Preorderings in Division Rings and the Intersection Theorem

Q: Why can a preordering fail to be an intersection of orderings in a general ring?

Because an intersection of orderings is always division-closed. Fix an ordering P and suppose at ∈ P and t ∈ P but a P; then -a ∈ P, so (-a)t ∈ P P ⊆ P, and adding gives 0 = at + (-a)t ∈ P, which is impossible. So every ordering — hence every intersection of orderings — is division-closed, and a preordering with T ≠ T cannot be such an intersection. (17.14) says this is the only obstruction, and (18.3) observes that in a division ring it never arises.

Q: What exactly is the difference between T_b and the subgroup generated by T and b?

T_b is required to be closed under addition as well as multiplication, so it is the smallest candidate *preordering* containing T b, not merely the smallest subgroup. Over a division ring the two happen to have the same description T + bT once one notes b^2 ∈ T, but T_b can contain 0 and then fails to be a preordering at all, whereas a subgroup never contains 0.

Q: Does (18.4) say that totally positive elements form a group?

Yes, when char D ≠ 2: they equal T(D), which by (18.1) is a normal subgroup of D^* closed under addition. In particular the inverse of a totally positive element is totally positive, and so is any conjugate of one — a fact used in the proof of Albert's theorem.

Q: If D has exactly one ordering, does that make T(D) an ordering?

Yes. By (18.3) the weak preordering is the intersection of all orderings, so with a unique ordering P we get T(D) = P. Equivalently, uniqueness of the ordering is the statement that T(D) already has index 2 in D^*. This is what happens for Q and for R.

Q: Is there a version of the intersection theorem for orderings of a domain that is not a division ring?

There is: (17.13) holds for any ring, with T replaced by T. And by (17.17), an ordering of a domain extends uniquely to any ring of quotients, so if a domain embeds in a division ring of quotients one can often transfer the division-ring statement back. The catch is that a noncommutative domain need not have a ring of quotients at all.

Q: Where is the noncommutativity actually felt in these proofs?

In two places. The description T_b = T + bT = T + Tb needs the normality of T in D^*, which comes from (18.1) and ultimately from the commutator identity. And (18.4) delivers *square-products* rather than squares, because a^2b^2 need not be a square when ab ≠ ba.

Executive Summary

In a general ring, a preordering need not be the intersection of the orderings that contain it; the obstruction is that $T$ may fail to be division-closed, and one must pass to the division closure $\bar{T}$ . In a division ring the obstruction evaporates. A preordering is a subgroup of $D^{*}$ by $(18.1)$ , so $a t \in T$ with $t \in T$ gives $a = (a t) t^{- 1} \in T$ at once.

The result is a clean two-part theorem $(18.3)$ : every preordering of a division ring is the intersection of the orderings containing it, and the adjunction $T_{b}$ of a new element $b$ collapses to $T + b T$ , which is a preordering exactly when $- b \notin T$ . The corollary $(18.4)$ identifies the totally positive elements with the sums of square-products, provided $char D \neq 2$ .

Intersection theorem

T = ⋂ {P : P an ordering, P \supseteq T}

Adjunction

T_{b} = T + b T = T + T b

Extension test

T_{b}

is a preordering

iff b \notin - T

Totally positive

= T (D)

when

char D \neq 2

Overview

The classical model is Artin's theorem on formally real fields: a nonzero element of a formally real field $F$ is a sum of squares if and only if it is positive in every ordering of $F$ . Artin used exactly that statement in his solution of Hilbert's 17th problem. Generalising it has two separate difficulties, and division rings dispose of both.

Which preorderings are intersections of orderings? In a general ring the answer is the division-closed ones, by $(17.14)$ . In a division ring every preordering qualifies.
Which single elements can be forced positive? One adjoins $b$ to $T$ and asks whether the enlarged set is still a preordering. The general construction $T_{b}$ is a set of sums of complicated permuted products; over a division ring it is just $T + b T$ .

T_{b} = T + b T = T + T b, b \in D^{*},

(18.3a)

Even powers of $b$ are squares and are absorbed into $T$ ; odd powers leave one factor $b$ on either side, which may be moved because $T$ is normal in $D^{*}$ .

The price of the noncommutative setting appears in $(18.4)$ : the answer is sums of square-products, not sums of squares. That the two genuinely differ is the content of $(18.7)$ in Formally Real Division Rings.

Learning Objectives

Prove that a preordering of a division ring is division-closed and deduce $\bar{T} = T$ .
State $(18.3)$ in full and identify $(17.8)$ and $(17.13)$ – $(17.14)$ as its two inputs.
Compute $T_{b} = T + b T$ and characterise when it is a preordering.
Prove $(18.4)$ : for $char D \neq 2$ , totally positive $iff$ sum of square-products.
Explain the failure of $(18.4)$ over a nonperfect field of characteristic $2$ .
Apply the theorem to determine the orderings of $ℚ (\sqrt{2})$ and the totally positive elements of $ℍ$ .

Definitions

Definition(17.12)Division closure

For a preordering $T$ in a ring $R$ , the division closure is

\bar{T} = {a \in R : a t \in T for some t \in T},

(17.12)

and the same set is obtained using $t a \in T$ , or $a b^{2} \in T$ for some $b \neq 0$ , or $b^{2} a \in T$ . One always has $T \subseteq \bar{T}$ and $0 \notin \bar{T}$ . $T$ is division-closed when $\bar{T} = T$ .

Definition(17.8)The adjunction

T_{b}

For a preordering $T$ in a ring $R$ and $b \neq 0$ , $T_{b}$ denotes the set of all sums of elements $per (b^{i} a_{1}^{2} \dots a_{m}^{2} t_{1} \dots t_{n})$ with $a_{j} \in R ∖ {0}$ , $t_{k} \in T$ and $i, m, n \geq 0$ . It always satisfies the two preordering axioms except possibly $0 \notin T_{b}$ .

$T (D)$: The weak preordering: all sums of square-products $a_{1}^{2} \dots a_{m}^{2}$ with $a_{i} \in D^{*}$ .
Totally positive: $a \in D^{*}$ lying in every ordering of $D$ . If $D$ has no ordering the condition is vacuous and every $a$ qualifies.
$Σ (D)$: The group of square-products, that is, the subgroup of $D^{*}$ generated by all squares.
Maximal preordering: A preordering maximal under inclusion; by $(17.10)$ these are exactly the orderings.

Throughout $D$ is a division ring and $T \subseteq D^{*}$ a preordering; by $(18.1)$ this means a subgroup of $D^{*}$ containing all squares and closed under addition.

Core Concepts

Division closure is automatic

Let $T$ be a preordering of $D$ and suppose $a t \in T$ for some $t \in T$ . By $(18.1)$ , $T$ is a subgroup of $D^{*}$ , so $t^{- 1} \in T$ and $a = (a t) t^{- 1} \in T \cdot T \subseteq T$ . Hence $\bar{T} = T$ for every preordering, and the criterion of $(17.14)$ — *division-closed $iff$ an intersection of orderings* — is satisfied by every preordering of a division ring.

This is the point at which division rings genuinely simplify the general theory. In a commutative ring the weak preordering $T (R)$ need not be division-closed; Exercise 17.8 exhibits $R = ℝ [x_{1}, \dots, x_{n}, y, z] / (x_{1}^{2} + \dots + x_{n}^{2} - y z)$ with $T (R) ⊊ \bar{T (R)}$ .

Why $T_{b}$ collapses to $T + b T$

Each generator of $T_{b}$ is a permuted product of $i$ copies of $b$ , paired squares, and elements of $T$ . Sandwich elimination — the identity $a u a = (a u)^{2} (u^{- 1})^{2} u$ from Ordered Division Rings — converts the paired squares into factors lying in $T$ . If $i$ is even, $b^{i} = (b^{i / 2})^{2}$ is itself a square and the whole generator lies in $T$ ; if $i$ is odd, one factor $b$ survives and the generator lies in $b T$ (equivalently in $T b$ , since $T$ is normal in $D^{*}$ ). Adding up, $T_{b} = T + b T$ .

The obstruction is a single equation

By $(17.8)$ , $T_{b}$ fails to be a preordering precisely when there is a relation $t' + b t = 0$ with $t, t' \in T$ . In a division ring that reads $b = - t' t^{- 1}$ , and since $T$ is a group the elements $- t' t^{- 1}$ run over exactly $- T$ . So the failure condition is the membership $b \in - T$ , and nothing more.

Key Results

Theorem(18.3)Intersection theorem for division rings

Let $T$ be any preordering in a division ring $D$ . Then:

$T$ is the intersection of all the orderings of $D$ containing $T$ ;
for $b \in D^{*}$ , the set $T_{b} = T + b T$ is a preordering of $D$ if and only if $b \notin - T$ .

Proof

(2). By $(17.8)$ , $T_{b}$ is a preordering unless there is an equation $t' + b t = 0$ with $t, t' \in T$ ; the set $T_{b}$ satisfies the closure axioms in all cases, so the only possible failure is $0 \in T_{b}$ . Suppose such an equation holds. Then $b = - t' t^{- 1}$ , and $t' t^{- 1} \in T$ because $T$ is a subgroup of $D^{*}$ by $(18.1)$ ; hence $b \in - T$ . Conversely if $b = - s$ with $s \in T$ , then $s + b \cdot 1 = 0$ with $s, 1 \in T$ , so $0 \in T_{b}$ and $T_{b}$ is not a preordering.

(1). Write $T'$ for the intersection of all orderings $P \supseteq T$ ; at least one such $P$ exists, since $T$ extends to a maximal preordering by Zorn's Lemma and maximal preorderings are orderings by $(17.10)$ . Clearly $T \subseteq T'$ . For the reverse inclusion, let $a \in D^{*}$ with $a \notin T$ . Then $- (- a) = a \notin T$ , so by part (2) applied to $b = - a$ the set $T_{- a} = T + (- a) T$ is a preordering. Enlarge it to an ordering $P$ ; then $P \supseteq T$ and $- a = (- a) \cdot 1 \in T_{- a} \subseteq P$ , so $a \notin P$ because $P \cap (- P) = \emptyset$ . Hence $a \notin T'$ .

This is the specialisation of $(17.13)$ : the division closure of $T$ equals the intersection of the orderings above $T$ , and here $\bar{T} = T$ because a preordering of a division ring is a subgroup.

Corollary(18.4)Totally positive elements

Let $D$ be a division ring with $char D \neq 2$ . An element $a \in D^{*}$ is totally positive — that is, positive with respect to every ordering of $D$ — if and only if $a$ is a sum of square-products.

Proof

**Case 1: $D$ is formally real.** Then $T (D)$ , the set of sums of square-products, is a preordering, and it is contained in every preordering, hence in every ordering. So the orderings containing $T (D)$ are all the orderings of $D$ , and $(18.3) (1)$ gives $T (D) = ⋂_{P} P$ , which is precisely the set of totally positive elements.

**Case 2: $D$ is not formally real.** Then $D$ has no ordering at all, by $(18.2)$ , and every $a \in D^{*}$ is vacuously totally positive; we must show every $a \in D^{*}$ is a sum of square-products. Since $char D \neq 2$ we may divide by $2$ , and since $1$ is central the two elements $(1 \pm a) / 2$ commute with each other, so

a = (\frac{1 + a}{2})^{2} - (\frac{1 - a}{2})^{2} = u^{2} + (- 1) v^{2}, u = \frac{1 + a}{2}, v = \frac{1 - a}{2} .

(18.4a)

By $(18.2)$ , $- 1$ is a sum of square-products, say $- 1 = s_{1} + \dots + s_{k}$ . If $v \neq 0$ then $(- 1) v^{2} = s_{1} v^{2} + \dots + s_{k} v^{2}$ , and each $s_{j} v^{2}$ is again a square-product; if $v = 0$ the term is absent. Likewise $u^{2}$ is a square-product when $u \neq 0$ and absent otherwise ( $u$ and $v$ cannot both vanish, since $u + v = 1$ ). So $a$ is a sum of square-products.

Counterexample—Characteristic

2

is genuinely excluded

Let $F = 𝔽_{2} (s)$ , a nonperfect field of characteristic $2$ . No ordering exists ( $(17.4)$ forces characteristic $0$ ), so every element of $F^{*}$ is vacuously totally positive. But in characteristic $2$ the Frobenius identity $(x + y)^{2} = x^{2} + y^{2}$ makes the set of sums of squares equal to the set of squares $F^{2} = 𝔽_{2} (s^{2})$ , and $s \notin F^{2}$ . So $s$ is totally positive but not a sum of square-products, and $(18.4)$ fails.

Corollary—Artin's theorem recovered

For a field $F$ of characteristic $\neq 2$ , a nonzero $a \in F$ is positive in every ordering of $F$ if and only if $a$ is a sum of squares. This is $(18.4)$ together with the fact that square-products in a commutative ring are squares.

Proof Techniques and Method

How these proofs work, and which move to reuse.

The pattern is Artin–Schreier's, executed three times in this section: to show an element is not in $T$ , build an ordering that excludes it.

Take the element you want to excludeGiven

a \notin T

, the goal is an ordering

P \supseteq T

with

a \notin P

Adjoin its negativeForm

T_{- a} = T + (- a) T

. Part (2) of

(18.3)

says this is a preordering as soon as

a \notin T

Maximalise with ZornA union of a chain of preorderings is a preordering, so a maximal one exists above

T_{- a}

Invoke maximality

(17.10)

: maximal preorderings are orderings. The resulting

P

contains

- a

, hence excludes

a

Two features of the division-ring case make the argument shorter than in $§ 17$ . Membership tests reduce to group membership because $T \leq D^{*}$ ; and the obstruction equation $t' + b t = 0$ can be solved for $b$ , which is impossible in a ring with non-invertible elements.

Worked Example

A commutative case with two orderings

Take $F = ℚ (\sqrt{2})$ and $T = T (F)$ , the sums of squares. The two field embeddings $ι_{\pm} : F \to ℝ$ with $\sqrt{2} \mapsto \pm 1.414 \dots$ give two orderings $P_{\pm}$ , and $(18.3)$ says $T = P_{+} \cap P_{-}$ .

Testing elements of $ℚ (\sqrt{2})$
Element $b$	$ι_{+} (b)$	$ι_{-} (b)$	$b \in T$ ?	$T_{b}$ a preordering?
$3 + 2 \sqrt{2} = (1 + \sqrt{2})^{2}$	$5.83 > 0$	$0.17 > 0$	yes	yes (equals $T$ )
$1 + \sqrt{2}$	$2.41 > 0$	$- 0.41 < 0$	no	yes; forces $P_{+}$
$- 1 - \sqrt{2}$	$- 2.41 < 0$	$0.41 > 0$	no	yes; forces $P_{-}$
$- 3 - 2 \sqrt{2}$	$- 5.83 < 0$	$- 0.17 < 0$	no	no: it lies in $- T$

The last row is the test of $(18.3) (2)$ in action: $- b = 3 + 2 \sqrt{2} = (1 + \sqrt{2})^{2} \in T$ , so $b \in - T$ and $T_{b}$ contains $0$ . Indeed $(1 + \sqrt{2})^{2} + b \cdot 1 = 0$ is exactly the forbidden relation $t' + b t = 0$ .

A noncommutative case where the corollary is vacuous but computable

Take $D = ℍ$ . It is not formally real ( $- 1 = i^{2}$ ), so every $q \in ℍ^{*}$ is vacuously totally positive, and Case 2 of $(18.4)$ must exhibit each $q$ as a sum of square-products. Follow the proof with $q = j$ :

u = \frac{1 + j}{2}, v = \frac{1 - j}{2}, u^{2} = \frac{1 + 2 j + j^{2}}{4} = \frac{j}{2}, v^{2} = \frac{1 - 2 j + j^{2}}{4} = - \frac{j}{2},

(E.1)

Using $j^{2} = - 1$ ; note $u$ and $v$ commute with each other and with $j$ .

j = u^{2} - v^{2} = u^{2} + i^{2} v^{2} = \frac{j}{2} + \frac{j}{2},

(E.2)

$u^{2}$ is a square and $i^{2} v^{2}$ is a product of two squares, so $j$ is a sum of two square-products — as $(18.4)$ predicts.

Process and Workflow

You are given a preordering $T$ of a division ring $D$ and an element $b \in D^{*}$ . Can $b$ be made positive?

$b \in T$ Already positive in every ordering above

T

. Nothing to do;

T_{b} = T

$b \notin T$ and $- b \notin T$ Both

T_{b}

and

T_{- b}

are preorderings. There are orderings above

T

making

b

positive and others making it negative —

b

is not decided by

T

$- b \in T$

b \in - T

, so

T_{b}

contains

0

and is not a preordering. Every ordering above

T

makes

b

negative.

The trichotomy is exhaustive because $T \cap (- T) = \emptyset$ , and it is exactly the statement that $T$ is the intersection of the orderings above it: an element is undecided by $T$ precisely when some ordering above $T$ puts it on each side.

Start from $T (D)$ Sums of square-products. If

0 \in T (D)

, stop:

D

has no ordering.

List undecided elementsThose

b

with

b \notin T

and

- b \notin T

. Each is a branch point in the ordering space.

Adjoin one at a timeReplace

T

T + b T

and repeat. Every choice sequence converges, by Zorn, to a maximal preordering.

Read off the orderingsMaximal preorderings are the orderings, and

T

is recovered as the intersection of all of them.

Comparison and Classification

What survives from the general theory of $§ 17$
	General ring	Commutative domain	Field	Division ring
Every preordering is division-closed	no	no	yes	yes
$T = ⋂$ orderings above $T$	partial	partial	yes	yes
$T_{b}$ has the simple form $T + b T$	no	no	yes	yes
Totally positive $=$ sums of squares	no	no	yes	no
Totally positive $=$ sums of square-products	no	no	yes	yes

What survives from the general theory of $§ 17$

The bottom two rows are the same statement for fields, because there a square-product is a square. The columns assume characteristic $\neq 2$ where the last two rows are concerned.

Statements and where they come from
Statement	General ring version	Division-ring version
Intersection of orderings	$\bar{T} = ⋂_{P \supseteq T} P$ $(17.13)$	$T = ⋂_{P \supseteq T} P$ $(18.3)$
Which $T$ are intersections	the division-closed ones $(17.14)$	all of them
Totally positive	$\exists b \neq 0$ with $a b^{2} \in T (R)$ $(17.15)$	$a \in T (D)$ $(18.4)$
Failure of $T_{b}$	$\exists t, t' \in T$ with $t' + b t = 0$ $(17.8)$	$b \in - T$

Relationship Map

The section is a two-step deduction: a structural fact about $T$ feeds a general theorem, and the general theorem yields the arithmetic corollary.

(18.1)

T \leq D^{*}

T

division-closed

(17.14)

(18.3)

T = ⋂ P

(18.4)

: totally positive

= T (D)

All subsets of $D^{*}$ no constraints

Subgroups containing $Σ (D)$ closed under multiplication and inverses

Preorderings…and closed under addition; each equals the intersection of the orderings above it

Orderings…and of index

2

D^{*}

; these are exactly the maximal preorderings

The bands classify *subsets of $D^{*}$ *, each class contained in the one outside it. The weak preordering $T (D)$ is the smallest member of the third band, and by $(18.3)$ it is the intersection of every member of the fourth.

Standards and Notation

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

Division closure

\bar{T}

(Lam); some authors write

T^{*}

\sqrt{T}

— the last clashes with the real radical of a preordering

Adjunction

T_{b}

for the preordering generated by

T

and

b

;

T [b]

appears in the quadratic-forms literature

Totally positive

a ≫ 0

in number theory;

a \in ⋂_{P} P

here

Space of orderings

X_{F}

Sper F

for fields; there is no standard notation for division rings

MarkupPresentation MathML per ISO/IEC 40314; relation symbols per ISO 80000-2

SoftwareSage and Magma expose real embeddings and totally positive tests for number fields; nothing analogous exists for division rings

Failure Modes and Common Mistakes

Do not carry $(18.3)$ back to general rings. There it holds only after replacing $T$ by its division closure $\bar{T}$ , and the gap is genuine — Exercise 17.8 gives a commutative example.
Do not confuse * $T_{b}$ is a preordering* with * $b$ is positive in some ordering above $T$ *: the two are equivalent here, but only because Zorn's Lemma then produces the ordering. The construction never exhibits it.
Do not assume the intersection of two orderings is an ordering. It is a preordering, generally of index $4$ or more, and the difference is exactly what $(18.3)$ measures.
Do not use $T + b T$ for a general ring $R$ ; the correct object there is the much larger set of sums of permuted products $per (b^{i} a_{1}^{2} \dots a_{m}^{2} t_{1} \dots t_{n})$ .

Quick Reference

Setting

D

a division ring,

T \subseteq D^{*}

a preordering, hence a subgroup by

(18.1)

Division closure

\bar{T} = T

always

Intersection theorem

T = ⋂ {P : P \supseteq T an ordering}

Adjunction

T_{b} = T + b T = T + T b

Test

T_{b}

is a preordering

iff b \notin - T

Totally positive

char D \neq 2

: totally positive

iff

sum of square-products

Field caseArtin: totally positive

iff

sum of squares

Excluded

char D = 2

; nonperfect fields break

(18.4)

Trichotomy for an element $b \in D^{*}$ against a preordering $T$
Condition	Meaning	Consequence
$b \in T$	forced positive	$b \in P$ for every ordering $P \supseteq T$
$b \in - T$	forced negative	$T_{b}$ is not a preordering
$b \notin T \cup (- T)$	undecided	orderings above $T$ exist on both sides

Frequently Asked Questions

Why can a preordering fail to be an intersection of orderings in a general ring?

Because an intersection of orderings is always division-closed. Fix an ordering $P$ and suppose $a t \in P$ and $t \in P$ but $a \notin P$ ; then $- a \in P$ , so $(- a) t \in P \cdot P \subseteq P$ , and adding gives $0 = a t + (- a) t \in P$ , which is impossible. So every ordering — hence every intersection of orderings — is division-closed, and a preordering with $\bar{T} \neq T$ cannot be such an intersection. $(17.14)$ says this is the only obstruction, and $(18.3)$ observes that in a division ring it never arises.

What exactly is the difference between

T_{b}

and the subgroup generated by

T

and

b

$T_{b}$ is required to be closed under addition as well as multiplication, so it is the smallest candidate preordering containing $T \cup {b}$ , not merely the smallest subgroup. Over a division ring the two happen to have the same description $T + b T$ once one notes $b^{2} \in T$ , but $T_{b}$ can contain $0$ and then fails to be a preordering at all, whereas a subgroup never contains $0$ .

Does

(18.4)

say that totally positive elements form a group?

Yes, when $char D \neq 2$ : they equal $T (D)$ , which by $(18.1)$ is a normal subgroup of $D^{*}$ closed under addition. In particular the inverse of a totally positive element is totally positive, and so is any conjugate of one — a fact used in the proof of Albert's theorem.

D

has exactly one ordering, does that make

T (D)

an ordering?

Yes. By $(18.3)$ the weak preordering is the intersection of all orderings, so with a unique ordering $P$ we get $T (D) = P$ . Equivalently, uniqueness of the ordering is the statement that $T (D)$ already has index $2$ in $D^{*}$ . This is what happens for $ℚ$ and for $ℝ$ .

Is there a version of the intersection theorem for orderings of a domain that is not a division ring?

There is: $(17.13)$ holds for any ring, with $T$ replaced by $\bar{T}$ . And by $(17.17)$ , an ordering of a domain extends uniquely to any ring of quotients, so if a domain embeds in a division ring of quotients one can often transfer the division-ring statement back. The catch is that a noncommutative domain need not have a ring of quotients at all.

Where is the noncommutativity actually felt in these proofs?

In two places. The description $T_{b} = T + b T = T + T b$ needs the normality of $T$ in $D^{*}$ , which comes from $(18.1)$ and ultimately from the commutator identity. And $(18.4)$ delivers square-products rather than squares, because $a^{2} b^{2}$ need not be a square when $a b \neq b a$ .

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §17–§18, pp. 281–288.
E. Artin, “Über die Zerlegung definiter Funktionen in Quadrate”, Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg 5 (1927), 100–115.
T. Y. Lam, The Algebraic Theory of Quadratic Forms, W. A. Benjamin, Reading, Massachusetts, 1973, Chapter 8.
T. Y. Lam, Orderings, Valuations and Quadratic Forms, CBMS Regional Conference Series in Mathematics 52, American Mathematical Society, 1983.
R. E. Johnson, “On ordered domains of integrity”, Proceedings of the American Mathematical Society 3 (1952).
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988.

AI Suggested Questions

Prove $(17.8)$ in full and identify precisely where invertibility is used to reach the form $b \in - T$ .
Give a commutative example where the weak preordering is not division-closed, and compute its division closure.
How many orderings does the rational function field $ℚ (t)$ have, and what is the intersection of all of them?
Is the set of totally positive elements of a formally real division ring ever of finite index in $D^{*}$ ?
Work out the analogue of $(18.3)$ for an ordered domain with a two-sided ring of quotients.
Explain how Artin's solution to Hilbert's 17th problem uses the field case of $(18.4)$ .
Does the space of orderings of a formally real division ring carry a useful topology, as the real spectrum does for fields?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Division closure is automatic

Why Tb collapses to T+bT

The obstruction is a single equation

06Key Results

07Proof Techniques and Method

08Worked Example

A commutative case with two orderings

A noncommutative case where the corollary is vacuous but computable

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Standards and Notation

13Failure Modes and Common Mistakes

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Why $T_{b}$ collapses to $T + b T$

Key Results

Proof Techniques and Method

Worked Example

Process and Workflow

Comparison and Classification

Relationship Map

Standards and Notation

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions