Preorderings in Rings

Q: Where exactly is Zorn's Lemma used, and can it be avoided?

It is used once, in (17.10a), to produce a maximal preordering above a given one. It cannot be avoided in general: the existence of an ordering on an arbitrary formally real field already needs a choice principle, and even for countable fields the resulting cone need not be computable. For specific rings one always constructs the ordering explicitly instead.

Q: Does (17.10) mean every ordering arises as a maximal preordering containing T(R)?

Yes. Every preordering contains T(R), so in particular every ordering does, and orderings are maximal among preorderings. Hence the orderings of R are exactly the maximal elements of the poset of preorderings containing T(R).

Q: Can a ring have a preordering but no ordering?

No — that is exactly the content of (17.10a) together with (17.10). Having a preordering, being formally real, and having an ordering are all equivalent for a nonzero ring; the equivalence is R. E. Johnson's theorem (17.11).

Executive Summary

Orderings are hard to construct directly: axiom $(17.3)$ demands a decision about the sign of every nonzero element at once. A preordering keeps the two easy closure conditions and drops the totality requirement. Preorderings are then cheap — every formally real ring has one — and the payoff is $(17.10)$ : a preordering is an ordering exactly when it is maximal.

Since the union of a chain of preorderings is a preordering, Zorn's Lemma converts any preordering into an ordering. That single move reduces the orderability problem to a finitary condition on the ring, and is the engine behind R. E. Johnson's theorem and the Artin–Schreier theorem it generalises.

Axioms

T + T \subseteq T

and

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

Excluded

0 \notin T

by definition

Forced

1 \in T

;

R

a domain,

char R = 0

Key theoremordering

=

maximal preordering,

(17.10)

Toolthe auxiliary set

T_{b}

(17.8)

Overview

Fix a ring $R$ . A preordering is a subset $T \subseteq R ∖ {0}$ satisfying two conditions:

(17.5) T + T \subseteq T, (17.6) per (a_{1}^{2} \dots a_{m}^{2} t_{1} \dots t_{n}) \in T

(17.5)–(17.6)

for all $a_{1}, \dots, a_{m} \in R ∖ {0}$ , all $t_{1}, \dots, t_{n} \in T$ and all $m, n \geq 0$ .

Here $per (a_{1}^{i_{1}} \dots a_{m}^{i_{m}})$ denotes any product of the $i_{1} + \dots + i_{m}$ listed factors, arranged in an arbitrary order. Condition $(17.6)$ therefore packages three separate facts at once: $T$ is closed under multiplication, $T$ contains every permuted product of doubled elements, and $T$ is closed under sandwiching, $a t a \in T$ .

Every ordering is a preordering, because the sign map of Ordered Rings and Positive Cones is multiplicative and assigns $+ 1$ to any product with all multiplicities even. Intersections of orderings are preorderings too — and Division-Closed Preorderings identifies precisely which preorderings arise that way.

Learning Objectives

Read and manipulate the notation $per (a_{1}^{i_{1}} \dots a_{m}^{i_{m}})$ .
Prove $(17.7)$ : a preordering forces $R$ to be a domain of characteristic $0$ with $1 \in T$ .
Construct $T_{b}$ and prove the trichotomy of $(17.8)$ .
Prove both directions of $(17.10)$ : ordering if and only if maximal preordering.
Verify that a union of a chain of preorderings is a preordering, and apply Zorn's Lemma.
Identify the preordering $T (ℝ [x])$ and two distinct orderings extending it.

Definitions

Definition(17.5)–(17.6)Preordering

$T \subseteq R ∖ {0}$ is a preordering if it is closed under addition and if, for all $a_{1}, \dots, a_{m} \in R ∖ {0}$ and $t_{1}, \dots, t_{n} \in T$ (with $m, n \geq 0$ ), every product of the list

a_{1}, a_{1}, a_{2}, a_{2}, \dots, a_{m}, a_{m}, t_{1}, \dots, t_{n}

(17.6)

taken in any order, lies in $T$ . Taking $m = 0$ shows $T \cdot T \subseteq T$ ; taking $n = 0$ shows every permuted doubled product lies in $T$ ; taking $m = n = 1$ gives $a t a \in T$ .

Definition(17.8)The auxiliary set

T_{b}

Let $T$ be a preordering and $b \in R ∖ {0}$ . Write $T_{b}$ for the set of all finite sums of elements

per (b^{i} a_{1}^{2} \dots a_{m}^{2} t_{1} \dots t_{n}), a_{j} \in R ∖ {0}, t_{k} \in T, i, m, n \geq 0 .

(T-b)

$T_{b}$ is the smallest candidate preordering containing both $T$ and $b$ : take $i = 0, m = 0, n = 1$ to get $T \subseteq T_{b}$ , and $i = 1, m = n = 0$ to get $b \in T_{b}$ .

$T (R)$: The weak preordering: all sums of permuted doubled products, with no elements of $T$ inserted. It is contained in every preordering of $R$ .
Proper: A subset of $R ∖ {0}$ closed under the operations is a preordering precisely when it does not contain $0$ ; failure of properness is the only way the definition can break.
Sandwiching: Replacing $t$ by $a t a$ . This is the noncommutative substitute for multiplying by a square, and is sign-preserving because $a$ occurs twice.

$T_{b}$ satisfies $(17.5)$ and $(17.6)$ by construction. The only question is whether $0 \in T_{b}$ — and that is precisely what $(17.8)$ answers.

Core Concepts

Parity is the invariant

Every argument in this section counts how many times a distinguished element $b$ occurs in a product. If $b$ occurs an even number of times, the product already lies in $T$ ; if odd, then multiplying once more by $b$ puts it in $T$ . Written out:

$i$ even: $per (b^{i} a_{1}^{2} \dots a_{m}^{2} t_{1} \dots t_{n}) \in T$ , since $b$ contributes $i / 2$ doubled factors.
$i$ odd: $b \cdot per (b^{i} a_{1}^{2} \dots a_{m}^{2} t_{1} \dots t_{n}) \in T$ , since the extra $b$ makes the count even.

So an element of $T_{b}$ splits as $t + r$ with $t \in T \cup {0}$ collecting the even terms and $r$ collecting the odd ones, where $b r \in T \cup {0}$ . That decomposition is the whole content of $(17.8)$ .

Sums of preorderings, and chains

Preorderings are closed under intersection (of a nonempty family) and under unions of chains. Chains are the important case: any instance of $(17.5)$ or $(17.6)$ involves finitely many elements, and in a chain those finitely many all lie in a single member. Hence the union satisfies both axioms, and it omits $0$ because each member does.

What maximality buys

If $T$ is maximal and $b \notin T$ , then $T_{b}$ cannot be a preordering: it contains $T$ and $b$ , so it would properly contain $T$ . This is a very strong tool, because $(17.8)$ converts " $T_{b}$ is not a preordering" into an explicit algebraic identity $t' + b t = 0$ with $t, t' \in T$ . Maximality is thus traded for equations.

Key Results

Proposition(17.7)What a preordering forces

Let $T$ be a preordering on a ring $R \neq 0$ . Then $T \cap (- T) = \emptyset$ , $1 \in T$ , and $R$ is a domain of characteristic zero.

Proof

If $a \in T \cap (- T)$ then $a$ and $- a$ both lie in $T$ , so $0 = a + (- a) \in T + T \subseteq T$ , contradicting $T \subseteq R ∖ {0}$ .

Taking $m = 1$ , $a_{1} = 1$ , $n = 0$ in $(17.6)$ gives $1 = 1 \cdot 1 \in T$ . Then $n \cdot 1 = 1 + \dots + 1 \in T$ by $(17.5)$ , and since $0 \notin T$ we get $n \cdot 1 \neq 0$ for all $n \geq 1$ ; hence $char R = 0$ .

Finally suppose $b c = 0$ with $b, c \in R ∖ {0}$ . The arrangement $b \cdot b \cdot c \cdot c$ is a legitimate instance of $per (b^{2} c^{2})$ , so it lies in $T$ . But $b b c c = b (b c) c = 0$ , giving $0 \in T$ — impossible. Hence $R$ is a domain.

Remark

The domain argument is a genuine improvement on the corresponding step for orderings: it uses only one admissible arrangement of the factors, and needs no sign discussion. It is also the first place where permuting the factors of $per$ is doing real work.

Lemma(17.8)When

T_{b}

fails

Let $T$ be a preordering on $R$ and $b \in R ∖ {0}$ . The following are equivalent:

$T_{b}$ is not a preordering in $R$ ;
there exist $t, t' \in T$ with $t' + b t = 0$ ;
there exist $t', t'' \in T$ with $t'' + t' b = 0$ .

Proof

**(2) $\Rightarrow$ (3).** Multiply $t' + b t = 0$ on the right by $b$ : $t' b + b t b = 0$ . Since $b t b = per (b^{2} t) \in T$ , statement (3) holds with $t'' : = b t b$ .

**(3) $\Rightarrow$ (2).** Multiply $t'' + t' b = 0$ on the left by $b$ : $b t'' + b t' b = 0$ , and $b t' b \in T$ , so (2) holds with the roles $t : = t''$ and the new $t' : = b t' b$ .

**(2) $\Rightarrow$ (1).** Both $t'$ and $b t$ are among the generators of $T_{b}$ (take $i = 0$ and $i = 1$ respectively), so $0 = t' + b t \in T_{b}$ . A preordering never contains $0$ .

**(1) $\Rightarrow$ (2).** $T_{b}$ satisfies $(17.5)$ and $(17.6)$ by construction, so if it is not a preordering the only possible failure is $0 \in T_{b}$ : there is an identity

0 = \sum_{k} per (b^{i_{k}} a_{k 1}^{2} \dots a_{k m_{k}}^{2} t_{k 1} \dots t_{k n_{k}}) .

(17.9)

Group the terms by the parity of $i_{k}$ . The even terms sum to some $t \in T$ (or to $0$ if there are none); the odd terms sum to some $r$ with $b r \in T$ (or $r = 0$ if there are none), because multiplying an odd term on the left by $b$ makes every multiplicity even. If there were no odd terms then $t = 0 \in T$ , absurd; if there were no even terms then $r = 0$ , whence $b r = 0 \in T$ , equally absurd. So both groups are non-empty, $t \in T$ and $b r \in T$ .

Now $(17.9)$ reads $0 = t + r$ . Multiplying on the left by $b$ gives $0 = b t + b r$ , that is $t' + b t = 0$ with $t' : = b r \in T$ . This is (2).

Theorem(17.10)Orderings are the maximal preorderings

Let $R \neq 0$ be a ring and $T \subseteq R ∖ {0}$ a preordering. Then $T$ is an ordering of $R$ if and only if $T$ is maximal among the preorderings of $R$ .

Proof

**Ordering $\Rightarrow$ maximal.** Let $T$ be an ordering and suppose $T' ⊋ T$ is a preordering. Pick $a \in T' ∖ T$ . Then $a \neq 0$ , so totality gives $- a \in T \subseteq T'$ , and $0 = a + (- a) \in T' + T' \subseteq T'$ — impossible.

**Maximal $\Rightarrow$ ordering.** Let $T$ be a maximal preordering and suppose it is not an ordering; then some $b \neq 0$ has $b \notin T$ and $- b \notin T$ . Since $T_{b} \supseteq T \cup {b}$ satisfies $(17.5)$ and $(17.6)$ , maximality forbids $T_{b}$ from being a preordering (it would strictly contain $T$ ). By $(17.8)$ there are $t_{1}, t_{2} \in T$ with

t_{1} + b t_{2} = 0 .

(P.1)

The same argument applied to $- b$ gives $t_{3}, t_{4} \in T$ with $t_{3} + (- b) t_{4} = 0$ , i.e.

t_{3} = b t_{4} .

(P.2)

Set $t_{5} : = (b t_{4}) (b t_{2})$ . Because $b$ occurs twice and $t_{4}, t_{2} \in T$ , this is an instance of $per (b^{2} t_{4} t_{2})$ and so $t_{5} \in T$ . On the other hand $(P . 1)$ gives $b t_{2} = - t_{1}$ , so using $(P . 2)$ ,

t_{3} t_{1} = (b t_{4}) (- b t_{2}) = - (b t_{4}) (b t_{2}) = - t_{5} .

(P.3)

Since $t_{3}, t_{1} \in T$ we have $t_{3} t_{1} \in T$ , and therefore $0 = t_{3} t_{1} + t_{5} \in T + T \subseteq T$ — contradicting $0 \notin T$ . Hence $T$ was an ordering after all.

Corollary(17.10a)Every preordering extends to an ordering

If $R \neq 0$ has a preordering $T$ , then $R$ has an ordering $P \supseteq T$ .

Proof. The preorderings of $R$ containing $T$ form a poset in which every chain has the union as an upper bound: each axiom involves only finitely many elements, all lying in a single member of the chain, and no member contains $0$ . Zorn's Lemma supplies a maximal element $P$ , which is an ordering by $(17.10)$ .

Proof Techniques and Method

How these proofs work, and which move to reuse.

Assume the enlargement failsTo show

b

- b

already lies in a maximal preordering

T

, assume both are missing and form

T_{b}

and

T_{- b}

Convert failure into an equation

(17.8)

turns "

T_{b}

is not a preordering" into

t_{1} + b t_{2} = 0

with

t_{1}, t_{2} \in T

— an identity you can multiply.

Multiply the two equationsCombine the identities for

b

and

- b

so that

b

occurs an even number of times. The result is an element of

T

Exhibit zeroThe combination is the negative of an element of

T

, so

0 \in T + T \subseteq T

. Contradiction; the assumption was false.

This four-step pattern recurs verbatim in $(17.13)$ for the division closure and in $(17.16)$ for extension to a larger ring. Learning it once covers most of §17.

Worked Example

A preordering that is already an ordering

Take $R = ℚ$ . Here $T (ℚ)$ is the set of nonzero sums of squares. Every positive rational $a / b$ (with $a, b$ positive integers) equals $a b / b^{2}$ , and $a b$ is a sum of four integer squares by Lagrange's theorem, so

T (ℚ) = ℚ_{> 0}, e.g. \frac{5}{3} = \frac{15}{9} = (\frac{3}{3})^{2} + (\frac{2}{3})^{2} + (\frac{1}{3})^{2} + (\frac{1}{3})^{2} .

(E.1)

Check: $9 + 4 + 1 + 1 = 15$ , and $15 / 9 = 5 / 3$ .

So the weak preordering of $ℚ$ is already total, hence maximal, hence the unique ordering of $ℚ$ by $(17.10)$ .

A preordering that is not an ordering

Take $R = ℝ [x]$ . Since $R$ is commutative, $T (R)$ is the set of nonzero sums of squares of polynomials, which in one variable is exactly

T (ℝ [x]) = {f \neq 0 : f (t) \geq 0 for all t \in ℝ},

(E.2)

A nonnegative real polynomial in one variable is a sum of two squares — factor it over $ℂ$ and split into real and imaginary parts.

This is a preordering but not an ordering: $x \notin T$ and $- x \notin T$ , so totality fails. Two orderings above it, both containing $T$ :

Two orderings of $ℝ [x]$ extending $T (ℝ [x])$
Ordering	Positivity rule	Sign of $x$	Sign of $x^{3} - 5 x$
$P_{\infty}$	$f (t) > 0$ for all large $t$ ; equivalently the leading coefficient is positive	positive	positive
$P_{0^{-}}$	$f (t) > 0$ for all $t < 0$ close to $0$	negative	positive

For $x^{3} - 5 x = x (x^{2} - 5)$ : as $t \to + \infty$ it is positive, and for small negative $t$ the factor $x$ is negative while $x^{2} - 5$ is negative, so the product is positive. Both orderings agree here, while they disagree on $x$ — consistent with the fact that $x$ lies in no intersection of all orderings.

The auxiliary set in this example

With $T = T (ℝ [x])$ and $b = x$ , the set $T_{x}$ consists of all sums $\sum_{i} x^{i} s_{i} (x)$ with each $s_{i}$ a nonzero sum of squares. Could $0$ lie in $T_{x}$ ? By $(17.8)$ that would need $t' + x t = 0$ with $t, t'$ nonnegative and nonzero — impossible, since $- x t$ changes sign at $0$ while $t'$ does not. So $T_{x}$ is a preordering, and any maximal preordering above it is an ordering in which $x > 0$ ; $P_{\infty}$ is one such.

Process and Workflow

You want to know whether a given $b \neq 0$ can be made positive, relative to a preordering $T$ .

No identity $t' + b t = 0$ exists

T_{b}

is a preordering by

(17.8)

. Enlarge it by Zorn to an ordering

P \supseteq T

with

b \in P

. Conclusion:

b

is positive in at least one ordering above

T

Such an identity existsThen

b t = - t'

with

t, t' \in T

, so

b

is negative relative to

T

: in every ordering

P \supseteq T

b t \in - P

with

t \in P

forces

b \in - P

. Conclusion:

- b

lies in the division closure of

T

Identities exist for both $b$ and $- b$ Impossible when

T

is a preordering — that is exactly the contradiction driving the proof of

(17.10)

Start from a candidate coneAny set closed under sums and permuted doubled products is a candidate; usually

T (R)

itself.

Test propernessThe only obstruction is

0 \in T

. If

0 \notin T

, you have a preordering.

Enlarge by ZornChains have unions; take a maximal preordering containing your candidate.

Read off the orderingBy

(17.10)

the maximal preordering is total: it is an ordering, and it restricts to any subring.

Comparison and Classification

Orderings, preorderings and their weak analogues
Notion	Closed under sums	Contains permuted doubled products	Total	Exists when
Ordering $P$	yes	yes	yes	$R$ formally real, via Zorn
Preordering $T$	yes	yes	no	$R$ formally real
Weak preordering $T (R)$	yes	yes	no	$R$ formally real; smallest of all
Auxiliary $T_{b}$	yes	yes	no	no identity $t' + b t = 0$ over $T$
Intersection of orderings	yes	yes	no	division-closed, $(17.14)$

Which properties survive which operation
	Intersection of a family	Union of a chain	Restriction to a subring	Extension to an overring
Preordering	yes	yes	yes	partial
Ordering	no	no	yes	partial
Maximality	no	yes	no	no

Which properties survive which operation

"Part" records that extension to an overring is governed by the criterion of $(17.16)$ and can fail, as it does for $ℚ \subseteq ℚ (i)$ .

Relationship Map

T (R)

preordering

T

maximal preorderingordering

P

Read left to right this is the construction; read right to left it is the fact that every ordering contains the weak preordering. The middle arrow is Zorn's Lemma, the right-hand identification is $(17.10)$ .

Preorderings of $R$ — ordered by inclusion
- smallest element
  - the weak preordering $T (R)$ , present exactly when $R$ is formally real
- maximal elements
  - the orderings of $R$ , by $(17.10)$
  - at least one above every preordering, by Zorn
- closure operations
  - intersection of orderings, always a preordering
  - division closure $\bar{T}$ , itself a preordering, $(17.13)$

Computational Notes

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

The passage from a preordering to an ordering uses Zorn's Lemma and is not effective: there is no procedure that outputs a cone element by element, and orderings of a countable ring can be non-computable.
Testing $(17.6)$ directly is an infinite family of conditions. In practice one exhibits a multiplicative sign invariant — the leading coefficient, the least support element, an evaluation homomorphism — and verifies closure once.
For a finitely generated commutative $ℚ$ -algebra, existence of an ordering is decidable: it amounts to the existence of a real point of the associated variety, settled by quantifier elimination over real closed fields. Cylindrical algebraic decomposition is doubly exponential in the number of variables; QEPCAD B and Redlog are the standard implementations.
For finitely presented noncommutative algebras nothing of the sort holds: even deciding whether a given word is zero is undecidable, so no general orderability test can exist.
Membership in the cone generated by squares in a commutative polynomial ring is testable by semidefinite programming with a fixed degree bound, which is the computational face of preorderings in real algebraic geometry.

Failure Modes and Common Mistakes

Do not assume a maximal preordering is unique; a ring with several orderings has several.
Do not assume the union of two preorderings is a preordering — it is not closed under addition. Only chains behave.
Do not forget the degenerate cases $m = 0$ and $n = 0$ in $(17.6)$ ; they are what give $T \cdot T \subseteq T$ and $1 \in T$ .
Do not conclude from $b t \in T$ that $b \in T$ . That implication is division-closedness, and it can genuinely fail — see Division-Closed Preorderings.

Best Practices

When constructing a preordering, describe its generators, then prove only that $0$ is not a sum of them.
Keep an explicit note of the parity bookkeeping: which elements are doubled, and which are inserted from $T$ .
State whether a claimed cone is maximal; without maximality, $(17.10)$ gives nothing.
When you need an ordering with a prescribed positive element $b$ , build $T_{b}$ rather than trying to extend a total order by hand.

Quick Reference

Axiom (17.5)

T + T \subseteq T

Axiom (17.6)

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

Degenerate cases

m = 0

gives

T \cdot T \subseteq T

;

n = 0

gives

1 \in T

(17.7)

T \cap (- T) = \emptyset

R

a domain,

char R = 0

(17.8)

T_{b}

fails

\Leftrightarrow

t' + b t = 0

for some

t, t' \in T

(17.10)ordering

\Leftrightarrow

maximal preordering

Zornevery preordering lies in an ordering

Smallest

T (R) \subseteq T

for every preordering

T

Instances of $(17.6)$ worth memorising
Choice of $m, n$	Consequence
$m = 0$ , $n = 2$	$t_{1} t_{2} \in T$ : closure under multiplication
$m = 1$ , $n = 0$	$a^{2} \in T$ for all $a \neq 0$
$m = 1$ , $a_{1} = 1$ , $n = 0$	$1 \in T$
$m = 1$ , $n = 1$	$a t a \in T$ : sandwiching
$m = 2$ , $n = 0$	$a b a b, a b^{2} a, a^{2} b^{2} \in T$ for all $a, b \neq 0$

Frequently Asked Questions

Why not define a preordering simply as a subset closed under addition and multiplication containing all squares?

Because that set need not be closed under sandwiching in a noncommutative ring, and sandwiching is what the proofs use. The condition $a t a \in T$ does not follow from $T \cdot T \subseteq T$ plus $a^{2} \in T$ unless the ring is commutative. Axiom $(17.6)$ builds all the needed closure into one statement.

Where exactly is Zorn's Lemma used, and can it be avoided?

It is used once, in $(17.10 a)$ , to produce a maximal preordering above a given one. It cannot be avoided in general: the existence of an ordering on an arbitrary formally real field already needs a choice principle, and even for countable fields the resulting cone need not be computable. For specific rings one always constructs the ordering explicitly instead.

Does

(17.10)

mean every ordering arises as a maximal preordering containing

T (R)

Yes. Every preordering contains $T (R)$ , so in particular every ordering does, and orderings are maximal among preorderings. Hence the orderings of $R$ are exactly the maximal elements of the poset of preorderings containing $T (R)$ .

Why does the proof of

(17.10)

need both

b

and

- b

Failing to be an ordering means some element is neither positive nor negative, which is two pieces of information. Each gives an identity through $(17.8)$ , and the contradiction requires multiplying the two together so that $b$ occurs an even number of times. One identity alone yields no contradiction — indeed $T_{b}$ failing for a single $b$ is perfectly normal and simply says $- b$ is forced positive.

T_{b}

the smallest preordering containing

T

and

b

When it is a preordering at all, yes: any preordering containing $T$ and $b$ must contain every permuted product of $b$ , doubled ring elements and elements of $T$ , hence every generator of $T_{b}$ , hence all their sums. When $0 \in T_{b}$ , no preordering contains $T$ and $b$ at all.

Can a ring have a preordering but no ordering?

No — that is exactly the content of $(17.10 a)$ together with $(17.10)$ . Having a preordering, being formally real, and having an ordering are all equivalent for a nonzero ring; the equivalence is R. E. Johnson's theorem $(17.11)$ .

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §17, results (17.5)–(17.10), pp. 277–280.
E. Artin and O. Schreier, Algebraische Konstruktion reeller Körper, Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg 5 (1927). The commutative prototype of the maximality argument.
T. Y. Lam, Orderings, Valuations and Quadratic Forms, CBMS Regional Conference Series in Mathematics 52, American Mathematical Society, 1983. Preorderings, fans and spaces of orderings in the field case.
A. Prestel, Lectures on Formally Real Fields, Lecture Notes in Mathematics 1093, Springer-Verlag, 1984.
M. Marshall, Positive Polynomials and Sums of Squares, Mathematical Surveys and Monographs 146, American Mathematical Society, 2008. Preorderings, quadratic modules and their computational use.

AI Suggested Questions

What is the exact relationship between Lam's preorderings and the quadratic modules used in the noncommutative Positivstellensatz?
For which noncommutative rings can a maximal preordering be described explicitly rather than obtained from Zorn's Lemma?
How does the poset of preorderings of a commutative ring relate to the topology of its real spectrum?
Are there natural weakenings of axiom $(17.6)$ that still force the ring to be a domain?
What does the set of orderings above a fixed preordering look like for a free algebra over a formally real field?
How do preorderings interact with Ore localisation, and when is the extension of a preordering still proper?
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01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Parity is the invariant

Sums of preorderings, and chains

What maximality buys

06Key Results

07Proof Techniques and Method

08Worked Example

A preordering that is already an ordering

A preordering that is not an ordering

The auxiliary set in this example

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Computational Notes

13Failure Modes and Common Mistakes

14Best Practices

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

Computational Notes

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions