How can uniqueness in (17.17) be reconciled with Q(2) having two orderings?

There is no conflict: Q(2) is not a ring of quotients of Q, because no nonzero rational multiple of 2 is rational. Condition (Q) is what removes the freedom, by forcing each new element's sign to be readable inside R.

Extending Orderings | KEVOS® Mathematics

Q: Does (17.17) need the Ore condition?

Not as stated — the hypothesis is only (Q), which is weaker and does not presuppose that R' is built from fractions. The Ore condition is how one usually *produces* a ring of quotients that happens to be a division ring; Goldie's theorem supplies it for Noetherian domains.

Q: Why does it matter that the proof avoids the identity element?

Because (17.19) applies to an ideal I of a domain, and a proper ideal is a ring without identity. Since the cone axioms never mention 1, orderings of such rings make sense, and the extension theorem transfers unchanged. This is exactly how Grätzer and Schmidt's result falls out.

Q: Is the extended ordering computable from the original one?

For a ring of quotients, yes in a useful sense: (17.18) reduces the sign of x to the sign of ax inside R, so any effective way of finding a and deciding signs in R gives an algorithm. For a general overring, (17.16) is only a criterion and the extension comes from Zorn's Lemma.

Q: Can an ordering extend in more than one way to a ring of quotients if the ring is noncommutative?

No. The uniqueness argument uses only that ax ∈ P with a ∈ P forces x positive in any compatible ordering, which is valid in any ring. Noncommutativity affects the two-sided bookkeeping in (17.18), not the conclusion.

Executive Summary

Restricting an ordering is free: if $P'$ is an ordering of $R'$ and $R \subseteq R'$ is a subring, then $P' \cap R$ is an ordering of $R$ . Extending is not. $(17.16)$ identifies the exact obstruction: an ordering $P$ of $R$ extends to $R'$ if and only if $0$ is not a sum of products in which the elements of $R'$ occur with even multiplicity and finitely many factors are drawn from $P$ .

For the most important enlargement — passing to a ring of quotients — the obstruction never occurs, and more is true: the extension is unique. That is the Albert–Neumann–Fuchs theorem $(17.17)$ , and since its proof never uses an identity element, it yields the Grätzer–Schmidt result $(17.19)$ that an ordering of a nonzero ideal extends uniquely to the whole domain.

Restrictionalways an ordering; no hypotheses

Extension criterion

0

not a sum of

per (a_{1}^{2} \dots a_{m}^{2} t_{1} \dots t_{n})

a_{i} \in R'

t_{j} \in P

Quotient ringsextension exists and is unique,

(17.17)

Idealsan ordering of

I \neq 0

extends uniquely to

R

(17.19)

Failure

ℚ \subseteq ℚ (i)

1 + i^{2} = 0

Overview

Let $R \subseteq R'$ be rings and let $P$ be an ordering of $R$ . To order $R'$ compatibly one must assign a sign to every new element while keeping the old assignments. The obstruction is visible already in $(17.4)$ : a sign assignment must make products with even multiplicities positive, so any identity in $R'$ writing $0$ as such a sum blocks the extension permanently.

0 = \sum_{k} per (a_{k 1}^{2} \dots a_{k m_{k}}^{2} t_{k 1} \dots t_{k n_{k}}), a_{k j} \in R' ∖ {0}, t_{k l} \in P .

(17.16a)

The obstruction. $(17.16)$ says its absence is also sufficient.

Once that is seen, the proof is three lines: if the generated set omits $0$ it is a preordering of $R'$ , Zorn plus $(17.10)$ turn it into an ordering, and its restriction to $R$ is an ordering containing $P$ , hence equal to $P$ because orderings are maximal preorderings.

For the special case of a ring of quotients, one does better than existence by abstract nonsense: an explicit formula for the extended cone, uniqueness, and no need for an identity element — the last point being what makes $(17.19)$ possible.

Learning Objectives

Show that $P' \cap R$ is always an ordering of $R$ when $P'$ is one of $R'$ .
State and prove the extension criterion $(17.16)$ .
Recognise the generated set as the preordering of $R'$ generated by $P$ .
Prove the two-sided description $(17.18)$ of the extended cone.
Prove existence and uniqueness of the extension to a ring of quotients $(17.17)$ .
Deduce $(17.19)$ and apply it to an ideal of a domain.

Definitions

Definition(17.17)Ring of quotients

Let $R \subseteq R'$ be domains. $R'$ is a ring of quotients of $R$ if for every $x \in R'$ there exist $a, b \in R ∖ {0}$ with

a x \in R and x b \in R .

(Q)

The definition is deliberately weak: it demands only that every element of $R'$ can be pushed into $R$ from each side, not that $R'$ consists of formal fractions. For a commutative domain the quotient field qualifies; for a two-sided Ore domain the division ring of fractions qualifies; and by $(17.19)$ a domain is a ring of quotients of any of its nonzero ideals.

Restricts / extends: $P'$ restricts to $P$ , and $P$ extends to $P'$ , when $P' \cap R = P$ .
Generated preordering: For $P$ an ordering of $R \subseteq R'$ , the set of sums of permuted products of doubled elements of $R'$ with factors from $P$ inserted. It is the smallest candidate preordering of $R'$ containing $P$ .
Ore condition: Any two nonzero elements have a common nonzero multiple on the given side; a Noetherian domain satisfies it on both sides by Goldie's theorem.
Ordering without identity: A cone satisfying the three axioms in a ring not assumed to have 1. All of (17.17) goes through in this setting.

Malcev's examples show that a noncommutative domain need not embed in any division ring, which is why the extension theorem is stated for rings of quotients rather than for fields of fractions.

Core Concepts

Restriction is automatic

If $P'$ is an ordering of $R'$ and $R \subseteq R'$ is a subring, then $P : = P' \cap R$ satisfies all three cone axioms: closure under sums and products is inherited, and for $0 \neq a \in R$ either $a$ or $- a$ lies in $P'$ , and both lie in $R$ . The same argument applies to preorderings. So the interesting direction is always upwards.

Extension is a properness question

Let $T'$ denote the set of all sums appearing in $(17.16 a)$ . It automatically satisfies $(17.5)$ and $(17.6)$ , and it contains $P$ (take $m = 0$ , $n = 1$ ). So the only question is whether $0 \in T'$ — and this is the same dichotomy that governs $T_{b}$ in $(17.8)$ and the division closure in $(17.13)$ .

Why a quotient ring never obstructs

For a ring of quotients the extension can be written down. Set

P' = {x \in R' : a x b \in P for some a, b \in P} .

(17.17a)

Sandwich $x$ between positives until it lands in $R$ , then read off its sign there.

The content of $(17.18)$ is that one-sided multiplication already suffices, which is what makes $P'$ closed under addition. The property $(Q)$ is used exactly once, to produce the element that pushes $x$ back into $R$ .

Key Results

Theorem(17.16)Extension criterion

Let $R \subseteq R'$ be rings and let $P$ be an ordering of $R$ . Then $P$ extends to an ordering of $R'$ if and only if, in $R'$ , $0$ is not a sum of elements of the form

per (a_{1}^{2} \dots a_{m}^{2} t_{1} \dots t_{n}), a_{1}, \dots, a_{m} \in R' ∖ {0}, t_{1}, \dots, t_{n} \in P .

(17.16)

Proof

Necessity. Suppose $P'$ is an ordering of $R'$ with $P' \cap R = P$ . Then $P \subseteq P'$ , and $P'$ is a preordering of $R'$ , so every displayed element lies in $P'$ by $(17.6)$ and every sum of them lies in $P'$ by $(17.5)$ . Since $0 \notin P'$ , no such sum is $0$ .

Sufficiency. Let $T'$ be the set of all such sums and assume $0 \notin T'$ . By construction $T'$ is closed under addition, and inserting doubled elements of $R'$ or elements of $T'$ into a permuted product again yields such a sum, so $T'$ satisfies $(17.6)$ . Hence $T'$ is a preordering of $R'$ , and it contains $P$ (take $m = 0$ , $n = 1$ ).

By Zorn's Lemma and $(17.10)$ , $T'$ is contained in an ordering $P'$ of $R'$ . Then $P \subseteq T' \cap R \subseteq P' \cap R$ , and $P' \cap R$ is an ordering of $R$ . Two orderings of $R$ with one contained in the other are equal: if $a \in (P' \cap R) ∖ P$ then $- a \in P \subseteq P' \cap R$ , giving $0 = a + (- a)$ in a cone. Hence $P' \cap R = P$ .

Theorem(17.17)Albert, Neumann, Fuchs

Let $R \subseteq R'$ be domains such that $R'$ is a ring of quotients of $R$ in the sense of $(Q)$ . Then every ordering $P$ of $R$ extends to an ordering $P'$ of $R'$ , and $P'$ is unique.

Proof

Define $P' = {x \in R' : a x b \in P for some a, b \in P}$ . We first prove the one-sided descriptions

(a) P' = {x \in R' : a x \in P for some a \in P}, (b) P' = {x \in R' : x b \in P for some b \in P} .

(17.18)

For (a), the inclusion $\supseteq$ is clear: if $a x \in P$ with $a \in P$ , pick any $b \in P$ ; then $(a x) b \in P \cdot P \subseteq P$ . For $\subseteq$ , let $a x b \in P$ with $a, b \in P$ . By $(Q)$ choose $c \in R ∖ {0}$ with $c (a x) \in R$ ; replacing $c$ by $- c$ if necessary we may take $c \in P$ . Then $c (a x b) \in P \cdot P \subseteq P$ , i.e. $(c a x) b \in P$ with $c a x \in R$ . If $c a x$ were $0$ this product would be $0 \notin P$ ; if $- c a x \in P$ then $- (c a x) b \in P$ , contradicting $P \cap (- P) = \emptyset$ . Hence $c a x \in P$ , and $c a \in P$ , which places $x$ in the right-hand set. Statement (b) is proved symmetrically, choosing $d \in P$ with $(x b) d \in R$ .

** $P'$ is an ordering.** For $x, y \in P'$ use (a) to get $a \in P$ with $a x \in P$ and (b) to get $b \in P$ with $y b \in P$ . Then

a (x + y) b = (a x) b + a (y b) \in P + P \subseteq P, a (x y) b = (a x) (y b) \in P \cdot P \subseteq P,

(17.18c)

so $x + y \in P'$ and $x y \in P'$ . For totality, let $0 \neq x \in R'$ and choose $a \in R ∖ {0}$ with $a x \in R$ , taking $a \in P$ after a sign change. Since $R'$ is a domain, $a x \neq 0$ , so $a x \in P$ or $a (- x) = - (a x) \in P$ ; correspondingly $x \in P'$ or $- x \in P'$ . Also $0 \notin P'$ , since $a \cdot 0 = 0 \notin P$ .

** $P'$ restricts to $P$ .** $P \subseteq P'$ by (a) with any $a \in P$ . Conversely if $x \in P' \cap R$ and $a x \in P$ with $a \in P$ , then $x \neq 0$ and $- x \in P$ would give $- (a x) = a (- x) \in P$ , impossible; so $x \in P$ .

Uniqueness. Let $Q$ be any ordering of $R'$ with $Q \cap R = P$ and let $x \in P'$ , say $a x \in P$ with $a \in P$ . Both $a$ and $a x$ lie in $P \subseteq Q$ . If $x \notin Q$ then $- x \in Q$ , so $a (- x) = - (a x) \in Q$ , contradicting $Q \cap (- Q) = \emptyset$ . Hence $P' \subseteq Q$ , and since both are orderings — maximal preorderings by $(17.10)$ — we get $P' = Q$ .

RemarkNo identity is used

The cone axioms never mention $1$ , and the proof above never multiplies by $1$ : every sandwich uses elements of $P$ itself. Consequently $(17.17)$ holds verbatim for rings possibly without identity, which is exactly what the next corollary needs.

Corollary(17.19)Grätzer–Schmidt

Let $R$ be a domain and $I \neq 0$ a two-sided ideal of $R$ . Then every ordering of $I$ , regarded as a ring possibly without identity, extends uniquely to an ordering of $R$ .

Proof. Fix $a \in I ∖ {0}$ . For any $x \in R$ we have $a x \in I$ and $x a \in I$ , so $R$ is a ring of quotients of $I$ in the sense of $(Q)$ , with the same element serving on both sides. Both $I$ and $R$ are domains. Apply the identity-free form of $(17.17)$ .

Corollary(17.17b)Ore domains and fraction fields

If $R$ is a commutative domain, every ordering of $R$ extends uniquely to its quotient field. If $R$ is a domain satisfying the Ore condition on both sides, with division ring of fractions $D$ , then every ordering of $R$ extends uniquely to $D$ : each $z \in D$ can be written $a b^{- 1}$ and $c^{- 1} d$ with $a, b, c, d \in R$ , $b, c \neq 0$ , so $z b \in R$ and $c z \in R$ , which is $(Q)$ .

Proof Techniques and Method

How these proofs work, and which move to reuse.

GenerateForm the smallest set in

R'

closed under the preordering axioms and containing

P

. No verification of totality is needed at this stage.

Test for zeroThe single obstruction is

0 \in T'

. This is the criterion of

(17.16)

, and it is a statement about identities in

R'

, not about orderings.

MaximiseZorn plus

(17.10)

turn the preordering into an ordering of

R'

Restrict and compareThe restriction is an ordering of

R

containing

P

; comparable orderings are equal, so nothing was lost.

Comparability of orderings — if $P \subseteq Q$ are orderings then $P = Q$ — is used in almost every proof on this page and is worth isolating: it is immediate from totality plus $0 \notin Q$ .

Worked Example

Polynomials to rational functions

Let $R = ℝ [x]$ with the ordering $P$ in which $f > 0$ means the leading coefficient of $f$ is positive, and let $R' = ℝ (x)$ . Every $z = f / g$ satisfies $z g \in R$ and $g z \in R$ , so $(Q)$ holds and $(17.17)$ applies. The extended cone is

P' = {\frac{f}{g} \neq 0 : lead (f) \cdot lead (g) > 0},

(E.1)

Obtained from $(17.18)$ (a): multiply $f / g$ by $g^{2} \in P$ to get $f g \in R$ , and read the sign there.

Concretely $\frac{x - 3}{- x^{2} + 1} \in - P'$ , since $lead (x - 3) = 1$ and $lead (- x^{2} + 1) = - 1$ . The resulting ordered field is nonarchimedean: $x > n$ for every integer $n$ , so $1 / x$ is infinitely small.

A noncommutative extension

Let $A_{1} (ℝ) = ℝ ⟨ x, y ⟩ / (x y - y x - 1)$ be the first Weyl algebra with the ordering of Ordered Rings and Positive Cones: $r > 0$ when the top $y$ -coefficient $r_{n} (x)$ has positive leading coefficient. $A_{1} (ℝ)$ is a Noetherian domain, hence Ore on both sides, so it has a division ring of fractions $D_{1} (ℝ)$ , and $(17.17 b)$ extends the ordering uniquely to $D_{1} (ℝ)$ .

Ordering an ideal, then the ring

Let $I = 2 ℤ$ , a ring without identity, ordered by $P_{I} = {2, 4, 6, \dots}$ . By $(17.19)$ this extends uniquely to $ℤ$ — necessarily to the usual ordering, since $n \in ℤ$ is positive exactly when $2 n \in P_{I}$ . The point of the corollary is that no compatibility had to be checked: the ideal already determines the ring's ordering.

A failure

Take $R = ℚ$ with its unique ordering and $R' = ℚ (i)$ . In $R'$ ,

0 = per (i^{2}) + 1 = i \cdot i + 1,

(E.2)

A sum of one permuted doubled product and one element of $P$ — precisely the obstruction in $(17.16)$ .

So the ordering does not extend, as expected. Note also that $ℚ (i)$ is not a ring of quotients of $ℚ$ : no nonzero rational multiple of $i$ is rational, so $(Q)$ fails and $(17.17)$ never applied.

Process and Workflow

You have an ordering $P$ of $R$ and an overring $R'$ . Does it extend?

$R'$ is a ring of quotients of $R$ Yes, uniquely, by

(17.17)

. The cone is given explicitly by

(17.18)

: multiply into

R

from either side and read the sign.

$R$ is a nonzero ideal of the domain $R'$ Yes, uniquely, by

(17.19)

— a special case of the previous branch, with the identity-free form of the theorem.

You can exhibit an identity as in $(17.16 a)$ No. The obstruction is permanent; no cleverness recovers an extension, since the identity would hold in any candidate ordering.

NeitherTest properness of the preordering generated by

P

R'

. If it omits

0

, extensions exist — possibly several, as for

ℚ \subseteq ℚ (\sqrt{2})

How many extensions?
Inclusion	Extensions of the ordering	Why
$ℤ \subseteq ℚ$	exactly one	$ℚ$ is a ring of quotients, $(17.17)$
$ℝ [x] \subseteq ℝ (x)$	exactly one	ring of quotients
$A_{1} (ℝ) \subseteq D_{1} (ℝ)$	exactly one	Ore domain, $(17.17 b)$
$2 ℤ \subseteq ℤ$	exactly one	ideal of a domain, $(17.19)$
$ℚ \subseteq ℚ (\sqrt{2})$	exactly two	criterion holds; genuine choice of the sign of $\sqrt{2}$
$ℝ \subseteq ℝ [x]$	infinitely many	one per cut of $ℝ$
$ℚ \subseteq ℚ (i)$	none	$1 + i^{2} = 0$ , criterion fails

Comparison and Classification

Behaviour of positivity structures under change of ring
	Restricts to a subring	Extends to an overring	Extends to a ring of quotients	Unique when it extends
Ordering	yes	partial	yes	partial
Preordering	yes	partial	yes	no
Formal reality	yes	no	yes	not applicable
Archimedean property	yes	no	no	not applicable

Behaviour of positivity structures under change of ring

"Part" means: governed by $(17.16)$ , and can fail. Uniqueness of an extension holds for rings of quotients and fails in general, as $ℚ \subseteq ℚ (\sqrt{2})$ shows.

Comparison with the commutative theory
Aspect	Commutative domain	Noncommutative domain
Field of fractions	always exists	may not exist (Malcev)
Correct hypothesis	quotient field	ring of quotients in the sense of $(Q)$
Extension of an ordering	unique, to the quotient field	unique, to any ring of quotients, $(17.17)$
Role of the identity	assumed throughout	never used; hence $(17.19)$
Sufficient condition in practice	any localisation at a multiplicative set	the two-sided Ore condition

Relationship Map

Extension problem for $(R, P) \subseteq R'$ — governed by one properness condition
- always solvable when
  - $R'$ is a ring of quotients of $R$ , $(17.17)$
  - $R$ is a nonzero ideal of the domain $R'$ , $(17.19)$
- solvable iff
  - the preordering of $R'$ generated by $P$ omits $0$ , $(17.16)$
- never solvable when
  - $R'$ has zero divisors or positive characteristic, $(17.4)$
  - $- 1$ becomes a sum of permuted doubled products in $R'$
- unique when
  - $(Q)$ holds: signs are forced by sandwiching into $R$

ordering of

I

ordering of

R

ordering of the Ore quotient

D

Read left to right, this is the standard route to an ordered division ring: order a small piece, extend to the ring by $(17.19)$ , then extend to the division ring of fractions by $(17.17 b)$ . Each step is unique, so the whole chain is.

Applications and Industry Use

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

Ring theory

Building ordered division rings

The standard supply of noncommutative ordered division rings comes from ordering an Ore domain — a Weyl algebra, a skew polynomial ring, a group algebra of an ordered group — and then extending along $(17.17)$ .

Symbolic computation

Skew polynomials and D-modules

Ore algebras such as $k [x] [\partial; d / d x]$ underlie holonomic function algorithms. Their fraction fields are exactly the rings of quotients of $(17.17)$ , and orderings extend to them without extra hypotheses.

Control theory

Linear time-varying systems

Skew polynomial rings model linear differential and difference operators with variable coefficients; localising at nonzero operators produces the transfer-function calculus, and any compatible ordering travels with it.

Ordered algebra

Detecting orderability of subrings

Because restriction is free, an ordering on a large ring certifies orderability of every subring at once. This is often the cheapest way to prove a small ring formally real.

Honestly stated: this page is a tool page. Its results are used to move an ordering to wherever it is needed, most often into a division ring where infinitely large and infinitely small elements can be inverted.

Design Considerations

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

Order the small ring. Because restriction is automatic and extension is not, define the cone on the largest ring you can order explicitly, then restrict.
Prefer rings of quotients. If the enlargement you need is a localisation satisfying $(Q)$ , the extension is free and unique; if not, be prepared to verify $(17.16)$ by hand.
Check the Ore condition early. Noetherian domains are Ore on both sides; free algebras of rank $\geq 2$ are not, and their fraction constructions behave differently.
Do not assume an identity. Working without $1$ costs nothing here and buys the ideal-to-ring extension $(17.19)$ .
Expect several extensions when the enlargement is algebraic. Adjoining a square root splits the ordering into as many extensions as there are compatible sign choices, or none at all.

Failure Modes and Common Mistakes

Do not expect the archimedean property to survive an extension: $ℚ$ is archimedean, $ℚ (x)$ with the leading-coefficient ordering is not.
Do not read $(17.16)$ as a decision procedure; it is an existence criterion, and the search for an obstructing identity is unbounded.
Do not forget that $R'$ must be a domain in $(17.17)$ ; the theorem says nothing about overrings with zero divisors, which cannot be ordered at all.
Do not confuse extending an ordering with extending a preordering — the latter may be possible when the former is not, and $(17.16)$ is about the ordering.

Quick Reference

Restriction

P' \cap R

is always an ordering of

R

(17.16)extend

\Leftrightarrow

0

is not a sum of

per (a_{1}^{2} \dots a_{m}^{2} t_{1} \dots t_{n})

t_{j} \in P

Ring of quotients

\forall x \in R'

\exists a, b \in R ∖ {0}

with

a x \in R

x b \in R

(17.17)extension exists and is unique for a ring of quotients

(17.18)

P' = {x : a x \in P, a \in P} = {x : x b \in P, b \in P}

(17.19)an ordering of a nonzero ideal

I

of a domain extends uniquely to

R

No identity neededthe cone axioms and the proof of

(17.17)

never use

1

Comparabilityorderings

P \subseteq Q

of the same ring are equal

Which theorem applies
Situation	Result	Outcome
Arbitrary overring	$(17.16)$	criterion; may fail
Ring of quotients	$(17.17)$	exists, unique
Two-sided Ore domain into $D$	$(17.17 b)$	exists, unique
Nonzero ideal into a domain	$(17.19)$	exists, unique
Subring	restriction	always an ordering

Frequently Asked Questions

Why is restriction of an ordering automatic but extension not?

Restriction only has to verify conditions on elements that already have signs; all three cone axioms are inherited. Extension has to invent signs for new elements, and those inventions must be consistent with every algebraic identity holding in the larger ring. A single identity of the shape in $(17.16 a)$ makes consistency impossible.

How can uniqueness in

(17.17)

be reconciled with

ℚ (\sqrt{2})

having two orderings?

There is no conflict: $ℚ (\sqrt{2})$ is not a ring of quotients of $ℚ$ , because no nonzero rational multiple of $\sqrt{2}$ is rational. Condition $(Q)$ is what removes the freedom, by forcing each new element's sign to be readable inside $R$ .

Does

(17.17)

need the Ore condition?

Not as stated — the hypothesis is only $(Q)$ , which is weaker and does not presuppose that $R'$ is built from fractions. The Ore condition is how one usually produces a ring of quotients that happens to be a division ring; Goldie's theorem supplies it for Noetherian domains.

Why does it matter that the proof avoids the identity element?

Because $(17.19)$ applies to an ideal $I$ of a domain, and a proper ideal is a ring without identity. Since the cone axioms never mention $1$ , orderings of such rings make sense, and the extension theorem transfers unchanged. This is exactly how Grätzer and Schmidt's result falls out.

Is the extended ordering computable from the original one?

For a ring of quotients, yes in a useful sense: $(17.18)$ reduces the sign of $x$ to the sign of $a x$ inside $R$ , so any effective way of finding $a$ and deciding signs in $R$ gives an algorithm. For a general overring, $(17.16)$ is only a criterion and the extension comes from Zorn's Lemma.

Can an ordering extend in more than one way to a ring of quotients if the ring is noncommutative?

No. The uniqueness argument uses only that $a x \in P$ with $a \in P$ forces $x$ positive in any compatible ordering, which is valid in any ring. Noncommutativity affects the two-sided bookkeeping in $(17.18)$ , not the conclusion.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §17, results (17.16)–(17.19), pp. 280–282.
A. A. Albert, On ordered algebras, Bulletin of the American Mathematical Society 46 (1940).
B. H. Neumann, On ordered division rings, Transactions of the American Mathematical Society 66 (1949). Orderings on division rings built from ordered groups.
L. Fuchs, Partially Ordered Algebraic Systems, Pergamon Press, 1963. Extension theorems for ordered rings and their quotient structures.
P. M. Cohn, Skew Fields: Theory of General Division Rings, Encyclopedia of Mathematics and its Applications 57, Cambridge University Press, 1995. Ore localisation and domains not embeddable in division rings.

AI Suggested Questions

For which non-Ore domains does an ordering extend to a universal field of fractions?
How does the extension criterion $(17.16)$ specialise to group rings of ordered groups over ordered coefficient rings?
What is the space of extensions of a fixed ordering along a finite algebraic extension of a formally real field?
How do orderings interact with Gabriel localisation and with rings of quotients in the sense of torsion theories?
Can the Grätzer–Schmidt corollary be extended to one-sided ideals, and if not, what fails?
Which skew polynomial rings used in control theory carry orderings, and do those orderings extend to their transfer-function fields?
Is there an effective bound on the size of an obstructing identity when an ordering fails to extend to a finitely generated overring?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Restriction is automatic

Extension is a properness question

Why a quotient ring never obstructs

06Key Results

07Proof Techniques and Method

08Worked Example

Polynomials to rational functions

A noncommutative extension

Ordering an ideal, then the ring

A failure

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Building ordered division rings

Skew polynomials and D-modules

Linear time-varying systems

Detecting orderability of subrings

13Design Considerations

14Failure Modes and Common Mistakes

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Design Considerations

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions