Polynomials over Division Rings: Evaluation, Roots and Division

Q: Why insist that t be central when the coefficients are not?

Because the two constructions answer different questions. A central t gives the ring of polynomial *functions-in-waiting* on D, and the resulting theory is about roots and conjugacy classes. A skew t, with td = (d)t + (d), gives a ring whose module theory encodes linear difference or differential operators. Both are studied in this collection, but only the central case supports the substitution f f(d) for arbitrary d ∈ D.

Q: Is the evaluation map ever multiplicative?

Yes, on useful sub-collections. If h has coefficients in Z(D) then (gh)(d) = g(d)h(d) for all d, because h(d) commutes with everything in sight; this is Lam's Exercise 16.7. It is also multiplicative whenever d commutes with all coefficients of g. Outside these cases it fails.

Q: If d is a root of f and f = gh, must d be a root of g or of h?

Not in general. What (16.3) gives is that d is a root of h, or else some *conjugate* of d is a root of g. The example (t-j)(t-i) shows the conjugate really is needed: i is a root of the product, is a root of the right factor, and no root of the left factor t-j equals i.

Q: Does D[t] have a good factorisation theory?

It is a principal left and right ideal domain, hence a non-commutative unique factorisation domain in Ore's sense: any two factorisations of a fixed element into irreducibles have the same length and isomorphic factors up to similarity. But the factors themselves are far from unique — t^2+1 factors as (t+q)(t-q) for every unit pure quaternion q.

Q: What replaces the derivative and multiplicity?

There is no formal derivative test that works unchanged, because the product rule fails at the level of evaluation. Multiplicity is normally tracked by divisibility in D[t] instead: one asks for the largest m with (t-d)^m a right divisor of f. That notion is well behaved because right divisibility, unlike evaluation, is multiplicative by construction.

Q: How do I get the left-handed version of every statement?

Apply the right-handed statement in D^op. Multiplication reverses, so right divisors become left divisors and (16.3) becomes f(d) = h(d),g(a^-1da) in the appropriate left-evaluation sense. Lam states everything on the right and invokes the opposite ring when a left statement is needed, as in the proof of (16.14).

Executive Summary

Over a commutative field $K$ , the substitution $f \mapsto f (c)$ is a $K$ -algebra homomorphism $K [t] \to K$ , and every elementary fact about roots — the factor theorem, the bound of $n$ roots for degree $n$ , the uniqueness of interpolation — is a consequence. Over a noncommutative division ring $D$ the substitution map is still additive, but it is not multiplicative, and all three consequences fail.

What survives is sharper than it first appears. The factor theorem survives verbatim for right roots $(16.2)$ . Multiplicativity is replaced by an exact conjugation formula $(16.3)$ : if $f = g h$ and $a : = h (d) \neq 0$ , then $f (d) = g (a d a^{- 1}) h (d)$ . That formula is the engine of the whole section — Gordon–Motzkin, Wedderburn's factorisation and the Bray–Whaples uniqueness theorem are all obtained by iterating it.

$t$ centralConvention in

D [t]

Additive onlyEvaluation map

$g (a d a^{- 1}) h (d)$ Replaces

g (d) h (d)

$\infty$ Possible root count, degree 2

Overview

Fix a ring $R$ with identity and let $R [t]$ be the polynomial ring in one indeterminate $t$ that commutes elementwise with $R$ . This is not a skew polynomial ring: the twisting constructions $R [t; σ]$ and $R [t; δ]$ discussed in Polynomial Rings and Laurent Series and Differential Polynomial Rings and the Weyl Algebra deform the commutation rule between $t$ and the coefficients. Here $t$ is genuinely central, and all the difficulty comes from the coefficients failing to commute with the element we substitute.

f (t) = \sum_{i = 0}^{n} a_{i} t^{i} \in R [t], f (r) : = \sum_{i = 0}^{n} a_{i} r^{i} \in R .

(16.0)

Evaluation. The coefficients must be written on the left of the powers of $t$ before substituting.

The parenthetical instruction in the caption is the whole subtlety. In $R [t]$ the identities $\sum a_{i} t^{i} = \sum t^{i} a_{i}$ hold, because $t$ is central; in $R$ the two elements $\sum a_{i} r^{i}$ and $\sum r^{i} a_{i}$ are usually different. So a polynomial determines two evaluation maps, and hence two notions of root.

Specialising to a division ring lets us divide by $h (d)$ when it is nonzero, and that is where conjugation enters. Because conjugation is what replaces substitution, conjugacy classes — not individual elements — become the natural carriers of root information. That is the theme picked up in The Gordon–Motzkin Theorem and Vanishing Polynomials.

Learning Objectives

Set up $D [t]$ with a central indeterminate and evaluate polynomials correctly on the left.
Produce the standard counterexample showing that $f \mapsto f (d)$ is not multiplicative.
State and prove the Remainder Theorem $(16.2)$ and identify the left ideal of polynomials killed by a fixed element.
Prove the conjugation formula $(16.3)$ and use it to transfer roots from a product to its left factor.
Explain why the roots of a right factor are roots of the product but the roots of a left factor need not be.
Find all roots of $(t - j) (t - i)$ in the real quaternions.

Definitions

Definition(16.1)Right root

Let $R$ be a ring and $f (t) = \sum_{i = 0}^{n} a_{i} t^{i} \in R [t]$ . An element $r \in R$ is a right root of $f$ if $f (r) = \sum_{i = 0}^{n} a_{i} r^{i} = 0$ . Following Lam, root without qualification means right root throughout §16.

$R [t]$: Polynomials with coefficients in $R$ in a central indeterminate $t$ : $t r = r t$ for all $r \in R$ . As an abelian group it is $⨁_{i \geq 0} R t^{i}$ .
Left root: $r$ with $\sum_{i} r^{i} a_{i} = 0$ . Equivalently a right root of the corresponding polynomial over $R^{op}$ . Left and right roots of the same $f$ need not coincide.
$\deg f$: The largest $i$ with $a_{i} \neq 0$ ; $\deg 0$ is undefined (or $- \infty$ ). Over a division ring $\deg (f g) = \deg f + \deg g$ , so $D [t]$ is a domain.
Monic: Leading coefficient $1$ . Division by a monic polynomial is possible over any ring, on either side.
Central coefficients: $F [t] \subseteq D [t]$ where $F = Z (D)$ . In fact $Z (D [t]) = F [t]$ , and polynomials with central coefficients behave far better than general ones.

$D$ always denotes a division ring, $F = Z (D)$ its centre, and $D^{*} = D ∖ {0}$ its multiplicative group. Conjugation means $d \mapsto a d a^{- 1}$ for $a \in D^{*}$ .

Core Concepts

Why evaluation is not multiplicative

Take $a, b \in R$ with $a b \neq b a$ and set $g (t) = t - a$ , $h (t) = t - b$ . Multiplying in $R [t]$ , where $t$ is central,

f (t) = g (t) h (t) = t^{2} - (a + b) t + a b, f (a) = a^{2} - (a + b) a + a b = a b - b a \neq 0,

(16.0a)

while $g (a) h (a) = 0 \cdot (a - b) = 0$ . Substitution destroys the factorisation.

The failure is not pathological: it is generic. It happens for every pair of noncommuting elements, and the size of the failure, $a b - b a$ , is exactly the additive commutator. So there is no hope of patching the definition of evaluation; the theory has to be rebuilt around a substitute for multiplicativity.

The master identity

The substitute costs one line. Write $g (t) = \sum_{i} b_{i} t^{i}$ and let $h$ be arbitrary. Because $t$ is central, $g (t) h (t) = \sum_{i} b_{i} h (t) t^{i}$ , and expanding $h$ and collecting powers of $t$ gives, for any $d \in R$ ,

(g h) (d) = \sum_{i} b_{i} h (d) d^{i} .

(16.0b)

Valid over any ring. Note how $h (d)$ is trapped between the coefficient $b_{i}$ and the power $d^{i}$ — that is precisely why $g (d)$ does not appear.

Two readings follow immediately. If $h (d) = 0$ then $(g h) (d) = 0$ : every root of a right factor is a root of the product. If $h (d) = a$ is invertible, insert $a^{- 1} a$ after each $b_{i}$ and the powers of $d$ turn into powers of $a d a^{- 1}$ ; that is $(16.3)$ .

Division on both sides

Division by a monic polynomial works over any ring by the usual leading-term subtraction, and it works on both sides: given $f$ and monic $g$ of degree $m$ , there are unique $q, r$ with $f = q g + r$ and unique $q', r'$ with $f = g q' + r'$ , all remainders of degree $< m$ . Over a division ring every nonzero polynomial can be scaled to be monic, so $D [t]$ is Euclidean on both sides. Consequently $D [t]$ is a principal left ideal domain and a principal right ideal domain.

f (d) = 0

t - d

right-divides

f

f \in D [t] (t - d)

The two-sided ideal structure is much thinner than the one-sided structure: the centre of $D [t]$ is $F [t]$ , and the ideal generated by $f$ is generated by the largest divisor of $f$ lying in $F [t]$ (Lam, Exercise 16.8). In particular $D [t]$ is far from simple, and when $\dim_{F} D < \infty$ it is neither left nor right primitive (Exercise 16.9).

Key Results

Lemma—Evaluation of a product

Let $R$ be any ring, $f = g h$ in $R [t]$ with $g (t) = \sum_{i} b_{i} t^{i}$ , and let $d \in R$ . Then $f (d) = \sum_{i} b_{i} h (d) d^{i}$ .

Proof

Write $h (t) = \sum_{j} c_{j} t^{j}$ . Since $t$ is central, $f (t) = \sum_{i, j} b_{i} c_{j} t^{i + j}$ , and the coefficient of $t^{k}$ in $f$ is $\sum_{i + j = k} b_{i} c_{j}$ . Hence $f (d) = \sum_{k} (\sum_{i + j = k} b_{i} c_{j}) d^{k} = \sum_{i} b_{i} (\sum_{j} c_{j} d^{j}) d^{i} = \sum_{i} b_{i} h (d) d^{i}$ , the regrouping being legitimate because it only reassociates a finite sum in $R$ .

Proposition(16.2)Remainder Theorem

Let $R$ be a ring, $f \in R [t]$ and $r \in R$ . Then $f (r) = 0$ if and only if $t - r$ is a right divisor of $f$ in $R [t]$ , that is $f = q \cdot (t - r)$ for some $q \in R [t]$ . Consequently ${f \in R [t] : f (r) = 0} = R [t] (t - r)$ , a left ideal of $R [t]$ .

Proof

Sufficiency. If $f = q \cdot (t - r)$ , apply the Lemma with $g = q$ and $h = t - r$ : since $h (r) = 0$ , every term of $\sum_{i} b_{i} h (r) r^{i}$ vanishes, so $f (r) = 0$ .

Necessity. The polynomial $t - r$ is monic, so right division gives $f = q \cdot (t - r) + s$ with $\deg s < 1$ , i.e. $s \in R$ . Evaluation is additive, and $(q \cdot (t - r)) (r) = 0$ by the first part, so $0 = f (r) = 0 + s$ , whence $s = 0$ .

The left ideal. The displayed set contains $R [t] (t - r)$ by sufficiency and is contained in it by necessity; it is a left ideal because it is a left multiple set, or directly because evaluation of $p f$ at $r$ is $\sum_{i} p_{i} f (r) r^{i} = 0$ whenever $f (r) = 0$ .

Proposition(16.3)Conjugation formula

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

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

In particular, if $d$ is a root of $f$ but not a root of $h$ , then $a d a^{- 1}$ is a root of $g$ — so $g$ has a root in the conjugacy class of $d$ .

Proof

Write $g (t) = \sum_{i} b_{i} t^{i}$ . By the Lemma, $f (d) = \sum_{i} b_{i} a d^{i}$ . Insert $a^{- 1} a$ between $a$ and $d^{i}$ and note $a d^{i} a^{- 1} = (a d a^{- 1})^{i}$ :

f (d) = \sum_{i} b_{i} a d^{i} a^{- 1} a = \sum_{i} b_{i} (a d a^{- 1})^{i} a = g (a d a^{- 1}) a .

Since $a = h (d)$ this is the stated identity. If moreover $f (d) = 0$ then $g (a d a^{- 1}) a = 0$ , and $D$ has no zero divisors with $a \neq 0$ , so $g (a d a^{- 1}) = 0$ .

Corollary—Asymmetry of factors

For $f = g h$ in $D [t]$ : every root of $h$ is a root of $f$ , but a root of $g$ need not be a root of $f$ . What is true is the weaker statement that if $d$ is a root of $f$ and not of $h$ , then some conjugate of $d$ is a root of $g$ ; the correspondence between the roots of $f$ and those of $g$ is only up to conjugacy.

Remark—Left roots

Everything above has a mirror image. Passing to $D^{op}$ converts left roots into right roots and reverses the order of the factors, so $(16.2)$ becomes: $r$ is a left root of $f$ iff $f \in (t - r) D [t]$ . The two notions are genuinely different — see the analysis of $(t - j) (t - i)$ below, where $j$ is a left root but not a right root. Niven's proof of the Fundamental Theorem of Algebra for quaternions, discussed in The Niven–Jacobson Theorem, moves between the two deliberately.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Conjugate, do not substitute

Whenever a factorisation $f = g h$ meets an element $d$ , do not evaluate the factors separately. Compute $a = h (d)$ and pass to $a d a^{- 1}$ . The conjugating element is always the value of the right factor.

Move 2

Split off a linear right factor

A root $c$ of $f$ yields $f = g \cdot (t - c)$ with $\deg g = \deg f - 1$ . Induction on degree is then available, and it is the standard opening of the proofs of $(16.4)$ , $(16.9)$ and $(16.13)$ .

Move 3

Divide with remainder, then evaluate

To prove that a set of polynomials is a left ideal $D [t] g$ , divide by $g$ on the right and show the remainder has more roots than its degree permits. This converts a divisibility statement into a counting statement.

The three moves compose. Move 2 reduces degree, Move 1 keeps track of how the root set is displaced under that reduction, and Move 3 converts the resulting count back into an algebraic conclusion. All of §16 is built from this loop.

Worked Example

Evaluation destroys a factorisation in $ℍ$

Let $ℍ = ℝ \oplus ℝ i \oplus ℝ j \oplus ℝ k$ be the real quaternions and put $f (t) = (t - i) (t - j)$ . Expanding with $t$ central, $f (t) = t^{2} - (i + j) t + i j = t^{2} - (i + j) t + k$ . Now evaluate at $i$ :

f (i) = i^{2} - (i + j) i + k = - 1 - (i^{2} + j i) + k = - 1 - (- 1 - k) + k = 2 k,

(E.1)

in agreement with the general formula $f (a) = a b - b a$ : here $i j - j i = k - (- k) = 2 k$ .

So $i$ is a root of the left factor $t - i$ but not of the product. Its right factor $t - j$ does contribute: $f (j) = j^{2} - (i + j) j + k = - 1 - (i j + j^{2}) + k = - 1 - k + 1 + k = 0$ .

All roots of $(t - j) (t - i)$

Reverse the factors: $f_{1} (t) = (t - j) (t - i) = t^{2} - (i + j) t + j i = t^{2} - (i + j) t - k$ . Its right factor gives one root, $d = i$ . Suppose $d \neq i$ is another root. Put $a : = (t - i) (d) = d - i \neq 0$ . By $(16.3)$ applied to $g = t - j$ , $h = t - i$ , the element $a d a^{- 1}$ is a root of $t - j$ , that is

a d a^{- 1} = j, d = a + i ⟹ a + a i a^{- 1} = j .

(E.2)

Set $m : = a i a^{- 1}$ , so $a = j - m$ . Rewriting $m a = a i$ with $a = j - m$ gives $m j - m^{2} = j i - m i$ . Since $m$ is conjugate to $i$ we have $m^{2} = - 1$ , and $j i = - k$ , so $m j + m i = - k - 1$ , i.e. $m (i + j) = - (1 + k)$ . As $(i + j)^{- 1} = - (i + j) / 2$ ,

m = - (1 + k) \cdot (- \frac{1}{2} (i + j)) = \frac{1}{2} (i + j + k i + k j) = \frac{1}{2} (i + j + j - i) = j .

(E.3)

using $k i = j$ and $k j = - i$ .

Then $a = j - m = 0$ , contradicting $a \neq 0$ . Hence ** $i$ is the only root of $f_{1}$ in $ℍ$ ** — a quadratic with exactly one root, even though $i$ and $j$ are conjugate in $ℍ$ (both have minimal polynomial $t^{2} + 1$ over $ℝ$ ). Note also that $j$ is a left root of $f_{1}$ , since $f_{1}$ has $t - j$ as a left factor.

Comparison and Classification

Polynomials over a field versus over a noncommutative division ring
Feature	$K [t]$ , $K$ a commutative field	$D [t]$ , $D$ noncommutative
Evaluation at $d$	ring homomorphism	additive and unital only
Value of a product	$f (d) = g (d) h (d)$	$f (d) = g (a d a^{- 1}) h (d)$ when $a = h (d) \neq 0$
Factor theorem	$f (d) = 0 \Leftrightarrow (t - d) ∣ f$	$f (d) = 0 \Leftrightarrow (t - d)$ is a right divisor
Roots of degree $n$	at most $n$	at most $n$ conjugacy classes; possibly infinitely many elements
Left and right roots	the same notion	different; $j$ is a left but not a right root of $(t - j) (t - i)$
Ideal structure	principal ideal domain	principal left and right ideal domain; two-sided ideals come from $F [t]$
Centre	$K [t]$	$F [t]$ with $F = Z (D)$
Splitting into linear factors	unique up to order when it exists	wildly non-unique — see Wedderburn's Factorisation Theorem

Relationship Map

The three results of this page sit at the base of the section; everything later is a consequence.

$(16.2)$ Remainder Theorem — roots ↔ linear right factors
- with $(16.3)$ gives
  - $(16.4)$ Gordon–Motzkin: roots lie in $\leq n$ conjugacy classes
  - $(16.9)$ Wedderburn: a minimal polynomial splits into linear factors from one class
  - $(16.13)$ Bray–Whaples: unique monic interpolant on nonconjugate points
- with the division algorithm gives
  - $(16.6)$ vanishing on a class $\Leftrightarrow$ divisible by the minimal polynomial
  - $D [t]$ is a principal left and right ideal domain

(16.0 b)

master identity

(16.3)

conjugation formulaconjugacy classes carry the root data§16 in full

Standards and Notation

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

This collection

D [t]

t

central; coefficients on the left; root means right root

Common variantCoefficients on the right, giving left roots as the default (Ore, and much of the quaternionic analysis literature)

Skew case

D [t; σ, δ]

with

t d = σ (d) t + δ (d)

— a different ring; do not conflate

Evaluation notation

f (d)

here; some authors write

f^{r} (d)

f (d)_{r}

to flag the side

MarkupPresentation MathML per ISO/IEC 40314; symbol conventions per ISO 80000-2

ImplementationsSage OrePolynomialRing, Magma twisted polynomial rings, GAP quaternion algebras via QuaternionAlgebra

Computational Notes

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

Right division of $f$ (degree $n$ ) by a monic $g$ (degree $m \leq n$ ) costs $O ((n - m + 1) m)$ multiplications in $D$ . Nothing is saved relative to the commutative case, but the side must be fixed before the loop is written: f = q*g + r and f = g*q + r produce different quotients.
Evaluation of a degree- $n$ polynomial by Horner's rule reads $f (d) = (\dots ((a_{n} d + a_{n - 1}) d + a_{n - 2}) d \dots) + a_{0}$ , which is correct for right roots because the coefficients stay on the left throughout. The mirrored Horner loop computes left-root values.
Greatest common right divisors exist and are computed by the Euclidean algorithm run on the right; $\gcd$ is then only well defined up to a unit of $D$ , i.e. up to a nonzero left scalar.
For $D$ finite-dimensional over $F$ , root-finding for $f \in D [t]$ reduces to solving a system of polynomial equations over $F$ in $\dim_{F} D$ unknowns; the eigenvalue reformulation via companion matrices (Lam, Exercise 16.9) is usually the practical route.
Computer algebra systems model $D [t]$ as an Ore polynomial ring with trivial twisting automorphism and trivial derivation; Sage's OrePolynomialRing and Magma's twisted polynomial rings both specialise correctly, but their roots methods generally assume a commutative base and should not be trusted here.

Failure Modes and Common Mistakes

Do not write $\sum t^{i} a_{i}$ and then substitute — you will have computed a left-root condition while believing you computed a right-root condition.
Do not cancel a common factor from an equation such as $g (a d a^{- 1}) a = 0$ on the wrong side; in $D$ cancellation is legitimate but the side matters for what remains.
Do not assume $D [t]$ is simple because $D$ is. It has a large centre $F [t]$ and correspondingly many two-sided ideals.
Do not identify $D [t]$ with the free product or with a skew polynomial ring; $t$ is central here by fiat, and every proof above uses that.

Best Practices

Declare the side once, at the top of any argument, and never switch silently.
When a factorisation is available, record which factor is on the right — that is the one whose roots you already know.
Track conjugacy classes rather than elements; the class is the invariant object, the element is a choice.
Before claiming a root set is finite, check whether two roots lie in a common conjugacy class.
Verify a computed factorisation by expanding it in $D [t]$ , not by comparing values at sample points.

Quick Reference

Ring

D [t]

t

central,

D

a division ring,

F = Z (D)

Evaluation

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

, coefficients on the left

Master identity

(g h) (d) = \sum_{i} b_{i} h (d) d^{i}

Remainder Theorem

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

Conjugation formula

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

a = h (d) \neq 0

Right factorsroots of

h

are roots of

g h

Left factorsroots of

g

move by conjugation

Ideal theoryprincipal left and right ideal domain;

Z (D [t]) = F [t]

The three foundational statements
Reference	Statement	Hypotheses
(16.1)	Definition of right root	any ring $R$
(16.2)	$f (r) = 0 \Leftrightarrow t - r$ is a right divisor; root set is $R [t] (t - r)$	any ring $R$
(16.3)	$f = g h$ , $a = h (d) \neq 0 \Rightarrow f (d) = g (a d a^{- 1}) a$	$D$ a division ring

Frequently Asked Questions

Why insist that

t

be central when the coefficients are not?

Because the two constructions answer different questions. A central $t$ gives the ring of polynomial functions-in-waiting on $D$ , and the resulting theory is about roots and conjugacy classes. A skew $t$ , with $t d = σ (d) t + δ (d)$ , gives a ring whose module theory encodes linear difference or differential operators. Both are studied in this collection, but only the central case supports the substitution $f \mapsto f (d)$ for arbitrary $d \in D$ .

Is the evaluation map ever multiplicative?

Yes, on useful sub-collections. If $h$ has coefficients in $Z (D)$ then $(g h) (d) = g (d) h (d)$ for all $d$ , because $h (d)$ commutes with everything in sight; this is Lam's Exercise 16.7. It is also multiplicative whenever $d$ commutes with all coefficients of $g$ . Outside these cases it fails.

d

is a root of

f

and

f = g h

, must

d

be a root of

g

or of

h

Not in general. What $(16.3)$ gives is that $d$ is a root of $h$ , or else some conjugate of $d$ is a root of $g$ . The example $(t - j) (t - i)$ shows the conjugate really is needed: $i$ is a root of the product, is a root of the right factor, and no root of the left factor $t - j$ equals $i$ .

Does

D [t]

have a good factorisation theory?

It is a principal left and right ideal domain, hence a non-commutative unique factorisation domain in Ore's sense: any two factorisations of a fixed element into irreducibles have the same length and isomorphic factors up to similarity. But the factors themselves are far from unique — $t^{2} + 1$ factors as $(t + q) (t - q)$ for every unit pure quaternion $q$ .

What replaces the derivative and multiplicity?

There is no formal derivative test that works unchanged, because the product rule fails at the level of evaluation. Multiplicity is normally tracked by divisibility in $D [t]$ instead: one asks for the largest $m$ with $(t - d)^{m}$ a right divisor of $f$ . That notion is well behaved because right divisibility, unlike evaluation, is multiplicative by construction.

How do I get the left-handed version of every statement?

Apply the right-handed statement in $D^{op}$ . Multiplication reverses, so right divisors become left divisors and $(16.3)$ becomes $f (d) = h (d) g (a^{- 1} d a)$ in the appropriate left-evaluation sense. Lam states everything on the right and invokes the opposite ring when a left statement is needed, as in the proof of $(16.14)$ .

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §16 (pp. 261–274).
O. Ore, “Theory of non-commutative polynomials”, Annals of Mathematics 34 (1933), 480–508.
P. M. Cohn, Skew Fields: Theory of General Division Rings, Encyclopedia of Mathematics and its Applications 57, Cambridge University Press, 1995, Chapters 1 and 2.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964.
B. Gordon and T. S. Motzkin, “On the zeros of polynomials over division rings”, Transactions of the American Mathematical Society 116 (1965), 218–226.

AI Suggested Questions

Work out the full proof that right division by a monic polynomial is possible and unique over an arbitrary ring.
Show that the two-sided ideals of $D [t]$ are exactly the ideals generated by polynomials in $F [t]$ , where $F = Z (D)$ .
Give an example of $f, g \in ℍ [t]$ where $f$ and $g$ have the same right roots but different left roots.
Explain how the companion-matrix construction of Lam's Exercise 16.9 turns root-finding in $D [t]$ into an eigenvalue problem.
Compare the untwisted ring $D [t]$ with the skew polynomial ring $D [t; σ]$ : which of $(16.2)$ and $(16.3)$ survive, and in what form?
How large can the set of monic degree-2 polynomials over $ℍ$ with exactly one root be, as a subset of $ℍ^{2}$ ?
Describe the notion of similarity of polynomials in $D [t]$ and its relation to conjugacy of roots.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why evaluation is not multiplicative

The master identity

Division on both sides

06Key Results

07Proof Techniques and Method

Conjugate, do not substitute

Split off a linear right factor

Divide with remainder, then evaluate

08Worked Example

Evaluation destroys a factorisation in ℍ

All roots of (t−j)(t−i)

09Comparison and Classification

10Relationship Map

11Standards and Notation

12Computational Notes

13Failure Modes and Common Mistakes

14Best Practices

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

Evaluation destroys a factorisation in $ℍ$

All roots of $(t - j) (t - i)$

Comparison and Classification

Relationship Map

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions