contradicting the choice of $b$ . So the sum is direct, and the partial sums $\sum_{i = 0}^{N} x^{i} (b x) R$ form a strictly ascending chain of right ideals. Hence $R$ is not right noetherian.

(4). $R x ⊋ R x^{2} ⊋ \dots$ and $x R ⊋ x^{2} R ⊋ \dots$ are strictly descending, since every nonzero element of $R x^{n}$ and of $x^{n} R$ has order at least $n$ while $x^{n}$ has order exactly $n$ . So $R$ satisfies neither descending chain condition. $□$

Theorem(1.26)Dieudonné's example

Let $R = ℤ ⟨ x, y ⟩ / (y^{2}, y x)$ , writing $x, y$ for the images of the generators. Then $R = ℤ [x] \oplus ℤ [x] y$ as an abelian group, $R$ is left noetherian, $R$ is not right noetherian, and $R$ is neither left nor right artinian.

Proof

Additive structure. In $R$ the relations $y^{2} = 0$ and $y x = 0$ annihilate every word in which $y$ is followed by another letter, so the monomials $x^{i}$ and $x^{i} y$ ( $i \geq 0$ ) span $R$ over $ℤ$ . They are independent: inside the triangular ring $(\begin{matrix} ℤ [x] & ℤ [x] \\ 0 & ℤ \end{matrix})$ the elements $X = (\begin{matrix} x & 0 \\ 0 & 0 \end{matrix})$ and $Y = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ satisfy $Y^{2} = 0$ and $Y X = 0$ , and generate over $ℤ$ a subring whose elements $1, X^{i}, X^{i} Y$ are visibly $ℤ$ -independent. So the spanning set is a basis and $R = ℤ [x] \oplus ℤ [x] y$ .

** $I = ℤ [x] y$ is an ideal.** It is a left ideal because $x \cdot x^{i} y = x^{i + 1} y$ and $y \cdot x^{i} y = (y x) x^{i - 1} y = 0$ for $i \geq 1$ while $y \cdot y = 0$ . It is a right ideal because $x^{i} y \cdot x = x^{i} (y x) = 0$ and $x^{i} y \cdot y = x^{i} y^{2} = 0$ . In fact $I \cdot x = I \cdot y = 0$ , so for $u \in I$ and $r \in R$ we get $u r = n u$ where $n \in ℤ$ is the constant term of $r$ .

Left noetherian. $ℤ [x]$ is a noetherian ring by the Hilbert Basis Theorem. As a left $ℤ [x]$ -module, $R = ℤ [x] \cdot 1 \oplus ℤ [x] \cdot y$ is finitely generated, hence a noetherian left $ℤ [x]$ -module. Every left ideal of $R$ is in particular a left $ℤ [x]$ -submodule of $R$ , so the left ideals satisfy the ACC and $R$ is left noetherian.

Not right noetherian. Suppose $I$ were generated as a right ideal by $u_{1}, \dots, u_{t} \in I$ . By the computation above, $\sum_{j} u_{j} R$ is just the additive subgroup generated by $u_{1}, \dots, u_{t}$ . But $I = ⨁_{i \geq 0} ℤ x^{i} y$ is a free abelian group of infinite rank, hence not finitely generated as an abelian group. So $I$ is not a finitely generated right ideal and $R$ is not right noetherian.

Not artinian. $R / I ≅ ℤ [x]$ , which is not artinian: $(x) ⊋ (x^{2}) ⊋ \dots$ is strictly descending. A quotient ring of a left (or right) artinian ring is left (or right) artinian, so $R$ is artinian on neither side. $□$

Corollary—One-sided hypotheses that do give two-sided conclusions

Every left noetherian ring $R$ is Dedekind-finite: if $a b = 1$ in $R$ then $b a = 1$ .

Proof

Let $a b = 1$ and define $φ :_{R} R \to_{R} R$ by $φ (r) = r b$ . This is a homomorphism of left $R$ -modules. It is surjective, since $φ (r a) = r a b = r$ for every $r$ . Because $_{R} R$ is a noetherian module, a surjective endomorphism of it is injective — see Chain Conditions. Finally

φ (b a - 1) = (b a - 1) b = b (a b) - b = b - b = 0,

so $b a - 1 \in \ker φ = 0$ , that is, $b a = 1$ . $□$

Remark—The triangular examples

Two further examples come free from Triangular Rings. With $S \subseteq R$ fields and $\dim_{S} R = \infty$ , the ring $(\begin{matrix} R & R \\ 0 & S \end{matrix})$ is left artinian and left noetherian but neither right noetherian nor right artinian; taking $R = ℝ$ , $S = ℚ$ makes this concrete. Small's example $(\begin{matrix} ℤ & ℚ \\ 0 & ℚ \end{matrix})$ is right noetherian and not left noetherian.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Compare leading coefficients across a forced identity

Reduce a purported relation to $b σ (c) = σ (a)$ and read off $b \in σ (k)$ . The whole failure of the right ACC in $(1.25)$ is this single comparison.

Move 2

Infinite direct sum instead of an infinite chain

A ring with an infinite direct sum of nonzero right ideals cannot be right noetherian, because the partial sums strictly ascend. Producing a direct sum is usually easier than producing a chain.

Move 3

Restrict to a noetherian subring

If $R$ is finitely generated as a left module over a noetherian subring $T$ , then $R$ is left noetherian: left ideals are $T$ -submodules of a noetherian $T$ -module. Dieudonné's example is exactly this with $T = ℤ [x]$ .

Move 3 is the most reusable and the most often forgotten. It requires $R$ to be finitely generated as a left $T$ -module, not merely as a $T$ -algebra: $ℤ ⟨ x, y ⟩$ is a finitely generated $ℤ$ -algebra and is neither left nor right noetherian.

A fourth move worth recording is the one used for the Dedekind-finite corollary: turn a ring-theoretic identity into a module endomorphism, then apply a chain condition to the module. Chain conditions are hypotheses about modules, and the shortest route to a ring-theoretic consequence is usually through a well-chosen endomorphism.

Worked Example

A concrete non-surjective twist

Take $k = ℚ (t)$ and let $σ$ be the field embedding determined by $σ |_{ℚ} = id$ and $σ (t) = t^{2}$ . It is injective, being a field homomorphism, and its image is $σ (k) = ℚ (t^{2}) ⊊ ℚ (t)$ . So $σ$ is an endomorphism of a field that is not an automorphism, and $R = ℚ (t) [x; σ]$ satisfies the hypotheses of $(1.25)$ .

Concretely, in $R$ one has $x t = t^{2} x$ , and more generally $x^{n} t = t^{2^{n}} x^{n}$ . Choose $b = t \notin ℚ (t^{2})$ . Then the right ideals $x^{i} (t x) R$ have direct sum, so

t x R ⊊ t x R + x (t x) R ⊊ t x R + x (t x) R + x^{2} (t x) R ⊊ \dots

(E.1)

A strictly ascending chain of right ideals. Meanwhile every left ideal of $R$ is principal.

Note what fails if $σ$ is an automorphism, say $σ (t) = t + 1$ : then $σ (k) = k$ , there is no admissible $b$ , and in fact $ℚ (t) [x; σ]$ is a principal right ideal domain as well — both Euclidean algorithms work.

Multiplying in Dieudonné's ring

In $R = ℤ ⟨ x, y ⟩ / (y^{2}, y x)$ take $u = 3 x^{2} y + 5 y \in I$ and $r = 7 - 2 x + 4 x y$ . Since $I x = I y = 0$ ,

u r = 7 u = 21 x^{2} y + 35 y, r u = 7 u - 2 x u + 4 x y u = 21 x^{2} y + 35 y - 6 x^{3} y - 10 x y,

(E.2)

Right multiplication only ever rescales by an integer; left multiplication moves elements up the $x$ -degree.

The asymmetry is visible in one line. On the left, $I$ is $ℤ [x]$ -free of rank $1$ on the generator $y$ ; on the right, $I$ is a trivial module, so its right submodules are exactly its subgroups, and it has infinitely many that are not finitely generated.

Reading off the opposite ring

$R^{op}$ for either example is right noetherian and not left noetherian. So each construction gives both asymmetries; there is never a need to build a separate example for the other side.

Frameworks and Models

It is worth keeping a mental list of which ring-theoretic properties survive passage to the opposite ring.

Left–right symmetry of standard properties
Property	Symmetric?	Comment
$rad R$	yes	characterised by $1 - x y z \in U (R)$ , a condition mentioning only units
Semisimple	yes	left semisimple and right semisimple coincide, via Wedderburn–Artin
Semiprimitive	yes	immediate from the symmetry of the radical
Prime, semiprime, simple	yes	defined by two-sided ideals
Von Neumann regular	yes	$a \in a R a$ is a side-neutral condition
Semiperfect	yes	idempotents lift and $R / rad R$ is semisimple; both symmetric
Dedekind-finite	yes	the condition $a b = 1 \Rightarrow b a = 1$ is itself symmetric
Noetherian	no	$(1.25)$ , $(1.26)$ and the triangular examples
Artinian	no	$(\begin{matrix} ℝ & ℝ \\ 0 & ℚ \end{matrix})$
Perfect	no	Bass; left perfect requires $rad R$ to be left T-nilpotent
Primitive	no	Bergman constructed a left primitive ring that is not right primitive
Goldie, hereditary, self-injective	no	each has standard counterexamples; global dimension can also differ on the two sides

Rule of thumb

Unit-theoretic conditions are symmetric

If a property can be phrased entirely in terms of invertibility, idempotents or two-sided ideals, it survives $R \mapsto R^{op}$ . The Jacobson radical is the model case.

Rule of thumb

Finiteness conditions on one-sided ideals are not

Anything defined by chains, generation or direct sums of one-sided ideals should be presumed asymmetric until a proof is supplied.

Comparison and Classification

Chain conditions in the standard one-sided examples
	left noeth.	right noeth.	left art.	right art.
$k [x; σ]$ , $k$ division ring, $σ$ not onto	yes	no	no	no
$ℤ ⟨ x, y ⟩ / (y^{2}, y x)$	yes	no	no	no
$(\begin{matrix} ℝ & ℝ \\ 0 & ℚ \end{matrix})$	yes	no	yes	no
$(\begin{matrix} ℤ & ℚ \\ 0 & ℚ \end{matrix})$	no	yes	no	no
$k ⟨ x, y ⟩$	no	no	no	no

Chain conditions in the standard one-sided examples

What each construction costs and buys
Construction	Input needed	Strength of the example
$k [x; σ]$ , $(1.25)$	a division ring with a non-surjective endomorphism	a principal left ideal domain that is not right noetherian — the strongest form, since domains resist most pathologies
$ℤ ⟨ x, y ⟩ / (y^{2}, y x)$ , $(1.26)$	nothing beyond $ℤ$	finitely presented on two generators and two relations; the cheapest example to write down
Triangular ring, $(1.23)$ – $(1.24)$	a bimodule lopsided over its two corner rings	the most flexible; also delivers left artinian with no right chain condition at all

Relationship Map

left artinianleft noetherianleft GoldieDedekind-finite

Every arrow is one-directional and every hypothesis is genuinely on the left. The first arrow is Hopkins–Levitzki; the second is elementary; the third is the corollary proved above.

Sources of one-sided behaviour
- a non-surjective injection
  - $σ : k \to k$ with $σ (k) ⊊ k$ , giving $k [x; σ]$ — this page
  - the same twist with negative powers is impossible: $k [x, x^{- 1}; σ]$ needs an automorphism
- a one-sided annihilator
  - $y x = 0$ with $x y \neq 0$ — Dieudonné's ring
  - the analogous phenomenon in $(\begin{matrix} ℤ & ℤ / 2 ℤ \\ 0 & ℤ \end{matrix})$ , a left zero-divisor that is not a right one
- a lopsided bimodule
  - $M$ finitely generated over $R$ and not over $S$ — Triangular Rings
  - the most flexible source, and the only one that delivers left artinian without right noetherian

In the other direction, the existence of these examples is why every theorem in this collection carries an explicit side. The Jacobson radical is the notable exception, and its symmetry is a theorem rather than a convention — see Jacobson Radical: Definition and Characterisations.

Design Considerations

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

Fix a side and keep it. Decide at the outset whether modules are left or right and never switch mid-argument. Most published errors involving chain conditions are side-slips, not mathematical mistakes.
**Use $R^{op}$ deliberately.** Every left theorem yields a right theorem for the opposite ring at no cost. This halves the work and makes it explicit when a claimed symmetry has actually been proved rather than assumed.
Pick the cheapest construction. For a pure noetherian asymmetry, Dieudonné's ring needs no input data at all. For an asymmetry involving artinian, only the triangular construction works, because $(1.25)$ and $(1.26)$ produce rings that are artinian on neither side.
Watch what the example loses. $k [x; σ]$ is a domain, which many other constructions cannot manage; Dieudonné's ring is finitely presented, which the triangular examples over $ℝ$ and $ℚ$ are not. Choose the example that keeps the properties your argument still needs.
Do not over-hypothesise. Requiring two-sided noetherian when only the left condition is used weakens every theorem that quotes yours. State the minimal side.

Failure Modes and Common Mistakes

Neither example on this page is artinian on either side; do not cite them for artinian asymmetry.
In $(1.25)$ the generators of the direct sum are $x^{i} (b x)$ , not $x^{i} b$ . With the generators $x^{i} b$ the sum is in general not direct — the degree comparison that yields the contradiction disappears.
The ideal $I = ℤ [x] y$ in Dieudonné's ring is finitely generated as a left ideal (by $y$ alone) and not as a right ideal. Finite generation of an ideal is itself a one-sided notion.
Being left Goldie does not imply being right Goldie; the triangular ring over $ℝ$ and $ℚ$ separates them.

Quick Reference

$(1.25)$ hypotheses

k

a division ring,

σ \in End (k)

not surjective

$(1.25)$ conclusion

k [x; σ]

is a principal left ideal domain, not right noetherian, not artinian on either side

The witness

b \in k ∖ σ (k)

; the right ideals

x^{i} (b x) R

have direct sum

$(1.26)$

R = ℤ ⟨ x, y ⟩ / (y^{2}, y x) = ℤ [x] \oplus ℤ [x] y

$(1.26)$ conclusionleft noetherian, not right noetherian, artinian on neither side

The witness

I = ℤ [x] y

with

I x = I y = 0

, so right submodules are just subgroups

Left artinian caseuse

(\begin{matrix} ℝ & ℝ \\ 0 & ℚ \end{matrix})

instead

What is symmetric

rad R

, semisimple, semiprimitive, von Neumann regular, semiperfect, Dedekind-finite

Which example to reach for
You want	Use	Bonus properties
Left noetherian, not right	$ℤ ⟨ x, y ⟩ / (y^{2}, y x)$	finitely presented over $ℤ$
The same, in a domain	$k [x; σ]$ , $σ$ not onto	principal left ideal domain
Left artinian, not right noetherian	$(\begin{matrix} ℝ & ℝ \\ 0 & ℚ \end{matrix})$	left composition series of length $3$ ; left Goldie not right Goldie
Right noetherian, not left	$(\begin{matrix} ℤ & ℚ \\ 0 & ℚ \end{matrix})$	Small's example; or take opposites of the rows above

Frequently Asked Questions

Why does the left Euclidean algorithm work in

k [x; σ]

but not the right one?

Dividing on the left by a monic $f$ requires subtracting $c x^{n - d} f$ , whose leading coefficient is $c σ^{n - d} (1) = c$ — no condition on $c$ . Dividing on the right requires subtracting $f x^{n - d} c'$ , whose leading coefficient is $σ^{n - d} (c')$ , so one must solve $σ^{n - d} (c') = c$ . That is possible for all $c$ exactly when $σ$ is surjective.

k [x; σ]

ever right noetherian when

σ

is not surjective?

No. Part (3) of $(1.25)$ produces an infinite direct sum of nonzero right ideals from any $b \in k ∖ σ (k)$ , and such a $b$ exists precisely when $σ$ fails to be surjective. Conversely, if $σ$ is an automorphism then the mirror-image argument makes $k [x; σ]$ a principal right ideal domain as well.

Does an example exist that is left noetherian, right noetherian, and left artinian but not right artinian?

No, and for a structural reason: by Hopkins–Levitzki a left artinian ring is left noetherian, but that gives nothing on the right. A ring that is left artinian and right noetherian is in fact right artinian — this is a theorem, since a right noetherian ring with nilpotent radical and semisimple quotient satisfies the right DCC. The genuinely available asymmetry is left artinian with neither right condition, as in the triangular example.

Why is the direct sum in

(1.25)

built from

x^{i} (b x)

rather than

x^{i} b

Because the contradiction comes from comparing leading coefficients of $b x f$ and $x g$ , both of which have all coefficients modified by $σ$ . With the generators $x^{i} b$ one obtains $b c_{r} = σ (a_{r - 1})$ instead, which only says $b \in σ (k) c_{r}^{- 1}$ and yields no contradiction. Concretely, over $k = ℚ (t)$ with $σ (t) = t^{2}$ and $b = t$ , one has $b (t x) = t^{2} x = x b$ , so $b R$ and $x b R$ already intersect nontrivially.

What replaces symmetry when a property is one-sided?

The opposite ring. Every left statement about $R$ is a right statement about $R^{op}$ , so results proved on one side transfer automatically to the other side of a different ring. What does not transfer is a left result to a right result about the same ring; that requires a genuine proof, and for chain conditions no such proof exists.

Are there natural rings, not built as counterexamples, that are noetherian on one side only?

They are rare but real. Rings of differential operators on non-smooth varieties, certain skew group rings over non-noetherian coefficient rings, and some enveloping algebras of infinite-dimensional Lie algebras exhibit one-sided behaviour. Nearly all rings arising in commutative algebra, algebraic geometry and finite group representation theory are two-sided noetherian, which is why the asymmetry is easy to forget.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §1, examples (1.25) and (1.26) (pp. 22–24).
T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics 189, Springer-Verlag, 1999, §2 and §3.
J. C. McConnell and J. C. Robson, Noncommutative Noetherian Rings, revised edition, Graduate Studies in Mathematics 30, American Mathematical Society, 2001, Chapters 1–2 and 7.
L. W. Small, “An example in noetherian rings”, Proceedings of the National Academy of Sciences of the USA 54 (1965).
H. Bass, “Finitistic dimension and a homological generalization of semi-primary rings”, Transactions of the American Mathematical Society 95 (1960), 466–488.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapters 1–3.

AI Suggested Questions

Prove that a left artinian, right noetherian ring is right artinian.
Describe all right ideals of $k [x; σ]$ when $σ$ is a non-surjective endomorphism of a division ring.
Construct a division ring admitting an endomorphism that is not surjective, other than a rational function field.
Give a left primitive ring that is not right primitive, following Bergman's construction.
Show that left perfect and right perfect are distinct conditions, and identify where T-nilpotence enters.
Is there a finitely presented ring that is left artinian but not right noetherian?
Compare the left and right global dimensions of the triangular ring built from $ℤ$ , $ℚ$ and $ℚ$ .

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Where the asymmetry comes from

The Euclidean algorithm survives on one side only

Dieudonné's mechanism: a one-sided annihilator

06Key Results

07Proof Techniques and Method

Compare leading coefficients across a forced identity

Infinite direct sum instead of an infinite chain

Restrict to a noetherian subring

08Worked Example

A concrete non-surjective twist

Multiplying in Dieudonné's ring

Reading off the opposite ring

09Frameworks and Models

Unit-theoretic conditions are symmetric

Finiteness conditions on one-sided ideals are not

10Comparison and Classification

11Relationship Map

12Design Considerations

13Failure Modes and Common Mistakes

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

Worked Example

Frameworks and Models

Comparison and Classification

Relationship Map

Design Considerations

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions