Criteria for Central Finiteness and the Double Centralizer Theorem

Q: Why is D automatically a simple module over R?

Because the subring D ⊗ 1 already acts transitively enough: an R-submodule of D is in particular closed under left multiplication by all of D, hence is a left ideal of the division ring D, hence is 0 or D. Faithfulness is the deeper half, and it comes from the simplicity of R established in (15.1).

Q: What exactly does the Density Theorem contribute?

It converts a module-theoretic fact into an approximation statement: R can prescribe the images of any finite L-independent set of vectors in D. Density becomes an isomorphism R ≅ End(D_L) exactly when (D_L) is finite, and that is what turns three loosely related finiteness conditions into a genuine equivalence.

Q: Is L = C_D(K) ever equal to K?

Precisely when K is a maximal subfield of D, provided K is commutative — that is the content of (15.7). For noncommutative K one always has Z(K) ⊆ L K, and L = K would force K commutative, so the case L = K is exactly the maximal-subfield case.

Q: Does (15.5) give a practical test for central finiteness?

It gives a clean structural criterion rather than an algorithm. Its real use is theoretical: it makes central finiteness a property of one associated ring, so that theorems about artinian rings can be applied to it. In concrete cases one still computes _F D directly.

Q: Why are the dimensions in the formula cardinal numbers rather than integers?

Because D need not be finite-dimensional over F. When _F K is finite but _F D is infinite, the product (_F K)(_F L) collapses to _F L by cardinal arithmetic, and the formula degenerates into the statement that L is as large as D.

Q: How does this compare with the double centralizer theorem for group representations?

They are instances of one principle. There, a group algebra and its commutant in End(V) determine each other; here, a division subring and its centralizer determine each other. Both proofs realise an abstract commuting operator inside a concrete ring and then evaluate on a distinguished vector — the identity element in the present case.

Executive Summary

Let $D$ be a division ring with centre $F$ , let $K$ be a division subring containing $F$ , and let $L = C_{D} (K)$ . The single construction that unlocks the whole section is to view $D$ as a left module over $R : = D \otimes_{F} K^{op}$ , with $D$ acting by left multiplication and $K$ by right multiplication. That module is simple and faithful, its endomorphism ring is $L$ , and the Density Theorem applies.

Three conditions then turn out to be equivalent — $\dim_{F} K < \infty$ , $\dim (D_{L}) < \infty$ , and $R$ artinian — and when they hold, $R ≅ M_{r} (L)$ , $C_{D} (L) = K$ , and $\dim_{F} D = (\dim_{F} K) (\dim_{F} L)$ . The last two statements are the Double Centralizer Theorem for division rings. Taking $K = D$ gives the clean criterion: $D$ is centrally finite exactly when $D \otimes_{F} D^{op}$ is artinian.

3Equivalent criteria in

(15.4)

$C_{D} (C_{D} (K)) = K$ Double centralizer

$M_{r} (L)$ What

R

becomes

$\dim_{F} K < \infty$ The hypothesis that cannot be dropped

Overview

A division ring has no ideals and no idempotents, so it carries no chain conditions worth exploiting. The device of this section is to build a bigger ring, $D \otimes_{F} K^{op}$ , that acts on $D$ ; the bigger ring has plenty of ideals, and the Jacobson Density Theorem measures how close its action comes to being everything.

(d \otimes a^{op}) \cdot v = d v a (d, v \in D, a \in K),

(15.3)

Left multiplication by $D$ and right multiplication by $K$ commute by associativity, so this is a well-defined left $R$ -module structure on $D$ .

Two dictionaries are set up at once. Structural facts about $R$ (simple, artinian, matrix ring) translate into dimension facts about $K$ and $L$ ; conversely, dimension counts inside $D$ decide the ring theory of $R$ . The results of the previous page — Tensor Products of Algebras and Their Centralizers — supply the simplicity of $R$ that makes the module faithful.

The Double Centralizer Theorem that falls out is the division-ring case of a general principle for central simple algebras: a subalgebra and its centralizer determine each other, and their dimensions multiply to the dimension of the whole. Finiteness is essential — the identity $C_{D} (C_{D} (K)) = K$ can fail for infinite-dimensional subfields.

Learning Objectives

Verify that $(15.3)$ defines a module structure and that the module is simple and faithful.
Compute $End (_{R} D)$ and identify it with $C_{D} (K)$ .
State the equivalences of $(15.4)$ and trace which implication uses which earlier result.
Prove $C_{D} (C_{D} (K)) = K$ under the hypothesis $\dim_{F} K < \infty$ .
Derive the criterion $D$ centrally finite $\Leftrightarrow$ $D \otimes_{F} D^{op}$ artinian.
Use the dimension formula on the rational quaternions and check every number.

Definitions

Standing setup: $D$ is a division ring with centre $F = Z (D)$ ; $K$ is a division subring with $F \subseteq K \subseteq D$ ; $L : = C_{D} (K)$ ; $R : = D \otimes_{F} K^{op}$ .
$K^{op}$: The opposite ring, with elements written $a^{op}$ and product $a^{op} b^{op} = (b a)^{op}$ . It is a division ring whenever $K$ is, and it is canonically anti-isomorphic to $K$ .
$D_{L}$ and $_{L} D$: $D$ regarded as a right, resp. left, vector space over the division ring $L$ . The two dimensions can differ in principle; $(15.6)$ says they agree once one is finite.
Density: $R \subseteq End (D_{L})$ is dense if for all $L$ -independent $v_{1}, \dots, v_{n} \in D$ and arbitrary $w_{1}, \dots, w_{n} \in D$ there is $ρ \in R$ with $ρ v_{i} = w_{i}$ for each $i$ .
Centrally finite: $\dim_{F} D < \infty$ where $F = Z (D)$ . Equivalently, by $(15.5)$ , $D \otimes_{F} D^{op}$ is artinian.

Both $K$ and $L$ contain $F$ ; $K$ and $L$ centralize each other by construction, and $K \subseteq C_{D} (L)$ always holds. The content of the Double Centralizer Theorem is that this inclusion is an equality when $\dim_{F} K$ is finite.

Core Concepts

Why the opposite ring appears

We want $D$ to be a left module on which $K$ acts by right multiplication. Right multiplications compose in the reverse order: $v \mapsto (v a) b$ is right multiplication by $a b$ , so $ρ_{a} ρ_{b} = ρ_{b a}$ . Writing the acting copy of $K$ as $K^{op}$ makes the assignment $a^{op} \mapsto ρ_{a}$ a ring homomorphism rather than an anti-homomorphism. When $K$ is commutative — the case of maximal subfields — $K^{op} = K$ and the distinction disappears.

The endomorphism ring is the centralizer

Let $f$ be an endomorphism of the left $R$ -module $D$ , written on the right. Since $f$ commutes in particular with left multiplication by every $d \in D$ , we get $(d \cdot 1) f = d \cdot (1) f$ , so $f$ is right multiplication by the single element $c : = (1) f$ . Since $f$ also commutes with right multiplication by every $a \in K$ , we get $v a c = v c a$ for all $v$ , hence $a c = c a$ : that is, $c \in C_{D} (K) = L$ . Conversely every $ρ_{c}$ with $c \in L$ is such an endomorphism.

End (_{R} D)

right multiplications

ρ_{c}

c \in C_{D} (K)

L

With that identification, $D$ becomes a right $L$ -vector space and $R$ maps into $End (D_{L})$ . The module is simple because $D$ is already simple over the subring $D \otimes 1$ , and faithful because $R$ is simple by $(15.1)$ and the annihilator is a proper ideal.

What density buys, and what artinian buys

Density alone gives approximation on finitely many vectors at a time. It becomes an isomorphism exactly when the space is finite-dimensional: a dense subring of $End (D_{L})$ is left artinian if and only if $\dim (D_{L}) < \infty$ , and in that case it is all of $End (D_{L}) ≅ M_{r} (L)$ . That equivalence, from the density chapter, is the hinge of $(15.4)$ .

Key Results

Theorem(15.3)The module that carries the section

Let $D$ be a division ring with centre $F$ , let $K$ be a division subring of $D$ with $F \subseteq K$ , and set $L : = C_{D} (K)$ , a division subring of $D$ containing $F$ . Then the rule $(d \otimes a^{op}) \cdot v = d v a$ makes $D$ into a faithful simple left module over $R : = D \otimes_{F} K^{op}$ , with $End (_{R} D) ≅ L$ acting by right multiplication. Consequently $R$ acts as a dense ring of linear transformations on the right $L$ -vector space $D_{L}$ .

Proof

Well-definedness: the map $D \times K^{op} \to End (D, +)$ , $(d, a^{op}) \mapsto (v \mapsto d v a)$ , is $F$ -bilinear, and left and right multiplications commute by associativity, so it induces a ring homomorphism on the tensor product.

Simplicity: an $R$ -submodule of $D$ is in particular a $D$ -submodule for the action of $D \otimes 1$ , i.e. a left ideal of $D$ , hence $0$ or $D$ . Faithfulness: the annihilator is a two-sided ideal of $R$ , and $R$ is simple by $(15.1)$ applied with $D' = K^{op}$ — legitimate because $F$ is exactly $Z (D)$ and $K^{op}$ is a division $F$ -algebra. Since $1 \in D$ is not annihilated by $1 \otimes 1$ , the annihilator is proper, hence zero.

Endomorphisms: write them on the right. If $f \in End (_{R} D)$ and $c : = (1) f$ , then $(v) f = (v \cdot 1) f = v \cdot (1) f = v c$ using $D$ -linearity, so $f = ρ_{c}$ . Compatibility with the action of $1 \otimes a^{op}$ reads $(v a) c = (v c) a$ for all $v \in D$ ; taking $v = 1$ gives $a c = c a$ for all $a \in K$ , so $c \in L$ . Conversely $ρ_{c}$ is $R$ -linear for every $c \in L$ , and $ρ_{c} ρ_{c'} = ρ_{c c'}$ with the right-hand convention, so $End (_{R} D) ≅ L$ . Density is now the Density Theorem applied to the simple module $_{R} D$ .

Theorem(15.4)Criteria for finiteness, and the Double Centralizer Theorem

In the setting of $(15.3)$ the following are equivalent:

$\dim_{F} K < \infty$ ;
$\dim (D_{L}) < \infty$ ;
the simple ring $R = D \otimes_{F} K^{op}$ is artinian.

If these hold and $r : = \dim (D_{L})$ , then $\dim_{F} K = r$ as well, $R ≅ M_{r} (L)$ , $C_{D} (L) = K$ , and

\dim_{F} D = (\dim_{F} K) \cdot (\dim_{F} L) (as cardinal numbers).

Proof

**(1) $\Rightarrow$ (3)** is $(15.1) (3)$ : $\dim_{F} K^{op} = \dim_{F} K < \infty$ makes $R$ artinian. **(2) $\Leftrightarrow$ (3)** is the density criterion: a dense subring of $End (D_{L})$ is artinian precisely when $\dim (D_{L}) < \infty$ , and then equals $End (D_{L})$ .

**(3) $\Rightarrow$ (1) and the numerology.** Assume $r = \dim (D_{L}) < \infty$ . Density gives $R ≅ End (D_{L}) ≅ M_{r} (L)$ . As a left module over itself, $M_{r} (L)$ is a direct sum of $r$ copies of its simple module, which is $D$ ; so $_{R} R ≅ r \cdot (_{R} D)$ . Restricting the scalars to $D \otimes 1$ , this says $_{D} R ≅ r \cdot (_{D} D)$ , i.e. $\dim (_{D} R) = r$ . But $(15.2)$ exhibits $R$ as a free left $D$ -module of rank $\dim_{F} K$ . Hence $\dim_{F} K = r < \infty$ , which is (1).

The dimension formula is transitivity of dimension along $F \subseteq L \subseteq D$ : $\dim_{F} D = \dim (D_{L}) \cdot \dim_{F} L = (\dim_{F} K) (\dim_{F} L)$ . When $D$ is centrally infinite this reduces to $\dim_{F} D = \dim_{F} L$ , since the factor $\dim_{F} K$ is finite.

The double centralizer identity. Certainly $K \subseteq C_{D} (L)$ . Conversely take $b \in C_{D} (L)$ . Right multiplication $ρ_{b}$ is then an endomorphism of $D_{L}$ , because $ρ_{b} (v c) = v c b = v b c = ρ_{b} (v) c$ for $c \in L$ . Since $R = End (D_{L})$ , there is $y \in R$ acting as $ρ_{b}$ . But $ρ_{b}$ also commutes with left multiplication by every element of $D$ , so $y \in C_{R} (D) = 1 \otimes K^{op}$ by $(15.1) (1)$ ; write $y = 1 \otimes a^{op}$ with $a \in K$ . Evaluating both descriptions at $1 \in D$ gives $b = ρ_{b} (1) = y \cdot 1 = a \in K$ . Hence $C_{D} (L) = K$ .

Corollary(15.5)Central finiteness detected by the tensor square

Let $D$ be a division ring with centre $F$ . Then $D$ is centrally finite if and only if the simple ring $D \otimes_{F} D^{op}$ is artinian. If $n = \dim_{F} D < \infty$ , then

D \otimes_{F} D^{op} ≅ End (D_{F}) ≅ M_{n} (F) .

Proof. Take $K = D$ in $(15.4)$ ; then $L = C_{D} (D) = Z (D) = F$ , so $D_{L} = D_{F}$ and $r = \dim_{F} D$ . Condition (1) reads $\dim_{F} D < \infty$ and condition (3) reads $D \otimes_{F} D^{op}$ artinian; the isomorphism is $R ≅ M_{r} (L) = M_{n} (F)$ , realised by sending $d \otimes e^{op}$ to $v \mapsto d v e$ .

Corollary(15.6)Left and right dimensions over the centralizer agree

In the setting of $(15.3)$ , assume $r = \dim_{F} K < \infty$ . Then $\dim (D_{L}) = \dim (_{L} D) = r$ .

Proof. Apply $(15.4)$ to $D^{op}$ , whose centre is again $F$ and which contains $K^{op}$ as a division subring of the same $F$ -dimension $r$ ; its centralizer is $C_{D^{op}} (K^{op}) = L^{op}$ . The right-hand dimension of $D^{op}$ over $L^{op}$ is the left-hand dimension of $D$ over $L$ , and $(15.4)$ evaluates it as $\dim_{F} K^{op} = r$ .

Remark—Finiteness is not decorative

Without the hypothesis $\dim_{F} K < \infty$ , the identity $C_{D} (C_{D} (K)) = K$ can fail: Lam's Exercise 15.4 supplies a division ring $D$ and a subfield $K$ of infinite $F$ -dimension with $C_{D} (C_{D} (K)) ⊋ K$ . Everything in $(15.4)$ beyond the equivalence of (1)–(3) should be read as conditional on that finiteness.

Proof Techniques and Method

The reusable moves behind these proofs.

Move 1

Turn a subring into a module structure

To study $K \subseteq D$ , let $D$ act on itself from the left and $K$ from the right. The two actions commute, so they assemble into an action of $D \otimes_{F} K^{op}$ and the subring problem becomes a module problem.

Move 2

Count one object two ways

$\dim (_{D} R)$ is $\dim_{F} K$ by freeness, and $r$ by the matrix description. Equating two computations of the same invariant is what produces $\dim_{F} K = \dim (D_{L})$ out of thin air.

Move 3

Realise an operator inside the ring

To show $b \in C_{D} (L)$ lies in $K$ : check $ρ_{b}$ is $L$ -linear, use surjectivity of $R \to End (D_{L})$ to realise it as an element of $R$ , then locate that element with the centralizer computation $C_{R} (D) = 1 \otimes K^{op}$ and evaluate at $1$ .

Move 3 is the pattern of every double-centralizer proof in the subject, including the version for finite-dimensional central simple algebras: an abstract operator is captured inside a concrete ring, and evaluation at the identity element reads off the answer.

Worked Example

The rational quaternions, subfield by subfield

Let $D = ℍ_{ℚ}$ with $i^{2} = j^{2} = - 1$ , $i j = - j i = k$ , so $F = Z (D) = ℚ$ and $\dim_{F} D = 4$ . Take $K = ℚ (i)$ .

First compute $L = C_{D} (K)$ . Writing $q = a + b i + c j + d k$ , the condition $q i = i q$ gives $c = d = 0$ , so $L = ℚ (i) = K$ . Now every claim of $(15.4)$ can be checked numerically.

Auditing $(15.4)$ for $K = ℚ (i) \subseteq ℍ_{ℚ}$
Quantity	Predicted by $(15.4)$	Computed directly
$\dim_{F} K$	$r$	$2$
$\dim (D_{L})$	$r$	$2$ , since $D = K \oplus j K$ as a right $K$ -space
$R = D \otimes_{F} K^{op}$	$M_{r} (L)$	$M_{2} (ℚ (i))$
$C_{D} (L)$	$K$	$ℚ (i)$
$\dim_{F} D$	$(\dim_{F} K) (\dim_{F} L)$	$4 = 2 \cdot 2$

The identification $R ≅ M_{2} (ℚ (i))$ is concrete: $D = K \oplus j K$ , and writing $v = x + j y$ with $x, y \in K$ , left multiplication by $i$ sends $x + j y \mapsto i x - j i y$ — note $i j = - j i$ forces the conjugate — while right multiplication by $i$ sends $x + j y \mapsto x i + j y i$ . In the ordered basis $(1, j)$ these are the matrices $(\begin{matrix} i & 0 \\ 0 & - i \end{matrix})$ and $i \cdot id$ respectively.

Two degenerate but instructive choices

$K = F = ℚ$ . Then $L = C_{D} (F) = D$ , $r = \dim_{F} K = 1$ , $D_{L} = D_{D}$ has dimension $1$ , $R = D \otimes_{F} F ≅ D ≅ M_{1} (D)$ , and $C_{D} (L) = Z (D) = ℚ = K$ . Every clause holds trivially.
$K = D = ℍ_{ℚ}$ . Then $L = Z (D) = ℚ$ , $r = 4$ , and $R = ℍ_{ℚ} \otimes_{ℚ} ℍ_{ℚ}^{op} ≅ M_{4} (ℚ)$ — the statement of $(15.5)$ for $n = 4$ .

Process and Workflow

Fix the dataIdentify

F = Z (D)

exactly, the division subring

K \supseteq F

, and form

R = D \otimes_{F} K^{op}

Compute the centralizerDetermine

L = C_{D} (K)

by solving the commutation equations in a basis. This is linear algebra whenever

\dim_{F} D < \infty

Decide finitenessAny one of

\dim_{F} K < \infty

\dim (D_{L}) < \infty

R

artinian settles all three; pick whichever is cheapest for the case at hand.

Read off the structureIf finite,

R ≅ M_{r} (L)

with

r = \dim_{F} K = \dim (D_{L})

, and

C_{D} (L) = K

Multiply the dimensions

\dim_{F} D = (\dim_{F} K) (\dim_{F} L)

constrains what centralizers are possible — often enough to determine

L

without computing it.

Is $\dim_{F} K$ finite?

Yes

R

is simple artinian,

R ≅ M_{r} (L)

, the Double Centralizer Theorem applies, and the dimension formula pins down

\dim_{F} L

R

is simple but neither left nor right artinian. Density still holds, but only as approximation;

C_{D} (C_{D} (K))

may be strictly bigger than

K

Unknown, but $D$ centrally finiteThen

\dim_{F} K \leq \dim_{F} D < \infty

automatically, and the finite branch applies without further checking.

Comparison and Classification

Specialisations of $(15.4)$
Choice of $K$	$L = C_{D} (K)$	$R = D \otimes_{F} K^{op}$	Statement obtained
$K = F$	$D$	$D$	Trivial case; $r = 1$
$K = D$	$F$	$M_{n} (F)$ if $n = \dim_{F} D < \infty$	$(15.5)$ : central finiteness $\Leftrightarrow$ artinian tensor square
$K$ a maximal subfield	$K$ itself	$M_{r} (K)$ if $r = \dim_{F} K < \infty$	$(15.8)$ : $\dim_{F} D = r^{2}$
$K$ any subfield, $\dim_{F} K < \infty$	$C_{D} (K) \supseteq K$	$M_{r} (L)$	Double Centralizer Theorem
$K$ a subfield with $\dim_{F} K = \infty$	$C_{D} (K)$	simple, not artinian	Only density survives

Which conclusions require which hypotheses
	$K$ arbitrary division subring	$\dim_{F} K < \infty$	$D$ centrally finite
$_{R} D$ simple and faithful	yes	yes	yes
$End (_{R} D) = L$	yes	yes	yes
$R$ dense in $End (D_{L})$	yes	yes	yes
$R ≅ M_{r} (L)$	no	yes	yes
$C_{D} (C_{D} (K)) = K$	no	yes	yes
$\dim_{F} D = (\dim_{F} K) (\dim_{F} L)$	no	yes	yes

Which conclusions require which hypotheses

Relationship Map

The nesting of subrings drives the nesting of conclusions.

$D$ — a division ringNo chain conditions of its own

$L = C_{D} (K)$ A division subring;

D

is a vector space over it on both sides

$K = C_{D} (L)$ Recovered from

L

when

\dim_{F} K < \infty

$F = Z (D)$ The ground field; contained in every centralizer

(15.1)

simplicity

(15.3)

faithful simple moduleDensity

(15.4)

criteria

(15.5)

(15.6)

(15.8)

Downstream, $(15.4)$ is the tool that makes the theory of maximal subfields work: applied to a maximal subfield $K$ , where $L = K$ , it gives the perfect-square dimension theorem, and applied inside a centralizer it gives the existence of separable maximal subfields.

Applications and Industry Use

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

Central simple algebras

The template for the general theorem

The finite-dimensional Double Centralizer Theorem — for a simple subalgebra $B$ of a central simple $F$ -algebra $A$ , $C_{A} (C_{A} (B)) = B$ and $\dim_{F} B \cdot \dim_{F} C_{A} (B) = \dim_{F} A$ — is proved by exactly this argument with $A \otimes_{F} B^{op}$ in place of $R$ .

Brauer group

Inverses and splitting

$(15.5)$ is the statement that $[D] [D^{op}] = 1$ in $Br (F)$ , restricted to the centrally finite case. It also explains why the Brauer group only sees centrally finite algebras: outside that class the tensor square is not artinian and no Wedderburn description exists.

Representation theory

Endomorphism algebras

Schur's Lemma produces a division ring $L$ of endomorphisms; the centralizer calculus here says how much of the ambient algebra $L$ determines. This is the mechanism behind the double centralizer pairings — Schur–Weyl duality is the same statement for group algebras acting on tensor space.

Coding and signal design

Degree bookkeeping

Cyclic division algebras used for space–time codes are specified by a maximal subfield $K$ with $\dim_{F} K = r$ and the guarantee $\dim_{F} D = r^{2}$ . Those numbers are the code's rate and delay parameters, and $(15.4)$ is what certifies them.

Failure Modes and Common Mistakes

Do not confuse $L = C_{D} (K)$ with $Z (K)$ : $L$ lives in $D$ and typically contains $K$ properly when $K$ is not maximal.
Do not conclude $D$ is centrally finite from $R$ being simple; $R$ is simple for every $K$ , and only the artinian property is informative.
Do not read the formula $\dim_{F} D = (\dim_{F} K) (\dim_{F} L)$ as ordinary arithmetic when $D$ is centrally infinite — it is an identity of cardinals, and reduces to $\dim_{F} D = \dim_{F} L$ .
Do not forget that $F$ must be the exact centre when invoking $(15.1)$ inside the proof; taking a smaller ground field silently invalidates the faithfulness argument.

Best Practices

Compute $L = C_{D} (K)$ first; almost every quantity in $(15.4)$ is expressed through it.
Use the dimension formula as a consistency check on any centralizer computation — the two factors must multiply to $\dim_{F} D$ .
When $K$ is commutative, drop the opposite-ring notation explicitly rather than silently, so that readers know the simplification was justified.
State which of the three equivalent conditions you verified; in examples one of them is usually far cheaper than the others.
For a centrally infinite $D$ , say so before quoting any conclusion beyond density.

Quick Reference

Setup

F = Z (D)

F \subseteq K \subseteq D

division subring,

L = C_{D} (K)

R = D \otimes_{F} K^{op}

Action

(d \otimes a^{op}) \cdot v = d v a

Module

_{R} D

is simple and faithful

Endomorphisms

End (_{R} D) ≅ L

, acting by right multiplication

Equivalences

\dim_{F} K < \infty \Leftrightarrow \dim (D_{L}) < \infty \Leftrightarrow R

artinian

Structure

R ≅ M_{r} (L)

with

r = \dim_{F} K = \dim (D_{L}) = \dim (_{L} D)

Double centralizer

C_{D} (C_{D} (K)) = K

when

\dim_{F} K < \infty

Dimensions

\dim_{F} D = (\dim_{F} K) (\dim_{F} L)

Tensor square

D

centrally finite

\Leftrightarrow

D \otimes_{F} D^{op}

artinian; then

≅ M_{n} (F)

Numbering map for this page
Result	Content
$(15.3)$	$D$ is a faithful simple $D \otimes_{F} K^{op}$ -module with endomorphism ring $C_{D} (K)$
$(15.4)$	Three equivalent finiteness criteria; $R ≅ M_{r} (L)$ ; $C_{D} (L) = K$ ; dimension formula
$(15.5)$	Central finiteness $\Leftrightarrow$ $D \otimes_{F} D^{op}$ artinian; then $≅ M_{n} (F)$
$(15.6)$	$\dim (D_{L}) = \dim (_{L} D)$ when $\dim_{F} K < \infty$

Frequently Asked Questions

Why is

D

automatically a simple module over

R

Because the subring $D \otimes 1$ already acts transitively enough: an $R$ -submodule of $D$ is in particular closed under left multiplication by all of $D$ , hence is a left ideal of the division ring $D$ , hence is $0$ or $D$ . Faithfulness is the deeper half, and it comes from the simplicity of $R$ established in $(15.1)$ .

What exactly does the Density Theorem contribute?

It converts a module-theoretic fact into an approximation statement: $R$ can prescribe the images of any finite $L$ -independent set of vectors in $D$ . Density becomes an isomorphism $R ≅ End (D_{L})$ exactly when $\dim (D_{L})$ is finite, and that is what turns three loosely related finiteness conditions into a genuine equivalence.

L = C_{D} (K)

ever equal to

K

Precisely when $K$ is a maximal subfield of $D$ , provided $K$ is commutative — that is the content of $(15.7)$ . For noncommutative $K$ one always has $Z (K) \subseteq L \cap K$ , and $L = K$ would force $K$ commutative, so the case $L = K$ is exactly the maximal-subfield case.

Does

(15.5)

give a practical test for central finiteness?

It gives a clean structural criterion rather than an algorithm. Its real use is theoretical: it makes central finiteness a property of one associated ring, so that theorems about artinian rings can be applied to it. In concrete cases one still computes $\dim_{F} D$ directly.

Why are the dimensions in the formula cardinal numbers rather than integers?

Because $D$ need not be finite-dimensional over $F$ . When $\dim_{F} K$ is finite but $\dim_{F} D$ is infinite, the product $(\dim_{F} K) (\dim_{F} L)$ collapses to $\dim_{F} L$ by cardinal arithmetic, and the formula degenerates into the statement that $L$ is as large as $D$ .

How does this compare with the double centralizer theorem for group representations?

They are instances of one principle. There, a group algebra and its commutant in $End (V)$ determine each other; here, a division subring and its centralizer determine each other. Both proofs realise an abstract commuting operator inside a concrete ring and then evaluate on a distinguished vector — the identity element in the present case.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §15, results (15.3)–(15.6) (pp. 252–255).
T. Y. Lam, A First Course in Noncommutative Rings, §11, the Jacobson Density Theorem (11.16) and its artinian refinement (11.19).
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapters IV–V.
R. S. Pierce, Associative Algebras, Graduate Texts in Mathematics 88, Springer-Verlag, 1982, §12 (the centralizer theorem).
P. K. Draxl, Skew Fields, London Mathematical Society Lecture Note Series 81, Cambridge University Press, 1983, Chapters 1–3.

AI Suggested Questions

Write out the proof that a dense subring of $End (V_{L})$ is left artinian if and only if $\dim (V_{L}) < \infty$ .
Find a division ring $D$ and a subfield $K$ with $\dim_{F} K$ infinite for which $C_{D} (C_{D} (K)) \neq K$ .
State and prove the double centralizer theorem for a simple subalgebra of a finite-dimensional central simple algebra.
For $D$ the rational quaternions, list all division subrings $K$ and their centralizers, and check the dimension formula in each case.
How does $(15.5)$ specialise when $D$ is commutative, and why is the resulting statement uninteresting?
Explain the relationship between $(15.4)$ and the Skolem–Noether theorem on extending isomorphisms of subalgebras.
What is the analogue of the double centralizer theorem for Azumaya algebras over a commutative ring?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why the opposite ring appears

The endomorphism ring is the centralizer

What density buys, and what artinian buys

06Key Results

07Proof Techniques and Method

Turn a subring into a module structure

Count one object two ways

Realise an operator inside the ring

08Worked Example

The rational quaternions, subfield by subfield

Two degenerate but instructive choices

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

The template for the general theorem

Inverses and splitting

Endomorphism algebras

Degree bookkeeping

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

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions