Skew Polynomial Rings as Primitive Rings

Q: Is k[x;] ever simple?

Not under the hypothesis of (11.12), since Rx is then a proper nonzero ideal. Simplicity in the purely twisted case fails for a structural reason: x is a normal element, so it always generates a proper ideal. Simple examples come from the differential side, where x is not normal — this is the content of Amitsur's theorem.

Q: How can a ring over a commutative field k have infinitely many non-isomorphic faithful simple modules?

The classification is by -conjugacy, a = (c)a'c^-1, which over a commutative k reduces to a = ((c)c^-1)a' — not to a = a'. The set of elements (c)c^-1 is a proper subgroup of k^× whenever preserves some invariant, and for R(t) with : t t+1 that invariant is the degree.

Q: What replaces polynomial evaluation in the twisted setting?

The action on M_a = k. Ordinary evaluation is not a ring homomorphism, but g g * 1 is exactly the projection of g onto the complement of R(x-a), and it is all that is needed: g ∈ R(x-a) precisely when g * 1 = 0. The formula x^m * 1 = ^m-1(a)…(a)a is the twisted analogue of a^m.

Q: Why does (11.14) require k not algebraic over its centre?

Because the ideals of k[x] are generated by central polynomials, and if a were algebraic over C its minimal polynomial over C would generate a nonzero ideal inside R(x-a), destroying faithfulness. The hypothesis guarantees a supply of a with no such polynomial. It also forces k to be noncommutative, since a field is algebraic over itself.

Executive Summary

Abstract theory says left primitive rings exist in abundance. This page supplies them explicitly. Take a division ring $k$ and a twist — either an endomorphism $σ$ or a derivation $δ$ — and form the skew polynomial ring. The left ideals $R (x - a)$ are maximal, and the resulting simple modules $M_{a} = R / R (x - a)$ are faithful under mild hypotheses on the twist.

Two features make these examples decisive for the theory. First, $(11.12)$ shows $R = k [x; σ]$ has ideals $R x^{m}$ and nothing else, so $R$ is left primitive without being simple. Second, the modules $M_{a}$ are classified by a conjugacy-like relation, and for $k = ℝ (t)$ with $σ : t \mapsto t + 1$ there are infinitely many classes — a left primitive ring with infinitely many pairwise non-isomorphic faithful simple left modules, hence zero socle by $(11.11)$ .

$R x^{m}$ All nonzero ideals

$a \neq 0$ Condition for faithfulness

$\infty$ Non-isomorphic faithful simples

$0$ Socle of these rings

Overview

Let $k$ be a division ring and $σ : k \to k$ a ring endomorphism; $σ$ is automatically injective, since its kernel is a proper ideal of a division ring. The skew polynomial ring $R = k [x; σ]$ consists of left polynomials $\sum_{i} a_{i} x^{i}$ with multiplication determined by

x b = σ (b) x (b \in k) .

(11.a)

Hilbert's twist. Setting $σ = id$ recovers the ordinary polynomial ring $k [x]$ .

The construction and its basic properties are developed on the Skew Polynomial Rings and Hilbert's Twist page. What matters here is one consequence: right division works. Given $g$ and $f \neq 0$ in $R$ there are unique $q, r$ with $g = q f + r$ and $r = 0$ or $\deg r < \deg f$ . Injectivity of $σ$ is what makes the leading coefficient invertible at every step. Consequently every left ideal of $R$ is principal, generated by any element of least degree in it.

Step (ii) is where a hypothesis on the twist is needed. If $σ$ has a power that is inner, $R$ acquires extra ideals and the argument breaks. Lam's hypothesis — $σ$ is not an automorphism of finite inner order — is exactly what rules that out, and it holds automatically whenever $σ$ is not surjective.

Learning Objectives

State the right division algorithm in $k [x; σ]$ and deduce that left ideals are principal.
Prove $(11.12)$ : the nonzero ideals of $k [x; σ]$ are exactly $R x^{m}$ , $m \geq 0$ .
Prove $(11.13)$ : $M_{a}$ is faithful and simple for $a \neq 0$ , and classify the $M_{a}$ up to isomorphism.
Explain why $M_{0}$ is excluded from the faithfulness statement.
Verify that $ℝ (t) [x; σ]$ with $σ : t \mapsto t + 1$ has infinitely many non-isomorphic faithful simple modules.
State $(11.14)$ with the correct hypothesis on $k$ and its centre.

Definitions

$k [x; σ]$: Left polynomials over the division ring $k$ with $x b = σ (b) x$ ; $σ$ an injective ring endomorphism of $k$ .
$k [x; δ]$: Differential polynomial ring: left polynomials with $x b = b x + δ (b)$ , where $δ$ is a derivation of $k$ .
Inner automorphism: $a \mapsto u a u^{- 1}$ for a fixed unit $u$ . An automorphism $σ$ has finite inner order if $σ^{d}$ is inner for some $d \geq 1$ .
Inner derivation: $a \mapsto u a - a u$ for a fixed $u \in k$ . Amitsur's criterion for simplicity of $k [x; δ]$ in characteristic $0$ requires $δ$ to be non-inner.
$σ$ -conjugacy: $a \sim a'$ on $k ∖ {0}$ iff $a = σ (c) a' c^{- 1}$ for some $c \neq 0$ . An equivalence relation.
$δ$ -conjugacy: $a \sim a'$ on $k$ iff $a = c a' c^{- 1} + δ (c) c^{- 1}$ for some $c \neq 0$ .

Throughout, $R$ denotes the skew polynomial ring under discussion, $C$ the centre of $k$ , and degrees are the usual polynomial degrees, which are additive because $k$ is a division ring and the twist is injective.

Core Concepts

The module $M_{a}$ concretely

By right division, every $g \in R$ is uniquely $q (x - a) + r$ with $r \in k$ . Hence $R = R (x - a) \oplus k$ as left $k$ -spaces, and $M_{a} = R / R (x - a)$ is identified with $k$ , one-dimensional over $k$ . Since any nonzero $R$ -submodule of $M_{a}$ is in particular a nonzero left $k$ -subspace of a one-dimensional space, $M_{a}$ is simple. No hypothesis on the twist is needed for this.

Under the identification $M_{a} = k$ the action of $x$ is read off from the twist. For $R = k [x; σ]$ , since $x b = σ (b) x = σ (b) (x - a) + σ (b) a$ ,

x * b = σ (b) a (b \in k = M_{a}) .

(11.b)

For the differential ring $k [x; δ]$ the same computation gives $x * b = b a + δ (b)$ .

Why $a = 0$ must be excluded

$M_{0} = R / R x$ is simple, but $R x$ is a two-sided ideal, so $ann (M_{0}) \supseteq R x \neq 0$ and $M_{0}$ is not faithful. The whole point of $(11.12)$ is that these ideals $R x^{m}$ are the only obstruction: for $a \neq 0$ none of them fits inside $R (x - a)$ .

The invariant that separates the modules

A left $R$ -module homomorphism $M_{a} \to M_{a'}$ is in particular left $k$ -linear, so it is right multiplication by some $c \in k$ . Compatibility with the action of $x$ then reads $a c = σ (c) a'$ , that is $a = σ (c) a' c^{- 1}$ . So the isomorphism classes of the $M_{a}$ correspond exactly to the $σ$ -conjugacy classes of $k ∖ {0}$ , and finding many classes is a matter of finding an invariant that $σ (c) c^{- 1}$ cannot change.

Key Results

Proposition(11.12)The ideals of a skew polynomial ring

Let $k$ be a division ring and $σ$ an endomorphism of $k$ that is not an automorphism of finite inner order — that is, no positive power of $σ$ is an inner automorphism of $k$ . This includes every case where $σ$ is not surjective. Then the nonzero two-sided ideals of $R = k [x; σ]$ are exactly the ideals $R x^{m}$ for $m \geq 0$ .

Proof

Each $R x^{m}$ is an ideal: $x^{m} b = σ^{m} (b) x^{m}$ shows $x^{m} R \subseteq R x^{m}$ , so $R x^{m} R \subseteq R x^{m}$ .

Conversely let $𝔄 \neq 0$ be an ideal. By the division algorithm $𝔄 = R f$ for an $f$ of least degree in $𝔄$ , and multiplying by the inverse of its leading coefficient we may take $f$ monic:

f = x^{m} + a_{m - 1} x^{m - 1} + \dots + a_{n} x^{n}, m \geq n \geq 0, a_{n} \neq 0 .

We show $m = n$ , which forces $f = x^{m}$ and $𝔄 = R x^{m}$ .

**Step 1: the coefficients are $σ$ -fixed.** Since $𝔄$ is two-sided, $f x - x f \in 𝔄$ . Writing $f = \sum_{i} a_{i} x^{i}$ with $a_{m} = 1$ , we get $f x - x f = \sum_{i} (a_{i} - σ (a_{i})) x^{i + 1}$ ; the top term vanishes because $a_{m} = 1$ is $σ$ -fixed, so this element has degree at most $m$ and lies in $R f$ . Hence $f x - x f = c f$ for some $c \in k$ . All terms of $f x - x f$ have degree at least $n + 1$ , so its coefficient at $x^{n}$ is $0$ , while that of $c f$ is $c a_{n}$ . As $a_{n} \neq 0$ we get $c = 0$ , so $f x = x f$ and $σ (a_{i}) = a_{i}$ for every $i$ .

Step 2: a conjugation identity. For $a \in k$ the element $f a - σ^{m} (a) f$ lies in $𝔄$ . Using $x^{i} a = σ^{i} (a) x^{i}$ ,

f a - σ^{m} (a) f = \sum_{i} (a_{i} σ^{i} (a) - σ^{m} (a) a_{i}) x^{i},

whose coefficient at $x^{m}$ is $σ^{m} (a) - σ^{m} (a) = 0$ . So it has degree less than $m$ and lies in $R f$ , hence is $0$ . Reading its coefficient at $x^{n}$ gives $a_{n} σ^{n} (a) = σ^{m} (a) a_{n}$ . By Step 1, $σ^{n} (a_{n}) = a_{n} = σ^{m} (a_{n})$ , so this says $σ^{n} (a_{n} a) = σ^{m} (a a_{n}) = σ^{n} (σ^{m - n} (a a_{n}))$ . Injectivity of $σ$ gives

a_{n} a = σ^{m - n} (a) a_{n} for all a \in k .

Step 3: conclusion. If $m > n$ , put $d = m - n \geq 1$ ; the identity reads $σ^{d} (a) = a_{n} a a_{n}^{- 1}$ for all $a$ , so $σ^{d}$ is the inner automorphism determined by $a_{n}$ . In particular $σ^{d}$ is surjective, hence so is $σ$ , and $σ$ is an automorphism of finite inner order — contrary to hypothesis. Therefore $m = n$ , $f = x^{m}$ , and $𝔄 = R x^{m}$ .

Proposition(11.13)Faithful simple modules over

k [x; σ]

Under the hypothesis of $(11.12)$ , for every $a \in k ∖ {0}$ the module $M_{a} = R / R (x - a)$ is a faithful simple left $R$ -module. In particular $R = k [x; σ]$ is a left primitive ring. Moreover $M_{a} ≅ M_{a'}$ as $R$ -modules if and only if $a = σ (c) a' c^{- 1}$ for some $c \in k ∖ {0}$ .

Proof

Simplicity. As explained above, $R = R (x - a) \oplus k$ by right division, so $M_{a}$ is one-dimensional as a left $k$ -space and therefore has no proper nonzero $R$ -submodule.

Faithfulness. Since $ann (M_{a})$ is an ideal contained in $R (x - a)$ , by $(11.12)$ it suffices to show $x^{m} \notin R (x - a)$ for every $m \geq 0$ . Under the identification $M_{a} = k$ with $x * b = σ (b) a$ , the class of $g \in R$ is $g * 1$ , and $g \in R (x - a)$ exactly when $g * 1 = 0$ . Now

x^{m} * 1 = σ^{m - 1} (a) σ^{m - 2} (a) \dots σ (a) a (m \geq 1),

a product of nonzero elements of the division ring $k$ , since $a \neq 0$ and $σ$ is injective; and $x^{0} * 1 = 1 \neq 0$ . So $x^{m} \notin R (x - a)$ for all $m \geq 0$ , no nonzero ideal lies inside $R (x - a)$ , and $ann (M_{a}) = 0$ .

Classification. Let $f : M_{a} \to M_{a'}$ be an $R$ -isomorphism, both modules identified with $k$ . Restricting to $k \subseteq R$ shows $f$ is left $k$ -linear, so $f (b) = b c$ with $c = f (1) \neq 0$ . Compatibility with $x$ gives, for all $b \in k$ ,

f (x * b) = σ (b) a c and x * f (b) = σ (b c) a' = σ (b) σ (c) a',

so $a c = σ (c) a'$ , i.e. $a = σ (c) a' c^{- 1}$ . Conversely, given such a $c$ , the map $b \mapsto b c$ reverses the computation and is an isomorphism $M_{a} \to M_{a'}$ .

Corollary—Left primitive without being simple

Under the hypothesis of $(11.12)$ the ring $R = k [x; σ]$ is left primitive and has the proper nonzero ideal $R x$ , hence is not simple. It is a domain that is not a division ring, so it has no minimal one-sided ideals and $soc (R) = 0$ .

Proposition(11.14)The untwisted case

Let $k$ be a division ring which is not algebraic over its centre $C$ , and let $R = k [x]$ be the ordinary polynomial ring, $x$ central. Then for every $a \in k$ that is not algebraic over $C$ , the module $M_{a} = R / R (x - a)$ is a faithful simple left $R$ -module, and $R / (x - a) R$ is a faithful simple right $R$ -module. In particular $R$ is both left and right primitive. The isomorphism classes of the modules $M_{a}$ , $a \in k$ , correspond bijectively to the conjugacy classes of $k$ .

Proof

Simplicity of $M_{a}$ is as before. For faithfulness, one uses the description of the ideals of $k [x]$ : every ideal of $k [x]$ is of the form $R g$ with $g \in C [x]$ . Suppose $0 \neq R g \subseteq R (x - a)$ with $g = \sum_{i = 0}^{n + 1} c_{i} x^{i}$ , $c_{i} \in C$ . Dividing on the right, write $g = (\sum_{i = 0}^{n} b_{i} x^{i}) (x - a)$ with $b_{i} \in k$ . Comparing coefficients of $x^{i}$ gives $c_{i} = b_{i - 1} - b_{i} a$ for $0 \leq i \leq n + 1$ , where $b_{- 1} = b_{n + 1} = 0$ . Since the $c_{i}$ are central,

\sum_{i = 0}^{n + 1} c_{i} a^{i} = \sum_{i = 0}^{n + 1} (b_{i - 1} - b_{i} a) a^{i} = \sum_{i = 0}^{n + 1} b_{i - 1} a^{i} - \sum_{i = 0}^{n + 1} b_{i} a^{i + 1} = 0,

the two sums telescoping. So $a$ satisfies the nonzero polynomial $g \in C [x]$ , contradicting the assumption that $a$ is not algebraic over $C$ . Hence $R (x - a)$ contains no nonzero ideal and $M_{a}$ is faithful. The classification $M_{a} ≅ M_{a'} \Leftrightarrow a = c a' c^{- 1}$ is the computation of $(11.13)$ with $σ = id$ .

Example—The differential analogue

Let $k$ be a division ring of characteristic $0$ and $δ$ a non-inner derivation of $k$ . By Amitsur's theorem $(3.16)$ the differential polynomial ring $R = k [x; δ]$ is a simple domain, so every nonzero module is faithful and each $M_{a} = R / R (x - a)$ , $a \in k$ , is a faithful simple left $R$ -module. Here $x * b = b a + δ (b)$ on $M_{a} = k$ , and $M_{a} ≅ M_{a'}$ exactly when $a = c a' c^{- 1} + δ (c) c^{- 1}$ for some $c \neq 0$ . The class of $0$ consists of the logarithmic derivatives $δ (c) c^{- 1}$ , so if $k$ contains an element that is not a logarithmic derivative then $R$ has at least two non-isomorphic faithful simple left modules. Being a domain that is not a division ring, $R$ has zero socle.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Classify the ideals first

Primitivity is the statement that a maximal left ideal swallows no nonzero two-sided ideal. Knowing all the two-sided ideals turns that into a finite check — here, whether $x^{m}$ lies in $R (x - a)$ .

Move 2

Commutators to constrain a generator

For an ideal $𝔄 = R f$ , the elements $f x - x f$ and $f a - σ^{m} (a) f$ lie in $𝔄$ and have degree below the leading one. Degree bounds then force them to be zero, which pins the coefficients of $f$ .

Move 3

Compute in the quotient, not the ideal

Membership in $R (x - a)$ is decided by evaluating in $M_{a} = k$ : $g \in R (x - a)$ iff $g * 1 = 0$ . Twisted evaluation replaces polynomial division entirely.

Move 3 deserves emphasis. Ordinary evaluation $g \mapsto g (a)$ is not a ring homomorphism in the twisted setting, but the module action is perfectly well behaved, and $x^{m} * 1 = σ^{m - 1} (a) \dots σ (a) a$ is the twisted substitute for $a^{m}$ . Every faithfulness computation on this page is that formula.

Fix the twistChoose

k

and

σ

(or

δ

), and check the hypothesis: no power of

σ

inner, or

δ

non-inner in characteristic

0

List the ideals

(11.12)

gives

R x^{m}

for

k [x; σ]

; Amitsur's theorem gives only

0

and

R

for

k [x; δ]

Pick $a \neq 0$ and testShow

x^{m} * 1 \neq 0

M_{a}

, so no ideal sits inside

R (x - a)

Count the classesFind an invariant preserved by

c \mapsto σ (c) c^{- 1}

to separate the

M_{a}

up to isomorphism.

Worked Example

A primitive ring with infinitely many faithful simple modules

Let $k = ℝ (t)$ , the field of rational functions in one variable, and let $σ$ be the $ℝ$ -automorphism with $σ (t) = t + 1$ . Put $R = k [x; σ]$ .

The hypothesis holds

$k$ is commutative, so the only inner automorphism of $k$ is the identity. If $σ^{d}$ were inner for some $d \geq 1$ we would need $σ^{d} = id$ , but $σ^{d} (t) = t + d \neq t$ . Hence $σ$ is not an automorphism of finite inner order and $(11.12)$ – $(11.13)$ apply: the nonzero ideals of $R$ are the $R x^{m}$ , and every $M_{a}$ with $a \neq 0$ is a faithful simple left $R$ -module.

Separating the modules by degree

For $0 \neq f / g \in ℝ (t)$ with $f, g \in ℝ [t] ∖ {0}$ set $\deg (f / g) = \deg f - \deg g$ , a well-defined group homomorphism from $k^{\times}$ to $ℤ$ . For any $c = c (t) \neq 0$ ,

σ (c) c^{- 1} = \frac{c (t + 1)}{c (t)}, \deg (σ (c) c^{- 1}) = 0,

(E.1)

Translation of the variable does not change degrees, so the twisted logarithmic derivative is always of degree $0$ .

If $a = σ (c) a' c^{- 1}$ then, $k$ being commutative, $a = (σ (c) c^{- 1}) a'$ and therefore $\deg a = \deg a'$ . So $σ$ -conjugate elements have equal degree, and the elements $t, t^{2}, t^{3}, \dots$ of degrees $1, 2, 3, \dots$ lie in pairwise distinct $σ$ -conjugacy classes.

M_{t}, M_{t^{2}}, M_{t^{3}}, \dots are pairwise non-isomorphic faithful simple left R -modules .

(E.2)

What this ring demonstrates

$R$ is left primitive but not simple: $R x$ is a proper nonzero ideal.
$R$ has infinitely many pairwise non-isomorphic faithful simple left modules, so by $(11.11)$ it can have no minimal left ideal: $soc (R) = 0$ .
$R$ is a noetherian domain, so it is prime; consistent with $(11.6)$ , since left primitive rings are prime.
$R$ is not left artinian — it contains the strictly descending chain $R x ⊋ R x^{2} ⊋ \dots$ of left ideals.

Comparison and Classification

Three families of primitive polynomial rings
Ring	Hypothesis	Ideals	Faithful simple modules
$k [x; δ]$ , $char k = 0$	$δ$ non-inner	only $0$ and $R$ — simple	every $M_{a}$ , $a \in k$ ; classes are $δ$ -conjugacy classes
$k [x; σ]$	$σ$ not an automorphism of finite inner order	$R x^{m}$ , $m \geq 0$	$M_{a}$ for $a \neq 0$ ; classes are $σ$ -conjugacy classes of $k ∖ {0}$
$k [x]$ , $x$ central	$k$ not algebraic over its centre $C$	$R g$ , $g \in C [x]$	$M_{a}$ for $a$ transcendental over $C$ ; classes are conjugacy classes
$k [x]$ , $k$ a field	none	$R g$ , $g \in k [x]$	none — $k [x]$ is commutative and not a field, so not primitive

Properties of the examples
	Simple	Left primitive	Zero socle	Left noetherian
$k [x; δ]$ , $δ$ non-inner, $char k = 0$	yes	yes	yes	yes
$ℝ (t) [x; σ]$ , $σ : t \mapsto t + 1$	no	yes	yes	yes
$k [x]$ , $k$ not algebraic over $C$	no	yes	yes	yes
$ℚ [x]$	no	no	yes	yes
$End (V_{k})$ , $\dim_{k} V$ infinite	no	yes	no	no

Properties of the examples

Relationship Map

Hypothesis on the twistIdeals of

R

classified

R (x - a)

swallows no ideal

M_{a}

faithful simple

R

left primitive

The chain is the same in all three families; only the middle step changes. For $k [x; δ]$ it is Amitsur's simplicity theorem, for $k [x; σ]$ it is $(11.12)$ , and for $k [x]$ over a division ring it is the description of ideals by central polynomials.

Twisted polynomial rings over a division ring — $k [x; σ, δ]$ and its specialisations
- $δ = 0$ , $σ$ non-trivial
  - ideals $R x^{m}$ under $(11.12)$
  - left primitive, not simple
  - controls linear difference operators
- $σ = id$ , $δ$ non-inner, characteristic $0$
  - simple domain by Amitsur $(3.16)$
  - Weyl algebra when $k$ is a rational function field with $δ = d / d t$
  - controls linear differential operators
- $σ = id$ , $δ = 0$
  - $k [x]$ ; primitive only when $k$ is not algebraic over its centre $(11.14)$
  - commutative case is never primitive unless $k$ is a field and $R$ is not

The related pages Skew Polynomial Rings and Hilbert's Twist, Simplicity Criteria for Differential Polynomial Rings and Differential Polynomial Rings and the Weyl Algebra develop the constructions themselves; this page uses them as a source of primitive rings.

Applications and Industry Use

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

Control theory

Linear time-varying systems

Skew polynomial rings are the operator algebras of linear differential and difference systems with non-constant coefficients. The twist encodes the shift or the derivative, and module-theoretic notions such as controllability translate into the structure theory used here.

Coding theory

Skew cyclic codes

Codes defined as left ideals of $k [x; σ] / (x^{n} - 1)$ over finite fields — introduced by Boucher and Ulmer — exploit exactly the right-division algorithm quoted in this page, and give codes with parameters unattainable by classical cyclic codes.

Symbolic computation

Ore algebras in CAS

Maple's Ore algebra tools, Sage's skew and Ore polynomial rings, and holonomic-function packages all implement $k [x; σ, δ]$ directly, with noncommutative Groebner bases for left ideals.

Representation theory

Counterexample supply

These rings are the standard source of primitive rings that are neither simple nor artinian, and the classification by $σ$ -conjugacy is the model for parametrising simple modules over Ore extensions generally.

Stated honestly: the primitivity of these rings is used inside algebra, as a source of examples with prescribed pathology. Their construction, on the other hand, is genuinely applied — every computer algebra treatment of linear functional equations lives in a skew polynomial ring.

Standards and Notation

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

This collection

k [x; σ]

with

x b = σ (b) x

, left coefficients

Common variant

b x = x σ (b)

with right coefficients; the two conventions are related by passing to

R^{op}

General Ore extension

R [x; σ, δ]

with

x b = σ (b) x + δ (b)

Differential case

k [x; δ]

, sometimes written

k [\partial; δ]

A_{1} (k)

for the Weyl algebra

SageOrePolynomialRing, k['x', sigma] for the twisted case

Maple / MagmaOre_algebra package; TwistedPolynomials for finite-field twists

Failure Modes and Common Mistakes

Do not evaluate twisted polynomials naively: $g \mapsto g (a)$ is not multiplicative in $k [x; σ]$ . The correct substitute is the module action, $g * 1$ , with $x^{m} * 1 = σ^{m - 1} (a) \dots σ (a) a$ .
Do not assume $σ$ -conjugacy is ordinary conjugacy. It reduces to conjugacy only when $σ = id$ ; the factor $σ (c)$ is what makes infinitely many classes possible over a commutative $k$ .
Do not expect these rings to have minimal one-sided ideals. They are domains that are not division rings, so their socles vanish and $(11.11)$ gives no information about them.
Do not apply $(11.14)$ to a commutative $k$ : a commutative division ring is a field, which is algebraic over its own centre, so the hypothesis is vacuous there.

Historical Notes and Lessons Learned

1933Ore extensionsOystein Ore develops the theory of noncommutative polynomial rings with a twist, including the division algorithm and the conditions under which a ring of fractions exists.
1937Jacobson's differential operatorsJacobson studies rings of differential polynomials over division rings, establishing the module-theoretic machinery later used to produce primitive examples.
1957Amitsur's simplicity criterionAmitsur characterises when a differential polynomial ring over a division ring is simple; in characteristic zero a non-inner derivation suffices.
1964–65One-sided primitivityBergman's example shows primitivity is genuinely one-sided; twisted polynomial constructions become the standard toolkit for building rings with prescribed module-theoretic behaviour.
2007Skew cyclic codesBoucher, Geiselmann and Ulmer introduce codes defined by left ideals of skew polynomial rings over finite fields, giving the construction a concrete engineering use.

The lesson is one of economy. A single twisted variable over a division ring generates enough noncommutativity to realise, in explicit form, phenomena that abstract theory can only assert: primitive rings that are not simple, and primitive rings whose faithful simple modules form an infinite family.

Quick Reference

Twist

x b = σ (b) x

k [x; σ]

;

x b = b x + δ (b)

k [x; δ]

DivisionRight division by any nonzero

f

; left ideals are principal

$(11.12)$

σ

not an automorphism of finite inner order

\Rightarrow

ideals are

R x^{m}

$(11.13)$

a \neq 0 \Rightarrow M_{a}

faithful simple;

R

left primitive

Action on $M_{a} = k$

x * b = σ (b) a

; twisted evaluation

x^{m} * 1 = σ^{m - 1} (a) \dots σ (a) a

Isomorphism

M_{a} ≅ M_{a'} \Leftrightarrow a = σ (c) a' c^{- 1}

$(11.14)$

k

not algebraic over

C

k [x]

is left and right primitive

Socle

0

— these are domains, not division rings

Choosing the twist for a desired property
Want	Take	Reference
A simple primitive domain	$k [x; δ]$ , $char k = 0$ , $δ$ non-inner	Amitsur $(3.16)$
Primitive but not simple	$k [x; σ]$ with $σ$ of infinite inner order	$(11.12)$ , $(11.13)$
Infinitely many faithful simples	$ℝ (t) [x; σ]$ , $σ : t \mapsto t + 1$	degree argument
Left and right primitive, untwisted	$k [x]$ , $k$ not algebraic over $C$	$(11.14)$
Not primitive at all	$k [x]$ with $k$ a field	$(11.8)$

Frequently Asked Questions

Why does the hypothesis of

(11.12)

mention inner order rather than plain order?

Because the obstruction produced by the proof is an identity $σ^{d} (a) = a_{n} a a_{n}^{- 1}$ , which says $σ^{d}$ is inner — not that it is the identity. Over a noncommutative $k$ an automorphism can have infinite order while some power is inner, and such a $σ$ really does create extra ideals, so the weaker-looking hypothesis is the correct one.

k [x; σ]

ever simple?

Not under the hypothesis of $(11.12)$ , since $R x$ is then a proper nonzero ideal. Simplicity in the purely twisted case fails for a structural reason: $x$ is a normal element, so it always generates a proper ideal. Simple examples come from the differential side, where $x$ is not normal — this is the content of Amitsur's theorem.

How can a ring over a commutative field

k

have infinitely many non-isomorphic faithful simple modules?

The classification is by $σ$ -conjugacy, $a = σ (c) a' c^{- 1}$ , which over a commutative $k$ reduces to $a = (σ (c) c^{- 1}) a'$ — not to $a = a'$ . The set of elements $σ (c) c^{- 1}$ is a proper subgroup of $k^{\times}$ whenever $σ$ preserves some invariant, and for $ℝ (t)$ with $σ : t \mapsto t + 1$ that invariant is the degree.

Do these rings have minimal one-sided ideals?

No. They are domains that are not division rings, and in such a ring a minimal left ideal $R a$ would force $a$ to be a unit, making $R a = R$ and hence $R$ a division ring. So the socle is zero, which is consistent with $(11.11)$ : a ring with several non-isomorphic faithful simple left modules cannot have a minimal left ideal.

What replaces polynomial evaluation in the twisted setting?

The action on $M_{a} = k$ . Ordinary evaluation is not a ring homomorphism, but $g \mapsto g * 1$ is exactly the projection of $g$ onto the complement of $R (x - a)$ , and it is all that is needed: $g \in R (x - a)$ precisely when $g * 1 = 0$ . The formula $x^{m} * 1 = σ^{m - 1} (a) \dots σ (a) a$ is the twisted analogue of $a^{m}$ .

Why does

(11.14)

require

k

not algebraic over its centre?

Because the ideals of $k [x]$ are generated by central polynomials, and if $a$ were algebraic over $C$ its minimal polynomial over $C$ would generate a nonzero ideal inside $R (x - a)$ , destroying faithfulness. The hypothesis guarantees a supply of $a$ with no such polynomial. It also forces $k$ to be noncommutative, since a field is algebraic over itself.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §11, (11.12)–(11.14) and the differential example on pp. 186–192.
T. Y. Lam, A First Course in Noncommutative Rings, §1 for the construction of skew polynomial rings and §3 for Amitsur's simplicity theorem (3.16).
O. Ore, “Theory of non-commutative polynomials”, Annals of Mathematics 34 (1933), 480–508.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapters III–IV.
J. C. McConnell and J. C. Robson, Noncommutative Noetherian Rings, Wiley, 1987, Chapter 1 on Ore extensions.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 1.

AI Suggested Questions

Give a division ring $k$ and an automorphism $σ$ of infinite order some power of which is inner, and describe the extra ideals of $k [x; σ]$ .
Classify the simple modules of the general Ore extension $k [x; σ, δ]$ over a division ring.
For which fields $k$ and automorphisms $σ$ does $k [x; σ]$ have only finitely many isomorphism classes of faithful simple modules?
Work out the delta-conjugacy classes for the Weyl algebra over a rational function field and count the faithful simple modules.
How does the right-division algorithm in a skew polynomial ring underpin the decoding of skew cyclic codes?
Compare the primitivity of the free algebra with that of the skew polynomial rings on this page, and identify what plays the role of the twist.
Which of these examples remain left primitive after passing to the Ore quotient division ring, and why does the question become trivial there?

Primitive Skew Polynomial Rings

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

The module $M_{a}$ concretely

Why $a = 0$ must be excluded

The invariant that separates the modules

Key Results

Proof Techniques and Method

Classify the ideals first

Commutators to constrain a generator

Compute in the quotient, not the ideal

Worked Example

A primitive ring with infinitely many faithful simple modules

The hypothesis holds

Separating the modules by degree

What this ring demonstrates

Comparison and Classification

Relationship Map

Applications and Industry Use

Linear time-varying systems

Skew cyclic codes

Ore algebras in CAS

Counterexample supply

Standards and Notation

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

The module Ma concretely

Why a=0 must be excluded

The invariant that separates the modules

06Key Results

07Proof Techniques and Method

Classify the ideals first

Commutators to constrain a generator

Compute in the quotient, not the ideal

08Worked Example

A primitive ring with infinitely many faithful simple modules

The hypothesis holds

Separating the modules by degree

What this ring demonstrates

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Linear time-varying systems

Skew cyclic codes

Ore algebras in CAS

Counterexample supply

12Standards and Notation

13Failure Modes and Common Mistakes

14Historical Notes and Lessons Learned

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

The module $M_{a}$ concretely

Why $a = 0$ must be excluded

Key Results

Proof Techniques and Method

Worked Example

Comparison and Classification

Relationship Map

Applications and Industry Use

Standards and Notation

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions