Wedderburn’s Factorisation Theorem

Q: How can an irreducible polynomial factor?

Irreducibility is relative to the coefficient ring. f is irreducible in F[t] and stays irreducible there; the factorisation happens in the strictly larger ring D[t], whose linear factors t - a have noncentral a. There is no contradiction, exactly as t^2+1 is irreducible over R but factors over C — except that here the enlargement is noncommutative and produces infinitely many factorisations rather than one.

Q: Is the factorisation unique if I fix a_1?

The theorem does not claim uniqueness even then, and no canonical choice of a_2, …, a_n is produced. For n = 2 the remaining factor is forced by coefficient matching, but for larger n the freedom in the intermediate steps is genuine.

Q: Why are only cyclic permutations of the factors allowed?

The commuting lemma applies to a split of f into exactly two blocks, f = f_1f_2, and yields f = f_2f_1. Taking f_2 to be the rightmost linear factor rotates it to the left end. Iterating generates the cyclic group of rotations and nothing more; arbitrary transpositions of adjacent factors are not justified and are generally false.

Q: Does Wedderburn's theorem need D to be finite-dimensional over its centre?

No. Only the class A is assumed algebraic over F. The division ring may be centrally infinite; what matters is that the elements of A satisfy a polynomial over the centre, which is what produces f and makes the degree bound (16.5) available.

Q: What is the relationship to reduced characteristic polynomials?

If D is centrally finite of degree n and a generates a maximal subfield F(a), then the minimal polynomial of a over F coincides with its reduced characteristic polynomial. In that case (16.10) says the reduced trace of a is a sum of n conjugates of a and the reduced norm is a product of n conjugates.

Q: Can Dickson's criterion be used to test conjugacy in practice?

Yes, and it is the standard method for centrally finite algebras: compute both minimal polynomials over F by linear algebra and compare. It is far cheaper than searching for a conjugating element. Note the criterion fails if the minimal polynomials are taken over a noncentral subfield.

Executive Summary

Let $D$ be a division ring with centre $F$ and let $f \in F [t]$ be irreducible of degree $n$ . Over a field, irreducible means unfactorable. Over $D$ , Wedderburn's theorem $(16.9)$ says the opposite: as soon as $f$ has one root in $D$ , it splits into $n$ monic linear factors in $D [t]$ , and all $n$ roots may be taken inside the single conjugacy class $A$ determined by that root.

More is true. The rightmost factor $t - a_{1}$ can be prescribed with $a_{1}$ any element of $A$ , and the factors may be cyclically permuted. Since a noncentral class is infinite, an irreducible polynomial of degree $n \geq 2$ that meets $D$ has infinitely many distinct complete factorisations. Uniqueness of factorisation, the backbone of the commutative theory, disappears entirely.

$n$ Linear factors, all from one class

$\infty$ Distinct factorisations when

n \geq 2

cyclicPermutations that stay valid

1921Wedderburn

Overview

Two classical results share this page because they share a proof technique. Dickson's theorem $(16.8)$ identifies conjugacy classes with minimal polynomials: two elements of $D$ algebraic over $F$ are conjugate precisely when they satisfy the same monic irreducible polynomial over $F$ . Wedderburn's theorem $(16.9)$ then says that this polynomial, viewed in $D [t]$ , is a product of linear factors drawn from the class.

f (t) = (t - a_{n}) (t - a_{n - 1}) \dots (t - a_{1}), a_{1}, \dots, a_{n} \in A, a_{1} arbitrary in A,

(16.9)

$f \in F [t]$ is the minimal polynomial of the algebraic conjugacy class $A$ , of degree $n$ ; the identity holds in $D [t]$ .

The direction of the product matters. Right roots are the ones the theory controls, and the rightmost factor is the one whose root is genuinely a root of $f$ . The remaining $a_{j}$ are elements of $A$ , not in general roots of $f$ — a distinction spelled out on The Gordon–Motzkin Theorem.

The proof needs only two ingredients: the degree bound for polynomials vanishing on a class, from Vanishing Polynomials, and the conjugation rule $(16.3)$ . Neither requires $D$ to be finite-dimensional over $F$ .

Learning Objectives

State $(16.8)$ and $(16.9)$ with the hypothesis algebraic over the centre in place.
Run the maximal-right-factorisation argument and see where $(16.5)$ closes it.
Prove that a central polynomial factoring as $f_{1} f_{2}$ in $D [t]$ also factors as $f_{2} f_{1}$ .
Explain why an irreducible $f \in F [t]$ with a root in $D$ has infinitely many factorisations when $\deg f \geq 2$ .
Derive $(16.10)$ and interpret it as a statement about reduced trace and reduced norm.
Give the explicit factorisation of a quadratic minimal polynomial over a quaternion division algebra.

Definitions

$F = Z (D)$: The centre of the division ring $D$ ; a field, and the coefficient field for every polynomial called central here.
Algebraic class $A$: A conjugacy class of $D$ all of whose elements are algebraic over $F$ ; they then share a single minimal polynomial $f \in F [t]$ , necessarily monic and irreducible over $F$ .
$F (b)$: For $b \in D$ algebraic over $F$ , the field $F [b] \subseteq D$ ; commutative because $F$ is central, of dimension $\deg f$ over $F$ .
Complete factorisation: An expression of a monic polynomial of degree $n$ as a product of $n$ monic linear factors in $D [t]$ .
Cyclic permutation of factors: Moving the leftmost factor of a valid factorisation to the far right; for a central polynomial this again yields a valid factorisation.

$D$ is an arbitrary division ring — no chain condition, no finite dimension over $F$ . Only the class $A$ is assumed algebraic.

Core Concepts

Why a central polynomial is so flexible

A polynomial $f \in F [t]$ is a central element of $D [t]$ : it commutes with every coefficient and with $t$ . Two consequences drive the theory. First, its root set is a union of full conjugacy classes, since $f (d a d^{- 1}) = d f (a) d^{- 1}$ . Second, the left ideal it generates is two-sided, so divisibility statements are unambiguous.

Lemmawithin (16.9)Factors of a central polynomial commute

Let $f \in F [t] ∖ {0}$ and suppose $f = f_{1} f_{2}$ in $D [t]$ . Then $f_{1} f_{2} = f_{2} f_{1}$ ; in particular $f = f_{2} f_{1}$ is again a valid factorisation.

Proof

Since $f$ is central in $D [t]$ , $f f_{1} = f_{1} f$ . Substituting $f = f_{1} f_{2}$ on both sides gives $f_{1} f_{2} f_{1} = f_{1} f_{1} f_{2}$ . The ring $D [t]$ is a domain and $f_{1} \neq 0$ , so cancelling $f_{1}$ on the left yields $f_{2} f_{1} = f_{1} f_{2}$ .

Applying the lemma with $f_{2} = t - a_{1}$ turns $f = (t - a_{n}) \dots (t - a_{2}) (t - a_{1})$ into $f = (t - a_{1}) (t - a_{n}) \dots (t - a_{2})$ , and iterating produces all $n$ cyclic rotations. Non-cyclic permutations are not generally valid.

The engine: push the factorisation as far as it will go

Suppose $f$ has been written as $g \cdot (t - a_{r}) \dots (t - a_{1})$ with all $a_{i} \in A$ and $r$ as large as possible. Either the accumulated right factor already vanishes on all of $A$ — in which case the degree bound $(16.5)$ forces $r \geq \deg f$ , so $g$ is a constant and the factorisation is complete — or some $a \in A$ escapes it, and the conjugation rule $(16.3)$ converts that escape into one more linear factor, contradicting maximality. There is no third possibility.

r

maximalright factor vanishes on

A

(16.5)

r \geq \deg f

r = n

, complete split

Key Results

Theorem(16.8)Dickson

Let $D$ be a division ring with centre $F$ and let $a, b \in D$ both be algebraic over $F$ . Then $a$ and $b$ are conjugate in $D$ if and only if they have the same minimal polynomial over $F$ .

Proof

( $\Rightarrow$ ) If $b = d a d^{- 1}$ and $g \in F [t]$ , then $g (b) = d g (a) d^{- 1}$ because conjugation fixes $F$ ; so $a$ and $b$ annihilate the same polynomials over $F$ and share a minimal polynomial.

( $\Leftarrow$ ) Let $f \in F [t]$ be the common minimal polynomial and $A$ the conjugacy class of $a$ . In the commutative ring $F (b) [t]$ we may divide: $f (b) = 0$ gives $f (t) = h (t) (t - b) = (t - b) h (t)$ with $h \in F (b) [t]$ of degree $\deg f - 1$ .

By $(16.5)$ , a nonzero polynomial vanishing on $A$ has degree at least $\deg f$ ; since $\deg h < \deg f$ , $h$ cannot vanish on $A$ , so choose $a' \in A$ with $α : = h (a') \neq 0$ . Now $f (a') = 0$ because $a'$ is conjugate to $a$ . Applying the conjugation rule $(16.3)$ to $f = (t - b) \cdot h$ at $d = a'$ gives $0 = f (a') = (α a' α^{- 1} - b) α$ , hence $b = α a' α^{- 1}$ . So $b$ is conjugate to $a'$ , hence to $a$ .

Theorem(16.9)Wedderburn's factorisation theorem

Let $D$ be a division ring with centre $F$ and let $A$ be a conjugacy class of $D$ which is algebraic over $F$ , with minimal polynomial $f \in F [t]$ of degree $n$ . Then there exist $a_{1}, \dots, a_{n} \in A$ with

f (t) = (t - a_{n}) (t - a_{n - 1}) \dots (t - a_{1}) in D [t] .

Moreover $a_{1}$ may be prescribed to be any preassigned element of $A$ , and $f$ is also the product of the same linear factors permuted cyclically.

Proof

Fix $a_{1} \in A$ . Since $f \in F [t]$ and $a_{1} \in A$ , $f (a_{1}) = 0$ , so $f \in D [t] (t - a_{1})$ by the remainder theorem $(16.2)$ . Among all factorisations

f (t) = g (t) (t - a_{r}) \dots (t - a_{1}), g \in D [t], a_{1}, \dots, a_{r} \in A,

choose one with $r$ maximal; $r$ is bounded by $n$ because degrees add, and $r \geq 1$ by the previous sentence. Put $h (t) = (t - a_{r}) \dots (t - a_{1})$ .

**Claim: $h$ vanishes on $A$ .** If not, pick $a \in A$ with $α : = h (a) \neq 0$ . Since $f (a) = 0$ , the conjugation rule $(16.3)$ applied to $f = g h$ gives $0 = g (α a α^{- 1}) α$ , so $a_{r + 1} : = α a α^{- 1}$ — again an element of $A$ — is a root of $g$ . By $(16.2)$ , $g = g_{1} \cdot (t - a_{r + 1})$ , producing a factorisation with $r + 1$ linear factors from $A$ and contradicting maximality.

So $h (A) = 0$ , and $(16.5)$ gives $r = \deg h \geq \deg f = n$ . Combined with $r \leq n$ this forces $r = n$ , so $g$ has degree $0$ ; comparing leading coefficients of the monic $f$ gives $g = 1$ . The cyclic permutation statement is the lemma above applied repeatedly, and $a_{1}$ was arbitrary in $A$ from the outset.

Corollary(16.10)Trace and norm as sums and products of conjugates

In the notation of $(16.9)$ , write $f (t) = t^{n} + d_{1} t^{n - 1} + \dots + d_{n} \in F [t]$ . Then $- d_{1}$ is a sum of $n$ elements of $A$ and $(- 1)^{n} d_{n}$ is a product of $n$ elements of $A$ .

Proof

Expand $f = (t - a_{n}) \dots (t - a_{1})$ . The coefficient of $t^{n - 1}$ collects one $- a_{j}$ from each factor, so $d_{1} = - (a_{1} + \dots + a_{n})$ . The constant term is the product of the constant terms in order, $d_{n} = (- a_{n}) (- a_{n - 1}) \dots (- a_{1}) = (- 1)^{n} a_{n} a_{n - 1} \dots a_{1}$ .

Corollary—Irreducible over the centre, split over

D

If $f \in F [t]$ is monic irreducible of degree $n$ and has at least one root in $D$ , then $f$ splits into $n$ monic linear factors in $D [t]$ . If moreover $n \geq 2$ , the root is noncentral, so by Herstein's theorem $(13.26)$ its class $A$ is infinite and $f$ admits infinitely many distinct complete factorisations — one for each choice of $a_{1} \in A$ .

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Maximise, then contradict

Do not build the factorisation step by step to the end. Assume a longest one exists and show that any element of $A$ escaping it manufactures another factor. Extremal choice replaces an induction that would be awkward to set up.

Move 2

Convert escape into a root

$(16.3)$ turns *the right factor does not kill $a$ * into *the left factor has a root conjugate to $a$ *. This is the only way roots migrate between factors in the noncommutative setting.

Move 3

Close with a degree bound

Once the accumulated right factor vanishes on the whole class, $(16.5)$ says its degree is at least $\deg f$ . The two inequalities meet exactly, which is why the factorisation is complete rather than merely long.

Dickson's proof is the same machine run in reverse: instead of building factors, it uses $(16.5)$ to guarantee that some element of the class escapes a given factor, and then reads the escape as a conjugacy. In both proofs the degree bound is used only as an existence device — it never produces the element explicitly.

Worked Example

Quadratic classes in a quaternion division algebra

Let $D = ℍ$ with $F = ℝ$ , and let $a = α + β i$ with $β \neq 0$ . Its minimal polynomial over $ℝ$ is

f (t) = t^{2} - 2 α t + (α^{2} + β^{2}) = t^{2} - tr (a) t + N (a),

(E.1)

$tr$ and $N$ are the quaternionic trace $a + \bar{a}$ and norm $\bar{a} a$ .

Wedderburn's theorem predicts $f = (t - a_{2}) (t - a_{1})$ with $a_{1}, a_{2}$ in the class $A$ of $a$ and $a_{1}$ arbitrary. Take $a_{1} = a$ . Expanding $(t - a_{2}) (t - a) = t^{2} - (a_{2} + a) t + a_{2} a$ and matching coefficients forces $a_{2} = 2 α - a = \bar{a}$ and $a_{2} a = \bar{a} a = α^{2} + β^{2}$ — both satisfied. So

f (t) = (t - \bar{a}) (t - a) = (t - a) (t - \bar{a}),

(E.2)

$\bar{a}$ has the same real part and norm as $a$ , hence lies in $A$ — Dickson's criterion in action. The second equality is the cyclic permutation.

Non-uniqueness, made explicit

For $a = i$ we get $t^{2} + 1 = (t + i) (t - i)$ . But every unit purely imaginary quaternion $q$ lies in the same class, so

t^{2} + 1 = (t + q) (t - q) for every q with q^{2} = - 1,

(E.3)

A two-sphere of distinct complete factorisations of a single quadratic.

Checking $(16.10)$ : $- d_{1} = 0 = q + (- q)$ , a sum of two elements of $A$ ; and $(- 1)^{2} d_{n} = 1 = (- q) (q)$ , a product of two elements of $A$ . Both conclusions hold for every $q$ simultaneously.

Dickson at work

The elements $i$ and $\frac{1}{5} (3 i + 4 j)$ both satisfy $t^{2} + 1$ over $ℝ$ and neither satisfies a linear polynomial, so $t^{2} + 1$ is the minimal polynomial of each. Dickson's theorem therefore asserts they are conjugate in $ℍ$ — as they are, both being unit purely imaginary quaternions.

Process and Workflow

The proof of $(16.9)$ is constructive except at one point, and that point is where a search is required.

Confirm the class is algebraicFind the minimal polynomial

f \in F [t]

of one element

a_{1}

A

. If no such polynomial exists, nothing on this page applies.

Divide out the prescribed factorRight-divide

f

t - a_{1}

using the monic division algorithm; the remainder is

f (a_{1}) = 0

, so the division is exact.

Find an element of $A$ that the current right factor missesBy

(16.5)

such an element exists while the accumulated right factor has degree

< \deg f

. Locating it is the search step.

Conjugate it into a root of the quotientWith

α

the value of the current right factor at that element,

α a α^{- 1}

is a root of the quotient by

(16.3)

; peel off the corresponding linear factor.

Repeat until the degree is exhaustedAfter

n

steps the quotient is the constant

1

and the factorisation is complete.

Comparison and Classification

Factorisation over a field versus over a division ring
Property	$F$ a field	$D$ a division ring
Irreducible $f$ of degree $n \geq 2$	no proper factorisation	splits into $n$ linear factors once it has one root $(16.9)$
Number of complete factorisations	one, up to order and units	infinitely many when $n \geq 2$
Roots of $f$ determine the factors	yes	no; only the rightmost factor carries a root of $f$
Permuting factors	always valid	cyclic permutations only
Same minimal polynomial implies	equal up to $F$ -isomorphism of $F (a)$	conjugate in $D$ $(16.8)$
Coefficient of $t^{n - 1}$	sum of roots	sum of $n$ elements of the class $(16.10)$

Where each hypothesis is needed
Hypothesis	Used for	Consequence if dropped
$f$ has coefficients in $Z (D)$	root set is a union of classes; factors commute	cyclic permutation fails; $f (a) = 0$ says nothing about conjugates of $a$
$A$ algebraic over $F$	existence of $f$ and of the degree bound $(16.5)$	no minimal polynomial exists and no vanishing polynomial exists at all
$D$ is a division ring	inverting $h (a)$ in $(16.3)$ ; $D [t]$ a domain	the conjugation rule and the cancellation in the commuting lemma both fail
$f$ monic	right division algorithm	division by $f$ is not generally possible

Relationship Map

This page consumes the vanishing-polynomial theory and feeds the applications at the end of §16.

(16.3)

conjugation rule

(16.5)

degree bound

(16.9)

Wedderburn

(16.10)

trace and norm

Vanishing Polynomials supplies $(16.5)$ , the only nontrivial input.
The Gordon–Motzkin Theorem supplies the complementary bound and, via Herstein's theorem, the infinitude of a noncentral class that makes the factorisation non-unique.
The Bray–Whaples Theorem is the interpolation counterpart: instead of prescribing one class and factoring, it prescribes $n$ pairwise nonconjugate points and constructs the unique monic polynomial vanishing on them.
The Niven–Jacobson Theorem uses the quadratic case as its main tool over quaternion algebras.
The corollary $(16.10)$ connects to reduced trace and reduced norm for centrally finite algebras, and thence to Splitting Fields for Finite-Dimensional Algebras.

Applications and Industry Use

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

Division algebras

Reduced trace and reduced norm

If $D$ is centrally finite of degree $n$ and $F (a)$ is a maximal subfield, then $f$ is the reduced characteristic polynomial of $a$ , so $(16.10)$ writes the reduced trace as a sum of $n$ conjugates of $a$ and the reduced norm as a product of $n$ conjugates.

Quaternionic computation

Factoring quaternion polynomials

Software that factors polynomials over $ℍ$ relies on the fact that any real irreducible quadratic dividing the polynomial splits as $(t - \bar{q}) (t - q)$ for a whole sphere of $q$ , which is exactly $(E . 2)$ and $(E . 3)$ .

Skew polynomial codes

Non-unique factorisation as a resource

In $𝔽_{q^{m}} [t; σ]$ the analogue of Wedderburn's theorem gives many factorisations of the same generator polynomial; code constructions exploit that freedom to design generators with prescribed root spaces.

Group theory

Conjugacy testing

Dickson's criterion reduces conjugacy of elements of a division algebra to equality of two minimal polynomials — a linear-algebra computation — rather than a search for a conjugating element.

Honest summary: the theorem is the reason noncommutative factorisation theory is not a copy of the commutative one. Its practical value is mostly as a certificate — it tells an algorithm that a complete split exists, so the search for one need not be abandoned.

Failure Modes and Common Mistakes

Do not conclude from $(16.9)$ that $f$ is reducible in $F [t]$ ; it is irreducible there, and that is the whole point.
Do not apply the theorem to a class that is not algebraic over $F$ — there is then no $f$ to factor, and no nonzero polynomial vanishes on the class.
Do not expect the factorisation to be computable without a search: the proof guarantees an escaping element of $A$ exists but does not exhibit it.
Do not transfer the result to $\deg f = 1$ expecting content; a central singleton class gives $f = t - a$ and the statement is vacuous.

Historical Notes and Lessons Learned

1914–23Dickson on algebrasDickson develops the arithmetic of linear associative algebras and establishes the conjugacy criterion in terms of minimal polynomials over the centre.
1921Wedderburn on division algebrasWedderburn proves that a minimal polynomial over the centre splits completely inside the division algebra, exposing a phenomenon with no commutative counterpart.
1941NivenNiven solves quaternionic polynomial equations, making the quadratic case of the factorisation theorem completely explicit.
1965Gordon and MotzkinThe class bound and the two-implies-infinite dichotomy explain quantitatively why complete factorisations come in infinite families.
1980s–Skew polynomial analoguesWedderburn polynomials and their factorisation theory are developed for Ore extensions, in work of Lam, Leroy and others, and imported into rank-metric coding theory.

The lesson worth keeping is about the meaning of irreducibility. Over a field it is an absolute property of a polynomial; over a division ring it is a property of the coefficient ring one happens to be working in. Wedderburn's theorem measures the collapse precisely: irreducible of degree $n$ over $F$ becomes maximally reducible over $D$ the moment a single root appears.

Quick Reference

Setting

D

a division ring,

F = Z (D)

A

a class algebraic over

F

Minimal polynomial

f \in F [t]

, monic irreducible,

\deg f = n

Wedderburn

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

with all

a_{j} \in A

Prescription

a_{1}

may be any element of

A

Symmetrycyclic permutations of the factors are valid

Dicksonconjugate

\Leftrightarrow

same minimal polynomial over

F

Trace

- d_{1} = a_{1} + \dots + a_{n}

Norm

(- 1)^{n} d_{n} = a_{n} a_{n - 1} \dots a_{1}

Statements and hypotheses
Reference	Hypotheses	Conclusion
$(16.8)$	$a, b \in D$ algebraic over $F = Z (D)$	$a \sim b$ in $D$ $\Leftrightarrow$ same minimal polynomial over $F$
$(16.9)$	$A$ a class algebraic over $F$ with minimal polynomial $f$ , $\deg f = n$	$f$ splits into $n$ linear factors with roots in $A$ ; $a_{1}$ prescribable
$(16.9)$ , cyclic part	$f \in F [t]$ , $f = f_{1} f_{2}$ in $D [t]$	$f = f_{2} f_{1}$ as well
$(16.10)$	Notation of $(16.9)$ , $f = t^{n} + d_{1} t^{n - 1} + \dots + d_{n}$	$- d_{1} \in$ sums of $n$ elements of $A$ ; $(- 1)^{n} d_{n} \in$ products
Corollary	$f \in F [t]$ irreducible of degree $n \geq 2$ with a root in $D$	infinitely many distinct complete factorisations in $D [t]$

Frequently Asked Questions

How can an irreducible polynomial factor?

Irreducibility is relative to the coefficient ring. $f$ is irreducible in $F [t]$ and stays irreducible there; the factorisation happens in the strictly larger ring $D [t]$ , whose linear factors $t - a$ have noncentral $a$ . There is no contradiction, exactly as $t^{2} + 1$ is irreducible over $ℝ$ but factors over $ℂ$ — except that here the enlargement is noncommutative and produces infinitely many factorisations rather than one.

Is the factorisation unique if I fix

a_{1}

The theorem does not claim uniqueness even then, and no canonical choice of $a_{2}, \dots, a_{n}$ is produced. For $n = 2$ the remaining factor is forced by coefficient matching, but for larger $n$ the freedom in the intermediate steps is genuine.

Why are only cyclic permutations of the factors allowed?

The commuting lemma applies to a split of $f$ into exactly two blocks, $f = f_{1} f_{2}$ , and yields $f = f_{2} f_{1}$ . Taking $f_{2}$ to be the rightmost linear factor rotates it to the left end. Iterating generates the cyclic group of rotations and nothing more; arbitrary transpositions of adjacent factors are not justified and are generally false.

Does Wedderburn's theorem need

D

to be finite-dimensional over its centre?

No. Only the class $A$ is assumed algebraic over $F$ . The division ring may be centrally infinite; what matters is that the elements of $A$ satisfy a polynomial over the centre, which is what produces $f$ and makes the degree bound $(16.5)$ available.

What is the relationship to reduced characteristic polynomials?

If $D$ is centrally finite of degree $n$ and $a$ generates a maximal subfield $F (a)$ , then the minimal polynomial of $a$ over $F$ coincides with its reduced characteristic polynomial. In that case $(16.10)$ says the reduced trace of $a$ is a sum of $n$ conjugates of $a$ and the reduced norm is a product of $n$ conjugates.

Can Dickson's criterion be used to test conjugacy in practice?

Yes, and it is the standard method for centrally finite algebras: compute both minimal polynomials over $F$ by linear algebra and compare. It is far cheaper than searching for a conjugating element. Note the criterion fails if the minimal polynomials are taken over a noncentral subfield.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §16, (16.8)–(16.10) (pp. 267–269).
J. H. M. Wedderburn, “On division algebras”, Transactions of the American Mathematical Society 22 (1921), 129–135.
L. E. Dickson, Algebras and Their Arithmetics, University of Chicago Press, 1923.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964.
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

Construct an explicit complete factorisation of a degree-three minimal polynomial in a cyclic division algebra of degree three.
Given $a_{1}$ in a class of degree $n \geq 3$ , how much freedom remains in choosing $a_{2}$ ?
State and prove the analogue of Wedderburn's factorisation theorem for Wedderburn polynomials in an Ore extension $D [t; σ, δ]$ .
Is there a canonical choice of factorisation when $D$ is a quaternion algebra over a number field?
How does $(16.10)$ compare with the classical expression of the reduced norm as a determinant after scalar extension to a splitting field?
Give an example showing that a non-cyclic permutation of the linear factors in a Wedderburn factorisation is invalid.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why a central polynomial is so flexible

The engine: push the factorisation as far as it will go

06Key Results

07Proof Techniques and Method

Maximise, then contradict

Convert escape into a root

Close with a degree bound

08Worked Example

Quadratic classes in a quaternion division algebra

Non-uniqueness, made explicit

Dickson at work

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Reduced trace and reduced norm

Factoring quaternion polynomials

Non-unique factorisation as a resource

Conjugacy testing

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

Key Results

Proof Techniques and Method

Worked Example

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions