What happens to (15.8) if D is centrally infinite?

The equivalences remain true — and are all false simultaneously. D ⊗_F K is still simple and still acts densely on D_K, but it is not artinian, (D_K) is infinite, and there is no degree, no square-dimension statement and no matrix description. Density is all that survives.

Maximal Subfields of Division Rings

Q: Do maximal subfields always exist?

Yes, in any division ring. The union of a chain of subfields is a subfield, so Zorn's Lemma applies, and every subfield — in particular F(a) for any element a — is contained in a maximal one. Existence is free; good behaviour, such as separability over the centre, is not.

Q: Why must the dimension of a central division algebra be a perfect square?

Because a maximal subfield K is its own centralizer, so the dimension formula of the Double Centralizer Theorem reads _F D = (_F K)(_F C_D(K)) = (_F K)^2. Equivalently, D ⊗_F K ≅ M_r(K) and dimension counting over K gives _F D = r^2.

Q: Are all maximal subfields of a given D isomorphic?

No. They all have the same degree r over the centre when D is centrally finite, but H_Q contains Q(i), Q(-2) and Q(-3), pairwise non-isomorphic. The Skolem–Noether theorem says that *isomorphic* subfields are conjugate by an inner automorphism, which is a different statement.

Q: Is a maximal subfield the same as a splitting field?

Every maximal subfield of a centrally finite D splits D, by D ⊗_F K ≅ M_r(K). The converse fails: splitting fields can be much larger, need not embed in D at all, and include for instance any algebraically closed field containing F.

Q: Can a maximal subfield equal the centre?

Only if D = F. If K = F were maximal then C_D(F) = D would have to equal F, forcing D commutative. This is the degenerate case r = 1.

Executive Summary

Maximal subfields are the commutative shadows of a division ring: each one is a field $K$ inside $D$ that cannot be enlarged, equivalently a subfield satisfying $C_{D} (K) = K$ . Every such $K$ automatically contains the centre $F$ , so the whole theory of §15 — tensor products, centralizers, density — applies to the pair $(K, D)$ with $L = C_{D} (K) = K$ .

For a centrally finite $D$ the payoff is decisive: $D \otimes_{F} K ≅ M_{r} (K)$ with $r = \dim_{F} K$ , and $\dim_{F} D = r^{2}$ . Central finiteness, finiteness of $\dim_{F} K$ , and finiteness of $\dim (D_{K})$ are all the same condition; and a subfield $E \supseteq F$ is maximal precisely when $\dim_{F} E = \sqrt{\dim_{F} D}$ .

Test for maximality

C_{D} (K) = K

Dimension theorem

\dim_{F} D = r^{2}

with

r = \dim_{F} K

Splitting

D \otimes_{F} K ≅ M_{r} (K)

Degree test

E

maximal

\Leftrightarrow

\dim_{F} E = \sqrt{\dim_{F} D}

Overview

A division ring is hard to look at directly, but it is full of fields: every element $a \in D$ generates a commutative subring $F [a]$ , which sits inside a subfield. Studying $D$ through its maximal subfields replaces a noncommutative object by a family of commutative ones together with the way they overlap — the strategy behind cyclic algebras, crossed products and the Brauer group.

Two facts make the strategy work. First, maximality has a purely internal characterisation that needs no tensor products: $K$ is maximal iff nothing outside $K$ commutes with $K$ . Second, once that is known, the machinery of Criteria for Central Finiteness and the Double Centralizer Theorem applies with $L = C_{D} (K) = K$ , and the resulting numerology is unusually rigid.

K maximal ⟺ C_{D} (K) = K ⟹ F = Z (D) \subseteq K,

(15.7)

Maximal subfields are the self-centralizing subfields, and they contain the centre for free.

Existence is never an issue: any chain of subfields has an upper bound (its union), so Zorn's Lemma provides maximal subfields in every division ring, centrally finite or not. What varies wildly is their size and their arithmetic.

Learning Objectives

Prove that a commutative subring of a division ring lies in a subfield, so maximal subfields and maximal commutative subrings coincide.
Prove $(15.7)$ in both directions and deduce $Z (D) \subseteq K$ .
Derive the five equivalences of $(15.8)$ from the results on centralizers.
Show $\dim_{F} D = r^{2}$ and interpret $r$ as the degree of $D$ .
Apply the degree criterion to decide whether a given subfield is maximal.
Compute the maximal subfields of the rational quaternions and of a twisted Laurent series division ring.

Definitions

Definition—Maximal subfield

A subfield $K$ of a ring $D$ is a maximal subfield if there is no subfield $K'$ of $D$ with $K ⊊ K'$ . In a division ring this is equivalent to $K$ being a maximal commutative subring, because every commutative subring of a division ring is contained in a subfield.

Why the equivalence holds. Let $T$ be a commutative subring of $D$ and, by Zorn's Lemma, let $S \supseteq T$ be a maximal commutative subring. For $0 \neq s \in S$ , the inverse $s^{- 1}$ commutes with everything that commutes with $s$ , hence with all of $S$ ; so $S [s^{- 1}]$ is commutative and maximality gives $s^{- 1} \in S$ . Thus $S$ is a commutative division ring, i.e. a field.

$C_{D} (K)$: The centralizer of $K$ in $D$ ; a division subring containing $Z (D)$ , and containing $K$ whenever $K$ is commutative.
Degree $r$: For $D$ centrally finite with $\dim_{F} D = n$ , the integer $r = \sqrt{n}$ . Also called the index of $D$ .
$D_{K}$ , $_{K} D$: $D$ viewed as a right, respectively left, vector space over a subfield $K$ .
Splitting field: A field $K \supseteq F$ with $D \otimes_{F} K ≅ M_{r} (K)$ . By $(15.8)$ every maximal subfield of a centrally finite $D$ is one.
Centrally infinite: $\dim_{F} D = \infty$ . By $(15.8)$ this is equivalent to every — hence some — maximal subfield having infinite degree over $F$ .

Core Concepts

Maximality is a centralizer condition

If $K$ is commutative then $K \subseteq C_{D} (K)$ , so $L : = C_{D} (K)$ is a division subring containing $K$ . Enlarging $K$ inside $D$ means finding a commuting element outside it, and the elements available are precisely those of $L$ . So $K$ is maximal exactly when $L$ has nothing new to offer, i.e. $L = K$ .

The subtlety is that $L$ need not be commutative when $K$ is not maximal. What the argument uses is weaker: given $c \in L$ , the ring $K [c]$ is commutative, hence lies in a subfield, so maximality of $K$ forces $c \in K$ .

Why the dimension is a perfect square

Feed $L = K$ into the dimension formula of the Double Centralizer Theorem. It reads $\dim_{F} D = (\dim_{F} K) (\dim_{F} L) = r \cdot r$ . The square is not an accident of the proof — it is the shadow of the isomorphism $D \otimes_{F} K ≅ M_{r} (K)$ , whose left side has $K$ -dimension $\dim_{F} D$ and whose right side has $K$ -dimension $r^{2}$ .

\dim_{K} (D \otimes_{F} K) = \dim_{F} D = r^{2} = \dim_{K} M_{r} (K) .

(15.8)

Counting dimensions on both sides of the splitting isomorphism.

Consequences of a fixed degree

Because all maximal subfields of a centrally finite $D$ have degree exactly $r$ , and every element lies in some maximal subfield, the degree over $F$ of any $a \in D$ divides $r$ . In particular the minimal polynomial of an element of a quaternion algebra has degree $1$ or $2$ , and a degree- $3$ division algebra contains no quadratic subfields at all.

Key Results

Proposition(15.7)Maximal subfields are self-centralizing

Let $D$ be a division ring and $K \subseteq D$ a subfield. Then $K$ is a maximal subfield of $D$ if and only if $C_{D} (K) = K$ . When this holds, $Z (D) \subseteq K$ .

Proof

Write $L = C_{D} (K)$ ; since $K$ is commutative, $K \subseteq L$ .

**( $\Leftarrow$ )** Suppose $L = K$ and let $K'$ be a subfield of $D$ with $K \subseteq K'$ . Every element of $K'$ commutes with every element of $K$ , so $K' \subseteq C_{D} (K) = K$ , whence $K' = K$ . Thus $K$ is maximal.

**( $\Rightarrow$ )** Suppose $K$ is maximal and let $c \in L$ . Then $K [c]$ is a commutative subring of $D$ , hence is contained in a subfield $K'$ of $D$ containing $K$ . Maximality gives $K' = K$ , so $c \in K$ and $L = K$ .

Finally, every $c \in Z (D)$ commutes with $K$ , so $Z (D) \subseteq C_{D} (K) = K$ .

Theorem(15.8)Maximal subfields of a division ring

Let $D$ be a division ring with centre $F$ and let $K$ be a maximal subfield of $D$ (so $F \subseteq K$ by $(15.7)$ ). Then $D \otimes_{F} K$ is a simple $F$ -algebra acting as a dense ring of linear transformations on the right $K$ -vector space $D_{K}$ , and the following are equivalent:

$D \otimes_{F} K$ is artinian;
$\dim (D_{K}) < \infty$ ;
$\dim (_{K} D) < \infty$ ;
$\dim_{F} K < \infty$ ;
$D$ is centrally finite.

If $r : = \dim_{F} K < \infty$ , then $\dim (D_{K}) = \dim (_{K} D) = r$ ,

D \otimes_{F} K ≅ End (D_{K}) ≅ M_{r} (K), \dim_{F} D = r^{2} .

Moreover, in this case a subfield $E$ of $D$ with $F \subseteq E$ is a maximal subfield if and only if $\dim_{F} E = \sqrt{\dim_{F} D}$ .

Proof

By $(15.7)$ we have $L : = C_{D} (K) = K$ , and since $K$ is commutative, $K^{op} = K$ ; so the ring $R = D \otimes_{F} K^{op}$ of $(15.3)$ is just $D \otimes_{F} K$ . Simplicity, the module structure and the density statement are therefore immediate from $(15.1)$ and $(15.3)$ .

The equivalence of (1), (2) and (4) is $(15.4)$ with $L = K$ ; the equivalence of (2) and (3) is $(15.6)$ . For (4) $\Rightarrow$ (5), the dimension formula gives $\dim_{F} D = (\dim_{F} K) (\dim_{F} L) = r \cdot r = r^{2} < \infty$ . For (5) $\Rightarrow$ (4), $K$ is an $F$ -subspace of $D$ , so $\dim_{F} K \leq \dim_{F} D < \infty$ . The isomorphism $D \otimes_{F} K ≅ End (D_{K}) ≅ M_{r} (K)$ is the conclusion of $(15.4)$ with $L = K$ .

Degree criterion, "only if". If $E$ is a maximal subfield then applying the above to $E$ gives $\dim_{F} D = (\dim_{F} E)^{2}$ , so $\dim_{F} E = \sqrt{\dim_{F} D}$ .

Degree criterion, "if". Let $E \supseteq F$ be a subfield with $\dim_{F} E = \sqrt{\dim_{F} D} = r$ . Any subfield of $D$ is contained in a maximal one — take a chain and pass to its union, then apply Zorn's Lemma — so choose a maximal subfield $E' \supseteq E$ . By the previous paragraph $\dim_{F} E' = r = \dim_{F} E$ , and since $E \subseteq E'$ are both finite-dimensional $F$ -spaces of the same dimension, $E = E'$ . Hence $E$ is maximal.

Corollary—Degrees of elements divide the degree of the algebra

Let $D$ be centrally finite of degree $r$ over $F$ , and let $a \in D$ . Then $F (a)$ is a finite field extension of $F$ contained in some maximal subfield $K$ , so $\dim_{F} F (a)$ divides $\dim_{F} K = r$ . In particular the minimal polynomial of $a$ over $F$ has degree dividing $r$ .

Proof. $F [a]$ is a commutative domain, finite-dimensional over $F$ , hence a field, and it lies in a maximal subfield by the Zorn argument above. The tower formula $\dim_{F} K = \dim_{F (a)} K \cdot \dim_{F} F (a)$ finishes the argument.

Corollary—Forbidden dimensions

There is no division ring $D$ with $\dim_{Z (D)} D$ equal to $2, 3, 5, 6, 7, 8, 10, \dots$ — any value that is not a perfect square. The smallest noncommutative possibilities are $4$ (quaternion algebras) and $9$ (degree-three algebras such as Dickson's cyclic example over $ℚ$ ).

Proof Techniques and Method

The reusable moves behind these proofs.

Reduce maximality to a centralizerReplace "cannot be enlarged" by "

C_{D} (K) = K

". The second form is checkable by solving linear commutation equations.

Feed $L = K$ into the general theoryThe centralizer results of

(15.4)

were stated for arbitrary division subrings; maximal subfields are the case where the centralizer collapses onto the subfield.

Count dimensions twice

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

becomes

r^{2}

, and independently

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

. Agreement of the two counts is the square-dimension theorem.

Upgrade a subfield to a maximal oneZorn's Lemma gives a maximal subfield above any subfield, so statements about maximal subfields transfer to statements about arbitrary ones by a dimension comparison.

The last step is the whole content of the degree criterion, and it is the standard way to turn an existence theorem about maximal objects into a computable test.

Worked Example

Maximal subfields of the rational quaternions

Let $D = ℍ_{ℚ}$ , $F = ℚ$ , $\dim_{F} D = 4$ , so $r = 2$ . A pure quaternion $q = a i + b j + c k$ with $(a, b, c) \neq 0$ satisfies

q^{2} = - (a^{2} + b^{2} + c^{2}) \in ℚ^{\times},

(E.1)

Pure quaternions square to negative rationals; the cross terms cancel because $i j = - j i$ , $j k = - k j$ , $k i = - i k$ .

Hence $ℚ (q) ≅ ℚ (\sqrt{- n})$ with $n = a^{2} + b^{2} + c^{2} > 0$ , a quadratic field, so $\dim_{ℚ} ℚ (q) = 2 = \sqrt{\dim_{ℚ} D}$ and $ℚ (q)$ is maximal by the degree criterion. Conversely every maximal subfield is of this shape: it is a quadratic extension $ℚ (a)$ , and after subtracting the rational part of $a$ we obtain a pure quaternion generator.

Which quadratic fields sit inside $ℍ_{ℚ}$
Field	Generator	Inside $ℍ_{ℚ}$ ?	Reason
$ℚ (i)$	$i$	yes	$1 = 1^{2} + 0 + 0$
$ℚ (\sqrt{- 2})$	$i + j$	yes	$(i + j)^{2} = - 2$
$ℚ (\sqrt{- 3})$	$i + j + k$	yes	$(i + j + k)^{2} = - 3$
$ℚ (\sqrt{- 7})$	—	no	$7$ is not a sum of three rational squares
$ℚ (\sqrt{2})$	—	no	pure quaternions square to negative rationals

So $ℍ_{ℚ}$ has infinitely many maximal subfields, all of degree $2$ , and they fall into infinitely many isomorphism classes — degree is an invariant of $D$ , the isomorphism type of a maximal subfield is not.

The splitting isomorphism in this case is $ℍ_{ℚ} \otimes_{ℚ} ℚ (i) ≅ M_{2} (ℚ (i))$ , consistent with $r = 2$ and $\dim_{ℚ} D = 4$ .

A maximal subfield of infinite degree

Let $k = ℚ (t)$ and let $σ$ be the automorphism $f (t) \mapsto f (2 t)$ of $k$ , which has infinite order. Form the twisted Laurent series division ring $D = k ((x; σ))$ , whose elements are series $\sum_{i \geq i_{0}} a_{i} x^{i}$ with $a_{i} \in k$ , multiplied using $x a = σ (a) x$ .

Centre. Commuting with $x$ forces every coefficient to satisfy $σ (a_{i}) = a_{i}$ , i.e. $a_{i} \in ℚ$ — the fixed field of $σ$ , since $f (2 t) = f (t)$ forces $f$ constant. Commuting with $t$ forces $a_{i} (2^{i} - 1) = 0$ , i.e. $a_{i} = 0$ for $i \neq 0$ . Hence $F = Z (D) = ℚ$ .
A maximal subfield. The same computation shows $C_{D} (k) = k$ , so by $(15.7)$ the field $K = ℚ (t)$ is a maximal subfield of $D$ .
Its degree. $\dim_{ℚ} ℚ (t)$ is infinite, so by $(15.8)$ $D$ is centrally infinite, $\dim (D_{K})$ is infinite, and $D \otimes_{ℚ} K$ is simple but neither left nor right artinian.

Comparison and Classification

Maximal subfields across standard examples
$D$	$F = Z (D)$	$\dim_{F} D$	Degree $r$	A maximal subfield
A field $F$	$F$	$1$	$1$	$F$ itself
$ℍ_{ℝ}$	$ℝ$	$4$	$2$	$ℝ (i) ≅ ℂ$
$ℍ_{ℚ}$	$ℚ$	$4$	$2$	$ℚ (i)$ , $ℚ (\sqrt{- 2})$ , $\dots$
Cyclic algebra $(K / F, σ, a)$ of degree $n$	$F$	$n^{2}$	$n$	$K$
Dickson's cubic example over $ℚ$	$ℚ$	$9$	$3$	a cyclic cubic field
$ℚ (t) ((x; σ))$ , $σ (t) = 2 t$	$ℚ$	infinite	—	$ℚ (t)$ , of infinite degree

Properties that do and do not transfer between maximal subfields of one $D$
	Same for all maximal subfields?	Centrally finite $D$	Centrally infinite $D$
Self-centralizing	yes	yes	yes
Contains $Z (D)$	yes	yes	yes
Degree over $F$	yes	yes	no
Isomorphism type	no	no	no
Splits $D$	yes	yes	no
Separable over $F$	no	partial	no

Properties that do and do not transfer between maximal subfields of one $D$

Relationship Map

Maximal subfields are where the abstract centralizer theory becomes arithmetic.

$(15.7)$ — $K$ maximal $\Leftrightarrow$ $C_{D} (K) = K$ — Elementary; no tensor products needed.
- $(15.8)$ — five equivalences plus $\dim_{F} D = r^{2}$ — Obtained by substituting $L = K$ into the double centralizer results.
  - Degree test: $E$ maximal $\Leftrightarrow$ $\dim_{F} E = \sqrt{\dim_{F} D}$
  - Splitting: $D \otimes_{F} K ≅ M_{r} (K)$ , so every maximal subfield is a splitting field
  - $(15.12)$ — a separable maximal subfield always exists in the centrally finite case
  - $(15.16)$ — the Brauer–Albert basis built from a separable maximal subfield

Given a subfield $E \supseteq F$ of a centrally finite $D$ , is it maximal?

$\dim_{F} E = \sqrt{\dim_{F} D}$ Yes — maximal, and

E

splits

D

D \otimes_{F} E ≅ M_{r} (E)

$\dim_{F} E < \sqrt{\dim_{F} D}$ No. Then

C_{D} (E) ⊋ E

and

\dim_{F} C_{D} (E) = \dim_{F} D / \dim_{F} E

by the dimension formula; enlarge

E

inside

C_{D} (E)

$\dim_{F} E$ does not divide $r$ Impossible: the degree of any subfield divides

r

, so such an

E

does not exist inside

D

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

Splitting fields and the index

The relative Brauer group $Br (K / F)$ collects the classes split by $K$ . Since every maximal subfield splits $D$ , the index $r$ bounds the degrees needed, and the crossed-product description of $Br (F)$ rests on choosing maximal subfields that are Galois over $F$ .

Number theory

Embedding fields into algebras

Deciding which quadratic fields embed in a quaternion algebra is the local-global embedding problem, solved by comparing ramification. The computation for $ℍ_{ℚ}$ above — sums of three rational squares — is its most elementary instance.

Wireless communication

Rate and delay of space–time codes

A cyclic division algebra of degree $r$ has maximal subfield $K$ with $\dim_{F} K = r$ ; the code transmits $r \times r$ matrices over $K$ obtained from $D \otimes_{F} K ≅ M_{r} (K)$ . The square-dimension theorem is the reason the resulting codeword matrices are square.

Symbolic computation

Presenting a division algebra

Computer algebra systems represent a central division algebra by a maximal subfield plus a twisting element, precisely because $\dim_{F} D = r^{2}$ makes the resulting basis ${u^{i} v^{j}}$ predictable in size.

Design Considerations

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

Which maximal subfield? They all have the same degree but not the same arithmetic. Choose one that is separable — or better, Galois — over $F$ if you intend to use crossed-product or cyclic-algebra descriptions.
Maximal subfield or maximal commutative subring? Inside a division ring the two notions agree, and the second is often easier to verify because it needs no inverses. Outside division rings they diverge sharply.
Left or right dimension? State the side while it still matters. For a maximal subfield of a centrally finite $D$ both are $r$ , but the equality is a theorem, not a definition.
Do not model with a subfield that is too small. If $\dim_{F} E$ properly divides $r$ then $D \otimes_{F} E$ is a matrix ring over a smaller division algebra, not over $E$ — the splitting is only partial.

Failure Modes and Common Mistakes

Do not assume a maximal subfield is separable over $F$ ; in characteristic $p$ purely inseparable maximal subfields exist. What is true is that a separable one also exists — that is $(15.12)$ .
Do not confuse $C_{D} (K) = K$ with $K$ being the centre. The centre is contained in every maximal subfield and is usually much smaller.
Do not expect $\dim_{F} D$ to be a square when $F$ is a proper subfield of the centre. The theorem is about the full centre only.
Do not conclude that a subfield of degree $r$ over $F$ inside a simple algebra is maximal; the square-dimension theorem here is stated for division rings.

Historical Notes and Lessons Learned

1878FrobeniusClassifies the finite-dimensional real division algebras: $ℝ$ , $ℂ$ and the quaternions. In hindsight this is the degree- $2$ case of the square-dimension theorem over a real closed field.
1906–1914Dickson and WedderburnCyclic algebras are constructed from a cyclic extension $K / F$ and a twisting element; $K$ appears as a maximal subfield, and the degree of the algebra is the degree of $K$ .
1929–1932Brauer, Noether, Hasse, AlbertThe Brauer group and crossed products systematise the study of central simple algebras through their splitting fields; maximal subfields become the primary structural handle.
1972AmitsurConstructs central division algebras that are not crossed products — no maximal subfield is Galois over the centre. Maximal subfields exist, but need not be as well behaved as the classical theory hoped.

The lesson is that maximality is cheap and structure is expensive: Zorn's Lemma hands over maximal subfields in any division ring, but arranging them to be separable requires $(15.12)$ and arranging them to be Galois can be impossible.

Quick Reference

Characterisation

K

maximal

\Leftrightarrow

C_{D} (K) = K

Contains the centre

Z (D) \subseteq K

for every maximal subfield

K

Equivalent to central finiteness

\dim_{F} K < \infty \Leftrightarrow \dim (D_{K}) < \infty \Leftrightarrow D \otimes_{F} K

artinian

\Leftrightarrow \dim_{F} D < \infty

Degree

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

Dimension

\dim_{F} D = r^{2}

, always a perfect square

Splitting

D \otimes_{F} K ≅ End (D_{K}) ≅ M_{r} (K)

Degree test

E \supseteq F

maximal

\Leftrightarrow

\dim_{F} E = \sqrt{\dim_{F} D}

Elements

\dim_{F} F (a)

divides

r

for every

a \in D

Checklist for a candidate maximal subfield
Check	How	If it fails
$K$ is a field	Commutative and closed under inverses	Enlarge to a subfield first
$F \subseteq K$	Automatic once $K$ is maximal	Adjoin the centre
$C_{D} (K) = K$	Solve the commutation equations in a basis	Any new element enlarges $K$
$\dim_{F} K = \sqrt{\dim_{F} D}$	Linear algebra over $F$	$K$ is not maximal, or $D$ is centrally infinite

Frequently Asked Questions

Do maximal subfields always exist?

Yes, in any division ring. The union of a chain of subfields is a subfield, so Zorn's Lemma applies, and every subfield — in particular $F (a)$ for any element $a$ — is contained in a maximal one. Existence is free; good behaviour, such as separability over the centre, is not.

Why must the dimension of a central division algebra be a perfect square?

Because a maximal subfield $K$ is its own centralizer, so the dimension formula of the Double Centralizer Theorem reads $\dim_{F} D = (\dim_{F} K) (\dim_{F} C_{D} (K)) = (\dim_{F} K)^{2}$ . Equivalently, $D \otimes_{F} K ≅ M_{r} (K)$ and dimension counting over $K$ gives $\dim_{F} D = r^{2}$ .

Are all maximal subfields of a given

D

isomorphic?

No. They all have the same degree $r$ over the centre when $D$ is centrally finite, but $ℍ_{ℚ}$ contains $ℚ (i)$ , $ℚ (\sqrt{- 2})$ and $ℚ (\sqrt{- 3})$ , pairwise non-isomorphic. The Skolem–Noether theorem says that isomorphic subfields are conjugate by an inner automorphism, which is a different statement.

Is a maximal subfield the same as a splitting field?

Every maximal subfield of a centrally finite $D$ splits $D$ , by $D \otimes_{F} K ≅ M_{r} (K)$ . The converse fails: splitting fields can be much larger, need not embed in $D$ at all, and include for instance any algebraically closed field containing $F$ .

What happens to

(15.8)

D

is centrally infinite?

The equivalences remain true — and are all false simultaneously. $D \otimes_{F} K$ is still simple and still acts densely on $D_{K}$ , but it is not artinian, $\dim (D_{K})$ is infinite, and there is no degree, no square-dimension statement and no matrix description. Density is all that survives.

Can a maximal subfield equal the centre?

Only if $D = F$ . If $K = F$ were maximal then $C_{D} (F) = D$ would have to equal $F$ , forcing $D$ commutative. This is the degenerate case $r = 1$ .

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §15, results (15.7)–(15.8) (pp. 254–256).
T. Y. Lam, A First Course in Noncommutative Rings, §14 (cyclic algebras, Dickson's examples and twisted Laurent series constructions).
A. A. Albert, Structure of Algebras, American Mathematical Society Colloquium Publications 24, 1939, Chapters III–V.
N. Jacobson, Basic Algebra II, 2nd edition, W. H. Freeman, 1989, §4.6 (maximal subfields and splitting fields).
S. A. Amitsur, “On central division algebras”, Israel Journal of Mathematics 12 (1972), 408–420.
M.-A. Knus, A. Merkurjev, M. Rost and J.-P. Tignol, The Book of Involutions, American Mathematical Society Colloquium Publications 44, 1998, Chapter I.

AI Suggested Questions

Prove that every commutative subring of a division ring is contained in a maximal subfield, and identify where Zorn's Lemma is needed.
Determine exactly which quadratic fields embed into the quaternion algebra $(- 1, - 1)$ over $ℚ$ , and relate the answer to ramification at $2$ and $\infty$ .
Give an example of a centrally finite division ring with two maximal subfields that are not isomorphic as field extensions of the centre.
Show that a subfield $E$ of a centrally finite $D$ with $\dim_{F} E = m$ has centralizer of dimension $r^{2} / m$ over $F$ , and explain when $C_{D} (E)$ is again a division algebra of square dimension over its own centre.
Explain how the crossed product construction reconstructs $D$ from a Galois maximal subfield and a factor set.
Describe Amitsur's non-crossed-product division algebras and what they say about maximal subfields.
Compute the centre and a maximal subfield of the twisted Laurent series ring $k ((x; σ))$ for a general automorphism $σ$ of infinite order.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Maximality is a centralizer condition

Why the dimension is a perfect square

Consequences of a fixed degree

06Key Results

07Proof Techniques and Method

08Worked Example

Maximal subfields of the rational quaternions

A maximal subfield of infinite degree

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Splitting fields and the index

Embedding fields into algebras

Rate and delay of space–time codes

Presenting a division algebra

12Design Considerations

13Failure Modes and Common Mistakes

14Historical Notes and Lessons Learned

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

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Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Comparison and Classification

Relationship Map

Applications and Industry Use

Design Considerations

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions