Orderings on Division Rings

Q: Is every subgroup of D^* of index 2 an ordering?

No. Index 2 gives axiom (17.3) and multiplicative closure, but additive closure P + P ⊆ P is an extra condition and usually fails. For example Q^* has many index-2 subgroups — one for each subgroup of Q^*/Q^*2 of index 2 — and only Q_>0 is additively closed.

Q: Can a division ring of characteristic p be ordered?

Never. An ordering contains 1 and is closed under addition, so it contains every n 1 with n ≥ 1; if p 1 = 0 this would put 0 in the cone. So (17.4) forces characteristic 0, and orderable division rings all contain Q in their centre.

Q: How many orderings can a division ring have?

As many as there are maximal preorderings above T(D), and for formally real D there is at least one. For Q there is exactly one; for Q(2) there are two; for the rational function field Q(t) there are infinitely many. The set of all orderings carries a natural topology in the field case (the real spectrum), but no comparable theory exists for division rings.

Executive Summary

For a general ring the ordering axioms of $§ 17$ are awkward: the multiplicative axiom has to be stated with permuted products $per (a_{1}^{2} \dots a_{m}^{2} t_{1} \dots t_{n})$ because there is no way to move a factor past its neighbours. In a division ring every nonzero element is invertible, and that single fact collapses the whole apparatus.

An ordering $P \subseteq D^{*}$ becomes a normal subgroup of index $2$ ; a preordering $T$ becomes a normal subgroup containing every square and every commutator, with $D^{*} / T$ elementary abelian. The orderability criterion becomes a single arithmetic statement: $D$ can be ordered if and only if $- 1$ is not a sum of square-products.

$[D^{*} : P] = 2$ An ordering as a subgroup

$3$ axiomsPreordering test (18.1)

Exponent $2$ The group

D^{*} / T

$- 1 \notin T (D)$ Orderability (18.2)

Overview

Throughout, $D$ is a division ring and $D^{*} = D ∖ {0}$ its multiplicative group. An ordering on $D$ is a subset $P \subseteq D^{*}$ satisfying the three axioms inherited from the theory of ordered rings.

P + P \subseteq P, P \cdot P \subseteq P, P \cup (- P) = D ∖ {0} .

(17.1)–(17.3)

Setting $a < b iff b - a \in P$ recovers a total order compatible with addition and with multiplication by positive elements.

Two consequences from the general theory carry over at once. First, $P \cap (- P) = \emptyset$ and $1 \in P$ , so $D$ has characteristic $0$ — an ordered division ring contains a copy of $ℚ$ . Second, $P$ is closed under multiplication, and because inverses are available it is closed under inversion as well: for $a \in P$ we have $a^{- 1} = a (a^{- 1})^{2}$ , and squares of nonzero elements always lie in $P$ .

Because $D^{*} = P ⊔ (- P)$ , the cone $P$ has index $2$ and is therefore automatically normal. With the induced order $P$ is itself a multiplicative ordered group, with positive cone ${a \in P : a > 1}$ — the link to the material of Ordered Groups and Group Rings and to the Mal'cev–Neumann construction that produces the standard examples.

Learning Objectives

Translate the ordering axioms into group-theoretic language for a division ring.
State $(18.1)$ and use its three closure conditions to test a candidate set.
Prove that any preordering contains $[D^{*}, D^{*}]$ , hence is normal with $D^{*} / T$ of exponent $2$ .
Distinguish a square-product from a square, and explain why the distinction is invisible for fields.
State the Szele–Pickert theorem $(18.2)$ and deduce the Artin–Schreier criterion for fields.
Decide whether $ℚ$ , $ℍ$ and a twisted Laurent series ring are orderable.

Definitions

Definition(17.5)–(17.6)Preordering, specialised

A preordering in a ring $R$ is a subset $T \subseteq R ∖ {0}$ with $T + T \subseteq T$ such that every permuted product $per (a_{1}^{2} \dots a_{m}^{2} t_{1} \dots t_{n})$ lies in $T$ , for all $a_{i} \in R ∖ {0}$ and $t_{j} \in T$ . Here $per$ denotes the product of the listed factors — each $a_{i}$ occurring twice — taken in any order.

Every ordering is a preordering, and any intersection of orderings is a preordering.

Definition—Square-product and the weak preordering

A square-product in $D$ is an element of the form $a_{1}^{2} a_{2}^{2} \dots a_{m}^{2}$ with all $a_{i} \in D^{*}$ ; these form the subgroup $Σ (D) \leq D^{*}$ generated by the squares. The weak preordering $T (D)$ is the set of all finite sums of square-products. $D$ is formally real when $0 \notin T (D)$ .

$D^{*}$: The multiplicative group $D ∖ {0}$ of a division ring $D$ .
$Σ (D)$: The subgroup of $D^{*}$ generated by ${a^{2} : a \in D^{*}}$ ; its elements are exactly the square-products.
$T (D)$: Sums of square-products. It is contained in every preordering of $D$ , which is why it is called weak.
Totally positive: Positive with respect to every ordering of $D$ . Characterised in Preorderings in Division Rings.
$[D^{*}, D^{*}]$: The commutator subgroup of $D^{*}$ , generated by all $a b a^{- 1} b^{- 1}$ with $a, b \in D^{*}$ .

Orderings and preorderings are sets of nonzero elements throughout: $0$ never belongs to a cone, and the sign convention is that $P$ is the set of strictly positive elements.

Core Concepts

Inverses do the work

The general definition of a preordering is complicated only because the two copies of each $a_{i}$ may be separated by arbitrary other factors, and there is no commutativity to bring them together. In a division ring the separating block $u$ is invertible, and one identity closes the gap.

a u a = (a u)^{2} (u^{- 1})^{2} u (a, u \in D^{*}),

(C.1)

Expand the right side: $a u a u u^{- 1} u^{- 1} u = a u a$ . Two squares are produced and the pair $a, a$ disappears.

So a block $a u a$ inside a long permuted product may be traded for a single element $(a u)^{2} (u^{- 1})^{2}$ — a product of two squares — followed by the shortened block $u$ . Iterating removes every matched pair. That is the whole content of $(18.1)$ .

Commutators are square-products

The same identity, applied with the roles rearranged, exhibits every multiplicative commutator as a product of three squares.

a b a^{- 1} b^{- 1} = a^{2} (a^{- 1} b)^{2} (b^{- 1})^{2} (a, b \in D^{*}) .

(C.2)

Expanding: $a a a^{- 1} b a^{- 1} b b^{- 1} b^{- 1} = a b a^{- 1} b^{- 1}$ .

Hence $[D^{*}, D^{*}] \subseteq Σ (D) \subseteq T$ for every preordering $T$ . This is the structural reason a preordering can never be a lopsided subset: it is forced to be normal, and $D^{*} / T$ is an abelian group killed by squaring.

[D^{*}, D^{*}]

Σ (D)

any preordering

T

an ordering

P

D^{*}

Key Results

Proposition(18.1)Preorderings in a division ring

Let $D$ be a division ring. A subset $T \subseteq D^{*}$ is a preordering of $D$ if and only if

$T + T \subseteq T$ ;
$T \cdot T \subseteq T$ ;
$a^{2} \in T$ for every $a \in D^{*}$ .

If $T$ is a preordering then $t \in T \Rightarrow t^{- 1} \in T$ , and $T \supseteq [D^{*}, D^{*}]$ . In particular $T$ is a normal subgroup of $D^{*}$ , and $D^{*} / T$ is an abelian group of exponent $2$ .

Proof

Necessity. Condition (1) is the axiom $(17.5)$ . Taking $m = 0$ in $(17.6)$ gives $t_{1} t_{2} \in T$ , which is (2); taking $m = 1$ , $n = 0$ gives $per (a^{2}) = a \cdot a = a^{2} \in T$ , which is (3).

Sufficiency. Assume (1)–(3). Only $(17.6)$ needs checking. Let $W$ be any arrangement of the list $a_{1}, a_{1}, \dots, a_{m}, a_{m}, t_{1}, \dots, t_{n}$ with $a_{i} \in D^{*}$ and $t_{j} \in T$ ; we show the value of $W$ lies in $T$ , by induction on $m$ .

If $m = 0$ then $W$ is a product of elements of $T$ , so $W \in T$ by (2) (and $W = 1 = 1^{2} \in T$ if $n = 0$ as well). If $m \geq 1$ , choose an index $i$ and the two positions of $a_{i}$ in $W$ , and write $W = w_{1} a_{i} u a_{i} w_{3}$ where $u$ is the (possibly empty) block between the two occurrences. Every letter is nonzero and $D$ is a division ring, so $u \in D^{*}$ . By $(C . 1)$ ,

W = w_{1} [(a_{i} u)^{2} (u^{- 1})^{2}] u w_{3}, (a_{i} u)^{2} (u^{- 1})^{2} \in T \cdot T \subseteq T

(18.1a)

using (3) and (2). The right-hand side is an arrangement of $a_{1}, a_{1}, \dots, a_{m - 1}, a_{m - 1}$ together with $t_{1}, \dots, t_{n}$ and the one extra element $(a_{i} u)^{2} (u^{- 1})^{2}$ of $T$ . The inductive hypothesis applies and gives $W \in T$ .

The remaining assertions. $1 = 1^{2} \in T$ ; for $t \in T$ , $t^{- 1} = t (t^{- 1})^{2} \in T \cdot T \subseteq T$ , so $T$ is a subgroup of $D^{*}$ . By $(C . 2)$ each commutator $a b a^{- 1} b^{- 1}$ is a product of three squares, hence lies in $T$ ; therefore $[D^{*}, D^{*}] \subseteq T$ , so $T ⊴ D^{*}$ and $D^{*} / T$ is abelian. Finally $a^{2} \in T$ for all $a$ , so every element of $D^{*} / T$ squares to the identity.

Corollary—Orderings are the index-two preorderings

For a subset $P \subseteq D^{*}$ the following are equivalent: (i) $P$ is an ordering of $D$ ; (ii) $P$ is a preordering of $D$ with $[D^{*} : P] = 2$ .

Proof

(i) $\Rightarrow$ (ii): $P + P \subseteq P$ and $P \cdot P \subseteq P$ are axioms, and for $a \in D^{*}$ one of $\pm a$ lies in $P$ , so $a^{2} = (\pm a) (\pm a) \in P$ ; thus $P$ is a preordering by $(18.1)$ . Since $D^{*} = P \cup (- P)$ and $P \cap (- P) = \emptyset$ , the coset decomposition $D^{*} = P ⊔ (- 1) P$ shows $[D^{*} : P] = 2$ .

(ii) $\Rightarrow$ (i): $- 1 \notin P$ , since otherwise $0 = 1 + (- 1) \in P + P \subseteq P$ , contradicting $0 \notin P$ . So $(- 1) P \neq P$ , and as $[D^{*} : P] = 2$ these are the only two cosets: $D^{*} = P ⊔ (- P)$ , which is axiom $(17.3)$ . Axioms $(17.1)$ and $(17.2)$ hold because $P$ is a preordering.

Lemma—Formal reality tested at

- 1

For a division ring $D$ : $0 \in T (D)$ if and only if $- 1 \in T (D)$ . Hence $D$ is formally real if and only if $- 1$ is not a sum of square-products.

Proof

If $- 1 \in T (D)$ then $0 = 1 + (- 1)$ is a sum of square-products, since $1 = 1^{2}$ is one. Conversely suppose $0 = s_{1} + \dots + s_{n}$ with each $s_{i}$ a square-product. As square-products are nonzero, $n \geq 2$ ; then $- s_{1} = s_{2} + \dots + s_{n}$ , and multiplying on the right by $s_{1}^{- 1}$ gives $- 1 = s_{2} s_{1}^{- 1} + \dots + s_{n} s_{1}^{- 1}$ . Each $s_{i} s_{1}^{- 1}$ lies in the group $Σ (D)$ , so $- 1 \in T (D)$ .

Theorem(18.2)Szele–Pickert orderability criterion

A division ring $D$ admits an ordering if and only if $- 1$ is not a sum of square-products in $D$ .

Proof

By the Lemma, the stated condition is exactly formal reality of $D$ . R. E. Johnson's theorem $(17.11)$ says that for any nonzero ring, formal reality, the existence of a preordering, and the existence of an ordering are equivalent: $T (D)$ is a preordering as soon as $0 \notin T (D)$ , and Zorn's Lemma enlarges any preordering to a maximal one, which by $(17.10)$ is an ordering. Conversely an ordering $P$ contains $T (D)$ and misses $0$ .

Corollary—Artin–Schreier for fields

A field $F$ admits an ordering if and only if $- 1$ is not a sum of squares in $F$ . Indeed in a commutative ring $a_{1}^{2} \dots a_{m}^{2} = (a_{1} \dots a_{m})^{2}$ , so square-products are squares and $(18.2)$ specialises to the 1927 criterion of Artin and Schreier.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Three techniques carry the section; all three reappear in the later results on formally real division rings.

Move 1

Sandwich elimination

A block $a u a$ becomes $(a u)^{2} (u^{- 1})^{2} u$ . Whenever a hypothesis is stated for permuted products, this identity reduces it to ordinary products of squares.

Move 2

Right-translate to normalise

From $0 = s_{1} + \dots + s_{n}$ , multiply through by $s_{1}^{- 1}$ to reach a statement about $- 1$ . Only available because $Σ (D)$ is a group — this is the step that fails in a general ring.

Move 3

Zorn plus maximality

Preorderings are closed under unions of chains, so a maximal one exists; $(17.10)$ identifies maximal preorderings with orderings. Existence proofs never construct an ordering explicitly.

Move 2 explains why $(18.2)$ can be phrased with $- 1$ rather than with $0$ . In a general ring the weak preordering $T (R)$ is closed under addition and permuted multiplication but not under division, and the passage from * $0$ is a sum* to * $- 1$ is a sum* is exactly the step that needs an inverse.

Worked Example

The rationals: one ordering, and $T (ℚ)$ is already it

Take $D = ℚ$ . Square-products are squares. Which positive rationals are sums of squares of rationals? Write $a / b$ with $a, b$ positive integers; then $a / b = a b / b^{2}$ , and by Lagrange's four-square theorem $a b = n_{1}^{2} + n_{2}^{2} + n_{3}^{2} + n_{4}^{2}$ , so

\frac{a}{b} = (\frac{n_{1}}{b})^{2} + (\frac{n_{2}}{b})^{2} + (\frac{n_{3}}{b})^{2} + (\frac{n_{4}}{b})^{2} .

(E.1)

Every positive rational is a sum of at most four rational squares; no negative rational is, since sums of squares are positive.

Hence $T (ℚ) = ℚ_{> 0}$ , which is already an index- $2$ subgroup of $ℚ^{*}$ and therefore, by the corollary above, an ordering. The weak preordering is the unique ordering, which is the group-theoretic form of the statement that $ℚ$ is orderable in exactly one way. Note $ℚ^{*} / T (ℚ) = {\pm 1}$ , of exponent $2$ as $(18.1)$ demands, even though $ℚ^{*} / ℚ^{* 2}$ is infinite.

The real quaternions: no ordering at all

Take $D = ℍ$ , the division ring of real quaternions. Then $i^{2} = - 1$ , so $- 1$ is a square, in particular a sum of square-products, and $(18.2)$ rules out any ordering of $ℍ$ .

- 1 = i^{2} \in Σ (ℍ) \subseteq T (ℍ) ⟹ 0 = 1 + i^{2} \in T (ℍ) .

(E.2)

This is instructive because $ℍ$ passes every necessary condition from the general theory: it is a domain of characteristic $0$ with no zero divisors. Orderability is strictly stronger than those conditions, and the obstruction is arithmetic, not structural.

Comparison and Classification

The general theory of $§ 17$ versus its division-ring form
Notion	General ring $R$	Division ring $D$
Preordering axiom	$per (a_{1}^{2} \dots a_{m}^{2} t_{1} \dots t_{n}) \in T$	$T \cdot T \subseteq T$ and $a^{2} \in T$
Algebraic type of $T$	additively closed subset	normal subgroup of $D^{*}$
Quotient	no group structure available	$D^{*} / T$ abelian of exponent $2$
Ordering	maximal preordering $(17.10)$	preordering of index $2$
Division closure $\bar{T}$	can be strictly larger than $T$	always $\bar{T} = T$
Orderability test	$0 \notin T (R)$	$- 1 \notin T (D)$
Basic positive elements	sums of $per (a_{1}^{2} \dots a_{m}^{2})$	sums of square-products

Formal reality of some standard division rings
	Formally real	Orderable	Commutative	$- 1$ a square
$ℚ$ , $ℝ$	yes	yes	yes	no
$ℂ$	no	no	yes	yes
$𝔽_{p}$	no	no	yes	partial
$ℍ$	no	no	no	yes
$ℚ ((y)) ((x; σ))$ , $σ (y) = 2 y$	yes	yes	no	no
$ℝ ((y)) ((x; σ))$ , $σ (y) = - y$	no	no	no	no

Formal reality of some standard division rings

The last row is the point of the whole section: a division ring in which $- 1$ is a square-product but is not a square, and not even a sum of squares.

Relationship Map

The logical layout of the section is a single chain of equivalences supported by one structural proposition.

- 1 \notin T (D)

D

formally real

D

has a preordering

D

has an ordering

$(18.1)$ Preordering $T \leq D^{*}$ — additively closed subgroup containing all squares
- contains
  - every square $a^{2}$
  - every square-product, i.e. $Σ (D)$
  - every commutator $a b a^{- 1} b^{- 1}$
  - every inverse $t^{- 1}$ of its own elements
- is contained in
  - every ordering $P \supseteq T$
  - $D^{*}$ , with elementary abelian quotient
- never contains
  - $0$
  - $- 1$ (else $0 = 1 + (- 1) \in T$ )

Downstream, $(18.1)$ is the reason the intersection theorem of Preorderings in Division Rings is so much simpler than its general-ring ancestor, and the normality of orderings is exactly what makes Albert's theorem work in Formally Real Division Rings.

Standards and Notation

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

Cone notation

P

for an ordering,

T

for a preordering (Lam, Serre);

Σ F^{2}

is common for the field case

Multiplicative group

D^{*}

here;

D^{\times}

in much of the arithmetic literature

Weak preordering

T (D)

(Lam);

Σ D^{2}

Σ D^{* 2}

elsewhere — the latter is unsafe in the noncommutative case

Square-productsNo universal symbol;

Σ (D)

is used on this page for the subgroup generated by squares

Order relation

a < b iff b - a \in P

; ISO 80000-2 symbols for

<

\leq

SoftwareRealField/ordering machinery in Sage and Magma covers fields only; no CAS implements orderings of noncommutative division rings

Failure Modes and Common Mistakes

Do not read $P \cup (- P) = D ∖ {0}$ as allowing $0 \in P$ ; the cone consists of strictly positive elements and $P \cap (- P) = \emptyset$ is forced.
Do not assume an ordering makes $D$ into an ordered group under addition only; the multiplicative compatibility $P \cdot P \subseteq P$ is a separate axiom and is what fails in most attempted constructions.
Do not confuse formally real with real closed: the first is the orderability criterion, the second requires in addition that every positive element is a square and every odd-degree polynomial has a root — a condition with no useful noncommutative analogue.
Do not expect uniqueness of the ordering. A formally real division ring generally has many, and the intersection of them all is the division closure of $T (D)$ .

Historical Notes and Lessons Learned

1903Hilbert's twisted Laurent seriesIn the second edition of the Grundlagen der Geometrie, Hilbert produces the first noncommutative ordered division ring, to separate the axioms of geometry — arithmetic of ends over a twisted series field.
1927Artin–SchreierA field is orderable exactly when $- 1$ is not a sum of squares. The theory of formally real fields is launched and, with it, Artin's solution of Hilbert's 17th problem.
1940Albert on ordered algebrasAlbert proves that the centre of an ordered division ring is algebraically closed in it, the first genuinely noncommutative theorem of the subject.
1948–49Mal'cev and NeumannThe Laurent series construction with well-ordered support generalises Hilbert's example to an arbitrary ordered group, producing ordered division rings in abundance.
early 1950sSzele, Pickert, R. E. JohnsonThe orderability criterion is extended from fields to division rings, and Johnson gives the version for arbitrary rings via preorderings — the form used here.
1983Scharlau–TschimmelThe level of a non-formally-real division ring is shown to take every positive integer value, in sharp contrast with Pfister's powers-of-two theorem for fields.

The methodological lesson is that the correct noncommutative generalisation was not found by weakening the field statement but by strengthening the object: replacing square by square-product, the smallest multiplicatively closed set that a cone is forced to contain. Once that substitution is made, the Artin–Schreier proof runs almost verbatim.

Quick Reference

Ordering

P \subseteq D^{*}

with

P + P \subseteq P

P \cdot P \subseteq P

P \cup (- P) = D ∖ {0}

Ordering, group formAdditively closed subgroup of index

2

D^{*}

Preordering $(18.1)$

T + T \subseteq T

T \cdot T \subseteq T

D^{* 2} \subseteq T

Automatic

T ⊴ D^{*}

[D^{*}, D^{*}] \subseteq T

D^{*} / T

of exponent

2

Square-product

a_{1}^{2} \dots a_{m}^{2}

; the group

Σ (D)

generated by squares

Weak preordering

T (D) =

sums of square-products; smallest preordering when

D

is formally real

Formally real

0 \notin T (D) iff - 1 \notin T (D)

Szele–Pickert $(18.2)$

D

orderable

iff

- 1

is not a sum of square-products

Which statement to quote
You want	Use	Reference
To verify a candidate cone	the three closure conditions	(18.1)
To promote a preordering to an ordering	maximality, via Zorn	(17.10)
To rule out any ordering	exhibit $- 1$ as a sum of square-products	(18.2)
To order a field	check $- 1$ is not a sum of squares	Artin–Schreier
To build an ordered division ring	Mal'cev–Neumann series over an ordered group	(18.5)

Frequently Asked Questions

Why does the definition of a preordering need permuted products at all?

Because in a noncommutative ring you cannot bring the two copies of $a$ together, and the axiom must be strong enough to force the cone to be closed under conjugation. In a division ring the identity $a u a = (a u)^{2} (u^{- 1})^{2} u$ does the bringing-together for you, so the permutations can be dropped — that is exactly $(18.1)$ . In a general ring they cannot.

Is every subgroup of

D^{*}

of index 2 an ordering?

No. Index $2$ gives axiom $(17.3)$ and multiplicative closure, but additive closure $P + P \subseteq P$ is an extra condition and usually fails. For example $ℚ^{*}$ has many index- $2$ subgroups — one for each subgroup of $ℚ^{*} / ℚ^{* 2}$ of index $2$ — and only $ℚ_{> 0}$ is additively closed.

Can a division ring of characteristic

p

be ordered?

Never. An ordering contains $1$ and is closed under addition, so it contains every $n \cdot 1$ with $n \geq 1$ ; if $p \cdot 1 = 0$ this would put $0$ in the cone. So $(17.4)$ forces characteristic $0$ , and orderable division rings all contain $ℚ$ in their centre.

How many orderings can a division ring have?

As many as there are maximal preorderings above $T (D)$ , and for formally real $D$ there is at least one. For $ℚ$ there is exactly one; for $ℚ (\sqrt{2})$ there are two; for the rational function field $ℚ (t)$ there are infinitely many. The set of all orderings carries a natural topology in the field case (the real spectrum), but no comparable theory exists for division rings.

Does

(18.2)

give an algorithm for deciding orderability?

Not in general. It converts orderability into a single arithmetic question, but deciding whether $- 1$ is a sum of square-products requires understanding the subgroup generated by squares, which for a presented division ring is not effectively computable. For a finitely generated field extension of $ℚ$ the question is decidable by real-algebraic methods.

Why is the quotient

D^{*} / T

of exponent 2 so useful?

It converts questions about orderings into questions about index- $2$ subgroups of an elementary abelian $2$ -group, that is, into linear algebra over $𝔽_{2}$ . Orderings containing a given preordering $T$ correspond to hyperplanes in $D^{*} / T$ that remain additively closed, which is how one counts orderings in practice.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §18, pp. 285–286.
E. Artin and O. Schreier, “Algebraische Konstruktion reeller Körper”, Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg 5 (1927), 85–99.
T. Szele, “On ordered skew fields”, Proceedings of the American Mathematical Society 3 (1952).
G. Pickert, Einführung in die höhere Algebra, Vandenhoeck & Ruprecht, Göttingen, 1951.
T. Y. Lam, The Algebraic Theory of Quadratic Forms, W. A. Benjamin, Reading, Massachusetts, 1973, Chapters 8 and 10.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964.

AI Suggested Questions

Give a full proof that a maximal preordering in a division ring is an ordering, following $(17.10)$ .
Which subgroups of index $2$ in $ℚ (\sqrt{2})^{*}$ are additively closed, and how do they correspond to the two orderings?
Construct a division ring in which the set of squares is not closed under multiplication.
How does the space of orderings of a formally real field relate to the real spectrum of its coordinate ring?
What replaces the Artin–Schreier theory of real closures for noncommutative division rings, if anything?
Show that the group $Σ (D)$ of square-products equals the smallest normal subgroup containing all squares, and compute it for the real quaternions.
Compare the orderability criterion $(18.2)$ with the criterion for a domain to be orderable in $(17.11)$ .

Ordered Division Rings

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Inverses do the work

Commutators are square-products

Key Results

Proof Techniques and Method

Sandwich elimination

Right-translate to normalise

Zorn plus maximality

Worked Example

The rationals: one ordering, and $T (ℚ)$ is already it

The real quaternions: no ordering at all

Comparison and Classification

Relationship Map

Standards and Notation

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Inverses do the work

Commutators are square-products

06Key Results

07Proof Techniques and Method

Sandwich elimination

Right-translate to normalise

Zorn plus maximality

08Worked Example

The rationals: one ordering, and T(ℚ) is already it

The real quaternions: no ordering at all

09Comparison and Classification

10Relationship Map

11Standards and Notation

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

Key Results

Proof Techniques and Method

Worked Example

The rationals: one ordering, and $T (ℚ)$ is already it

Comparison and Classification

Relationship Map

Standards and Notation

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions