The Gordon–Motzkin Theorem | KEVOS® Mathematics

Q: If f = (t-a_1)…(t-a_n), are all the a_j roots of f?

Only a_n, the root of the rightmost factor, is guaranteed. The theorem says every root is *conjugate* to some a_j, not that each a_j is a root. The example (t-j)(t-i) over H has i as its only root; j is not a root.

Q: Is the bound of n classes attained?

Yes, and trivially so when D is a field, where classes are singletons and a separable polynomial of degree n has n roots. Over H a product (t-c_1)…(t-c_n) with pairwise nonconjugate real c_i realises n distinct classes. Bray–Whaples shows such configurations are exactly those with a unique monic interpolating polynomial.

Q: What replaces the theorem for skew polynomial rings D[t;]?

The evaluation map is twisted, f(r) = a_i N_i(r) with N_i the -twisted norms, and conjugacy is replaced by -conjugacy, r (d) r d^-1. With those substitutions the class bound and the dichotomy both survive; this is what underlies the Vandermonde and Gabidulin machinery used for rank-metric codes.

Q: Why does the proof of (16.11) need Wedderburn's little theorem?

To pass from *noncommutative* to *infinite*. Two roots in one class means the class is not a singleton, so the chosen a is noncentral and D is noncommutative; Wedderburn's little theorem then rules out D being finite, which is what allows (13.10) to make the centraliser infinite and (13.22) to make the projective space infinite.

Executive Summary

Over a field, degree bounds the number of roots. Over a division ring that fails immediately: in the real quaternions $ℍ$ the polynomial $t^{2} + 1$ is killed by every purely imaginary unit quaternion, a whole two-sphere of roots for a quadratic. The Gordon–Motzkin theorem $(16.4)$ repairs the count by changing what is counted: a nonzero $f \in D [t]$ of degree $n$ has all of its roots inside at most $n$ conjugacy classes of $D$ .

A second theorem of the same authors $(16.11)$ shows why classes are the right unit. Within one conjugacy class the root set obeys a rigid dichotomy: two roots already force infinitely many. Combining the two gives $(16.12)$ — a polynomial of degree $n$ has at most $n$ roots or infinitely many, and nothing in between.

$\leq n$ Classes meeting the root set

$2 \Rightarrow \infty$ Roots inside one class

$n$ or $\infty$ Size of the root set

1965Gordon and Motzkin

Overview

Let $D$ be a division ring and $D [t]$ the polynomial ring in a central indeterminate $t$ . A polynomial is written with its coefficients on the left, $f (t) = \sum_{i = 0}^{n} a_{i} t^{i}$ , and evaluated at $r \in D$ by substituting on the right: $f (r) = \sum_{i} a_{i} r^{i}$ . Such an $r$ is a right root. The conventions and the evaluation pathology are developed on Polynomials over Division Rings: Evaluation, Roots and Division; this page takes them as given.

The single fact that drives everything here is that evaluation is not a ring homomorphism. From $f = g h$ one cannot conclude $f (d) = g (d) h (d)$ . What one can conclude — result $(16.3)$ — is that the value of $g$ has to be taken at a conjugate of $d$ , twisted by $h (d)$ . Conjugation therefore enters the theory at the ground floor, and the invariant that survives is the conjugacy class.

f \in D [t], \deg f = n ⟹ {d \in D : f (d) = 0} \subseteq A_{1} \cup \dots \cup A_{m}, m \leq n,

(16.4)

The $A_{i}$ are conjugacy classes of $D$ . No bound on the size of the individual classes is possible.

The two theorems pull in opposite directions and meet exactly. Result $(16.4)$ says the root set is spread over few classes; result $(16.11)$ says that inside a class the root set is tiny or enormous. The corollary $(16.12)$ is the sharpest statement about $| Γ |$ available without further hypotheses on $D$ .

Learning Objectives

State $(16.4)$ with its exact hypotheses and explain why it does not bound the number of roots.
Apply the conjugation rule $(16.3)$ to transfer a root of $f = g h$ into a root of $g$ .
Reproduce the induction on degree that proves the class bound.
Prove $(16.11)$ by identifying $Γ \cap A$ with a projective space over $C_{D} (a)$ .
Deduce the dichotomy $(16.12)$ and recognise Herstein's theorem as the case $g = 0$ .
Compute the complete root set of a quadratic in $ℍ [t]$ by hand.

Definitions

Definition(16.1)Right root

Let $R$ be a ring and $f (t) = \sum_{i = 0}^{n} a_{i} t^{i} \in R [t]$ , written with all coefficients on the left. An element $r \in R$ is a right root of $f$ if $f (r) : = \sum_{i = 0}^{n} a_{i} r^{i} = 0$ . Throughout this page root means right root; the left-handed theory is the mirror image, obtained by passing to $D^{op}$ .

Conjugacy class $A$: For $a \in D$ , the set $A = {d a d^{- 1} : d \in D^{*}}$ . Classes partition $D$ ; the central elements are exactly those with singleton classes.
$C_{D} (a)$: The centraliser ${x \in D : x a = a x}$ , a division subring of $D$ containing the centre $F = Z (D)$ .
$Γ$: The root set ${d \in D : g (d) = 0}$ of a fixed polynomial $g \in D [t]$ . For $g = 0$ we have $Γ = D$ .
$P (V_{K})$: The projective space of a right $K$ -vector space $V$ : the set of one-dimensional $K$ -subspaces, equivalently the orbits of $V ∖ {0}$ under right multiplication by $K^{*}$ .
Algebraic class: A conjugacy class $A$ is algebraic over $F = Z (D)$ if one — hence every — element of $A$ is algebraic over $F$ ; all elements then share one minimal polynomial, the minimal polynomial of $A$ .

$D$ is a division ring with centre $F$ , and $t$ commutes elementwise with $D$ . No finiteness or chain condition is assumed anywhere on this page.

Core Concepts

Two tools, both cheap

Everything on this page is built from a division algorithm and a one-line computation.

Proposition(16.2)Noncommutative remainder theorem

Let $R$ be any ring and $f \in R [t]$ nonzero. Then $r \in R$ is a root of $f$ if and only if $t - r$ is a right divisor of $f$ in $R [t]$ . The set of polynomials having $r$ as a root is the left ideal $R [t] \cdot (t - r)$ .

Proposition(16.3)Conjugation rule for products

Let $D$ be a division ring, $f = g h$ in $D [t]$ , and $d \in D$ with $a : = h (d) \neq 0$ . Then

f (d) = g (a d a^{- 1}) h (d) .

In particular, if $d$ is a root of $f$ but not of $h$ , then the conjugate $a d a^{- 1}$ is a root of $g$ .

Proof

Write $g (t) = \sum_{i} b_{i} t^{i}$ . Since $t$ is central, $f (t) = g (t) h (t) = \sum_{i} b_{i} h (t) t^{i}$ , so $f (d) = \sum_{i} b_{i} h (d) d^{i} = \sum_{i} b_{i} a d^{i}$ . Insert $a^{- 1} a$ and use $a d^{i} a^{- 1} = (a d a^{- 1})^{i}$ to get $\sum_{i} b_{i} (a d a^{- 1})^{i} a = g (a d a^{- 1}) h (d)$ . The last sentence follows because $D$ has no zero divisors.

d

is a root of

f = g h

a : = h (d) \neq 0

a d a^{- 1}

is a root of

g

Why the class, and not the element, is the invariant

Stripping a linear factor from $f$ costs one degree, and $(16.3)$ says the surviving roots reappear in the quotient only after conjugation. Iterating the strip therefore produces a list of classes, one per degree consumed, and no finer record of the roots survives the process. That is precisely the shape of $(16.4)$ .

Key Results

Theorem(16.4)Gordon–Motzkin

Let $D$ be a division ring and let $f \in D [t]$ be a polynomial of degree $n \geq 1$ . Then the roots of $f$ in $D$ lie in at most $n$ conjugacy classes of $D$ . If moreover $f (t) = (t - a_{1}) \dots (t - a_{n})$ with $a_{1}, \dots, a_{n} \in D$ , then every root of $f$ is conjugate to some $a_{j}$ .

Proof

Induct on $n$ . For $n = 1$ , $f = a_{1} t + a_{0}$ with $a_{1} \neq 0$ has the single root $- a_{1}^{- 1} a_{0}$ , one class. Let $n \geq 2$ . If $f$ has no root there is nothing to prove, so fix a root $c$ and use $(16.2)$ to write $f (t) = g (t) (t - c)$ with $\deg g = n - 1$ .

Let $d$ be any root of $f$ with $d \neq c$ . Putting $h (t) = t - c$ we have $a : = h (d) = d - c \neq 0$ , so $(16.3)$ makes $a d a^{- 1}$ a root of $g$ . By the inductive hypothesis the roots of $g$ lie in at most $n - 1$ conjugacy classes; since $d$ and $a d a^{- 1}$ lie in the same class, $d$ lies in one of those $n - 1$ classes. Adjoining the class of $c$ gives at most $n$ classes in all.

For the second statement run the same induction on the given factorisation: $c$ is conjugate to $a_{n}$ — indeed the rightmost factor supplies the root $a_{n}$ — and the roots of $g (t) = (t - a_{1}) \dots (t - a_{n - 1})$ are conjugate to some $a_{j}$ with $j \leq n - 1$ by induction, hence so are their conjugates.

Theorem(16.11)Gordon–Motzkin dichotomy inside a class

Let $D$ be a division ring, let $g (t) = \sum_{i} c_{i} t^{i} \in D [t]$ (the zero polynomial is allowed), let $Γ$ be the set of roots of $g$ in $D$ , and let $A$ be a conjugacy class of $D$ . If $| Γ \cap A | \geq 2$ , then $Γ \cap A$ is infinite.

Proof

Fix $a \in A$ , so $A = {d a d^{- 1} : d \in D^{*}}$ . For $d \in D^{*}$ ,

g (d a d^{- 1}) = \sum_{i} c_{i} (d a d^{- 1})^{i} = (\sum_{i} c_{i} d a^{i}) d^{- 1},

so $d a d^{- 1} \in Γ$ exactly when $L (d) = 0$ , where $L : D \to D$ is the additive map $L (d) = \sum_{i} c_{i} d a^{i}$ . Put $K = C_{D} (a)$ , a division subring. For $k \in K$ we have $L (d k) = \sum_{i} c_{i} d k a^{i} = \sum_{i} c_{i} d a^{i} k = L (d) k$ , so $L$ is right $K$ -linear and $V : = \ker L$ is a right $K$ -subspace of $D_{K}$ .

The map $λ : V ∖ {0} \to Γ \cap A$ , $λ (d) = d a d^{- 1}$ , is onto, and $λ (d) = λ (d')$ holds iff $d^{- 1} d' \in C_{D} (a)^{*} = K^{*}$ , i.e. iff $d' \in d K^{*}$ . Hence $λ$ induces a bijection between $P (V_{K})$ and $Γ \cap A$ .

Now assume $| Γ \cap A | \geq 2$ . Then $P (V_{K})$ has at least two points, so $\dim V_{K} \geq 2$ . Also $A$ has at least two elements, so $a$ is noncentral and $D$ is noncommutative, hence infinite by Wedderburn's little theorem. Lam's $(13.10)$ then gives that $K = C_{D} (a)$ is infinite, and $(13.22)$ gives that $P (V_{K})$ is infinite for a $K$ -space of dimension at least two. Transporting along the bijection, $Γ \cap A$ is infinite.

Corollary(16.12)The size dichotomy

Let $g \in D [t]$ be of degree $n \geq 1$ and let $Γ$ be its set of roots in $D$ . Then either $| Γ | \leq n$ or $Γ$ is infinite.

Proof

Suppose $| Γ | > n$ and pick distinct roots $r_{1}, \dots, r_{n + 1}$ . By $(16.4)$ these lie in at most $n$ conjugacy classes, so by the pigeonhole principle two of them lie in a common class $A$ . Then $| Γ \cap A | \geq 2$ , and $(16.11)$ makes $Γ \cap A$ — hence $Γ$ — infinite.

Remark—Herstein as a degenerate case

Taking $g = 0$ in $(16.11)$ gives $Γ = D$ , $L = 0$ and $V = D$ , so $P (V_{K})$ becomes the coset space $D^{*} / C_{D} (a)^{*}$ and the bijection is the usual parametrisation of a conjugacy class by cosets of a centraliser. The theorem then reads: a class with at least two elements is infinite — Herstein's theorem $(13.26)$ , that a noncentral element of a division ring has infinitely many conjugates.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Peel a linear factor

A root $c$ gives $f = g \cdot (t - c)$ by $(16.2)$ , dropping the degree by one. Every induction in this section is an induction on how many times this can be done.

Move 2

Twist by the value of the right factor

The leftover roots do not survive into $g$ unchanged; $(16.3)$ conjugates them by $h (d)$ . Working modulo conjugacy makes the twist invisible, which is why classes are the natural bookkeeping unit.

Move 3

Linearise the class

Rewriting $g (d a d^{- 1}) = L (d) d^{- 1}$ converts a nonlinear root condition into the kernel of a right $C_{D} (a)$ -linear map. Counting roots in a class becomes counting lines in a vector space.

Move 3 is the reusable idea. It replaces a question about a conjugacy class — a set with no linear structure — by a question about a projective space over the centraliser, where dimension counting is available. The same device recurs whenever one has to count elements of a class satisfying a coefficientwise condition.

Worked Example

A quadratic with a sphere of roots

In $ℍ = ℝ \oplus ℝ i \oplus ℝ j \oplus ℝ k$ take $f (t) = t^{2} + 1$ . Writing $q = x_{0} + v$ with $v$ purely imaginary, $q^{2} = x_{0}^{2} - | v |^{2} + 2 x_{0} v$ , so $q^{2} = - 1$ forces $x_{0} = 0$ and $| v | = 1$ .

Γ = {x_{1} i + x_{2} j + x_{3} k : x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = 1},

(E.1)

A two-sphere of roots for a quadratic — and a single conjugacy class, namely that of $i$ .

This is consistent with $(16.4)$ : one class, and $1 \leq 2$ . Because $f$ has central coefficients, its root set is a union of full classes, which is why the whole sphere appears at once.

A quadratic whose two linear factors contribute only one root

Now take $f (t) = (t - j) (t - i)$ . Expanding, and using $j i = - k$ ,

f (t) = t^{2} - (i + j) t - k .

(E.2)

The right-hand factor supplies the root $i$ : indeed $f (i) = i^{2} - (i + j) i + j i = - 1 - i^{2} - j i + j i = 0$ . What about $j$ ? Evaluating, $f (j) = j^{2} - (i + j) j + j i = - 1 - i j - j^{2} - k = - 1 - k + 1 - k = - 2 k \neq 0$ . The root of the left factor is not a root of the product.

In fact $i$ is the only root. Suppose $d \neq i$ is a root. Applying $(16.3)$ with $g = t - j$ , $h = t - i$ and $a = d - i \neq 0$ , we get $a d a^{- 1} = j$ . Conjugation in $ℍ$ preserves the real part and the norm, so $Re d = 0$ and $| d | = 1$ , whence $d^{2} = - 1$ . Substituting into $(E . 2)$ :

0 = f (d) = - 1 - (i + j) d - k ⟹ d = - (i + j)^{- 1} (1 + k) = \frac{1}{2} (i + j) (1 + k) = \frac{1}{2} (i - j + j + i) = i,

(E.3)

Using $(i + j)^{- 1} = - \frac{1}{2} (i + j)$ , $i k = - j$ and $j k = i$ .

contradicting $d \neq i$ . So $Γ = {i}$ : a quadratic that splits into linear factors over $ℍ$ , whose roots occupy one class, and which meets that infinite class in a single point. Compare $(E . 1)$ : same class, same degree, radically different intersection — exactly the two extremes permitted by $(16.11)$ .

Comparison and Classification

Roots of a degree- $n$ polynomial: field versus division ring
Question	Field $F$	Division ring $D$
Bound on number of roots	$\leq n$	none; $\leq n$ or infinite $(16.12)$
Bound on conjugacy classes met	$\leq n$ (classes are points)	$\leq n$ $(16.4)$
Is evaluation multiplicative?	yes	no; corrected by $(16.3)$
Root set of $f$ with central coefficients	a finite set	a union of full conjugacy classes
Two roots in one class	impossible (classes are singletons)	forces infinitely many $(16.11)$
Factorisation into linear factors	essentially unique	wildly non-unique; see Wedderburn's Factorisation Theorem

What each hypothesis buys
	$(16.2)$	$(16.3)$	$(16.4)$	$(16.11)$
Needs $t$ central	yes	yes	yes	yes
Needs $D$ to be a division ring	no	yes	yes	yes
Needs $f \neq 0$	yes	no	yes	no
Needs coefficients in $Z (D)$	no	no	no	no
Uses Wedderburn's little theorem	no	no	no	yes

What each hypothesis buys

Relationship Map

The logical dependencies in §16 are short and almost linear; the two Gordon–Motzkin results sit at the junction.

(16.2)

remainder

(16.3)

conjugation rule

(16.4)

class bound

(16.12)

dichotomy

$(16.4)$ and $(16.11)$ — the two counting theorems
- feed directly into
  - $(16.12)$ : $| Γ | \leq n$ or $| Γ |$ infinite
  - $(16.13)$ Bray–Whaples uniqueness of the interpolating polynomial
  - $(16.18)$ Niven's criterion for infinitely many quaternionic roots
- specialise to
  - Herstein's theorem $(13.26)$ , the case $g = 0$
  - the classical root bound when $D$ is a field
- depend on
  - $(13.10)$ : centralisers in an infinite division ring are infinite
  - $(13.22)$ : projective spaces over infinite division rings are infinite
  - Wedderburn's little theorem

Applications and Industry Use

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

Inside algebra

The engine of §16

Dickson's conjugacy criterion, Wedderburn's factorisation theorem and Bray–Whaples interpolation are all proved by combining $(16.3)$ with the class bound. The class count is the substitute for the degree count that a field would supply.

Quaternionic computation

Root finding over $ℍ$

Algorithms that solve quaternionic polynomial equations — used in rigid-body kinematics, attitude estimation and rotation interpolation — rely on the fact that the solution set is a finite union of isolated points and two-spheres, which is exactly $(16.4)$ plus $(16.11)$ .

Quaternionic analysis

Zero sets of slice-regular functions

The theory of slice-regular quaternionic functions takes its structure theorem for zero sets — isolated zeros together with isolated spheres — from the polynomial case treated here.

Coding and skew polynomials

Contrast with twisted evaluation

Codes built from skew polynomial rings $D [t; σ]$ use a twisted evaluation and a twisted notion of conjugacy; the class bound survives in that setting, which is what makes the corresponding Vandermonde and Gabidulin constructions work.

The honest summary: this is infrastructure. Outside quaternionic computation the theorems are rarely the object of interest, but almost every statement about polynomial equations over a division ring is proved by invoking one of them.

Computational Notes

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

For $D = ℍ$ the structure theorems make root finding effective. Given $f \in ℍ [t]$ of degree $n$ , form $f \bar{f} \in ℝ [t]$ of degree $2 n$ , where the bar is quaternionic conjugation applied coefficientwise. Every root of $f$ is conjugate to a root of $f \bar{f}$ lying in $ℝ (i) ≅ ℂ$ , so the problem reduces to a complex root computation followed by a linear correction.

Compute the real polynomial $f \bar{f}$ of degree $2 n$ — $O (n^{2})$ quaternion multiplications, or $O (n \log n)$ with a transform.
Find the complex roots of $f \bar{f}$ by any standard method; each conjugate pair $α, \bar{α}$ is one candidate conjugacy class.
For each candidate class, test whether $f$ is right divisible by the real quadratic $t^{2} - 2 Re (α) t + | α |^{2}$ . If yes, the entire two-sphere is a root set; if no, the class contributes at most one root, recovered by solving one linear equation over $ℍ$ .
Deflate by the linear or quadratic factor found and repeat.

Over a general division ring nothing so clean is available: exhibiting even one root requires an effective model of $D$ , and for finitely presented division rings the word problem is already an obstruction. Computer algebra support is therefore restricted to quaternion algebras and to matrix models of centrally finite algebras, where Magma, Sage and GAP can solve the associated linear systems.

Failure Modes and Common Mistakes

Do not assume the $a_{j}$ appearing in a factorisation $f = (t - a_{1}) \dots (t - a_{n})$ are roots of $f$ ; only the rightmost one is guaranteed to be.
Do not confuse right roots with left roots. They are genuinely different sets, related by passing to $D^{op}$ , and $(16.2)$ is one-sided in exactly this way.
Do not apply $(16.11)$ to conclude that a class meets the root set; it says only that if the meeting has two points it has infinitely many.
Do not expect the count in $(16.4)$ to be sharp for every $f$ ; a degree- $n$ polynomial may meet far fewer than $n$ classes, or none.

Best Practices

When counting, always say whether you are counting roots or classes; the two statements have different content.
Record the side: right root, right divisor, *left ideal $D [t] (t - r)$ *. A one-sided statement quoted without its side is unusable.
To show a class contributes exactly one root, exhibit the root and rule out a second via $(16.11)$ or via right divisibility by the class's minimal polynomial.
When a polynomial has central coefficients, exploit it immediately: the root set is then a union of classes and the whole analysis simplifies.

Historical Notes and Lessons Learned

1921Wedderburn on division algebrasWedderburn studies factorisation of minimal polynomials over division algebras, establishing the first structural results about roots in a conjugacy class.
1941–44Niven and Eilenberg–NivenNiven solves polynomial equations over the real quaternions; Eilenberg and Niven prove the fundamental theorem of algebra for quaternionic polynomials by a topological degree argument.
1953Herstein on conjugatesHerstein proves that a noncentral element of a division ring has infinitely many conjugates — the degenerate case of the later dichotomy.
1965Gordon and MotzkinGordon and Motzkin publish the class bound and the two-implies-infinite dichotomy, giving the first uniform account of root sets over arbitrary division rings.
1983–86Bray–Whaples and Vandermonde theoryBray and Whaples establish interpolation over division rings; Lam develops the general theory of Vandermonde matrices, in which the class bound reappears as an invertibility criterion.

The methodological lesson is a familiar one. The naive invariant — the number of roots — is not controlled by the degree, and no amount of effort will make it so. Replacing it with an invariant that is controlled, the number of conjugacy classes, converts a false analogy into a theorem, and simultaneously identifies the correct object for every later result in the section.

Quick Reference

Evaluation

f (r) = \sum_{i} a_{i} r^{i}

for

f = \sum_{i} a_{i} t^{i}

Remainder theorem

f (r) = 0 \Leftrightarrow f \in D [t] (t - r)

Conjugation rule

f = g h

a = h (d) \neq 0 \Rightarrow f (d) = g (a d a^{- 1}) h (d)

Class bound

\deg f = n \Rightarrow

roots meet

\leq n

classes

Dichotomy in a class

| Γ \cap A | \geq 2 \Rightarrow | Γ \cap A | = \infty

Global dichotomy

| Γ | \leq n

| Γ | = \infty

Central coefficients

Γ

is a union of full conjugacy classes

Degenerate case

g = 0

recovers Herstein's theorem

(13.26)

The results of this page at a glance
Reference	Hypotheses	Conclusion
$(16.3)$	$D$ a division ring, $f = g h$ , $h (d) \neq 0$	$f (d) = g (a d a^{- 1}) h (d)$ with $a = h (d)$
$(16.4)$	$D$ a division ring, $0 \neq f \in D [t]$ , $\deg f = n$	Roots lie in at most $n$ conjugacy classes
$(16.4)$ , second part	$f = (t - a_{1}) \dots (t - a_{n})$ in $D [t]$	Every root is conjugate to some $a_{j}$
$(16.11)$	$g \in D [t]$ arbitrary, $A$ a class, $\| Γ \cap A \| \geq 2$	$Γ \cap A$ is infinite
$(16.12)$	$\deg g = n \geq 1$	$\| Γ \| \leq n$ or $Γ$ infinite

Frequently Asked Questions

Does the Gordon–Motzkin theorem say a degree-

n

polynomial has at most

n

roots?

No, and this is the point of the theorem. It says the roots occupy at most $n$ conjugacy classes. Since a noncentral class of a division ring is always infinite, a quadratic such as $t^{2} + 1$ over $ℍ$ has a two-sphere of roots while meeting only one class. The correct statement about cardinality is $(16.12)$ : at most $n$ roots, or infinitely many.

Why are right roots preferred over left roots?

Only convention, but it must be fixed before anything can be said. With coefficients written on the left and substitution on the right, the remainder theorem $(16.2)$ becomes a statement about right divisibility and the set of polynomials vanishing at $r$ becomes a left ideal. Choosing the other convention mirrors every statement; mixing the two produces false theorems.

f = (t - a_{1}) \dots (t - a_{n})

, are all the

a_{j}

roots of

f

Only $a_{n}$ , the root of the rightmost factor, is guaranteed. The theorem says every root is conjugate to some $a_{j}$ , not that each $a_{j}$ is a root. The example $(t - j) (t - i)$ over $ℍ$ has $i$ as its only root; $j$ is not a root.

Is the bound of

n

classes attained?

Yes, and trivially so when $D$ is a field, where classes are singletons and a separable polynomial of degree $n$ has $n$ roots. Over $ℍ$ a product $(t - c_{1}) \dots (t - c_{n})$ with pairwise nonconjugate real $c_{i}$ realises $n$ distinct classes. Bray–Whaples shows such configurations are exactly those with a unique monic interpolating polynomial.

What replaces the theorem for skew polynomial rings

D [t; σ]

The evaluation map is twisted, $f (r) = \sum a_{i} N_{i} (r)$ with $N_{i}$ the $σ$ -twisted norms, and conjugacy is replaced by $σ$ -conjugacy, $r \mapsto σ (d) r d^{- 1}$ . With those substitutions the class bound and the dichotomy both survive; this is what underlies the Vandermonde and Gabidulin machinery used for rank-metric codes.

Why does the proof of

(16.11)

need Wedderburn's little theorem?

To pass from noncommutative to infinite. Two roots in one class means the class is not a singleton, so the chosen $a$ is noncentral and $D$ is noncommutative; Wedderburn's little theorem then rules out $D$ being finite, which is what allows $(13.10)$ to make the centraliser infinite and $(13.22)$ to make the projective space infinite.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §16, especially (16.4) and (16.11)–(16.12) (pp. 264–271).
B. Gordon and T. S. Motzkin, “On the zeros of polynomials over division rings”, Transactions of the American Mathematical Society 116 (1965), 218–226.
I. Niven, “Equations in quaternions”, American Mathematical Monthly 48 (1941), 654–661.
S. Eilenberg and I. Niven, “The fundamental theorem of algebra for quaternions”, Bulletin of the American Mathematical Society 50 (1944), 246–248.
P. K. Draxl, Skew Fields, London Mathematical Society Lecture Note Series 81, Cambridge University Press, 1983.
T. Y. Lam, “A general theory of Vandermonde matrices”, Expositiones Mathematicae 4 (1986), 193–215.

AI Suggested Questions

Work through the proof that a noncentral conjugacy class of a division ring is infinite, and identify exactly where finiteness of $D$ would break it.
Give an example of a cubic over $ℍ$ whose root set is one isolated point together with a two-sphere.
How does the class bound $(16.4)$ change for skew polynomial rings $D [t; σ, δ]$ with $σ$ -conjugacy replacing conjugacy?
Is there a bound on the number of conjugacy classes meeting the root set of a polynomial in two noncommuting variables over a division ring?
Compare the Gordon–Motzkin dichotomy with the Eilenberg–Niven fundamental theorem of algebra for quaternions: what does each say that the other does not?
Explain why $\ker L$ in the proof of $(16.11)$ is only a one-sided subspace, and what fails if one tries to make it two-sided.
Describe an algorithm that decides, for $f \in ℍ [t]$ , whether the root set is finite, and estimate its cost.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Two tools, both cheap

Why the class, and not the element, is the invariant

06Key Results

07Proof Techniques and Method

Peel a linear factor

Twist by the value of the right factor

Linearise the class

08Worked Example

A quadratic with a sphere of roots

A quadratic whose two linear factors contribute only one root

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

The engine of §16

Root finding over ℍ

Zero sets of slice-regular functions

Contrast with twisted evaluation

12Computational Notes

13Failure Modes and Common Mistakes

14Best Practices

15Historical Notes and Lessons Learned

16Quick Reference

17Frequently Asked Questions

18Related KEVOS Topics

19References

20AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Comparison and Classification

Relationship Map

Applications and Industry Use

Root finding over $ℍ$

Computational Notes

Failure Modes and Common Mistakes

Best Practices

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions