Tensor Products of Algebras and Their Centralizers

Q: Why insist that F be the whole centre of D rather than just a central subfield?

Because the coefficient argument concludes c_i ∈ Z(D), and only the equality Z(D) = F lets those coefficients be absorbed as scalars into the second tensor factor. With F strictly smaller, the surviving coefficients are genuine elements of D and the centralizer of D is larger than D'. The failure is not subtle: C ⊗_R C is not even simple.

Q: Does (15.1) say anything when D' is not a division algebra?

Yes — parts (1) and (2) hold for an arbitrary F-algebra D', since their proof uses nothing but the basis decomposition. Only the simplicity and artinian statements need D' to be a division algebra, and simplicity in fact survives the weaker hypothesis that D' is simple.

Q: If both factors are division algebras, when is the product one?

Rarely, and there is no elementary criterion. Over a number field the answer is given by local invariants: D ⊗_F D' is a division algebra exactly when the sum of the local Hasse invariants of D and D' has the same order at every place as the degree predicts. The useful heuristic is that any common splitting field forces zero divisors, so two algebras split by a common quadratic extension will not tensor to a division ring.

Q: How does the artinian clause interact with the density theorem?

Directly. Once R acts faithfully and densely on a module over a division ring, artinian is equivalent to that module being finite-dimensional and to the density being an isomorphism onto the full endomorphism ring. That is how (15.4) upgrades the one-way implication here to a list of equivalences.

Q: What is the relation between this result and central simple algebras?

A central simple F-algebra is by definition simple with centre exactly F and finite-dimensional over F. (15.1) says that tensoring a central division algebra with any simple algebra keeps simplicity, and with any F-algebra keeps the centralizer relationship. Restricting to finite dimensions gives the standard facts: the tensor product of central simple algebras is central simple, and ⊗_F descends to a group law on Brauer classes.

Q: Is D ⊗_F D' ever commutative?

Only if both factors are. The centre is Z(D'), so commutativity of the product would force D' = Z(D') and, via the symmetric role of the factors when both are central, D = F as well.

Executive Summary

Fix a division ring $D$ and let $F = Z (D)$ be its centre — exactly the centre, not some smaller field inside it. Then tensoring $D$ with any $F$ -algebra $D'$ produces a ring in which the two factors sit as mutually centralizing subalgebras, and the pair $(D, D')$ can be recovered from the product: $C_{R} (D) = D'$ and $Z (R) = Z (D')$ .

If $D'$ is a division algebra too, $R = D \otimes_{F} D'$ is a simple ring, and it is artinian as soon as $\dim_{F} D' < \infty$ . The converse fails, and the failure is instructive: $ℍ_{ℚ} \otimes_{ℚ} ℝ$ is the real quaternions, artinian while $\dim_{ℚ} ℝ$ is infinite.

$D'$ Centralizer of

D

R

$Z (D')$ Centre of

R

SimpleIf both factors are division

$F = Z (D)$ Hypothesis that carries everything

Overview

The structure theory of division rings is hard because division rings resist decomposition: there are no idempotents to split off and no proper one-sided ideals to filter by. The tensor product is the standard way to manufacture a decomposable ring out of an indecomposable one — a ring with enough ideals, enough modules and enough chain conditions to be attacked by Wedderburn–Artin methods — while keeping the original division ring visible inside it.

R = D \otimes_{F} D', D ≅ D \otimes 1, D' ≅ 1 \otimes D',

(15.0)

The two factors embed as subalgebras of $R$ that commute elementwise and together generate $R$ .

Everything on this page rests on one hypothesis: the ground field $F$ must be the full centre of $D$ . Read it as a rigidity condition. It says $D$ has no scalars beyond $F$ , so the only way an element of $R$ can commute with all of $D$ is by living in the other factor. That is exactly result $(15.1) (1)$ , and results $(15.3)$ onwards — the criteria for central finiteness, the Double Centralizer Theorem, the theory of maximal subfields — are all downstream of it.

Lam develops these results for division rings only; the first half of the section generalises verbatim to simple rings, and the finite-dimensional case reappears as the theory of central simple algebras and the Brauer group. The page Centrally Finite and Centrally Infinite Division Rings supplies the dimension-theoretic vocabulary used here.

Learning Objectives

State $(15.1)$ with its hypotheses, distinguishing the parts that need $D'$ to be a division algebra from the parts that do not.
Reproduce the normal-form argument computing $C_{R} (D)$ and $Z (R)$ .
Prove simplicity of $D \otimes_{F} D'$ by the minimal-length method.
Explain why $\dim_{F} D' < \infty$ gives a left and right artinian ring, and exhibit an artinian product with $\dim_{F} D'$ infinite.
Show by example that $F ⊊ Z (D)$ destroys every conclusion.
Compute $ℍ_{ℚ} \otimes_{ℚ} K$ for $K = ℝ$ , $ℚ (i)$ and $ℍ_{ℚ}$ .

Definitions

Definition(15.0)Tensor product of

F

-algebras

Let $A$ and $B$ be algebras over a field $F$ . Their tensor product $A \otimes_{F} B$ is the $F$ -vector space $A \otimes_{F} B$ equipped with the multiplication determined by $(a \otimes b) (a' \otimes b') = a a' \otimes b b'$ . It is an $F$ -algebra containing $A \otimes 1 ≅ A$ and $1 \otimes B ≅ B$ as subalgebras which commute with each other elementwise.

$C_{R} (S)$: The centralizer ${r \in R : r s = s r for all s \in S}$ ; a subring of $R$ , and a subalgebra when $S$ is an $F$ -subalgebra.
$Z (R)$: The centre $C_{R} (R)$ ; a commutative subring, and a field when $R$ is a division ring.
$D^{op}$: The opposite algebra: same underlying set and addition, with $a^{op} \cdot b^{op} = (b a)^{op}$ . If $D$ is a division ring so is $D^{op}$ , and $D^{op} ≅ End (D_{D})$ .
Central $F$ -algebra: An $F$ -algebra $A$ with $Z (A) = F$ . The hypothesis in $(15.1)$ is that $D$ is central over $F$ ; $D'$ is not assumed central.
Artinian: Satisfying the descending chain condition on left (resp. right) ideals. For algebras finite-dimensional over a field this is automatic.

All rings have an identity, all algebras are associative and unital, and $\dim$ without a subscript always names a dimension over the field or division ring shown as a subscript on the module.

Core Concepts

The normal form

Choose an $F$ -basis ${d_{i}' : i \in I}$ of $D'$ . Because tensoring is additive in each variable, $R$ decomposes as a left $D$ -module:

R = D \otimes_{F} (⨁_{i \in I} F d_{i}') = ⨁_{i \in I} (D \otimes d_{i}'),

(15.2)

$R$ is free as a left $D$ -module on the set ${1 \otimes d_{i}'}$ ; every element is uniquely $\sum_{i} c_{i} \otimes d_{i}'$ with $c_{i} \in D$ almost all zero.

Uniqueness of the coefficients $c_{i}$ is the whole content. It converts an equation between elements of $R$ into a family of equations in $D$ , one per basis vector, and both proofs below are nothing more than that conversion applied to the equation $d y = y d$ .

Why the centre must be the ground field

Suppose $y = \sum_{i} c_{i} \otimes d_{i}'$ commutes with every $d \in D$ . Comparing coefficients in $(15.2)$ gives $d c_{i} = c_{i} d$ for every $i$ and every $d$ , so each $c_{i} \in Z (D)$ . If $Z (D) = F$ , the scalars $c_{i}$ can be absorbed into the second tensor slot and $y = 1 \otimes (\sum_{i} c_{i} d_{i}') \in D'$ . If $Z (D)$ were strictly larger than $F$ , the $c_{i}$ would be genuine non-scalar elements and $y$ would escape $D'$ .

F = Z (D)

C_{R} (D) = D'

Z (R) = Z (D')

The second implication is a one-line consequence of the first: an element of $Z (R)$ certainly centralizes $D$ , so it lies in $D'$ ; and an element of $D'$ automatically centralizes $D$ , so it is central in $R$ exactly when it centralizes $D'$ , that is, when it lies in $Z (D')$ .

Simplicity by minimal length

To show a nonzero ideal $𝔄 \subseteq R$ is everything, take $0 \neq z \in 𝔄$ written with the fewest nonzero terms $z = \sum_{j = 1}^{m} d_{j} \otimes d_{j}'$ , the $d_{j}'$ being $F$ -independent. Because $D$ is a division ring we may left-multiply by $d_{1}^{- 1}$ and assume $d_{1} = 1$ . The commutator $d z - z d$ then lies in $𝔄$ and has shorter length, so it vanishes; minimality has been converted into commutativity of the remaining coefficients.

Key Results

Theorem(15.1)Centralizer, centre and simplicity of a tensor product

Let $D$ and $D'$ be algebras over a field $F$ , and assume $F = Z (D)$ is exactly the centre of $D$ . Put $R : = D \otimes_{F} D'$ , and identify $D$ with $D \otimes 1$ and $D'$ with $1 \otimes D'$ . Then:

$C_{R} (D) = D'$ ;
$Z (R) = Z (D')$ ;
if in addition $D$ and $D'$ are both division algebras, then $R$ is a simple $F$ -algebra, and $R$ is left and right artinian whenever $\dim_{F} D' < \infty$ .

The artinian implication in (3) is not reversible: $R$ can be artinian with $\dim_{F} D'$ infinite.

Proof

(1). Fix an $F$ -basis ${d_{i}'}$ of $D'$ and use the decomposition $(15.2)$ . Let $y = \sum_{i} c_{i} \otimes d_{i}' \in C_{R} (D)$ with $c_{i} \in D$ . For $d \in D$ the relation $d y = y d$ reads $\sum_{i} d c_{i} \otimes d_{i}' = \sum_{i} c_{i} d \otimes d_{i}'$ , and uniqueness of coefficients gives $d c_{i} = c_{i} d$ for all $i$ . Hence $c_{i} \in Z (D) = F$ , so $y = 1 \otimes (\sum_{i} c_{i} d_{i}') \in D'$ . The reverse inclusion $D' \subseteq C_{R} (D)$ holds by construction, so $C_{R} (D) = D'$ .

(2). Since $Z (R) \subseteq C_{R} (D) = D'$ , an element of $Z (R)$ is of the form $1 \otimes b$ with $b \in D'$ . Such an element commutes with all of $D$ automatically, so it is central in $R$ precisely when it commutes with $D'$ , that is, when $b \in Z (D')$ . Hence $Z (R) = Z (D')$ .

(3), simplicity. Let $𝔄 \neq 0$ be a two-sided ideal of $R$ and choose $0 \neq z \in 𝔄$ of the form $z = \sum_{j = 1}^{m} d_{j} \otimes d_{j}'$ with the $d_{j}'$ linearly independent over $F$ and $m$ minimal among all nonzero elements of $𝔄$ . As $D$ is a division ring, replacing $z$ by $(d_{1}^{- 1} \otimes 1) z$ — still in $𝔄$ , still of length $m$ — lets us assume $d_{1} = 1$ . For any $d \in D$ ,

d z - z d = \sum_{j = 2}^{m} (d d_{j} - d_{j} d) \otimes d_{j}' \in 𝔄

has length at most $m - 1$ , so by minimality it is zero, and independence of the $d_{j}'$ forces $d d_{j} = d_{j} d$ for all $d \in D$ . Thus $d_{j} \in Z (D) = F$ for $j \geq 2$ (and $d_{1} = 1 \in F$ ), so $z = 1 \otimes (\sum_{j} d_{j} d_{j}')$ is a nonzero element of $1 \otimes D'$ . Since $D'$ is a division ring, $z$ is invertible in $1 \otimes D'$ and hence in $R$ , so $𝔄 = R$ .

(3), artinian. If $\dim_{F} D' = n < \infty$ then $(15.2)$ exhibits $R$ as a free left $D$ -module of rank $n$ . Every left ideal of $R$ is in particular a left $D$ -subspace of $R$ , and left $D$ -subspaces of an $n$ -dimensional left $D$ -space satisfy the descending chain condition. So $R$ is left artinian; the same argument on the right gives right artinian.

Remark(15.1)The artinian converse fails

Let $D = ℍ_{ℚ}$ be the rational quaternions, so $F = Z (D) = ℚ$ , and take $D' = ℝ$ . Then $D \otimes_{ℚ} ℝ ≅ ℍ_{ℝ}$ is the real quaternion division ring, which is artinian — indeed $4$ -dimensional over $ℝ$ — while $\dim_{ℚ} ℝ$ is uncountably infinite. Finiteness of $\dim_{F} D'$ is sufficient, never necessary.

Corollary(15.1)Scalar extension of a central division algebra

Let $D$ be a division algebra with centre $F$ and let $K \supseteq F$ be any field extension. Then $D \otimes_{F} K$ is a simple $F$ -algebra with centre $K$ ; it is artinian if $\dim_{F} K < \infty$ .

Proof. Apply $(15.1)$ with $D' = K$ , a commutative division algebra over $F$ ; part (2) gives $Z (R) = Z (K) = K$ and part (3) gives simplicity and the artinian conclusion.

Remark—Simple factors suffice

The simplicity proof used $D'$ only to invert an element of $𝔄 \cap (1 \otimes D')$ . The same argument shows more: if $F = Z (D)$ with $D$ a division algebra and $B$ is any $F$ -algebra, then every nonzero ideal $𝔄$ of $D \otimes_{F} B$ meets $1 \otimes B$ nontrivially, and $𝔄 \mapsto 𝔄 \cap (1 \otimes B)$ is injective on ideals. In particular $D \otimes_{F} B$ is simple whenever $B$ is simple. This is the form in which the result reappears in the theory of central simple algebras.

Proof Techniques and Method

The reusable moves behind these proofs.

Move 1

Expand in a basis of the second factor

Any statement about $R = D \otimes_{F} D'$ becomes a statement about coefficient families in $D$ once a basis of $D'$ is fixed. Uniqueness of coefficients is the only tool needed for parts (1) and (2).

Move 2

Minimise the length inside an ideal

Choose an element of the ideal with the fewest tensor terms, normalise the leading coefficient using invertibility, then commute with the first factor: the commutator is shorter, hence zero. This is the standard proof that central simple algebras have no ideals.

Move 3

Chain conditions from freeness

If $R$ is a free module of finite rank over a division ring sitting inside it, its one-sided ideals are subspaces and the DCC is automatic. Freeness over $D$ — not finite dimension over $F$ — is what does the work.

Move 2 deserves a warning label: it needs the $d_{j}'$ to be $F$ -independent, otherwise "length" is not well defined and the induction collapses. In practice one fixes a basis first and defines length as the size of the support.

Worked Example

Three tensor products of the rational quaternions

Let $D = ℍ_{ℚ} = ℚ \oplus ℚ i \oplus ℚ j \oplus ℚ k$ with $i^{2} = j^{2} = - 1$ and $i j = - j i = k$ . Then $Z (D) = ℚ$ , so $F = ℚ$ is legitimately the ground field for $(15.1)$ , and $\dim_{ℚ} D = 4$ .

ℍ_{ℚ} \otimes_{ℚ} ℝ ≅ ℍ_{ℝ}, Z = ℝ = Z (ℝ) .

(E.1)

Simple, artinian, and still a division ring — although $\dim_{ℚ} ℝ = \infty$ .

ℍ_{ℚ} \otimes_{ℚ} ℚ (i) ≅ M_{2} (ℚ (i)), Z = ℚ (i) .

(E.2)

Adjoining a square root of $- 1$ splits the algebra: a tensor product of two division algebras that is not a division algebra.

ℍ_{ℚ} \otimes_{ℚ} ℍ_{ℚ} ≅ M_{4} (ℚ), Z = ℚ .

(E.3)

Quaternion algebras satisfy $ℍ ≅ ℍ^{op}$ via conjugation, so this is the special case $D \otimes_{F} D^{op} ≅ M_{4} (F)$ .

Check $(E . 2)$ by hand. In $ℍ_{ℚ} \otimes ℚ (i)$ write $ε = \frac{1}{2} (1 \otimes 1 - i \otimes i)$ , where the left $i$ is the quaternion and the right $i$ is the scalar. Then $ε^{2} = \frac{1}{4} (1 \otimes 1 - 2 i \otimes i + i^{2} \otimes i^{2}) = \frac{1}{4} (2 \otimes 1 - 2 i \otimes i) = ε$ , so $ε$ is a nontrivial idempotent. A division ring has no idempotent other than $0$ and $1$ , so the product is not a division ring; being simple, artinian, of dimension $4$ over its centre $ℚ (i)$ , Wedderburn–Artin leaves only $M_{2} (ℚ (i))$ .

What goes wrong when $F$ is not the whole centre

Take $D = D' = ℂ$ and $F = ℝ$ . Here $Z (D) = ℂ \neq ℝ$ , so $(15.1)$ does not apply — and indeed

ℂ \otimes_{ℝ} ℂ ≅ ℂ \times ℂ, z \otimes w ⟼ (z w, z \bar{w}) .

(E.4)

Not simple: the idempotent $\frac{1}{2} (1 \otimes 1 - i \otimes i)$ splits it. Both factors are fields, both are division algebras, and the conclusion still fails.

The same phenomenon at higher degree: $ℚ (\sqrt[3]{2}) \otimes_{ℚ} ℚ (\sqrt[3]{2}) ≅ ℚ (\sqrt[3]{2}) \times ℚ (\sqrt[3]{2}, ω)$ , of dimensions $3 + 6 = 9$ , where $ω$ is a primitive cube root of unity. Whenever $D$ is commutative and $F ⊊ D$ , the hypothesis $F = Z (D)$ is violated at once.

Comparison and Classification

Tensor products computed
$D \otimes_{F} D'$	$F$	Result	Simple?	Artinian?	Division?
$ℍ_{ℚ} \otimes ℝ$	$ℚ$	$ℍ_{ℝ}$	yes	yes	yes
$ℍ_{ℚ} \otimes ℚ (i)$	$ℚ$	$M_{2} (ℚ (i))$	yes	yes	no
$ℍ_{ℚ} \otimes ℍ_{ℚ}$	$ℚ$	$M_{4} (ℚ)$	yes	yes	no
$ℍ_{ℝ} \otimes ℂ$	$ℝ$	$M_{2} (ℂ)$	yes	yes	no
$D \otimes K$ , $K / F$ a field extension	$Z (D)$	simple with centre $K$	yes	if $\dim_{F} K < \infty$	sometimes
$ℂ \otimes ℂ$	$ℝ$ — not $Z (ℂ)$	$ℂ \times ℂ$	no	yes	no
$ℚ (\sqrt[3]{2})^{\otimes 2}$	$ℚ$ — not the centre	$ℚ (\sqrt[3]{2}) \times ℚ (\sqrt[3]{2}, ω)$	no	yes	no

Which hypotheses buy which conclusions in $(15.1)$
	$C_{R} (D) = D'$	$Z (R) = Z (D')$	$R$ simple	$R$ artinian
$F = Z (D)$ , $D'$ an arbitrary $F$ -algebra	yes	yes	no	no
$F = Z (D)$ , $D$ division, $D'$ simple	yes	yes	yes	no
$F = Z (D)$ , both division	yes	yes	yes	partial
$F = Z (D)$ , both division, $\dim_{F} D' < \infty$	yes	yes	yes	yes
$F ⊊ Z (D)$	no	no	no	partial

Which hypotheses buy which conclusions in $(15.1)$

Relationship Map

$(15.1)$ is the root of the section; everything later in §15 is an application of it to a specially chosen second factor.

$(15.1)$ — centralizer, centre, simplicity — $D$ central over $F$ ; $R = D \otimes_{F} D'$ .
- $D' = K^{op}$ for a division subring $K \supseteq F$ — Gives $(15.3)$ : $D$ is a faithful simple $R$ -module with endomorphism ring $C_{D} (K)$ , so the Density Theorem applies.
  - $(15.4)$ — three equivalent criteria for $\dim_{F} K < \infty$ , plus $C_{D} (C_{D} (K)) = K$
  - $(15.5)$ — $D$ is centrally finite iff $D \otimes_{F} D^{op}$ is artinian
  - $(15.8)$ — maximal subfields and $\dim_{F} D = r^{2}$
- $D' = K$ a field extension of $F$ — Gives scalar extension: $D \otimes_{F} K$ is simple with centre $K$ ; when it is $M_{r} (K)$ we call $K$ a splitting field.
  - Splitting fields and the Brauer group $Br (F)$
  - The Schur index of a representation

(15.1)

(15.3)

(15.4)

(15.8)

(15.12)

Applications and Industry Use

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

Brauer group

Tensor product as a group law

Classes of finite-dimensional central simple $F$ -algebras form a group under $\otimes_{F}$ , with $[D]^{- 1} = [D^{op}]$ because $D \otimes_{F} D^{op} ≅ M_{n} (F)$ . $(15.1)$ is what makes the product well defined: it keeps the centre equal to $F$ and the algebra simple.

Representation theory

Schur indices and splitting fields

A simple component of a group algebra $k G$ is $M_{m} (D)$ ; the Schur index is $\sqrt{\dim_{Z (D)} D}$ . Determining which field extension $K$ makes $D \otimes_{Z (D)} K$ a matrix ring is precisely the scalar-extension corollary above.

Wireless communication

Space–time block codes

Codes for multiple-antenna channels are built from cyclic division algebras: transmitted matrices are the images of $D$ under an embedding $D ↪ D \otimes_{F} K ≅ M_{r} (K)$ . Non-vanishing determinant, the design criterion, is exactly the invertibility of nonzero elements of $D$ .

Symbolic computation

Recognising an algebra

Given structure constants for an $F$ -algebra, deciding whether it is a matrix algebra amounts to finding a splitting field or a zero divisor. Tensoring with a candidate extension and looking for idempotents is the standard constructive test.

The honest summary: $(15.1)$ is infrastructure. It is not usually the theorem one wants, but it licenses the manipulation — tensoring up to a bigger field until the algebra becomes matrices — on which the applied uses depend.

Design Considerations

Design considerations here means the choices made when modelling a problem with these algebraic structures.

Fix the ground field to be the centre. Almost every failure in this area traces to tensoring over a field strictly smaller than $Z (D)$ . If you must work over a smaller field, expect the product to decompose and plan for it.
**Second factor: $K$ or $K^{op}$ ?** Tensoring with $K^{op}$ makes $D$ into a left module by $d \otimes a^{op}$ acting as $v \mapsto d v a$ ; tensoring with $K$ makes it a bimodule-flavoured object. Choose $K^{op}$ when you want a module and $K$ when $K$ is commutative and the distinction evaporates.
Which side carries the dimension? $\dim (D_{L})$ and $\dim (_{L} D)$ agree once one of them is finite, but they are not a priori the same. State the side while the finiteness is still in question.
Finite dimension versus artinian. If you only need chain conditions, do not over-assume $\dim_{F} D' < \infty$ ; the weaker hypothesis " $R$ artinian" is what the theorems actually consume, and it is strictly weaker.

Standards and Notation

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

Tensor over a field

A \otimes_{F} B

; the subscript is never optional in this subject

Centralizer

C_{R} (S)

(Lam, Rowen);

Z_{R} (S)

and

{Cent}_{R} (S)

also occur

Opposite algebra

A^{op}

; older texts write

A^{*}

A^{0}

Central simple"CSA over

F

" always means

Z (A) = F

and

\dim_{F} A < \infty

Markup

\otimes

is U+2297; ISO 80000-2 fixes the symbol, MathML Core the rendering

ImplementationsMagma and Sage both construct quaternion algebras directly (QuaternionAlgebra); scalar extension is ChangeRing / base_extend

Computational Notes

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

Let $\dim_{F} D = m$ and $\dim_{F} D' = n$ , both finite, each given by structure constants.

The tensor product has dimension $m n$ and its structure-constant table has $(m n)^{2}$ entries, each a vector of length $m n$ — so $O (m^{3} n^{3})$ storage in the dense representation. This cubic blow-up is why implementations prefer matrix representations to structure constants whenever a splitting field is known.
Deciding whether $D \otimes_{F} D'$ is a division ring is the same as deciding whether it has a zero divisor. Over a number field this is decidable by computing local Hasse invariants and adding them; over a general field it is not a finite computation.
Detecting simplicity is cheap once $(15.1)$ applies — no computation is needed. Detecting it in general costs a radical computation followed by a Wedderburn decomposition.
Producing the isomorphism $D \otimes_{F} D^{op} ≅ End (D_{F}) ≅ M_{n} (F)$ is explicit and linear-algebraic: send $d \otimes e^{op}$ to the map $v \mapsto d v e$ and write it in a chosen $F$ -basis of $D$ .

Failure Modes and Common Mistakes

Do not confuse $C_{R} (D)$ with $Z (R)$ . The first is all of $D'$ ; the second is only $Z (D')$ . They coincide exactly when $D'$ is commutative.
Do not omit the $F$ -independence of the $d_{j}'$ when running the minimal-length argument; without it the commutator need not be shorter and the induction fails.
Do not assume $\dim_{F} (D \otimes_{F} D') = \dim_{F} D \cdot \dim_{F} D'$ carries over to dimensions over the centre of the product; the relevant centre is $Z (D')$ , which may be much larger than $F$ .
Do not expect $D \otimes_{F} D'$ to remember which factor was which if both are isomorphic — the identification $C_{R} (D) = D'$ depends on the chosen embedding.

Quick Reference

Standing hypothesis

F = Z (D)

exactly;

R = D \otimes_{F} D'

Centralizer

C_{R} (D) = D'

Centre

Z (R) = Z (D')

Simplicity

D

D'

division

\Rightarrow

R

simple

Chain condition

\dim_{F} D' < \infty \Rightarrow R

left and right artinian

ConverseFalse:

ℍ_{ℚ} \otimes_{ℚ} ℝ ≅ ℍ_{ℝ}

Scalar extension

D \otimes_{F} K

simple with centre

K

for every field

K \supseteq F

Free module form

R = ⨁_{i} (D \otimes d_{i}')

over an

F

-basis

{d_{i}'}

D'

The three hypotheses and what each one buys
Hypothesis	Used for	If dropped
$F = Z (D)$	$C_{R} (D) = D'$ , $Z (R) = Z (D')$ , $d_{j} \in F$ in the ideal argument	$ℂ \otimes_{ℝ} ℂ$ : every conclusion fails
$D$ a division ring	Normalising $d_{1} = 1$ inside an ideal	Simplicity argument breaks; $D$ simple artinian still works
$D'$ a division ring	Inverting the surviving element of $1 \otimes D'$	Simplicity survives if $D'$ is merely simple
$\dim_{F} D' < \infty$	Freeness of finite rank, hence DCC	Product may still be artinian, as $(E . 1)$ shows

Frequently Asked Questions

Why insist that

F

be the whole centre of

D

rather than just a central subfield?

Because the coefficient argument concludes $c_{i} \in Z (D)$ , and only the equality $Z (D) = F$ lets those coefficients be absorbed as scalars into the second tensor factor. With $F$ strictly smaller, the surviving coefficients are genuine elements of $D$ and the centralizer of $D$ is larger than $D'$ . The failure is not subtle: $ℂ \otimes_{ℝ} ℂ$ is not even simple.

Does

(15.1)

say anything when

D'

is not a division algebra?

Yes — parts (1) and (2) hold for an arbitrary $F$ -algebra $D'$ , since their proof uses nothing but the basis decomposition. Only the simplicity and artinian statements need $D'$ to be a division algebra, and simplicity in fact survives the weaker hypothesis that $D'$ is simple.

If both factors are division algebras, when is the product one?

Rarely, and there is no elementary criterion. Over a number field the answer is given by local invariants: $D \otimes_{F} D'$ is a division algebra exactly when the sum of the local Hasse invariants of $D$ and $D'$ has the same order at every place as the degree predicts. The useful heuristic is that any common splitting field forces zero divisors, so two algebras split by a common quadratic extension will not tensor to a division ring.

How does the artinian clause interact with the density theorem?

Directly. Once $R$ acts faithfully and densely on a module over a division ring, artinian is equivalent to that module being finite-dimensional and to the density being an isomorphism onto the full endomorphism ring. That is how $(15.4)$ upgrades the one-way implication here to a list of equivalences.

What is the relation between this result and central simple algebras?

A central simple $F$ -algebra is by definition simple with centre exactly $F$ and finite-dimensional over $F$ . $(15.1)$ says that tensoring a central division algebra with any simple algebra keeps simplicity, and with any $F$ -algebra keeps the centralizer relationship. Restricting to finite dimensions gives the standard facts: the tensor product of central simple algebras is central simple, and $\otimes_{F}$ descends to a group law on Brauer classes.

D \otimes_{F} D'

ever commutative?

Only if both factors are. The centre is $Z (D')$ , so commutativity of the product would force $D' = Z (D')$ and, via the symmetric role of the factors when both are central, $D = F$ as well.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §15, results (15.1)–(15.2) (pp. 250–252).
T. Y. Lam, A First Course in Noncommutative Rings, §11 (the Density Theorem, (11.16) and (11.19)), which supplies the machinery used to exploit (15.1).
N. Jacobson, Basic Algebra II, 2nd edition, W. H. Freeman, 1989, Chapter 4 (central simple algebras and the Brauer group).
R. S. Pierce, Associative Algebras, Graduate Texts in Mathematics 88, Springer-Verlag, 1982, Chapters 9–12.
L. H. Rowen, Ring Theory, Volume II, Academic Press, 1988, Chapter 7 (simple algebras and centralizers).
B. A. Sethuraman, B. Sundar Rajan and V. Shashidhar, “Full-diversity, high-rate space-time block codes from division algebras”, IEEE Transactions on Information Theory 49 (2003), 2596–2616.

AI Suggested Questions

Prove that $D \otimes_{F} B$ is simple whenever $D$ is a central division $F$ -algebra and $B$ is a simple $F$ -algebra, and identify where the argument uses that $D$ is a division ring.
Give an explicit isomorphism $ℍ_{ℝ} \otimes_{ℝ} ℍ_{ℝ} ≅ M_{4} (ℝ)$ in terms of left and right multiplication operators.
For which quadratic fields $K$ is $ℍ_{ℚ} \otimes_{ℚ} K$ a division algebra?
Work out $Z (A \otimes_{F} B) = Z (A) \otimes_{F} Z (B)$ for finite-dimensional algebras and explain why $(15.1) (2)$ is the special case $Z (A) = F$ .
What replaces $(15.1)$ when $F$ is a commutative ring rather than a field, as in the theory of Azumaya algebras?
Show that a tensor product of two centrally infinite division algebras can be artinian, and characterise when.
How is the space–time code design criterion of non-vanishing determinant expressed in terms of the embedding $D ↪ M_{r} (K)$ coming from a splitting field?

Tensor Products and Centralizers

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

The normal form

Why the centre must be the ground field

Simplicity by minimal length

Key Results

Proof Techniques and Method

Expand in a basis of the second factor

Minimise the length inside an ideal

Chain conditions from freeness

Worked Example

Three tensor products of the rational quaternions

What goes wrong when $F$ is not the whole centre

Comparison and Classification

Relationship Map

Applications and Industry Use

Tensor product as a group law

Schur indices and splitting fields

Space–time block codes

Recognising an algebra

Design Considerations

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

The normal form

Why the centre must be the ground field

Simplicity by minimal length

06Key Results

07Proof Techniques and Method

Expand in a basis of the second factor

Minimise the length inside an ideal

Chain conditions from freeness

08Worked Example

Three tensor products of the rational quaternions

What goes wrong when F is not the whole centre

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Tensor product as a group law

Schur indices and splitting fields

Space–time block codes

Recognising an algebra

12Design Considerations

13Standards and Notation

14Computational Notes

15Failure Modes and Common Mistakes

16Quick Reference

17Frequently Asked Questions

18Related KEVOS Topics

19References

20AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

What goes wrong when $F$ is not the whole centre

Comparison and Classification

Relationship Map

Applications and Industry Use

Design Considerations

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions