Is k[x;] ever commutative when ≠ id?

No. If (b) ≠ b for some b, then xb = (b)x ≠ bx. Commutativity of k[x;] is therefore equivalent to k commutative and = id.

How does k[x;] differ from a group ring or a crossed product?

The skew Laurent polynomial ring k[x,x^-1;] *is* the skew group ring k * Z for the action of Z generated by . The polynomial ring k[x;] is its 'positive half' — a skew semigroup ring over N — and it is precisely by dropping negative exponents that one is allowed to weaken 'automorphism' to 'endomorphism'.

Does k[x;] have a division algorithm on both sides?

Left division works with no hypothesis beyond k being a division ring, because killing a leading term uses ^m-n(1) = 1. Right division requires solving ^m-n(c) = a_m for c, so it is available exactly when is surjective. This is the source of the standard example of a principal left ideal domain that is not right noetherian.

Skew Polynomial Rings and Hilbert’s Twist

Q: Why are the coefficients written on the left rather than the right?

Because the rule xb = (b)x lets you push every x to the right of every coefficient, so left polynomials are automatically a normal form: each element has a unique expression a_i x^i. Right polynomials are not a normal form — they only span _i ^i(k)x^i, which is all of the ring precisely when is surjective.

Q: What is the centre of k[x;]?

Let k be a field. An element f = a_i x^i is central iff it commutes with every b ∈ k and with x. The first condition reads a_i ^i(b) = b a_i for all b, forcing ^i = id whenever a_i ≠ 0; the second reads (a_i) = a_i. So if has finite order n with fixed field F, then Z(k[x;]) = F[x^n]; if has infinite order, only the constant term survives and Z(k[x;]) = F.

Q: Why does the Laurent construction need an automorphism when the polynomial one does not?

Negative powers of x have to move past coefficients too, and x^-1b = ^-1(b)x^-1 is the only rule compatible with x x^-1 = 1. Without ^-1 there is no consistent multiplication, so k((x;)) simply does not exist for a non-surjective .

Executive Summary

Ordinary polynomials over a commutative ring are commutative, which makes them useless as a source of examples in noncommutative ring theory. Hilbert's twist repairs this in a single stroke: keep the additive structure $⨁_{i \geq 0} k x^{i}$ , but declare that moving $x$ past a coefficient applies a fixed ring endomorphism $σ$ of $k$ . The result, $k [x; σ]$ , is noncommutative whenever $σ \neq id$ , even for commutative $k$ .

Almost every asymmetry in this collection can be produced by choosing $σ$ badly on purpose. If $σ$ fails to be injective, $x$ becomes a left zero-divisor that is not a right zero-divisor. If $σ$ fails to be surjective, $k [x; σ]$ is a principal left ideal domain that is not right noetherian. If $σ$ is an automorphism of a division ring, the Laurent series ring $k ((x; σ))$ is a new division ring — the mechanism behind the first examples of noncommutative ordered division rings.

$x b = σ (b) x$ Defining rule

$U (k)$ Units,

k

a domain

PLID

k

a division ring

1899Hilbert's example

Overview

Let $k$ be a ring and let $σ : k \to k$ be a ring endomorphism (unital, so $σ (1) = 1$ ). Take the free left $k$ -module on symbols $1, x, x^{2}, \dots$ and impose one rule.

x b = σ (b) x (b \in k),

(1.7a)

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

Iterating gives $x^{i} b = σ^{i} (b) x^{i}$ , and hence a closed formula for the product of two left polynomials.

(\sum_{i} a_{i} x^{i}) (\sum_{j} b_{j} x^{j}) = \sum_{i, j} a_{i} σ^{i} (b_{j}) x^{i + j} .

(1.7b)

Associativity is a short check on monomials and follows from $σ$ being multiplicative; distributivity from $σ$ being additive. The same recipe applied to formal power series gives $k [[x; σ]]$ , and — when $σ$ is invertible — to formal Laurent series gives $k ((x; σ))$ .

The construction is the degree-one case of the general Ore extension $k [x; σ, δ]$ , in which the rule is $x b = σ (b) x + δ (b)$ . Setting $δ = 0$ gives this page; setting $σ = id$ gives the Differential Polynomial Rings page. Both are needed, and neither subsumes the other.

Learning Objectives

State the rule $x b = σ (b) x$ and derive the product formula $(1.7 b)$ for left polynomials.
Explain why the right polynomials form a proper subset of $k [x; σ]$ exactly when $σ$ is not surjective.
Show that $x$ is a left zero-divisor but not a right zero-divisor when $σ$ is not injective.
Compute $U (k [x; σ])$ for $k$ a domain and $U (k [[x; σ]])$ for an arbitrary ring $k$ .
Prove that $k ((x; σ))$ is a division ring whenever $k$ is one and $σ \in Aut (k)$ .
Realise $ℍ$ as $ℂ [x; σ] / (x^{2} + 1)$ for $σ$ complex conjugation.

Definitions

Definition(1.7)Skew polynomial ring

Let $k$ be a ring and $σ$ a ring endomorphism of $k$ . The skew polynomial ring $k [x; σ]$ has underlying additive group $⨁_{i \geq 0} k x^{i}$ , with multiplication determined by $(1.7 b)$ ; equivalently, by $k$ -bilinearity from the rule $x b = σ (b) x$ . Its elements are called left polynomials. The skew power series ring $k [[x; σ]]$ is defined identically on $\prod_{i \geq 0} k x^{i}$ .

Definition(1.8)Skew Laurent series ring

Let $k$ be a ring and $σ$ an automorphism of $k$ . The skew Laurent series ring $k ((x; σ))$ consists of the formal series $\sum_{i \in ℤ} a_{i} x^{i}$ with $a_{i} = 0$ for all sufficiently negative $i$ , multiplied using $x^{i} b = σ^{i} (b) x^{i}$ for every $i \in ℤ$ . The subring of series with only finitely many nonzero terms is the skew Laurent polynomial ring $k [x, x^{- 1}; σ]$ .

Invertibility of $σ$ is not a convenience here: the rule $x^{- 1} b = σ^{- 1} (b) x^{- 1}$ has no meaning otherwise.

$\deg f$: For $f = \sum a_{i} x^{i} \neq 0$ in $k [x; σ]$ , the largest $i$ with $a_{i} \neq 0$ . The leading coefficient is that $a_{i}$ .
$ord F$: For $F \neq 0$ in $k [[x; σ]]$ or $k ((x; σ))$ , the smallest $i$ with $a_{i} \neq 0$ . The order replaces the degree in the series setting.
Left zero-divisor: An element $a \neq 0$ with $a b = 0$ for some $b \neq 0$ . A right zero-divisor satisfies $b a = 0$ for some $b \neq 0$ .
Principal left ideal domain: A domain in which every left ideal has the form $R f$ for a single $f$ . Abbreviated PLID; the right-handed notion is PRID.
Ore extension: The common generalisation $k [x; σ, δ]$ with $x b = σ (b) x + δ (b)$ , where $δ$ is a sigma-derivation.

Throughout, endomorphisms are unital and $k$ is a ring with identity. Left polynomials are the default; the phrase 'polynomial' with no qualifier always means left polynomial.

Core Concepts

Left polynomials versus right polynomials

The definition privileges one side, and the privilege is real. A right polynomial $c_{0} + x c_{1} + \dots + x^{n} c_{n}$ can always be rewritten as a left polynomial, because $x^{i} c_{i} = σ^{i} (c_{i}) x^{i}$ :

c_{0} + x c_{1} + \dots + x^{n} c_{n} = c_{0} + σ (c_{1}) x + \dots + σ^{n} (c_{n}) x^{n} .

(1.7c)

Right polynomials are the left polynomials whose $i$ -th coefficient lies in $σ^{i} (k)$ .

So the right polynomials form the additive subgroup $⨁_{i} σ^{i} (k) x^{i}$ . If $σ$ is onto this is everything; if not, it is a proper subgroup, and the two notions genuinely differ. This is the first place where non-surjectivity of $σ$ leaves a visible mark, and it is the seed of the one-sided chain conditions discussed on the One-Sided Chain Conditions page.

Failure of injectivity: a lopsided zero-divisor

Suppose $σ (b) = 0$ for some $b \neq 0$ in $k$ . Then $x b = σ (b) x = 0$ , so $x$ is a left zero-divisor. But $x$ is not a right zero-divisor: for $f = \sum a_{i} x^{i} \neq 0$ we have $f x = \sum a_{i} x^{i + 1} \neq 0$ , since right multiplication by $x$ merely shifts the coefficients without touching them.

Degrees and orders

If $f$ has degree $m$ with leading coefficient $a_{m}$ and $g$ has degree $n$ with leading coefficient $b_{n}$ , then by $(1.7 b)$ the coefficient of $x^{m + n}$ in $f g$ is $a_{m} σ^{m} (b_{n})$ . Degrees therefore add precisely when this product is nonzero — which is guaranteed if $k$ is a domain and $σ$ is injective, and can fail otherwise.

Conjugation in the Laurent setting

In $k ((x; σ))$ the variable is invertible, so the rule $x b = σ (b) x$ can be rewritten as

σ (b) = x b x^{- 1} (b \in k) .

(1.8a)

The twist becomes an inner automorphism of the larger ring, restricted to $k$ .

This is the conceptual payoff of the Laurent construction: an arbitrary automorphism of $k$ is realised as conjugation inside a ring containing $k$ . It is the same idea that underlies cyclic algebras, where a generator of a Galois group is made inner by adjoining one element.

Key Results

Proposition(1.7)(a)Domains and degrees

Let $k$ be a domain and $σ$ an injective ring endomorphism of $k$ . Then $\deg (f g) = \deg f + \deg g$ for all nonzero $f, g \in k [x; σ]$ ; in particular $k [x; σ]$ is a domain, and so is $k [[x; σ]]$ .

Proof

Write $f = \sum_{i \leq m} a_{i} x^{i}$ with $a_{m} \neq 0$ and $g = \sum_{j \leq n} b_{j} x^{j}$ with $b_{n} \neq 0$ . By $(1.7 b)$ the coefficient of $x^{m + n}$ in $f g$ is $a_{m} σ^{m} (b_{n})$ , and no higher power occurs. Injectivity of $σ$ gives $σ^{m} (b_{n}) \neq 0$ , and $k$ being a domain gives $a_{m} σ^{m} (b_{n}) \neq 0$ . Hence $f g \neq 0$ and $\deg (f g) = m + n$ .

For power series replace degree by order: if $ord F = m$ and $ord G = n$ then the coefficient of $x^{m + n}$ in $F G$ is again $a_{m} σ^{m} (b_{n}) \neq 0$ .

Corollary(1.7)(b)Units of the polynomial ring

Under the hypotheses of $(1.7) (a)$ — $k$ a domain, $σ$ injective — we have $U (k [x; σ]) = U (k)$ .

Proof

If $f g = 1$ then $\deg f + \deg g = 0$ , so both are constants; the statement reduces to $U (k) = U (k)$ . Conversely a unit of $k$ is visibly a unit of $k [x; σ]$ .

The hypothesis that $k$ be a domain cannot be dropped: in $(ℤ / 4 ℤ) [x]$ one has $(1 + 2 x) (1 - 2 x) = 1 - 4 x^{2} = 1$ , so $1 + 2 x$ is a unit of degree $1$ .

Proposition(1.7)(c)Units of the power series ring

Let $k$ be any ring and $σ$ any ring endomorphism of $k$ . Then $F = \sum_{i \geq 0} a_{i} x^{i} \in k [[x; σ]]$ is a unit if and only if $a_{0} \in U (k)$ .

Proof

Necessity is clear: the constant term of a product is the product of the constant terms, so $F G = G F = 1$ forces $a_{0} \in U (k)$ .

For sufficiency, suppose $a_{0} \in U (k)$ and look for $G = \sum_{j} b_{j} x^{j}$ with $F G = 1$ . Comparing coefficients of $x^{n}$ in $(1.7 b)$ gives $\sum_{i + j = n} a_{i} σ^{i} (b_{j}) = δ_{n 0}$ , i.e. $a_{0} b_{0} = 1$ and $a_{0} b_{n} = - \sum_{i = 1}^{n} a_{i} σ^{i} (b_{n - i})$ for $n \geq 1$ . Since $a_{0}$ is invertible these equations determine $b_{0}, b_{1}, b_{2}, \dots$ recursively, so $F$ is right-invertible.

For a left inverse, solve $G F = 1$ instead: the equations are $b_{0} a_{0} = 1$ and $b_{n} σ^{n} (a_{0}) = - \sum_{j = 0}^{n - 1} b_{j} σ^{j} (a_{n - j})$ . These are solvable because $σ^{n} (a_{0}) \in U (k)$ — a unital ring homomorphism carries units to units, with $σ^{n} (a_{0})^{- 1} = σ^{n} (a_{0}^{- 1})$ . Having both a left and a right inverse, $F$ is a unit.

Theorem(1.8)New division rings from old

Let $k$ be a division ring and $σ$ an automorphism of $k$ . Then $k ((x; σ))$ is a division ring. Consequently the construction may be iterated to produce division rings of iterated skew Laurent series.

Proof

Let $F \neq 0$ have order $t$ , so $F = \sum_{i \geq t} a_{i} x^{i}$ with $a_{t} \neq 0$ . Then $F x^{- t} = \sum_{i \geq t} a_{i} x^{i - t}$ has nonzero constant term $a_{t}$ , which is a unit because $k$ is a division ring. By $(1.7) (c)$ , $F x^{- t}$ is a unit of $k [[x; σ]]$ , hence of $k ((x; σ))$ . Since $x^{- t}$ is invertible, so is $F$ .

The argument uses invertibility of $σ$ only to make $k ((x; σ))$ a ring at all; once that is granted, the unit computation is the one already performed for power series.

Proposition—Left division and principal left ideals

Let $k$ be a division ring and $σ$ any ring endomorphism of $k$ (automatically injective, since $\ker σ$ is a proper ideal of $k$ ). Then $k [x; σ]$ admits a left division algorithm: for $f, g \in k [x; σ]$ with $g \neq 0$ there exist $q, r$ with $f = q g + r$ and $r = 0$ or $\deg r < \deg g$ . Consequently every left ideal of $k [x; σ]$ is principal, so $k [x; σ]$ is a principal left ideal domain.

Proof

Normalise $g$ to be monic by replacing it with $b_{n}^{- 1} g$ , where $b_{n}$ is its leading coefficient; this changes neither $g k [x; σ]$ nor the left ideal $k [x; σ] g$ . Induct on $\deg f$ . If $\deg f = m < n = \deg g$ take $q = 0$ , $r = f$ . If $m \geq n$ and $f$ has leading coefficient $a_{m}$ , then $a_{m} x^{m - n} g$ has leading term $a_{m} σ^{m - n} (1) x^{m} = a_{m} x^{m}$ , so $f - a_{m} x^{m - n} g$ has strictly smaller degree; apply the inductive hypothesis to it.

Now let $I \neq 0$ be a left ideal and choose $f \in I ∖ {0}$ of least degree. For $h \in I$ write $h = q f + r$ ; then $r = h - q f \in I$ , and minimality forces $r = 0$ . Hence $I = k [x; σ] f$ . That $k [x; σ]$ is a domain is $(1.7) (a)$ .

Note where the argument breaks on the other side: killing the leading term of $f$ from the right would require solving $σ^{\cdot} (c) = a_{m}$ , which needs $σ$ to be surjective.

Remark—Ordered noncommutative division rings

Suppose $k$ is an ordered division ring and $σ$ an order-preserving automorphism. Declare $F \in k ((x; σ))$ positive when its lowest nonzero coefficient is positive. This is a total order compatible with addition, and compatible with multiplication because the lowest coefficient of $F G$ is $a_{m} σ^{m} (b_{n})$ , a product of positives. Taking $k = ℚ (t)$ ordered by the sign at $+ \infty$ and $σ (t) = 2 t$ gives a noncommutative ordered division ring — Hilbert's original point.

Proof Techniques and Method

How these arguments work, and which move to reuse.

Four techniques carry essentially all of the theory of twisted polynomial and series rings.

Move 1

Compare extreme coefficients

Degrees for polynomials, orders for series. The extreme coefficient of a product is $a_{m} σ^{m} (b_{n})$ , and every statement about domains, units and zero-divisors is a statement about when that expression vanishes.

Move 2

Solve triangular recursions

Inverting a power series means solving $a_{0} b_{n} = (known)$ for each $n$ in turn. The recursion is solvable precisely when $a_{0}$ — and hence every $σ^{n} (a_{0})$ — is a unit.

Move 3

Multiply by a power of $x$

In the Laurent ring, shifting by $x^{- t}$ converts a general series into one with nonzero constant term. Every Laurent statement reduces to a power-series statement this way.

Move 4

Kill the leading term

The Euclidean step $f \mapsto f - a_{m} x^{m - n} g$ works on the left with no hypothesis on $σ$ beyond $σ (1) = 1$ ; the mirrored step needs $σ$ surjective. That single observation explains every left-right asymmetry of $k [x; σ]$ .

Move 4 is the one to internalise. It shows that the failure of surjectivity is not a technical nuisance but the exact obstruction to running the theory on the other side, and it predicts in advance that a non-surjective $σ$ will produce a left-but-not-right noetherian ring.

Worked Example

The quaternions as a twisted polynomial quotient

Take $k = ℂ$ and let $σ$ be complex conjugation, $σ (z) = \bar{z}$ , an automorphism of order $2$ . Form $A = ℂ [x; σ]$ , so that

x z = \bar{z} x (z \in ℂ), x^{2} z = \bar{\bar{z}} x^{2} = z x^{2} .

(E.1)

$x^{2}$ commutes with every scalar and with $x$ , so $x^{2}$ is central in $A$ .

Because $x^{2}$ is central, so is $x^{2} + 1$ , and $A (x^{2} + 1)$ is a two-sided ideal. Write $B = A / A (x^{2} + 1)$ and let $j$ denote the image of $x$ , $i$ the image of $\sqrt{- 1} \in ℂ$ . Then

i^{2} = - 1, j^{2} = - 1, j i = \bar{i} j = - i j .

(E.2)

Setting $ℓ = i j$ gives $ℓ^{2} = i j i j = - i^{2} j^{2} = - 1$ , so $1, i, j, ℓ$ satisfy exactly Hamilton's relations. As a left $ℂ$ -module $A$ is free on $1, x$ modulo $(x^{2} + 1)$ , so $\dim_{ℝ} B = 4$ and

ℂ [x; σ] / (x^{2} + 1) ≅ ℍ,

(E.3)

The real quaternions, produced from a commutative field by one twist and one quotient.

A twist with infinite order

Let $k = ℚ (t)$ and let $σ$ be the $ℚ$ -automorphism with $σ (t) = 2 t$ . In $k [x; σ]$ the basic relation is $x t = 2 t x$ , and $σ$ has infinite order, so $x^{n} t = 2^{n} t x^{n}$ . Since $σ$ is an automorphism, $k ((x; σ))$ is a division ring by $(1.8)$ , and $ℚ (t)$ carries the ordering in which $t$ exceeds every rational; $σ$ preserves it, so $k ((x; σ))$ is an ordered division ring that is not commutative.

A twist that is injective but not surjective

Let $k = 𝔽_{p} (t)$ and $σ (f) = f^{p}$ , the Frobenius. Then $σ$ is injective (as $k$ is a field) with image $𝔽_{p} (t^{p}) ⊊ k$ . Here $k [x; σ]$ is a principal left ideal domain, but $t \notin σ (k)$ , and the One-Sided Chain Conditions page uses exactly this element to exhibit an infinite direct sum of right ideals. Note also that $𝔽_{p} (t)$ has degree $p$ over $σ (k)$ , so the failure of surjectivity here is as small as it can be and still fatal.

Comparison and Classification

One construction, four hypotheses on $σ$
Hypothesis on $σ$	$k [x; σ]$ a domain?	Right polys $=$ left polys?	Typical pathology
$σ = id$	iff $k$ is	yes	none — the ring is $k [x]$
$σ$ not injective	no	no	$x$ is a left but not a right zero-divisor
$σ$ injective, not onto	iff $k$ is	no	left noetherian, not right noetherian
$σ \in Aut (k)$	iff $k$ is	yes	none of the above; $k ((x; σ))$ exists

Which twisted ring has which property ( $k$ a division ring, $σ$ an automorphism unless stated)
	Domain	Division ring	PLID	Noetherian both sides
$k [x]$	yes	no	yes	yes
$k [x; σ]$ , $σ$ onto	yes	no	yes	yes
$k [x; σ]$ , $σ$ not onto	yes	no	yes	no
$k [[x; σ]]$	yes	no	yes	yes
$k [x, x^{- 1}; σ]$	yes	no	yes	yes
$k ((x; σ))$	yes	yes	yes	yes

Which twisted ring has which property ( $k$ a division ring, $σ$ an automorphism unless stated)

The table's last column is the interesting one: the only entry that fails is the one where $σ$ is not surjective, and it fails on exactly one side.

Relationship Map

The twisted constructions form a tower, each obtained from the previous by completing or inverting.

k [x; σ]

k [x, x^{- 1}; σ]

k ((x; σ))

Sideways, the twist by $σ$ sits alongside the twist by a derivation, both special cases of the Ore extension.

$k [x; σ, δ]$ — Ore extension: $x b = σ (b) x + δ (b)$
- $δ = 0$
  - $k [x; σ]$ — this page
  - $k [x]$ when also $σ = id$
- $σ = id$
  - $k [x; δ]$ — differential polynomial rings
  - the Weyl algebra $A_{1} (k)$
- both nontrivial
  - quantised Weyl algebras
  - $q$ -difference operator rings

Downstream, $k [x; σ]$ and its Laurent versions supply the standard examples for three later topics: simplicity criteria (Simplicity of Skew Laurent Polynomial Rings), primitivity (Skew Polynomial Rings as Primitive Rings), and division-ring construction (Twisted Laurent Series Division Rings).

Rings

k [x; σ]

for arbitrary

k

σ

Domains

k

a domain,

σ

injective

Principal left ideal domains

k

a division ring,

σ

any endomorphism

Division rings

k ((x; σ))

σ

an automorphism

Applications and Industry Use

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

Coding theory

Skew cyclic codes

Codes defined as left ideals of $𝔽_{q^{m}} [x; θ] / (x^{n} - 1)$ with $θ$ a power of Frobenius. Because $σ$ is not central, many more left ideals exist than in the commutative case, which yields codes with better minimum distance than any classical cyclic code of the same length.

Control theory

Linear time-varying systems

A linear system with time-dependent coefficients is a module over an Ore algebra; the shift or differentiation operator does not commute with the coefficient functions, and $k [x; σ]$ is exactly the algebra of difference operators with variable coefficients.

Symbolic computation

Ore algebras in CAS

Systems that manipulate recurrences and $q$ -difference equations represent operators in $k [x; σ, δ]$ and rely on the left division algorithm; noncommutative Groebner bases over these rings underpin creative-telescoping algorithms.

Division algebras

Building noncommutative fields

Iterating $k \mapsto k ((x; σ))$ produces division rings of prescribed centre and prescribed transcendence behaviour. This is the standard machine for counterexamples about ordered rings, and Hilbert's original motivation.

The honest summary: this construction is a factory. It rarely models a physical system directly, but it manufactures the algebras that other subjects — coding, symbolic computation, systems theory — then take as their ground rings.

Standards and Notation

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

Preferred notation

k [x; σ]

with

x b = σ (b) x

(Lam, McConnell–Robson)

Common variant

R [x; σ, δ]

for the full Ore extension;

θ

in place of

σ

in coding papers

Opposite conventionSome authors write

b x = x σ (b)

, which is this ring built over

k^{op}

— check before quoting

Series rings

k [[x; σ]]

and

k ((x; σ))

; the Laurent polynomial ring is

k [x, x^{- 1}; σ]

SageSkewPolynomialRing(k, sigma) and the Ore polynomial constructor k['x', sigma]

Other systemsMaple's Ore algebra package and Singular:Plural implement left division and noncommutative Groebner bases

Failure Modes and Common Mistakes

Do not assume $k [x; σ]$ is free as a right $k$ -module. It is free on ${x^{i}}$ as a left $k$ -module; on the right, $x^{i} k = σ^{i} (k) x^{i}$ is generally a proper subgroup of $k x^{i}$ .
Do not expect the centre to be large. For $k$ a field, $Z (k [x; σ]) = F [x^{n}]$ when $σ$ has finite order $n$ with fixed field $F$ , and $Z (k [x; σ]) = F$ when $σ$ has infinite order — in the latter case the ring is not even a finite module over its centre.
Do not confuse $k [x; σ]$ with the group ring or with $k \otimes k [x]$ ; the underlying additive group agrees with $k [x]$ but no ring map between them exists in general.
Do not transport a right-handed theorem by symmetry. The mirror of $k [x; σ]$ is $k^{op} [x; σ]$ , and 'the same argument on the other side' is available only when $σ$ is onto.

Quick Reference

Rule

x b = σ (b) x

, so

x^{i} b = σ^{i} (b) x^{i}

Product

(\sum a_{i} x^{i}) (\sum b_{j} x^{j}) = \sum a_{i} σ^{i} (b_{j}) x^{i + j}

Leading coefficient

a_{m} σ^{m} (b_{n})

for degrees

m

n

Domain

k

a domain and

σ

injective

Units (polynomials)

U (k)

, when

k

is a domain and

σ

injective

Units (power series)

a_{0} \in U (k)

, no hypotheses

Division ring

k ((x; σ))

for

k

a division ring,

σ \in Aut (k)

Ideal theoryPLID when

k

is a division ring; principal on the right only if

σ

is onto

Which ring to reach for
Want	Take	Because
A noncommutative domain	$k [x; σ]$ , $σ \neq id$ injective	degrees add, so no zero-divisors
A left-right asymmetric ring	$k [x; σ]$ , $σ$ not onto	left division works, right division does not
A left but not right zero-divisor	$k [x; σ]$ , $σ$ not injective	$x b = 0$ while $f x \neq 0$
A new division ring	$k ((x; σ))$ , $σ \in Aut (k)$	lowest-order coefficient is invertible
An ordered noncommutative division ring	$k ((x; σ))$ , $σ$ order-preserving	sign of the lowest coefficient is multiplicative
A local ring	$k [[x; σ]]$ , $k$ a division ring	non-units are exactly the series of positive order

Frequently Asked Questions

Why are the coefficients written on the left rather than the right?

Because the rule $x b = σ (b) x$ lets you push every $x$ to the right of every coefficient, so left polynomials are automatically a normal form: each element has a unique expression $\sum a_{i} x^{i}$ . Right polynomials are not a normal form — they only span $⨁_{i} σ^{i} (k) x^{i}$ , which is all of the ring precisely when $σ$ is surjective.

k [x; σ]

ever commutative when

σ \neq id

No. If $σ (b) \neq b$ for some $b$ , then $x b = σ (b) x \neq b x$ . Commutativity of $k [x; σ]$ is therefore equivalent to $k$ commutative and $σ = id$ .

What is the centre of

k [x; σ]

Let $k$ be a field. An element $f = \sum a_{i} x^{i}$ is central iff it commutes with every $b \in k$ and with $x$ . The first condition reads $a_{i} σ^{i} (b) = b a_{i}$ for all $b$ , forcing $σ^{i} = id$ whenever $a_{i} \neq 0$ ; the second reads $σ (a_{i}) = a_{i}$ . So if $σ$ has finite order $n$ with fixed field $F$ , then $Z (k [x; σ]) = F [x^{n}]$ ; if $σ$ has infinite order, only the constant term survives and $Z (k [x; σ]) = F$ .

Why does the Laurent construction need an automorphism when the polynomial one does not?

Negative powers of $x$ have to move past coefficients too, and $x^{- 1} b = σ^{- 1} (b) x^{- 1}$ is the only rule compatible with $x x^{- 1} = 1$ . Without $σ^{- 1}$ there is no consistent multiplication, so $k ((x; σ))$ simply does not exist for a non-surjective $σ$ .

How does

k [x; σ]

differ from a group ring or a crossed product?

The skew Laurent polynomial ring $k [x, x^{- 1}; σ]$ is the skew group ring $k * ℤ$ for the action of $ℤ$ generated by $σ$ . The polynomial ring $k [x; σ]$ is its 'positive half' — a skew semigroup ring over $ℕ$ — and it is precisely by dropping negative exponents that one is allowed to weaken 'automorphism' to 'endomorphism'.

Does

k [x; σ]

have a division algorithm on both sides?

Left division works with no hypothesis beyond $k$ being a division ring, because killing a leading term uses $σ^{m - n} (1) = 1$ . Right division requires solving $σ^{m - n} (c) = a_{m}$ for $c$ , so it is available exactly when $σ$ is surjective. This is the source of the standard example of a principal left ideal domain that is not right noetherian.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §1, Examples (1.7)–(1.8) (pp. 9–10); see also §3 and §14.
O. Ore, “Theory of non-commutative polynomials”, Annals of Mathematics 34 (1933), 480–508.
J. C. McConnell and J. C. Robson, Noncommutative Noetherian Rings, revised edition, Graduate Studies in Mathematics 30, American Mathematical Society, 2001, Chapter 1.
P. M. Cohn, Free Rings and Their Relations, 2nd edition, London Mathematical Society Monographs 19, Academic Press, 1985.
D. Boucher and F. Ulmer, “Coding with skew polynomial rings”, Journal of Symbolic Computation 44 (2009), 1644–1656.
N. Jacobson, The Theory of Rings, American Mathematical Society Mathematical Surveys 2, 1943, Chapter 3.

AI Suggested Questions

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01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Left polynomials versus right polynomials

Failure of injectivity: a lopsided zero-divisor

Degrees and orders

Conjugation in the Laurent setting

06Key Results

07Proof Techniques and Method

Compare extreme coefficients

Solve triangular recursions

Multiply by a power of x

Kill the leading term

08Worked Example

The quaternions as a twisted polynomial quotient

A twist with infinite order

A twist that is injective but not surjective

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Skew cyclic codes

Linear time-varying systems

Ore algebras in CAS

Building noncommutative fields

12Standards and Notation

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

Multiply by a power of $x$

Worked Example

Comparison and Classification

Relationship Map

Applications and Industry Use

Standards and Notation

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions