
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
5.3Cauchycompleteness 9
6Completion 9
6.1The completion theorem 10
6.2An analogy 11
Bibliography 11
Index 11
Glossary 11
Let us start with some conventions regarding linear (i.e. total) orders. My orderings are always strict. Given a linear order , the binary relation on is defined by
Given an element and subsets , I write
Note that all above statements are vacuously true when , or for the two last ones.
A subset of a linear order is dense in if
and that it is cofinal in if it has no strict upper bound in , i.e. if
Definition
.
.
.
An ordered field is an ordered domain which is a field.
Definition
If is a (cofinal, dense) embedding, then I say that is a (cofinal, dense) extension of .
Given an ordered domain , I write
Moreover, for all , I write for the absolute value of .
Definition
Note that if is an ordered domain, then is a positive cone on . We have the following converse.
Lemma
is an order on , and is an ordered domain with .
Sketch of Proof. The antireflexivity 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 and follow from Definition 1.3(a) and Definition 1.3(b) respectively.
Proposition
Sketch of Proof. We define a positive cone on as follows:
We immediately have and , as well as Definition 1.3(e) and Definition 1.3(b). For and in , 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 welldefined map . This functor is left adjoint to the forgetfull functor.
Remark
I will say that an extension of ordered fields is algebraic if it is algebraic as a field extension.
Lemma
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
By a famous result of Artin and Schreier, realclosed 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 .
We have the following famous theorem of Tarski:
Tarski's theorem. An ordered field is realclosed if and only if it has the same first order properties as the ordered field of real numbers.
This means that in realclosed 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 ArtinSchreier, we have a real closure functor:
Theorem
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 .
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 pseudoCauchy sequences. I will use the following definition of cuts:
Definition
Lemma
is an unfilled cut over .
Sketch of Proof. We have by definition. Since , Any must satisfy or , 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 nonzero polynomial, then has finitely many roots in the real closure of . Since is realclosed, Tarski's theorem implies that it has the same firstorder 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 if and and . Annihilator polynomials of of minimal degree are called minimal polynomials of .
Remark
In this subsection, we assume that has no annihilator polynomial. Then for , we define as if is nonempty, and as if is nonempty. Note that this is welldefined. 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
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 in . So is an ordered field extension of . Now for and , we have . We deduce that .
Lemma
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 realclosed, all roots of in lie in , so there are an and an such that contains no root of and with . But then by the IVT for in we have a root of in : a contradiction.
Theorem
Assume that has no maximum and has no minimum. Then there is a unique ordered field embedding over with .
Sketch of Proof. Since is realclosed, 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 to . 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 realclosed, so has no zero in and must have constant, strictly positive sign on this set. In particular, we get since . So is strictly increasing.
Theorem
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 welldefined 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 realclosed, as in the previous proof, we must have for all in 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 realclosed 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:
Lemma
Sketch of Proof. If , then in by definition, so the extension is not cofinal. Likewise in if . Conversely, assume that and that . So we have for a certain . It follows that for all , there is an
with in . Now let be positive. Our goal will be to prove that there are a and an with for all , which would mean that in .
Assume for contradiction that for all , there is an , such that and that the interval in 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 of . 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:
There is an such that for all . In that case, by Lemma 2.1, we may assume that . By Tarski's theorem, given , the polynomial function takes a strictly positive minimal value on the set
But by our assumption, we may choose so that , whence for all .
There is an such that contains no root of . In that case, the polynomial function takes a strictly positive minimal value (in ) on , whence for all .
Now for where , we may choose and thus obtain a with and an with , whence . This shows that is cofinal.
In this subsection, we assume that has an annihilator polynomial. In particular, and are nonempty. Pick a minimal polynomial of .
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 by 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 in with positive sign. We readily see that has positive sign, so satisfies Definition 1.3(a).
We next deal with Definition 1.3(b). Write
as above, and also consider the Euclidean division of by :
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
Lemma
Sketch of Proof. This is a corollary of Lemma 2.1.
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 of 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
Remark
Definition
4.1. those are sometimes called Veronese cuts [2] or Dedekind cuts or...
Note that any filled cut is good, and that a good cut must satisfy in particular and .
Lemma
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 in .
By similar arguments, we have:
Lemma
Proposition
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.
In this section, we fix an orderd field .
A sequence in is a map where is a nonzero limit ordinal.
Definition
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:
Lemma
Definition
Lemma
Sketch of Proof. Pick such that for all , then use the triangle inequality.
Conversely, some Cauchy sequences are not convergent: take a Cauchy sequence of rationnal numbers in the usual sense, whose limit in is irrational.
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
the unique regular ordinal which embeds cofinally (without strict upper bound) into ;
the least ordinal which embeds cofinally into ;
the least cardinal of a cofinal subset of .
Moreover, two linear orders and have the same cofinality if and only if there is a nondecreasing map whose range is cofinal in .
Lemma
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 nondecreasing. 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 .
If 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 .
Definition
Proposition
Sketch of Proof. Fix a cofinal embedding . Assume that is Cauchycomplete and assume for contradiction that there is a good, unfilled cut in . 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
(5.1) 
Assume that is defined on and that is a limit. Assume for contradiction that is cofinal in . Note that has a strict lower bound in since its cardinality is . Consider with for all with . We find a with , so : a contradiction. So has a strict upper bound in . We pick with and for all with and extend the sequence accordingly. For successor , we extend directly.
So 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 in . Set , and fix with for all and in . It is easy to see that for all . So , whence : a contradiction.
So is filled: a contradiction.
Conversely, assume that every good cut in is filled and that is not Cauchycomplete. Let be a nonconvergent 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 in . 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
So 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
Sketch of Proof. This follows from Propositions 5.7 and 4.6.
Proposition
There are several distinct ways to prove of this:
Use Dedekind's construction of the reals by means of operations on cuts, where instead of cuts of rational numbers, one considers good cuts in . This directly yields the extension, but it is somewhat tedious.
Use Cauchy's construction of the reals by means of Cauchy sequence, where instead of standard sequences, one considers indexed Cauchy sequences. This is also tedious.
Use Zorn's lemma and some abstract nonsense. This is shorter, which is why I will proceed thus.
Sketch of Proof. So let be a set of cardinality . Let denote the set of dense ordered field extensions of where is a subset of . We endow with the extension ordering: if 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
Sketch of Proof. Fix such an extension . Let . We have a corresponding unfilled cut in by Lemma 3.2, which as we saw in the proof of Proposition 4.6 is a good cut.
I claim that there is a unique element with
(6.1) 
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 on , we need only prove that and for all .
For and , we have and , so and are both the unique element of with
It follows that . Now let . Again the element is unique with . But by the previous argument, we also have
So , 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.
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 realclosed fields respectively. They are also both left adjoints to the forgetful functors.
Moreover, an ordered field is realclosed if it has no proper algebraic extension, complete if it has no proper dense extension. We also have the following analogy:
Completion: 
Real closure: 
minimal cofinal and complete extension 
minimal realclosed extension 
maximal dense extension 
maximal algebraic extension 
Finally, there is a nontrivial interaction between those two closure operators, which is that the completion of a realclosed field is realclosed, 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 realclosed fields, using Tarski's theorem.
N. Bourbaki. Topologie générale (chapitres 1 à 4). Eléments de mathématique. Springer Berlin Heidelberg, 1971.
P. Ehrlich. The Rise of nonArchimedean Mathematics and the Roots of a Misconception I: The Emergence of nonArchimedean Systems of Magnitudes. Archive for History of Exact Sciences, 60(1):1–121, 2006.
I. Kaplansky. Maximal fields with valuations, ii. Duke Mathematical Journal, 12(2):243–248, 06 1945.
absolute value 2
algebraic extension 3
annihilator polynomial 4
Cauchy sequence 8
Cauchycomplete 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
realclosed field 3
sequence 8
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