Cuts over ordered fields

1Ordered domains and fields 1

2Real closure 3

3Cuts and simple extensions 3

3.1The transcendental case 4

3.2The algebraic case 6

4Good cuts and completeness 7

5Cauchy sequences and completeness 8

5.1Cauchy sequences 8

5.2About cofinality 8

5.3Cauchy-completeness 9

6Completion 9

6.1The completion theorem 10

6.2An analogy 11

Bibliography 11

Index 11

Glossary 11

1Ordered domains and fields

Let us start with some conventions regarding linear (i.e. total) orders. My orderings are always strict. Given a linear order

, the binary relation

is defined by

Note that all above statements are vacuously true when

, or

for the two last ones.

Definition 1.1. An ordered domain is a domain equipped with a linear order on , with

Definition 1.2. An embedding between ordered domains is a morphism of rings which is strictly increasing. A embedding is said cofinal if is cofinal in , dense if is dense in (as a linear order).

If is a (cofinal, dense) embedding, then I say that is a (cofinal, dense) extension of .

Definition 1.3. If is a domain, then a positive cone on is a subset of such that for all , we have

Note that if

is an ordered domain, then

is a positive cone on

. We have the following converse.

Sketch of Proof. The anti-reflexivity of

follows from Definition 1.3(c), its transitivity follows from Definition 1.3(a), its linearity follows from Definition 1.3(e),

follows from Definition 1.3(d) and the compatibilities between

and

follow from Definition 1.3(a) and Definition 1.3(b) respectively.

Proposition 1.5. Let be an ordered domain. There is a unique ordering of its fraction field such that is an extension of .

We immediately have

and

, as well as Definition 1.3(e) and Definition 1.3(b). For

and

, where

, we have

where

and

are strictly positive, whence

. So

is a positive cone. Since

contains

(as naturally embedded into

), the corresponding ordered domain is an extension of

Now if

is an order on

such that

is an ordered ring extension of

, then

contains

, whence

by Definition 1.3(b). We deduce that

is the order derived from

So in the sequel, I will extend orderings to fraction fields without mention. The fraction field operation

induces a functor from the category of ordered domains with embeddings, to the gategory of ordered fields with embeddings, by sending an embedding

to the well-defined map

. This functor is left adjoint to the forgetfull functor.

Remark 1.6. Mind that the extension

might not be cofinal. For instance, consider the ordered domain

whose positive cone is the set of polynomials with strictly positive leading coefficient. Now consider the ordered domain

contained in

. We have

but the element

satisfies

2Real closure

I will say that an extension of ordered fields is algebraic if it is algebraic as a field extension.

Sketch of Proof. Let

be an algebraic extension of ordered fields. We will prove by induction on

that for all

, if there is a polynomial

of degree

with

, then

has an upper bound in

This is immediate if

. Let

such that the result holds at

, and let

and

with

and

. Set

, so that

We have

. If

, then we conclude by our induction hypothesis that

has an upper bound in

. If

, then we deduce that

, so

has an upper bound in

By induction, the result holds in general. Now since

is algebraic, for all

, there is a

with degree

and

, so any such

has an upper bound in

. We deduce that

is cofinal.

Definition 2.2. A real-closed field is an ordered field which has no proper extension which is algebraic as a field extension.

By a famous result of Artin and Schreier, real-closed fields are exactly the fields

whose absolute Galois group has finite order (i.e. with

), or, equivalently, order

. They are ordered according to the positive cone

Tarski's theorem. An ordered field is real-closed if and only if it has the same first order properties as the ordered field of real numbers.

This means that in real-closed fields, elementary statements regarding polynomials or more complicated definable functions and sets which are satisfied in

, are valid. We will use this in the next section. Moreover, again by Artin-Schreier, we have a real closure functor:

Theorem 2.3. For every ordered field , there is an algebraic extension where is real-closed. If is an extension and is real-closed, then there is a unique with .

The extension

is called the real closure of

. By the above initial property, it is unique up to unique isomorphism, so we consider that

is naturally contained in

3Cuts and simple extensions

In this section, we construct simple extensions of ordered fields. The method is inspired from Kaplansky's work in his paper Maximal fields with valuation II [3], except I use cuts instead of pseudo-Cauchy sequences. I will use the following definition of cuts:

Definition 3.1. A cut over an ordered field is an ordered pair where , and has no maximum. We say that the cut is filled if has a minimum, unfilled otherwise.

Lemma 3.2. Let be a cofinal proper extension of ordered fields. For all , the pair where

Sketch of Proof. We have

by definition. Since

, Any

must satisfy

, whence

. Now for

, there is an

with

. But

is cofinal, so there is a

with

, whence

which shows that

has no maximum. Likewise

has no minimum, so

is an unfilled cut.

In what follows, we will see how to construct simple cofinal ordered field extensions using unfilled cuts. We start with important remarks on the sign of polynomials near unfilled cuts. Let

be an unfilled cut in

. If

is a non-zero polynomial, then

has finitely many roots in the real closure

. Since

is real-closed, Tarski's theorem implies that it has the same first-order properties as

. In particular

satisfies the intermediate value theorem on

. So

has constant sign between roots in

(since between any sign shift there ought to be a root by the IVT). Assume that

. Then since

has finitely many roots in

and

has no maximum, there is a final segment of

(in

) where

has constant sign

. Likewise, if

, since

has no minimum, there is an initial segment of

where

has constant sign

. I will leave

undefined.

We say that

is an annihilator polynomial of

and

. Annihilator polynomials of

of minimal degree are called minimal polynomials of

Remark 3.3. Cuts may have distinct unitary minimal polynomials. Consider for instance, in the ordered field

where

, the cut

where

is the set of finite fractions and

is the set of positive infinite fractions. Then

and

are both minimal polynomials of

3.1The transcendental case

In this subsection, we assume that has no annihilator polynomial. Then for

, we define

is non-empty, and as

is non-empty. Note that this is well-defined. We also set

For

with

, we have

. It is easy to see consequently that the set of polynomials

with

is a positive cone on

. We write

for the corresponding ordered domain, and

for its fraction field, which is thus a field extension of

Proposition 3.4. If the unfilled cut has no annihilator polynomial, then is an ordered field extension of where .

Sketch of Proof. It is easy to see that

is an ordered domain extension of

, since the sign

of any constant polynomial

is that of the constant

. So

is an ordered field extension of

. Now for

and

, we have

. We deduce that

and assume that has no maximum and has no minimum. Then is an unfilled cut over and it has no annihilator polynomial.

Sketch of Proof. That

is an unfilled cut over

is immediate by definition. Assume for contradiction that

has an annihilator polynomial

over

. In particular, neither

nor

are empty. Since

is real-closed, all roots of

lie in

, so there are an

and an

such that

contains no root of

and with

. But then by the IVT for

we have a root of

: a contradiction.

Assume that has no maximum and has no minimum. Then there is a unique ordered field embedding over with .

Sketch of Proof. Since

is real-closed, the extension

is transcendental, and

has no annihilator polynomial by Lemma 3.5. So there is a field isomorphism

which is the unique ring morphism

over

which sends

. Let us show that

is strictly increasing. It is enough to show that for

with

, we have

. Now, for such

, there are

with

such that we have

for all

. So

has no zero in

. But

is real-closed, so

has no zero in

and must have constant, strictly positive sign on this set. In particular, we get

since

. So

is strictly increasing.

If has a maximum , then there is a unique ordered field embedding over with . If has a minimum , then there is a unique ordered field embedding over with .

Sketch of Proof. Again the fields embeddings are well-defined and unique, so there remains only to show that they are strictly increasing, hence that they preserve the sign of elements in

and

respectively.

We treat the first case. Let

be strictly positive. So

is strictly positive for large enough

. So there is an

with

for all

. Since

is real-closed, as in the previous proof, we must have

for all

as well, whence

. This concludes the proof. The case when

has a minimum is symmetric.

So we have a characterization of orderings on simple extensions of real-closed fields via cuts and sets of the form

where

has a maximum or

has a minimum. One final question is: when is the extension

cofinal? Using the chracterizations of orderings using cuts, this reduces to the following:

Sketch of Proof. If

, then

by definition, so the extension is not cofinal. Likewise

. Conversely, assume that

and that

. So we have

for a certain

. It follows that for all

, there is an

with

. Now let

be positive. Our goal will be to prove that there are a

and an

with

for all

, which would mean that

Assume for contradiction that for all

, there is an

, such that

and that the interval

contains a root of

. Choosing

smaller than the distances between distinct roots of

, we may fix

and

such that

and that

contains a unique root

. Let

denote the minimal polynomial of

over

. Recall that

is a perfect field since it has characteristic zero. So

, which implies by Tarski's theorem that the function

takes opposite signs on small intervals

and

for small enough

. Our assumption implies that there is an

such that

and

. But then we have

, so

is an annihilator polynomial of

: a contradiction. We deduce that one of the following holds:

Now for

where

, we may choose

and thus obtain a

with

and an

with

, whence

. This shows that

is cofinal.

3.2The algebraic case

In this subsection, we assume that has an annihilator polynomial. In particular,

and

are non-empty. Pick a minimal polynomial

Claim:

is irreducible in

. Indeed, for strictly lower degree polynomials

, we have

and

by minimality, whence

So the quotient ring

is a field extension of

. We define a positive cone for

as follows. Given

, the Euclidean division of

yields a unique

with

and

By minimality of

, we may set

and define

as the set of elements

with

. Note that Definition 1.3(c,d,e) are satisfied. Consider

and

with positive sign. We readily see that

has positive sign, so

satisfies Definition 1.3(a).

Since

, we have

. Assume for contradiction that

. So

is negative for large enough

and

is negative for small enough

. In particular

is strictly negative for large enough

and small enough

. Since

is an anihilator polynomial of

, this implies that

, whence that

. But then

: a contradiction. We deduce that in fact

, which proves that

is a positive cone on

I write

for the corresponding ordered field. Similar arguments as in Proposition 3.4 yield:

Proposition 3.9. If is unfilled and has a minimal polynomial , then is an ordered field extension of where .

4Good cuts and completeness

Any linearly oredered set enjoys a topology, called the order topology, whose open sets are arbitrary unions of open intervals. An ordered field

, equipped with the order topology, is a topological field, i.e. the sum and product functions are continuous for the product topology on

, whereas the additive inverse and multiplicative inverse functions are continuous on their respective domains. A subset

is dense in

as a linear order if and only if it is dense in

as a topological space, i.e. if and only if for all

and all

, there is an

with

Definition 4.1. An ordered field is said complete if it has no proper dense extension.

Remark 4.2. We can also endow

with a uniform structure as per [1, Chapter 2], derived from its ordering. Then the notions of Cauchy sequence, completeness and Cauchy-completion have equivalents in the language of uniform structures, which are more general but more involved.

Note that any filled cut is good, and that a good cut

must satisfy in particular

and

Lemma 4.4. Let be an ordered field, and let be a good, unfilled cut. If has no annihilator polynomial, then is a dense extension.

Sketch of Proof. Assume that

has no annihilator polynomial. So

is a proper ordered field extension. We claim that it is dense. It is cofinal by Lemma 3.8, so we need only prove that for all

and

, there is an

with

. Write

where

and

. In

, the rational function

is uniformly continuous on any bounded interval not containing its poles (indeed this is true in

). In particular, given

, there is a

such that for all

with

, we have

. Since

is cofinal, we may assume that

. Pick

with

. So for all

and with

, we have

. Setting

, this implies that

Lemma 4.5. Let be an ordered field, and let be a good, unfilled cut. If is a minimal polynomial of , then is a dense extension.

Proposition 4.6. An ordered field is complete if and only if every good cut in is filled.

Sketch of Proof. Assume that

is complete and assume for contradiction that

is a good unfilled cut. So by Lemmas 4.4 and 4.5, we have a proper dense extension of

: a contradiction. Conversely, consider a proper dense extension

and fix a

. Since

is cofinal, we have an unfilled cut

of Lemma 3.2. By density, given

, there are

with

. So

and

. This proves that

is good, hence the result.

5Cauchy sequences and completeness

5.1Cauchy sequences

Definition 5.1. A convergent sequence in is a sequence such that there is an with

Saying that

is a limit of

is the same as saying that the map

extending

with

is continuous. As a consequence of the fact that the topological space

is Hausdorff, we have:

Conversely, some Cauchy sequences are not convergent: take a Cauchy sequence of rationnal numbers in the usual sense, whose limit in

is irrational.

5.2About cofinality

One can skip this subsection since it is not used in the sequel. However the result written as Lemma 5.5 is quite useful when working with Cauchy sequences. Recall that the cofinality of a linear order

is equivalently

Moreover, two linear orders

and

have the same cofinality if and only if there is a non-decreasing map

whose range is cofinal in

Lemma 5.5. If is a Cauchy sequence in and is not eventually constant, then .

Sketch of Proof. By not eventually constant, I mean that for all

, there is a

with

. We will define a cofinal embedding

. First pick cofinal embeddings

and

. For all

, we write

for the least ordinal in

such that

for some

. This exists because

is not eventually constant and the range of

is cofinal in

. Note that

is non-decreasing. Now since

is Cauchy and

is a cofinal embedding, for all

, there is an

with

for all

, whence

. But

is strictly increasing, so

. This proves that the range of is cofinal in

, hence

is a strictly increasing and cofinal map (i.e. a map whose range is cofinal), then say that

is a subsequence of

. Note that subseqences of convergent/Cauchy sequences are convergent/Cauchy, and that the limit is preserved in the convergent case. The previous lemma shows that some properties of Cauchy sequences can be reduced to those of Cauchy sequences indexed by

5.3Cauchy-completeness

Definition 5.6. An ordered field is said Cauchy-complete if every Cauchy sequence in converges.

Proposition 5.7. An ordered field is Cauchy-complete if and only if it its good cuts are filled.

Sketch of Proof. Fix a cofinal embedding

. Assume that

is Cauchy-complete and assume for contradiction that there is a good, unfilled cut

. For all

, there is an

with

. Since

has no maximum, there is

with

, whence

. Using this, we define a strictly increasing sequence

such that for all

and

with

, we have

Assume that

is defined on

and that

is a limit. Assume for contradiction that

is cofinal in

. Note that

has a strict lower bound

since its cardinality is

. Consider

with

for all

with

. We find a

with

, so

: a contradiction. So

has a strict upper bound

. We pick

with

and

for all

with

and extend the sequence accordingly. For successor

, we extend

directly.

is strictly increasing, ranges in

and satisfies (5.1) by definition. Note that this implies that

is a Cauchy sequence. Let

be its limit in

. We claim that

is the maximum of

. Indeed assume that

has a strict upper bound

. Set

, and fix

with

for all

and

. It is easy to see that

for all

. So

, whence

: a contradiction.

Conversely, assume that every good cut in

is filled and that

is not Cauchy-complete. Let

be a non-convergent Cauchy sequence. Write

I claim that this is a cut. We have

by definition. For

and corresponding

, the element

is a strict upper bound of

. So

has no maximum, and likewise

has no minimum. Now consider

. Since

is an initial segment of

whereas

is a final segment of

, we in fact have

. Let

, and let

with

for all

. Since

and

, there are

with

. We deduce that for all

, we have

is the limit of

, which contradicts our assumption on

. We deduce that

, hence

is an unfilled cut. The fact that

is Cauchy implies that

is good: a contradiction.

Proposition 5.8. An ordered field Cauchy-complete if and only if it is complete.

6Completion

6.1The completion theorem

Proposition 6.1. Let be an ordered field. There is a dense, complete extension of .

Sketch of Proof. So let

be a set of cardinality

. Let

denote the set of dense ordered field extensions

where

is a subset of

. We endow

with the extension ordering:

and the operations and ordering on

extend those on

. This set contains

, and it is inductive since we can take increasing unions. By Zorn's lemma, the set

has a maximal element

. I claim that

is complete. Assume that it is not complete. So it has a proper dense extension

. Now as we've seen every element of

injectively yields an unfilled good cut in

, and a (Cauchy) sequence

. So we have an injection

. Likewise, there is an injection

. In particular,

, so there is an injection

with

. Using this, we obtain an isomorphic copy of

into

which extends the inclusion

. This contradicts the maximality of

. So

is in fact complete.

Theorem 6.2. Let be a dense embedding into a complete extension. If is a cofinal extension and is complete, then there is a unique embedding with .

Sketch of Proof. Fix such an extension

. Let

. We have a corresponding unfilled cut

by Lemma 3.2, which as we saw in the proof of Proposition 4.6 is a good cut.

The unicity follows from the fact that

is cofinal, indeed for

with

, we have

since

is good. But since

is a cofinal embedding, it follows that

, whence that

. Assume for contradiction that no such element

exists. Write

(resp.

) for the set of lower bounds (resp. upper bounds) of elements of

. We will prove that

is a good unfilled cut in

, contradicting the completeness of

. It is clear that

. Since

has no minimum, neither do

and thus

. Likewise

has no maximum. It remains to show that

. Assume that there is a

. By definition of

and

, we must have

. It follows that

: a contradiction. So

, which concludes our proof that

is an unfilled cut in

and thus our proof by contradiction of the existence and unicity of

We complete our definition of the map

by setting

for all

. For

with

, there is an

with

. It follows from the definition of

that

, whence

is strictly increasing. Note that

is the only strictly increasing map

with

, since any such map must satisfy the defining inequality (6.1). So it is enough to conclude to prove that

is a morphism of rings. Since

coincides with

, we need only prove that

and

for all

It follows that

. Now let

. Again the element

is unique with

. But by the previous argument, we also have

, i.e.

is a group morphism. It follows that in order to deal with the product, we need only consider the case of products of strictly positive elements. For those, similar arguments as those used for the sum yield the result. This concludes the (sketch of) proof.

The extension

is called the completion (or Cauchy completion) of

. It is unique up to unique isomorphism.

6.2An analogy

So you can see a similarity between the completion operator

and the real closure operator

. First of all those are both functors, from the category of ordered fields with cofinal embeddings and the category of ordered fields with embeddings respectively, to their full subcategory of complete and real-closed fields respectively. They are also both left adjoints to the forgetful functors.

Moreover, an ordered field is real-closed if it has no proper algebraic extension, complete if it has no proper dense extension. We also have the following analogy:

Finally, there is a non-trivial interaction between those two closure operators, which is that the completion of a real-closed field is real-closed, or equivalently

for any ordered field

. This can be shown by lifting the theorem of continuity of roots (for polynomials), which is valid in

, to arbitrary real-closed fields, using Tarski's theorem.

Bibliography

[1]: N. Bourbaki. Topologie générale (chapitres 1 à 4). Eléments de mathématique. Springer Berlin Heidelberg, 1971.
[2]: P. Ehrlich. The Rise of non-Archimedean Mathematics and the Roots of a Misconception I: The Emergence of non-Archimedean Systems of Magnitudes. Archive for History of Exact Sciences, 60(1):1–121, 2006.
[3]: I. Kaplansky. Maximal fields with valuations, ii. Duke Mathematical Journal, 12(2):243–248, 06 1945.

Index

absolute value 2

algebraic extension 3

annihilator polynomial 4

Cauchy sequence 8

Cauchy-complete ordered field 9

(cofinal, dense) embedding 2

cofinal subset 1

(cofinal,dense) extension 2

complete ordered field 7

completion 10

convergent sequence 8

cut 4

dense subset 1

filled cut, unfilled cut 4

good cut 7

limit of a sequence 8

minimal polynomial 4

ordered domain 2

ordered field 2

positive cone 2

real closure 3

real-closed field 3

sequence 8

Glossary

set of strictly positive elements in 2

ordered fraction field of 2

real closure of 7 3

simple extension for without annihilator polynomial 4

simple extension of the cut with minimal polynomial 6

Completion:	Real closure:
minimal cofinal and complete extension	minimal real-closed extension
maximal dense extension	maximal algebraic extension

Table of contents

1Ordered domains and fields

2Real closure

3Cuts and simple extensions

3.1The transcendental case

3.2The algebraic case

4Good cuts and completeness

5Cauchy sequences and completeness

5.1Cauchy sequences

5.2About cofinality

5.3Cauchy-completeness

6Completion

6.1The completion theorem

6.2An analogy

Bibliography

Index

Glossary