Polynomial Rings, Power Series and Laurent Series over a Ring

Q: Why does the unit criterion for k[[x]] not need k to be commutative?

Because the recursion only ever divides by the constant term, and it can be run on either side. Solving FG = 1 from the left gives a right inverse; solving HF = 1 from the right gives a left inverse; the two coincide by associativity. Nowhere is a coefficient moved past another coefficient. The convention that x commutes with k is used, however, and dropping it changes the answer — see the skew case.

Q: Is k((x)) the field of fractions of k[[x]]?

For k a field, yes: k[[x]] is a discrete valuation ring and k((x)) is its fraction field, obtained by inverting x alone. For k a division ring the same statement holds with *fraction field* replaced by *Ore quotient ring*, since x is central the localisation is unproblematic.

Q: Why is rad k[[x]] not nil?

The radical is x,k[[x]] for k a field, and no nonzero element of it is nilpotent because ord(F^n) = n ord(F) is finite for every n. This is the cleanest counterexample to the false implication *radical implies nil*; the true general statement runs the other way, every nil one-sided ideal lies in the radical.

Q: What breaks if I allow infinitely many negative exponents?

Multiplication stops being defined. The coefficient of x^n in a product is _p+q=n f_p g_q, and with unbounded negative support this is an infinite sum with no meaning in a bare ring. Any repair requires an ordering on the exponents together with a well-ordering condition on supports, which is precisely the Malcev–Neumann series construction.

Q: Does k[[x]] inherit the noetherian property from k?

Yes for one variable, and for finitely many: if k is left noetherian then so is k[[x_1,…,x_n]]. For infinitely many variables it fails, exactly as the Hilbert Basis Theorem fails for infinitely many variables. Artinian is never inherited by any of these constructions.

Q: How much of this survives when x no longer commutes with k?

The unit criterion survives in the form U(k[[x;]]) = a_0 + a_1 x + … : a_0 ∈ U(k) for any endomorphism , and the Laurent construction still yields a division ring provided is an automorphism. What changes is the ideal theory and the chain conditions: k[x;] can be left noetherian without being right noetherian.

Executive Summary

Four constructions build new rings from a given ring $k$ by adjoining a variable: the polynomial ring $k [x]$ , the formal power series ring $k [[x]]$ , the Laurent polynomial ring $k [x, x^{- 1}]$ and the Laurent series ring $k ((x))$ . They differ in exactly one respect that matters — which elements become invertible.

Adjoining $x$ to $k$ polynomially adds nothing to the unit group when $k$ is a domain. Completing to power series makes every series with unit constant term invertible. Inverting $x$ as well makes every nonzero series invertible when $k$ is a division ring — which is why the Laurent construction is a machine for producing new division rings, and why Hilbert used its twisted version to build the first noncommutative ordered division ring.

4Constructions

$f_{0} \in U (k)$ Unit test in

k [[x]]

$k ((x))$ Division ring from division ring

(1.5)–(1.6)Lam's numbering

Overview

Let $k$ be a ring with identity, not assumed commutative, and let ${x_{i} : i \in I}$ be a family of independent variables. The standing convention in this collection is that the variables commute with the coefficients: $x_{i} b = b x_{i}$ for every $b \in k$ . Relaxing exactly that convention is what produces the twisted constructions treated in Skew Polynomial Rings and Hilbert's Twist.

With the convention in force, a formal power series is a formal sum of homogeneous components, one in each degree:

F = f_{0} + f_{1} + f_{2} + \dots, f_{n} a homogeneous polynomial of degree n in the x_{i} over k,

(1.5)

Multiplication is formal: the degree- $n$ component of $F G$ is $\sum_{p + q = n} f_{p} g_{q}$ , a finite sum.

In one variable this is the familiar $F = \sum_{n \geq 0} f_{n} x^{n}$ with $f_{n} \in k$ . The Laurent series ring loosens the bound on the left rather than the right: negative exponents are permitted, but only finitely many of them.

These rings are the standard supply of examples throughout the subject: $k [x]$ supplies a noetherian ring with zero radical, $k [[x]]$ supplies a local ring whose radical is nonzero and not nil, and $k ((x))$ supplies division rings that are not fields once the twist of the next page is added.

Learning Objectives

Construct $k [x]$ , $k [[x]]$ , $k [x, x^{- 1}]$ and $k ((x))$ over an arbitrary ring $k$ and identify the inclusions between them.
Prove the unit criterion $(1.5)$ for formal power series and explain why solving one-sided equations suffices.
Prove that $k ((x))$ is a division ring when $k$ is, using the order function.
Compute $rad k [[x]]$ and contrast it with $rad k [x]$ .
Give a ring $R$ and a unit of $R [x]$ that does not lie in $U (R)$ .
State the chain conditions inherited by each construction from $k$ .

Definitions

Definition(1.5)Formal power series ring

Let $k$ be a ring and ${x_{i} : i \in I}$ a family of independent variables, taken pairwise commuting or not, but in all cases commuting with the elements of $k$ . The ring $k [[x_{i} : i \in I]]$ has as elements the formal sums $F = f_{0} + f_{1} + f_{2} + \dots$ in which each $f_{n}$ is a *homogeneous polynomial of degree $n$ * in the $x_{i}$ with coefficients in $k$ . Addition is componentwise and multiplication is the formal one, which is well defined because each degree of $F G$ receives contributions from only finitely many pairs.

Definition(1.6)Laurent series ring

For a ring $k$ , $k ((x))$ denotes the set of formal series $F = \sum_{i \in ℤ} f_{i} x^{i}$ with $f_{i} \in k$ such that $f_{i} = 0$ for all but finitely many $i < 0$ . Multiplication is formal, with $x$ central. The condition that the negative part is finite is what makes each coefficient of a product a finite sum.

$\deg F$: For a nonzero polynomial, the largest $n$ with $f_{n} \neq 0$ . Undefined (or $- \infty$ ) for $F = 0$ .
$ord F$: For a nonzero series, the least $n$ with $f_{n} \neq 0$ ; set $ord 0 = \infty$ . On $k ((x))$ this takes values in $ℤ$ .
$k [x, x^{- 1}]$: The Laurent polynomial ring: series in $k ((x))$ with only finitely many nonzero coefficients. It is $ℤ$ -graded.
Constant term: The degree-zero component $f_{0}$ of $F$ . The unit criterion $(1.5)$ says invertibility of $F$ in $k [[x]]$ depends on $f_{0}$ alone.
$x$ -adic filtration: The descending chain $k [[x]] \supseteq (x) \supseteq (x^{2}) \supseteq \dots$ with intersection zero, which makes $k [[x]]$ a complete filtered ring.

Throughout, ring means ring with identity, and $U (R)$ denotes the group of two-sided invertible elements.

Core Concepts

Degree and order run in opposite directions

A polynomial ring is graded from below and bounded above; a power series ring is bounded below and unbounded above. That single asymmetry drives everything. Degree gives $k [x]$ a $ℤ_{\geq 0}$ -grading and hence a wealth of homogeneous components to compare; order gives $k [[x]]$ a filtration whose associated topology is complete, so infinite processes converge.

ord (F + G) \geq \min {ord F, ord G}, ord (F G) = ord F + ord G (k a domain) .

(1.5a)

The second identity requires $k$ to be a domain; it fails in $(ℤ / 4) [[x]]$ , where $ord (2 x \cdot 2 x) = \infty$ .

Why the constant term decides everything

To invert $F = f_{0} + f_{1} x + \dots$ one must solve $\sum_{p + q = n} f_{p} g_{q} = δ_{n, 0}$ for the unknowns $g_{0}, g_{1}, g_{2}, \dots$ . The equation in degree $n$ reads $f_{0} g_{n} = - \sum_{p \geq 1} f_{p} g_{n - p}$ : the unknown of highest index appears only once, multiplied on the left by $f_{0}$ . So the system is triangular, and it is solvable by recursion precisely when $f_{0}$ can be cancelled — that is, when $f_{0} \in U (k)$ .

f_{0} \in U (k)

triangular system solvable

F

right-invertible

F \in U (k [[x]])

The last arrow is the only one requiring care in a noncommutative ring: right-invertibility does not imply invertibility in general. Here it does, because the same recursion run on the other side produces a left inverse, and an element with a left and a right inverse is a unit.

Inverting $x$

In $k [[x]]$ the element $x$ is a non-unit — its constant term is $0$ . The passage to $k ((x))$ formally inverts it, and this is the entire content of the Laurent construction: every nonzero $F \in k ((x))$ factors as $F = x^{v} G$ with $v = ord F$ and $G \in k [[x]]$ having nonzero constant term. If $k$ is a division ring that constant term is a unit, so $G$ is a unit, and $x^{v}$ is a unit by fiat.

Key Results

Proposition(1.5)Units of a formal power series ring

Let $k$ be a ring and $R = k [[x_{i} : i \in I]]$ the formal power series ring, the variables commuting with $k$ . For $F = f_{0} + f_{1} + f_{2} + \dots \in R$ , one has $F \in U (R)$ if and only if $f_{0} \in U (k)$ .

Proof

Necessity. The map $R \to k$ sending $F$ to $f_{0}$ is a surjective ring homomorphism (set every variable to $0$ ). Ring homomorphisms carry units to units, so $F \in U (R)$ forces $f_{0} \in U (k)$ .

Sufficiency. Suppose $f_{0} \in U (k)$ and look for $G = g_{0} + g_{1} + g_{2} + \dots$ with $F G = 1$ . Comparing homogeneous components of degree $0, 1, 2, \dots$ gives

f_{0} g_{0} = 1, f_{0} g_{1} + f_{1} g_{0} = 0, f_{0} g_{2} + f_{1} g_{1} + f_{2} g_{0} = 0, \dots

Set $g_{0} = f_{0}^{- 1}$ and, recursively, $g_{n} = - f_{0}^{- 1} (f_{1} g_{n - 1} + f_{2} g_{n - 2} + \dots + f_{n} g_{0})$ . Each $g_{n}$ is a homogeneous element of degree $n$ , and each is determined by finitely many earlier ones, so $G$ is a well-defined element of $R$ with $F G = 1$ .

The same argument applied to the equations $\sum_{p + q = n} h_{p} f_{q} = δ_{n, 0}$ — now solving from the left, using $h_{n} = - (h_{n - 1} f_{1} + \dots + h_{0} f_{n}) f_{0}^{- 1}$ — produces $H$ with $H F = 1$ . Then $H = H (F G) = (H F) G = G$ , so $G$ is a two-sided inverse and $F \in U (R)$ . $□$

Theorem(1.6)Laurent series over a division ring

Let $k$ be a division ring. Then the Laurent series ring $R = k ((x))$ is a division ring. Consequently the construction may be iterated, producing division rings $k ((x_{1})) ((x_{2})) \dots$ of iterated Laurent series.

Proof

$R \neq 0$ since $1 \neq 0$ in $k$ . Let $F \in R$ be nonzero and put $v = ord F \in ℤ$ , so that $F = \sum_{i \geq v} f_{i} x^{i}$ with $f_{v} \neq 0$ . Because $x$ is central and invertible in $R$ ,

F x^{- v} = g_{0} + g_{1} x + g_{2} x^{2} + \dots \in k [[x]], g_{0} = f_{v} \neq 0 .

Since $k$ is a division ring, $g_{0} \in U (k)$ , so $F x^{- v} \in U (k [[x]])$ by $(1.5)$ ; a unit of the subring $k [[x]]$ is in particular a unit of $R$ . As $x^{- v} \in U (R)$ as well, $F = (F x^{- v}) x^{v}$ is a product of two units, hence a unit. Every nonzero element of $R$ is therefore invertible. $□$

Proposition—Radical of a power series ring

For any ring $R$ , the formal power series ring $R [[x]]$ in one central variable satisfies

rad R [[x]] = rad R + x R [[x]] = {F : f_{0} \in rad R} .

In particular $R [[x]]$ is a local ring if and only if $R$ is, and for $R = k$ a division ring, $rad k [[x]] = x k [[x]]$ .

Proof

**( $\supseteq$ ).** Let $F$ have $f_{0} \in rad R$ and let $G \in R [[x]]$ be arbitrary. The constant term of $1 - G F$ is $1 - g_{0} f_{0}$ , and $g_{0} f_{0} \in rad R$ because the radical is a two-sided ideal, so $1 - g_{0} f_{0} \in U (R)$ . By $(1.5)$ , $1 - G F \in U (R [[x]])$ . Since this holds for every $G$ , the element $F$ lies in $rad R [[x]]$ .

**( $\subseteq$ ).** Let $F \in rad R [[x]]$ and let $r \in R$ be arbitrary. Then $1 - r F \in U (R [[x]])$ , so by $(1.5)$ its constant term $1 - r f_{0}$ lies in $U (R)$ . As $r$ was arbitrary, $f_{0} \in rad R$ . $□$

Proposition—Units of a polynomial ring over a domain

If $R$ is a domain then $U (R [x]) = U (R)$ , and $R [x]$ is again a domain. If instead $R$ is commutative, then $f = a_{0} + a_{1} x + \dots + a_{n} x^{n}$ is a unit of $R [x]$ if and only if $a_{0} \in U (R)$ and $a_{1}, \dots, a_{n}$ are nilpotent.

Proof

For the first claim, if $R$ is a domain the leading coefficient of a product is the product of the leading coefficients, so $\deg (f g) = \deg f + \deg g$ . From $f g = 1$ we get $\deg f = \deg g = 0$ , so $f, g \in R$ and $f \in U (R)$ . The same additivity shows $R [x]$ has no zero divisors.

The commutative statement is the classical criterion. Sufficiency: a nilpotent element of a commutative ring added to a unit stays a unit, and $a_{1} x + \dots + a_{n} x^{n}$ is nilpotent in $R [x]$ when all $a_{i}$ are. Necessity: reduce modulo an arbitrary prime ideal $𝔭$ of $R$ ; the image of $f$ is a unit of the polynomial ring over the domain $R / 𝔭$ , so by the first part $a_{1}, \dots, a_{n} \in 𝔭$ . As $𝔭$ was arbitrary, the $a_{i}$ lie in the nilradical. $□$

Remark—Chain conditions

If $R$ is left noetherian then so are $R [x]$ , $R [[x]]$ and $R [x, x^{- 1}]$ — the first by the Hilbert Basis Theorem, the second by its power series analogue, the third because $x$ is a central unit so the ring is a central localisation. None of them is ever artinian for $R \neq 0$ : the chain $(x) ⊋ (x^{2}) ⊋ \dots$ never stabilises. See Chain Conditions.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Three moves carry all the arguments above, and all three recur in the twisted constructions.

Move 1

Triangular recursion on components

Comparing homogeneous components turns one equation in $R$ into a countable triangular system over $k$ . The unknown of highest index appears once, so a single invertible coefficient unlocks the whole system.

Move 2

Run the recursion on both sides

A one-sided inverse is not enough in a noncommutative ring. Solve on the left as well, then use $H = H (F G) = (H F) G = G$ to identify the two solutions.

Move 3

Normalise by a power of the variable

Factor out $x^{ord F}$ to reduce a Laurent question to a power series question with nonzero constant term. This is the standard reduction for any valuation-like function.

Move 3 is where completeness is quietly used: the reduction is legitimate only because the order is a genuine integer, which is exactly the finitely many negative terms clause in the definition of $k ((x))$ .

Worked Example

Inverting a quadratic in $ℚ [[x]]$

Take $k = ℚ$ and $F = 1 - x - x^{2}$ . Its constant term is $1 \in U (ℚ)$ , so $F$ is a unit. The recursion of the proof gives $g_{0} = 1$ , $g_{1} = - f_{1} g_{0} = 1$ , and $g_{n} = g_{n - 1} + g_{n - 2}$ for $n \geq 2$ .

(1 - x - x^{2})^{- 1} = 1 + x + 2 x^{2} + 3 x^{3} + 5 x^{4} + 8 x^{5} + 13 x^{6} + \dots = \sum_{n \geq 0} F_{n + 1} x^{n},

(E.1)

$F_{m}$ the Fibonacci numbers with $F_{1} = F_{2} = 1$ . Check the coefficient of $x^{3}$ : $3 - 2 - 1 = 0$ .

Note that $1 - x - x^{2}$ is not a unit of $ℚ [x]$ : it has degree $2$ , and $U (ℚ [x]) = ℚ^{\times}$ . Completion has genuinely enlarged the unit group.

A unit of $R [x]$ outside $U (R)$

Take $R = ℤ / 4 ℤ$ , which is not reduced: $2^{2} = 0$ . Then

(1 + 2 x) (1 + 2 x) = 1 + 4 x + 4 x^{2} = 1 in (ℤ / 4 ℤ) [x],

(E.2)

So $1 + 2 x \in U (R [x]) ∖ U (R)$ , in agreement with the commutative unit criterion: $2$ is nilpotent.

The order function at work in $ℚ ((x))$

Let $F = 3 x^{- 2} + x^{- 1} + 0 + x + \dots$ , so $ord F = - 2$ . Then $F x^{2} = 3 + x + x^{3} + \dots$ has constant term $3 \in U (ℚ)$ , hence is a unit of $ℚ [[x]]$ ; multiplying its inverse by $x^{2}$ inverts $F$ . Explicitly $F^{- 1} = x^{2} (\frac{1}{3} - \frac{1}{9} x + \frac{1}{27} x^{2} - \dots)$ , with $ord F^{- 1} = 2 = - ord F$ .

Process and Workflow

Which one-variable extension of $k$ do you need?

Finite data and degree bounds

k [x]

. Graded, noetherian when

k

is, and

U (k [x]) = U (k)

for

k

a domain — nothing new becomes invertible.

Local behaviour near $x = 0$

k [[x]]

. Complete, local when

k

is, with

rad k [[x]] = rad k + x k [[x]]

. The standard source of a radical that is not nil.

Negative exponents without completion

k [x, x^{- 1}]

. Still

ℤ

-graded and noetherian; the natural home for the skew Laurent polynomials

k [x, x^{- 1}; σ]

A new division ring

k ((x))

. A division ring whenever

k

is, and the construction iterates. With a twist

σ

it produces noncommutative — and orderable — division rings.

Fix the conventionDecide whether the variable commutes with the coefficients. If not, you are in the skew world and

(1.5)

needs restating.

Compute the order or degreeThis is the cheapest invariant and it decides membership in every ideal generated by powers of

x

Normalise by $x^{v}$ Reduce to a series with unit constant term.

Invert by recursionSolve the triangular system, then verify on the other side or invoke

(1.5)

Comparison and Classification

The four constructions in one variable
Ring	Typical element	Units when $k$ is a domain	$rad$ when $k$ is a field
$k [x]$	$\sum_{i = 0}^{n} a_{i} x^{i}$	$U (k)$	$0$
$k [[x]]$	$\sum_{i \geq 0} a_{i} x^{i}$	${F : f_{0} \in U (k)}$	$x k [[x]]$
$k [x, x^{- 1}]$	$\sum_{i = - m}^{n} a_{i} x^{i}$	${u x^{n} : u \in U (k), n \in ℤ}$	$0$
$k ((x))$	$\sum_{i \geq - m} a_{i} x^{i}$	all $F \neq 0$ when $k$ is a division ring	$0$

Structural properties, $k$ a field
	$k [x]$	$k [[x]]$	$k [x, x^{- 1}]$	$k ((x))$
$ℤ$ -graded	yes	no	yes	no
Complete for the $x$ -adic filtration	no	yes	no	yes
Local ring	no	yes	no	yes
Every nonzero element invertible	no	no	no	yes
Noetherian	yes	yes	yes	yes
Artinian	no	no	no	yes

Structural properties, $k$ a field

Relationship Map

The four rings sit in a square of inclusions, with completion running one way and localisation the other.

k [x]

k [[x]]

k ((x))

k [x]

k [x, x^{- 1}]

k ((x))

$k ((x))$ Laurent series — a division ring when

k

$k [[x]]$ power series — local, complete, radical

x k [[x]]

$k [x]$ polynomials — graded, radical zero for

k

a field

$k$ the coefficient ring

Adjoining a variable to $k$ — four routes
- keep $x$ commuting with $k$
  - $k [x]$ , $k [[x]]$ , $k [x, x^{- 1}]$ , $k ((x))$ — this page
  - the radical of $k [x]$ is described by Amitsur's theorem, treated in Radical of Polynomial Rings
- twist by an endomorphism $σ$
  - $k [x; σ]$ and $k [[x; σ]]$ — Skew Polynomial Rings and Hilbert's Twist
  - $k [x, x^{- 1}; σ]$ and $k ((x; σ))$ , requiring $σ$ to be an automorphism
- twist by a derivation $δ$
  - $k [x; δ]$ — differential polynomial rings, with the Weyl algebra as the leading case

Standards and Notation

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

Polynomials

k [x]

; several variables

k [x_{1}, \dots, x_{n}]

k [x_{i} : i \in I]

Noncommuting variables

k ⟨ x, y ⟩

— angle brackets, never square

Power series

k [[x]]

; Lam writes

k [[x_{i} : i \in I]]

for the general case

Laurent series

k ((x))

; the twisted version is

k ((x; σ))

Laurent polynomials

k [x, x^{- 1}]

, twisted

k [x, x^{- 1}; σ]

Order

ord

v

ν

depending on the source; all mean the least nonzero index

SagePowerSeriesRing, LaurentSeriesRing, LaurentPolynomialRing

Magma / Macaulay2PowerSeriesRing, LaurentSeriesRing; Macaulay2 works with truncations only

Computational Notes

Computational notes cover algorithms, cost and library behaviour rather than manufacturing process.

Formal series are infinite objects, so every implementation works with truncations $k [[x]] / (x^{N})$ and tracks the precision $N$ explicitly.

Naive inversion by the recursion of $(1.5)$ costs $O (N^{2})$ coefficient operations to precision $N$ .
Newton iteration $G_{2 m} = G_{m} (2 - F G_{m})$ doubles the correct precision at each step and costs $O (M (N))$ , where $M (N)$ is the cost of one multiplication; with FFT multiplication this is near-linear. The identity requires the coefficients to commute with the series, so it is a commutative technique.
Composition, reversion and exponentiation of series all reduce to multiplication and inversion; the Brent–Kung algorithm gives composition in $O (\sqrt{N} M (N))$ for commutative coefficient rings.
Laurent series are stored as a valuation together with a power series; arithmetic normalises the valuation after every operation, which is the practical form of Move 3.
Equality of two formal series is not decidable from truncations alone. Systems compare to a declared precision and warn rather than assert.

Failure Modes and Common Mistakes

$ord (F G) = ord F + ord G$ requires $k$ to be a domain. In $(ℤ / 4) [[x]]$ the product $2 x \cdot 2 x$ vanishes.
$k [[x]]$ is never artinian for $k \neq 0$ , so results proved for artinian rings do not transfer. Its radical is nonzero and contains no nonzero nilpotent — the standard counterexample to radical implies nil.
For finitely many commuting variables $k [[x_{1}, \dots, x_{n}]] = k [[x_{1}]] \dots [[x_{n}]]$ , but this iterated description is unavailable for infinitely many variables, where the two candidate definitions genuinely differ.
The unit criterion $(1.5)$ says nothing about $k [x]$ . The radical of a polynomial ring is a genuinely hard invariant — see Radical of Polynomial Rings.

Quick Reference

Power series unit test

F \in U (k [[x]]) \Leftrightarrow f_{0} \in U (k)

Laurent series

k

a division ring

\Rightarrow k ((x))

a division ring

Radical

rad k [[x]] = rad k + x k [[x]]

Order

ord (F G) = ord F + ord G

for

k

a domain

Degree

\deg (f g) = \deg f + \deg g

for

k

a domain

Polynomial units

U (k [x]) = U (k)

for

k

a domain

Chain conditionsleft noetherian is inherited; artinian never is

Factorisation

F = x^{ord F} G

with

G \in U (k [[x]])

, for

k

a division ring

Which ring makes which element invertible
Element of $k [x]$	$k [x]$	$k [[x]]$	$k ((x))$
$u \in U (k)$	unit	unit	unit
$x$	not a unit	not a unit	unit
$1 - x$	not a unit	unit	unit
$x - x^{2}$	not a unit	not a unit	unit
$0$	never	never	never

Frequently Asked Questions

Why does the unit criterion for

k [[x]]

not need

k

to be commutative?

Because the recursion only ever divides by the constant term, and it can be run on either side. Solving $F G = 1$ from the left gives a right inverse; solving $H F = 1$ from the right gives a left inverse; the two coincide by associativity. Nowhere is a coefficient moved past another coefficient. The convention that $x$ commutes with $k$ is used, however, and dropping it changes the answer — see the skew case.

k ((x))

the field of fractions of

k [[x]]

For $k$ a field, yes: $k [[x]]$ is a discrete valuation ring and $k ((x))$ is its fraction field, obtained by inverting $x$ alone. For $k$ a division ring the same statement holds with fraction field replaced by Ore quotient ring, since $x$ is central the localisation is unproblematic.

Why is

rad k [[x]]

not nil?

The radical is $x k [[x]]$ for $k$ a field, and no nonzero element of it is nilpotent because $ord (F^{n}) = n ord (F)$ is finite for every $n$ . This is the cleanest counterexample to the false implication radical implies nil; the true general statement runs the other way, every nil one-sided ideal lies in the radical.

What breaks if I allow infinitely many negative exponents?

Multiplication stops being defined. The coefficient of $x^{n}$ in a product is $\sum_{p + q = n} f_{p} g_{q}$ , and with unbounded negative support this is an infinite sum with no meaning in a bare ring. Any repair requires an ordering on the exponents together with a well-ordering condition on supports, which is precisely the Malcev–Neumann series construction.

Does

k [[x]]

inherit the noetherian property from

k

Yes for one variable, and for finitely many: if $k$ is left noetherian then so is $k [[x_{1}, \dots, x_{n}]]$ . For infinitely many variables it fails, exactly as the Hilbert Basis Theorem fails for infinitely many variables. Artinian is never inherited by any of these constructions.

How much of this survives when

x

no longer commutes with

k

The unit criterion survives in the form $U (k [[x; σ]]) = {a_{0} + a_{1} x + \dots : a_{0} \in U (k)}$ for any endomorphism $σ$ , and the Laurent construction still yields a division ring provided $σ$ is an automorphism. What changes is the ideal theory and the chain conditions: $k [x; σ]$ can be left noetherian without being right noetherian.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §1, examples (1.5) and (1.6) (pp. 8–10).
T. Y. Lam, Exercises in Classical Ring Theory, 2nd edition, Problem Books in Mathematics, Springer-Verlag, 2003, Chapter 1.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 1.
M. F. Atiyah and I. G. Macdonald, Introduction to Commutative Algebra, Addison-Wesley, 1969, Chapter 1 and Chapter 10 (completions).
P. M. Cohn, Free Rings and Their Relations, 2nd edition, London Mathematical Society Monographs 19, Academic Press, 1985 (Malcev–Neumann series).

AI Suggested Questions

Derive the Newton iteration for power series inversion and show it doubles precision at each step.
For which commutative rings $R$ is $R [[x]]$ a principal ideal ring?
Prove that $k [[x]]$ is the $(x)$ -adic completion of $k [x]$ and describe the induced topology on $k ((x))$ .
Compare the Malcev–Neumann series ring over an ordered group with the iterated Laurent construction.
Give a ring $R$ for which $rad R [x]$ is nonzero, and identify it using Amitsur's theorem.
Why does the Hilbert Basis Theorem hold for $R [[x]]$ but fail for $R [x_{1}, x_{2}, \dots]$ ?
Describe all ring endomorphisms of $k ((x))$ that fix $k$ pointwise.

Polynomial and Laurent Series Rings

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Degree and order run in opposite directions

Why the constant term decides everything

Inverting $x$

Key Results

Proof Techniques and Method

Triangular recursion on components

Run the recursion on both sides

Normalise by a power of the variable

Worked Example

Inverting a quadratic in $ℚ [[x]]$

A unit of $R [x]$ outside $U (R)$

The order function at work in $ℚ ((x))$

Process and Workflow

Comparison and Classification

Relationship Map

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Degree and order run in opposite directions

Why the constant term decides everything

Inverting x

06Key Results

07Proof Techniques and Method

Triangular recursion on components

Run the recursion on both sides

Normalise by a power of the variable

08Worked Example

Inverting a quadratic in ℚ[[x]]

A unit of R[x] outside U(R)

The order function at work in ℚ((x))

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Standards and Notation

13Computational Notes

14Failure Modes and Common Mistakes

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Inverting $x$

Key Results

Proof Techniques and Method

Worked Example

Inverting a quadratic in $ℚ [[x]]$

A unit of $R [x]$ outside $U (R)$

The order function at work in $ℚ ((x))$

Process and Workflow

Comparison and Classification

Relationship Map

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions