Herstein’s Lemma and Jacobson’s Commutativity Theorem

Q: Why must the eigenvalue b_0 be nonzero?

The factorisation (13.8b) isolates _a itself as the factor corresponding to b = 0. If _a were the non-injective factor there would be nothing to conclude — its kernel is just the centralizer of a. The argument therefore uses _a ≠ 0, i.e. noncentrality of a, to push the failure of injectivity onto some _a - b_0 with b_0 ∈ K^*, and b_0 ≠ 0 is exactly what makes xax^-1 = a - b_0 differ from a.

Q: Why does the lemma bother to say y can be chosen to be an additive commutator?

Because (13.9) needs it. The hypothesis of (13.9) constrains additive commutators only, so to conclude that y is torsion one must know y is a commutator. Replacing the eigenvector x by _a(x) = b_0 x costs nothing — it is a scalar multiple, so it conjugates a identically — and buys exactly that.

Q: Does (13.10) say anything when the centre is infinite?

It is then trivial: F(a) F is already infinite, so take K = F(a). The content is entirely in the case of a finite centre, where the ring is infinite but its centre is not, and Herstein's Lemma is needed to manufacture an element of infinite order commuting with a.

Executive Summary

Herstein's Lemma is a rigidity statement. In a division ring of characteristic $p > 0$ , a noncentral element $a$ of finite multiplicative order cannot sit in isolation: some $y \in D^{*}$ conjugates it to a different power of itself, and $y$ may be taken to be an additive commutator $a x - x a$ .

The proof is a piece of linear algebra. The inner derivation $δ_{a}$ is linear over the finite field $K = 𝔽_{p} [a]$ , satisfies $δ_{a}^{| K |} = δ_{a}$ , and therefore factors as $\prod_{b \in K} (δ_{a} - b) = 0$ . Since $δ_{a} \neq 0$ , one factor is non-injective — and its kernel supplies the required eigenvector.

$y a y^{- 1} = a^{i}$ The conclusion

$p > 0$ Required characteristic

$δ_{a}^{p^{n}} = δ_{a}$ Key operator identity

3Applications, (13.9)–(13.11)

Overview

Chapter 5 of Lam leaves one debt outstanding. The Jacobson–Herstein theorem $(12.9)$ — if for all $a, b$ in a ring $R$ there is $n = n (a, b) > 1$ with $(a b - b a)^{n} = a b - b a$ , then $R$ is commutative — was reduced to the case of division rings, but that case was not proved. Herstein's Lemma is the tool that settles it.

The mechanism is a collision between two finiteness statements. Herstein's Lemma produces a torsion element $y$ conjugating $a$ nontrivially; the hypothesis of $(13.9)$ makes $y$ torsion as well; then $⟨ a ⟩ ⟨ y ⟩$ is a finite subgroup of $D^{*}$ , which by Wedderburn's Little Theorem — through $(13.3)$ — must be cyclic, hence abelian. But $y a y^{- 1} \neq a$ . Contradiction.

a

noncentral torsion

(13.8)

y a y^{- 1} = a^{i} \neq a

y

torsion too

⟨ a ⟩ ⟨ y ⟩

finite

(13.3)

: cyclic, so abelianContradiction

The same lemma also proves $(13.10)$ : in an infinite division ring, every element lies in an infinite subfield, so no centralizer is finite. The mechanism there is the opposite reading of the same collision — since a finite subgroup would be abelian, the conjugating element $y$ is forced to have infinite order.

Learning Objectives

State $(13.8)$ with the hypotheses noncentral, torsion, and $char D = p > 0$ .
Show $δ_{a}$ is $K$ -linear for $K = 𝔽_{p} [a]$ and prove $δ_{a}^{| K |} = δ_{a}$ .
Extract an eigenvector from the factorisation $t^{| K |} - t = \prod_{b \in K} (t - b)$ .
Convert the eigenvector equation into $x a x^{- 1} = a - b_{0}$ and then into $a^{i}$ .
Prove $(13.9)$ , the division-ring case of the Jacobson–Herstein theorem.
Prove $(13.10)$ and deduce $(13.11)$ , Jacobson's theorem on algebraic division algebras over finite fields.

Definitions

$δ_{a}$: The inner derivation $x \mapsto a x - x a$ of $D$ . It is zero exactly when $a \in Z (D)$ .
$λ, ρ$: Left and right multiplication by $a$ : $λ (x) = a x$ , $ρ (x) = x a$ . They commute, and $δ_{a} = λ - ρ$ .
$K = 𝔽_{p} [a]$: The subring generated by the prime field and a torsion element $a$ ; a finite field, since it is a finite commutative domain.
$E = End (_{K} D)$: The ring of endomorphisms of $D$ regarded as a left $K$ -vector space. It is a $K$ -algebra of characteristic $p$ , and contains $λ$ , $ρ$ and $δ_{a}$ .
$F (a)$: For $F$ a subfield of $D$ , the division subring generated by $F \cup {a}$ ; a field whenever $F$ is central.

In (13.8) the element a is assumed noncentral and of finite multiplicative order; both are used, and neither can be dropped.

Core Concepts

Turning a ring question into an operator question

The obstacle in $(13.8)$ is that we must produce an element $y$ out of nothing. The trick is to stop looking inside $D$ and look instead inside the endomorphism algebra $E = End (_{K} D)$ , where $δ_{a}$ becomes an operator with computable minimal behaviour.

Two observations make this legitimate. First, $δ_{a}$ is $K$ -linear: for $z \in K$ we have $a z = z a$ , so $δ_{a} (z x) = a z x - z x a = z (a x - x a) = z δ_{a} (x)$ . Second, left multiplication by $a$ is the scalar action of $a \in K$ on $_{K} D$ , so $λ$ is the image of $a$ under $K \to E$ and may be treated as a scalar.

The Frobenius identity for $δ_{a}$

Write $| K | = p^{n}$ , so $a^{p^{n}} = a$ . Since $λ$ and $ρ$ commute and $E$ has characteristic $p$ , the freshman's dream applies to $δ_{a} = λ - ρ$ :

δ_{a}^{p^{n}} = (λ - ρ)^{p^{n}} = λ^{p^{n}} - ρ^{p^{n}}, so δ_{a}^{p^{n}} (x) = a^{p^{n}} x - x a^{p^{n}} = a x - x a = δ_{a} (x) .

(13.8a)

$δ_{a}$ satisfies the same equation $t^{p^{n}} = t$ that the elements of $K$ satisfy.

Now use the factorisation of that polynomial over $K$ , which is complete because $K$ has exactly $p^{n}$ elements and each is a root:

t^{p^{n}} - t = \prod_{b \in K} (t - b) \in K [t], hence 0 = δ_{a}^{p^{n}} - δ_{a} = (\prod_{b \in K^{*}} (δ_{a} - b)) \circ δ_{a} in E .

(13.8b)

From eigenvector to conjugation

An eigenvector is an $x \in D^{*}$ with $a x - x a = b_{0} x$ . Rearranged, $(a - b_{0}) x = x a$ , so $x a x^{- 1} = a - b_{0}$ . Because $b_{0} \neq 0$ , the conjugate differs from $a$ ; because $b_{0} \in K$ , the conjugate still lies in $K$ . Conjugate elements of $D^{*}$ have equal order, and in the cyclic group $K^{*}$ two elements of equal order generate the same subgroup — so $x a x^{- 1}$ is a power of $a$ .

Key Results

Lemma(13.8)Herstein's Lemma

Let $D$ be a division ring with $char D = p > 0$ , and let $a \in D^{*}$ be noncentral and of finite multiplicative order. Then there exists $y \in D^{*}$ with

y a y^{- 1} = a^{i} \neq a for some integer i > 0 .

(13.8c)

Moreover $y$ may be chosen to be an additive commutator of $D$ .

Proof

Setup. Since $a$ is torsion, $K : = 𝔽_{p} [a]$ is a finite field; write $| K | = p^{n}$ , so $a^{p^{n}} = a$ and $K^{*}$ is cyclic. Let $δ = δ_{a}$ , which is nonzero because $a \notin Z (D)$ . As shown above, $δ$ is $K$ -linear on the left $K$ -vector space $D$ , so $δ \in E = End (_{K} D)$ .

An operator identity. Writing $δ = λ - ρ$ with $λ, ρ$ commuting and $char E = p$ , identity $(13.8 a)$ gives $δ^{p^{n}} = δ$ in $E$ .

An eigenvector. Substituting $δ$ into $t^{p^{n}} - t = \prod_{b \in K} (t - b)$ — legitimate because $δ$ lies in the $K$ -algebra $E$ and commutes with the scalars — yields $(13.8 b)$ . If every factor $δ - b$ with $b \in K^{*}$ were injective, their composite would be injective, and composing with $δ$ to get $0$ would force $δ = 0$ , contradicting noncentrality of $a$ . So there is $b_{0} \in K^{*}$ and $x \in D^{*}$ with $δ (x) = b_{0} x$ , that is $a x - x a = b_{0} x$ .

Conjugation. Rearranging, $(a - b_{0}) x = x a$ , hence $x a x^{- 1} = a - b_{0} \in K ∖ {a}$ . Conjugation is an automorphism of $D^{*}$ , so $x a x^{- 1}$ and $a$ have the same finite order; two elements of the same order in the cyclic group $K^{*}$ generate the same cyclic subgroup, so $x a x^{- 1} = a^{i}$ for some $i > 0$ , and $a^{i} \neq a$ since $b_{0} \neq 0$ .

**Making $y$ a commutator.** Put $y : = δ (x) = a x - x a = b_{0} x$ , which is nonzero and is by construction an additive commutator. Since $b_{0} \in K$ commutes with $a$ ,

y a y^{- 1} = b_{0} (x a x^{- 1}) b_{0}^{- 1} = b_{0} a^{i} b_{0}^{- 1} = a^{i},

(13.8d)

so $y$ has the required property.

Theorem(13.9)Jacobson–Herstein, division-ring case

Let $D$ be a division ring such that for all $a, b \in D$ there exists an integer $n = n (a, b) > 1$ with $(a b - b a)^{n} = a b - b a$ . Then $D$ is a field.

Proof

Commutators are torsion. If $u \neq 0$ is an additive commutator then $u^{n} = u$ with $n > 1$ gives $u^{n - 1} = 1$ , so $u$ has finite order in $D^{*}$ .

A noncentral commutator exists. Suppose $D \neq F : = Z (D)$ . By $(13.5)$ , not all additive commutators are central, so choose additive commutators $a = b b' - b' b \notin F$ .

The centre has positive characteristic. For $c \in F^{*}$ , centrality gives $c a = (c b) b' - b' (c b)$ , again a nonzero additive commutator, hence torsion. Let $k$ be a common multiple of the orders of $a$ and of $c a$ . Then $1 = (c a)^{k} = c^{k} a^{k} = c^{k}$ , so every element of $F^{*}$ is torsion. A field of characteristic zero contains $ℚ$ , whose nonzero elements other than $\pm 1$ have infinite order; therefore $char F = char D = p > 0$ .

Apply Herstein's Lemma. $a$ is noncentral and torsion, so $(13.8)$ supplies an additive commutator $y \in D^{*}$ with $y a y^{- 1} = a^{i} \neq a$ . By hypothesis $y$ is itself torsion. Since $y$ normalises the finite cyclic group $⟨ a ⟩$ , the set $⟨ a ⟩ ⟨ y ⟩$ is a subgroup of $D^{*}$ , and it is finite because both $⟨ a ⟩$ and $⟨ y ⟩$ are.

Contradiction. By $(13.3)$ , a finite subgroup of $D^{*}$ in characteristic $p$ is cyclic, hence abelian. But $y a y^{- 1} = a^{i} \neq a$ . Therefore $D = F$ is a field.

Theorem(13.10)Infinite subfields and infinite centralizers

Let $D$ be an infinite division ring with centre $F$ . Then for every $a \in D$ , the field $F (a)$ is contained in an infinite subfield $K$ of $D$ . In particular the centralizer $C_{D} (a) \supseteq K$ is infinite.

Proof

If $F = D$ then $D$ itself is an infinite field and $K = D$ works, so assume $F ⊊ D$ . If $F (a)$ is infinite take $K = F (a)$ ; so assume $F (a)$ is finite, which forces $F$ finite and hence $char D = p > 0$ . If $a \in F$ , replace $a$ by any element of $D ∖ F$ : an infinite subfield containing $F (a')$ for such an $a'$ contains $F = F (a)$ . So we may assume $a \notin F$ .

Now $a$ lies in the finite field $F (a)$ , so $a$ is torsion, and $a$ is noncentral. Herstein's Lemma gives $y \in D^{*}$ with $y a y^{- 1} = a^{i} \neq a$ .

** $y$ has infinite order.** If $⟨ y ⟩$ were finite then, as in $(13.9)$ , $⟨ a ⟩ ⟨ y ⟩$ would be a finite subgroup of $D^{*}$ , hence cyclic by $(13.3)$ and therefore abelian — contradicting $y a y^{- 1} \neq a$ .

Producing the field. Conjugation by $y$ permutes the finite group $⟨ a ⟩$ , so the induced homomorphism $⟨ y ⟩ \to Aut (⟨ a ⟩)$ has finite image; some $y^{m}$ with $m > 0$ acts trivially, i.e. $y^{m}$ commutes with $a$ . Then $K : = F (a, y^{m})$ is generated over the central field $F$ by two commuting elements, hence is a subfield of $D$ containing $F (a)$ . Since $y$ has infinite order, so does $y^{m}$ , and $K \supseteq ⟨ y^{m} ⟩$ is infinite.

Theorem(13.11)Jacobson: algebraic division algebras over finite fields

Let $F$ be a finite field and let $D$ be a division algebra over $F$ that is algebraic over $F$ — every element of $D$ satisfies a nonzero polynomial over $F$ . Then $D$ is commutative, and hence an algebraic field extension of $F$ .

Proof

Let $p = char F$ . For $d \in D$ , $F [d]$ is a commutative domain, finite-dimensional over the finite field $F$ because $d$ is algebraic; so $F [d]$ is a finite field, say $| F [d] | = p^{k}$ with $k \geq 1$ . Every element of a field of order $p^{k}$ satisfies $t^{p^{k}} = t$ , so $d^{p^{k}} = d$ with $p^{k} > 1$ .

Applying this to $d = a b - b a$ shows that the hypothesis of $(13.9)$ holds for $D$ , with $n (a, b) = p^{k (a, b)}$ . Hence $D$ is a field.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Promote an element to an operator

$a$ is hard to analyse; $δ_{a} \in End (_{K} D)$ satisfies a known polynomial. The move is to change categories until a finiteness statement becomes available.

Move 2

Split a polynomial identity into linear factors

$t^{p^{n}} - t$ factors completely over $K$ . A vanishing product of commuting operators, none of which may be assumed injective, hands you a kernel.

Move 3

Play two finiteness statements against each other

Torsion of $a$ and torsion of $y$ build a finite subgroup; $(13.3)$ says it is abelian; the conjugation relation says it is not.

Move 3 has a dual use that is easy to miss. In $(13.9)$ the finite subgroup is constructed and the contradiction kills the assumption $D \neq Z (D)$ . In $(13.10)$ the same construction is run in reverse: since no contradiction is permitted, $y$ must fail to be torsion, and the infinitude of $⟨ y ⟩$ is the desired output.

Build $K = 𝔽_{p} [a]$ Finite field because

a

is torsion and

D

has no zero divisors.

Make $D$ a $K$ -vector space

δ_{a}

K

-linear because

a

commutes with

K

Prove $δ_{a}^{| K |} = δ_{a}$

λ

and

ρ

commute; characteristic

p

makes the binomial collapse.

Factor and find a kernel

\prod_{b \in K} (δ_{a} - b) = 0

with

δ_{a} \neq 0

gives

b_{0} \in K^{*}

with

\ker (δ_{a} - b_{0}) \neq 0

Read off the conjugation

x a x^{- 1} = a - b_{0}

, a power of

a

by cyclicity of

K^{*}

Normalise $y = δ_{a} (x)$

y = b_{0} x

conjugates

a

the same way and is visibly an additive commutator.

Worked Example

A skew Laurent division ring in characteristic 2

Let $K = 𝔽_{4} = {0, 1, ω, ω^{2}}$ with $ω^{2} = ω + 1$ and $ω^{3} = 1$ , and let $σ$ be the Frobenius $c \mapsto c^{2}$ , which generates $Gal (𝔽_{4} / 𝔽_{2})$ . Form the skew Laurent polynomial ring $R = K [y, y^{- 1}; σ]$ , with multiplication determined by

y c = σ (c) y = c^{2} y (c \in K) .

(E.1)

$R$ is a noetherian domain, hence an Ore domain, so it embeds in its division ring of fractions $D$ . This $D$ is infinite, noncommutative, and of characteristic $2$ — a legitimate arena for $(13.8)$ .

Herstein's Lemma made explicit

Take $a = ω$ . It has multiplicative order $3$ , so it is torsion, and it is noncentral because $y ω y^{- 1} = ω^{2} \neq ω$ . Here $K = 𝔽_{2} [ω] = 𝔽_{4}$ , so $| K | = 4 = 2^{2}$ and $n = 2$ . Compute the inner derivation on $y$ :

δ_{ω} (y) = ω y - y ω = ω y - ω^{2} y = (ω + ω^{2}) y = 1 \cdot y = y,

(E.2)

Using $ω + ω^{2} = ω + ω + 1 = 1$ in characteristic $2$ .

So $y$ is an eigenvector of $δ_{ω}$ with eigenvalue $b_{0} = 1 \in K^{*}$ — and, remarkably, $y$ is its own additive commutator $δ_{ω} (y)$ . The conclusion of $(13.8)$ reads

y ω y^{- 1} = ω^{2} = ω^{i} with i = 2 \neq 1,

(E.3)

exactly as predicted, with $y$ an additive commutator as the last clause of the lemma promises.

Checking (13.10) and (13.11) on the same object

The three results evaluated on $D$ , the fraction division ring of $𝔽_{4} [y, y^{- 1}; σ]$
Result	Prediction	Verification in $D$
(13.8)	Some additive commutator conjugates $ω$ to a power	$δ_{ω} (y) = y$ and $y ω y^{- 1} = ω^{2}$
(13.9)	$D$ noncommutative, so its hypothesis must fail	$y$ is an additive commutator of infinite order, so no $n > 1$ has $y^{n} = y$
(13.10)	$C_{D} (ω)$ is infinite	$y^{2} ω y^{- 2} = ω^{4} = ω$ , so $𝔽_{4} (y^{2}) \subseteq C_{D} (ω)$ is an infinite subfield
(13.11)	$D$ cannot be algebraic over a finite field	$y$ is transcendental over $𝔽_{2}$ ; $D$ has infinite-dimensional centre $𝔽_{2} (y^{2})$

Comparison and Classification

Hypotheses consumed by the four results
	$char p > 0$	Torsion element	Uses $(13.3)$	Conclusion is commutativity
(13.8) Herstein's Lemma	yes	yes	no	no
(13.9) Jacobson–Herstein	derived, not assumed	derived	yes	yes
(13.10) Infinite subfields	derived in the hard case	derived	yes	no
(13.11) Jacobson	yes, from finite $F$	derived	yes	yes
(13.12) Frobenius	no — base field is $ℝ$	no	no	no

Hypotheses consumed by the four results

The last row is the contrast worth holding on to. Over $ℝ$ , the analogous classification — Frobenius' theorem, treated in Division Rings and the Real Quaternions — does not conclude commutativity: $ℍ$ survives. Commutativity theorems of the type on this page are a characteristic- $p$ phenomenon.

Commutativity criteria in the neighbourhood of $(13.9)$
Hypothesis	Scope	Result
$a^{n (a)} = a$ , $n (a) > 1$ , for all $a$	any ring	Jacobson $(12.10)$
$(a b - b a)^{n (a, b)} = a b - b a$ , $n > 1$	any ring	Jacobson–Herstein $(12.9)$
same, for division rings	division rings	$(13.9)$ — the case proved here
algebraic over a finite field	division algebras	$(13.11)$
reduced algebraic over a finite field	any algebra	Exercise 13.11 in Lam

Relationship Map

Herstein's Lemma $(13.8)$ — $char p$ , $a$ noncentral torsion
- depends on
  - $(13.2)$ : finite subrings of $D$ are fields, giving $K = 𝔽_{p} [a]$ finite
  - cyclicity of $K^{*}$ for a finite field $K$
  - linear algebra over $K$ inside $End (_{K} D)$
- proves
  - $(13.9)$ Jacobson–Herstein for division rings, closing $(12.9)$
  - $(13.10)$ infinite subfields and infinite centralizers
  - $(13.11)$ Jacobson: algebraic division algebras over finite fields are commutative
- used jointly with
  - $(13.3)$ finite subgroups of $D^{*}$ are cyclic in characteristic $p$
  - $(13.5)$ existence of a noncentral additive commutator

The dependency on Wedderburn's Little Theorem is not decorative: $(13.3)$ is used in both $(13.9)$ and $(13.10)$ , and $(13.2)$ is what makes $𝔽_{p} [a]$ a field in the first place. Wedderburn's Little Theorem is therefore a prerequisite for this page in the strong sense.

(13.1)

(13.2)

(13.3)

(13.8)

(13.9)

(13.11)

Computational Notes

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

The proof of $(13.8)$ is effective whenever $D$ is presented concretely enough to compute in.

Finding $K = 𝔽_{p} [a]$ costs one order computation for $a$ ; the field is then $𝔽_{p^{n}}$ where $n$ is the degree of the minimal polynomial of $a$ over $𝔽_{p}$ .
Finding $b_{0}$ and the eigenvector requires a kernel computation for $δ_{a} - b$ over $K$ , one $b$ at a time — at most $p^{n} - 1$ nullspace computations, and in practice the first nonzero eigenvalue is found immediately.
In the skew Laurent setting of the worked example the eigenvectors are the monomials $c y^{m}$ , so $δ_{a}$ is diagonal in the monomial basis and no linear algebra is needed at all.
Determining the exponent $i$ is a discrete logarithm in the cyclic group $K^{*}$ of order $p^{n} - 1$ ; for the sizes arising in practice this is trivial, and in the skew case $i$ is simply the power of $p$ recording the twist.

Failure Modes and Common Mistakes

Do not assume $δ_{a}^{p^{n}} = δ_{a}$ without noting that $λ$ and $ρ$ commute — the collapse of the binomial coefficients requires it.
Do not conclude from $(13.10)$ that maximal subfields of an infinite division ring are infinite-dimensional; they are infinite as sets, which for a finite centre is the same thing and for an infinite centre is automatic.
Do not read $(13.11)$ as covering algebraic division algebras over $ℚ$ . Over $ℚ$ noncommutative examples abound, including infinite-dimensional ones built as unions of tensor products of quaternion-like algebras.

Historical Notes and Lessons Learned

1905WedderburnFinite division rings are commutative. Via $(13.2)$ and $(13.3)$ this becomes the engine of every argument on this page.
1945Jacobson's commutativity theoremIf every element of a ring satisfies $a^{n (a)} = a$ with $n (a) > 1$ , the ring is commutative — a vast generalisation of Wedderburn's theorem, proved by structure theory and reduction to division rings.
1953–54Herstein's commutator conditionsHerstein weakens the hypothesis from all elements to all additive commutators, yielding the Jacobson–Herstein theorem $(12.9)$ .
1968Herstein's lemma in printThe eigenvector argument for $(13.8)$ appears in Herstein's Carus monograph Noncommutative Rings, and is the proof reproduced by Lam.

The methodological lesson: an element of a division ring is hard to constrain directly, but the operator it induces on the ring lives in an endomorphism algebra where polynomial identities are available. Herstein's argument is the standard example of that translation, and the same idea reappears wherever derivations are used to study skew structures.

Quick Reference

(13.8) hypotheses

char D = p > 0

;

a

noncentral;

a

torsion

(13.8) conclusion

\exists y

an additive commutator with

y a y^{- 1} = a^{i} \neq a

Operator identity

δ_{a}^{p^{n}} = δ_{a}

where

p^{n} = | 𝔽_{p} [a] |

Eigenvector step

\prod_{b \in K} (δ_{a} - b) = 0

δ_{a} \neq 0

Conjugation formula

δ_{a} (x) = b_{0} x \Rightarrow x a x^{- 1} = a - b_{0}

(13.9)

(a b - b a)^{n (a, b)} = a b - b a

n > 1

\Rightarrow D

a field

(13.10)

D

infinite

\Rightarrow

every

F (a)

sits in an infinite subfield

(13.11)

D

algebraic over a finite field

\Rightarrow D

commutative

What each proof needs from earlier results
Proof	Imports
(13.8)	(13.2) via finiteness of $𝔽_{p} [a]$ ; cyclicity of $K^{*}$
(13.9)	(13.5), (13.8), (13.3)
(13.10)	(13.8), (13.3)
(13.11)	(13.9)

Frequently Asked Questions

Why must the eigenvalue

b_{0}

be nonzero?

The factorisation $(13.8 b)$ isolates $δ_{a}$ itself as the factor corresponding to $b = 0$ . If $δ_{a}$ were the non-injective factor there would be nothing to conclude — its kernel is just the centralizer of $a$ . The argument therefore uses $δ_{a} \neq 0$ , i.e. noncentrality of $a$ , to push the failure of injectivity onto some $δ_{a} - b_{0}$ with $b_{0} \in K^{*}$ , and $b_{0} \neq 0$ is exactly what makes $x a x^{- 1} = a - b_{0}$ differ from $a$ .

Why does the lemma bother to say

y

can be chosen to be an additive commutator?

Because $(13.9)$ needs it. The hypothesis of $(13.9)$ constrains additive commutators only, so to conclude that $y$ is torsion one must know $y$ is a commutator. Replacing the eigenvector $x$ by $δ_{a} (x) = b_{0} x$ costs nothing — it is a scalar multiple, so it conjugates $a$ identically — and buys exactly that.

Does

(13.10)

say anything when the centre is infinite?

It is then trivial: $F (a) \supseteq F$ is already infinite, so take $K = F (a)$ . The content is entirely in the case of a finite centre, where the ring is infinite but its centre is not, and Herstein's Lemma is needed to manufacture an element of infinite order commuting with $a$ .

(13.11)

a special case of Wedderburn's Little Theorem?

No, it is a strict generalisation in one direction. Wedderburn assumes finitely many elements; $(13.11)$ allows $D$ to be infinite, requiring only that each element be algebraic over a finite base field. A finite division ring is algebraic over its prime field, so $(13.1)$ follows from $(13.11)$ — but $(13.11)$ 's proof runs through $(13.9)$ , which itself uses $(13.3)$ , so the logical order is the reverse of the generality order.

What replaces these theorems over

ℚ

Nothing this strong is available. Algebraic division algebras over $ℚ$ are plentiful and noncommutative: the finite-dimensional ones are classified by the Albert–Brauer–Hasse–Noether theorem as cyclic algebras over their centres, and infinite-dimensional algebraic examples exist, obtained as unions of ascending tensor products of division algebras of coprime degrees. No classification of the infinite-dimensional case is known.

Where is the Jacobson–Herstein theorem for general rings completed?

In Lam's §12. The reduction from arbitrary rings to division rings is carried out there using subdirect decomposition and the structure theory of primitive rings; $(13.9)$ supplies the division-ring case that the reduction leaves open, so the two sections together constitute the full proof of $(12.9)$ .

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §13, (13.8)–(13.11), pp. 218–220; the reduction to division rings is §12, (12.9)–(12.10).
I. N. Herstein, Noncommutative Rings, Carus Mathematical Monographs 15, Mathematical Association of America, 1968, Chapter 3.
N. Jacobson, “Structure theory for algebraic algebras of bounded degree”, Annals of Mathematics 46 (1945), 695–707.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapter VII.
L. H. Rowen, Ring Theory, Volume II, Academic Press, 1988 (commutativity theorems and division algebras).

AI Suggested Questions

Give a complete proof that $(λ - ρ)^{p^{n}} = λ^{p^{n}} - ρ^{p^{n}}$ and identify where commutativity of $λ$ and $ρ$ is used.
Construct explicitly an infinite-dimensional algebraic division algebra over $ℚ$ and explain why $(13.11)$ does not apply.
Work out Exercise 13.8 in Lam: for $a$ with $a^{p^{n}}$ central, find $b$ with $a b a^{- 1} = 1 + b$ .
Prove the general Jacobson theorem: a ring in which every element satisfies $a^{n (a)} = a$ with $n (a) > 1$ is commutative.
How does Herstein's Lemma compare with the Skolem–Noether theorem as a source of conjugating elements?
Determine the centre of the skew Laurent division ring of fractions of $𝔽_{p^{n}} [y, y^{- 1}; σ]$ for $σ$ the Frobenius.
Is there a characteristic-zero analogue of $(13.8)$ under an additional hypothesis such as central finiteness?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Turning a ring question into an operator question

The Frobenius identity for δa

From eigenvector to conjugation

06Key Results

07Proof Techniques and Method

Promote an element to an operator

Split a polynomial identity into linear factors

Play two finiteness statements against each other

08Worked Example

A skew Laurent division ring in characteristic 2

Herstein's Lemma made explicit

Checking (13.10) and (13.11) on the same object

09Comparison and Classification

10Relationship Map

11Computational Notes

12Failure Modes and Common Mistakes

13Historical Notes and Lessons Learned

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

The Frobenius identity for $δ_{a}$

Key Results

Proof Techniques and Method

Worked Example

Comparison and Classification

Relationship Map

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions