The Real Holomorphy Ring and Sums ofPowers in Rational and Elliptic FunctionFields over The Field of Real Numbers

ByNatea Hunde

Advisor: Prof. Dr. Eberhard BeckerCo-advisor: Dr. Berhanu Bekele

A Thesis Submitted toThe Department of Mathematics

Presented in Fulfilment of The Requirements for Degree ofDoctor of Philosophy (Mathematics)

Addis Ababa UniversityAddis Ababa, Ethiopia

March 2017

Abstract

This thesis is devoted to a study of real holomorphy rings in R(x) and ellipticfunction fields R(x)(

√f(x)), f is a monic polynomial of degree 3 with no

multiple roots in C over R and sums of powers in R(x). The objects of studyare real valuation rings, real holomorphy rings and real places in the rationalfunction fields R(x) and then investigate their extension to elliptic functionfields which are quadratic extension of R(x). We write down the full list ofreal valuation rings to give the description of holomorphy ring in both fieldsand finally we study applications to sums of powers.

Acknowledgements

Praise be to the LORD!I believe that we are the sum total of all the individuals who have in

some way, small or great, contributed to our lives. I learned almost every-thing I know from some one and my life depend on the labours of othermen, living and dead. The likes of Serge Lang, T.Y.Lam, Bruce Reznick,Irving Kaplansky, David Eisenbud, William Fulton, Robin Hartshorne andmany others who wrote on commutative algebra and algebraic geometry I amhighly grateful for all of them having learned from their papers and books.

My sincere thanks to my Principal Advisor, Prof.Dr. Eberhard Becker,for introducing me to the fascinating topics of algebraic curves, real holomor-phy rings, higher level orderings and sums of powers. I am highly favouredor privileged in that he was very patient with me in introducing me to newconcepts and mathematical research. I appreciate and thank him for what Ilearnt from him during our conversation be it formal or informal that whatresearch in mathematics and the unity of mathematics is. His meticulousedition gave life to my scratch works and to my surprise helped much inunderstanding sometimes even what I wrote. His edition also helped me tolearn writing skills in mathematics. In fact it is advantageous to learn fromthe masters. Professor Becker invited me to his home university TU of Dort-mund, Germany for eight months in three separate times by covering myexpenses. He also showed me ancient churches or cathedrals in Cologne andDortmund cities, St. Goarshausen with the famous rock ” The LORELEY”,VATER RHEIN, the nice city of ”Limburg an der Lahn” and Museum atGOETHE university in FRANKFURT AM MAIN. He also took me to thegame in the famous stadium of Borussia Dortmund.

I would like also to thank my co-advisor Dr. Berhanu Bekele for hehelped me in changing my principal advisor to Prof. Becker and he invitedhim to give a lecture on algebraic curves at Addis Ababa University. Hiscontribution to the department’s graduate committee decision to invite Prof.Becker on my defense date of this thesis from ISP fund of the departmentis great. In my first trip to Germany he helped me much for which I amgrateful. I also acknowledge the discussions made during seminar times.

I enjoyed the time I spent with my friends and also instructors in algebrastream and want to say thank you all.

No less important is Professor H.V.Kumbhojkar who taught me algebracourses, challenged and introduced me to fuzzy sets and logic. He encouragesme by giving advice not only on algebra but also things helpful in the walkof life something for which I am grateful.

The prayer of my parents is always with me and God bless you all!Last, but definitely not least, I would like to thank AMA for your writings

kept me strong and inspired. JL I could never forget the way you told meeverything by saying nothing.

I dedicated this thesis to Hayilamariyam Gammadaa Masqalee who pavedthe way and every generation after him who followed his way.

1

Contents

1 A Sketch of The General Scenario 51.1 Hilbert’s 17th Problem . . . . . . . . . . . . . . . . . . . . . 51.2 Ordered Fields, Real Closed Fields . . . . . . . . . . . . . . 101.3 The Space of Orderings of a Field . . . . . . . . . . . . . . . 171.4 Valuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2 Real Holomorphy Rings of Rational Function Fields 212.1 Algebraic Function Fields of One Variable . . . . . . . . . . 212.2 Real Valuation Rings and Description of The Real Holomor-

phy Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.2.1 Real Valuation Rings of R(x) . . . . . . . . . . . . . 222.2.2 Real Holomorphy Rings . . . . . . . . . . . . . . . . 25

2.3 Real Places . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.4 Topological Representation of H(R(x)) . . . . . . . . . . . . 302.5 H(R(x)) a Dedekind Ring . . . . . . . . . . . . . . . . . . . 32

3 Real Holomorphy Rings of Elliptic Function Fields 383.1 Quadratic Extension of Fields . . . . . . . . . . . . . . . . . 383.2 Extension of Valuation Rings from R(x) to Elliptic Function

Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.3 The Valuation Theoretic Approach to Elliptic Function Fields 453.4 The Geometric Approach to Elliptic Function Fields . . . . . 493.5 Topological Representation . . . . . . . . . . . . . . . . . . . 51

4 Totally Positive Units and Sums of Powers 534.1 The Units of H . . . . . . . . . . . . . . . . . . . . . . . . . 534.2 The Valuation Criterion for Sums of Powers . . . . . . . . . 554.3 Sums of Mixed Powers . . . . . . . . . . . . . . . . . . . . . 554.4 Topics of Complexity . . . . . . . . . . . . . . . . . . . . . . 58

2

Introduction

This thesis is devoted to a study of real holomorphy rings, in R(x) and ellipticfunction fields R(x)(

√f(x) ), f any monic polynomial of degree 3 with no

multiple roots in C and sums of powers in these fields, mainly in R(x). Itis the central objective to interpret the general theory of real holomorphyrings and sums of powers in arbitrary formally real fields by providing a moreconcrete approach and add additional results based on special properties ofthe fields in questions.

As in the general theory, the central objects are the so called real valu-ation rings, i.e. valuation rings with a formally real valuation residue fieldwhich in the cases of study will be just R. Also they turn out to be discretevaluation rings. We list them completely as concretely as possible.

The next object to study is the real holomorphy ring, by definition theintersection of all real valuation rings. In the context of this thesis these ringsare Dedekind domains. These real holomorphy rings show one astonishingfeature known from the general theory: their totally positive units are sumsof 2n− th powers for every n.

Connected to the real holomorphy ring is another central object of thetheory: the so called real places of a formally real field which form a compactHausdorff spaces M . The real holomorphy ring H(K) of a formally real fieldK is isomorphic, in a natural way, to a dense subring of C(M,R), by virtue ofthe Stone-Weierstraß approximation theorem. In the present situation, thespace of real places can be realized as compact smooth real algebraic curvesassigned to our fields of study. Regarding the rational function field we aredealing with the affine line, in the elliptic function fields we are concernedwith the famous elliptic curves. In either case, the space of real places ishomeomorphic to the space of real points on the projective closure of thesecurves. This fact allows to draw information about the real holomorphy ringsand real valuation rings.

Methodologically, we first study these objects(real valuation rings, realholomorphy rings and real places) in the rational function field R(x) andthen investigate the extension to elliptic function fields which are quadraticextension of R(x). Finally, we are using these results to draw conclusions onsums of powers.

The first chapter describes the mathematical scenario to which the topicof this thesis belongs. It is justified to state that the topic had started toevolve into a substantial theory with Hilbert’s 17th Problem and its solutionby Artin roughly 25 years later. In addition to this historical sketch weoutline the concepts and general theorems we are going to use throughoutthis thesis.

3

Chapter 2 is devoted to the study of the rational function field R(x). It ispossible to list all real valuation rings which turn out to be discrete valuationrings with residue field R. As a consequence, a concrete description of thereal holomorphy ring H := H(R(x)) can be derived and it will be proventhat this ring is a Dedekind ring with field of fraction equal to R(x). Nomaximal ideal is a principal ideal, but their squares are principal. In bothcases explicit generators can be presented.Finally, the space of real places is shown to be homeomorphic to the real pro-jective line P1. We end up with a topological representation H → C(P1,R)and show, by virtue of the Stone-Weierstraß approximation theorem, that itadmits a dense image.

Chapter 3 treats the case of an elliptic function field L in an analoguemanner: finding a complete list of its real valuation rings, a description ofH(L), generators for its maximal ideals and their squares, proving that H(L)is a Dedekind ring, presenting a geometric interpretation of its space of realplaces and the topological representation of H(L) . In this case, the spaceof real places turns out to be homeomorphic to the 1-point compactificationof the set of real points on the affine elliptic curve given by the equationY 2 = f(X). In order to achieve all this, methods and results from commuta-tive ring theory will be applied in combination with the detailed informationon valuation rings we can provide.

In the final chapter 4 we apply the structural results of the previouschapters to the study of sums of powers. Originally, in the early work of E.Becker, sums of powers were studied by the so called orderings of higher level.Later on, it was observed that the real holomorphy ring is suitable to assumea central role. In fact, results on its groups of units and its ideal structureimmediately lead to results on sums of powers. It is the objective to displaythis approach in the case of our fields by representing a few striking results.Whenever it comes to really concrete statements we confine ourselves to therational function field case where the arguments can be given in a very clearmanner.

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Chapter 1

A Sketch of The GeneralScenario

1.1 Hilbert’s 17th Problem

The material used as background theory is mainly drawn from [12], [18],[19], [21], [25] and lectures given to me in Dortmund and texts sentthrough e-mail to me by Professor Becker.The following result from linear algebra is used in what follows.

Proposition 1 For a real symmetric n×n matrix A, the following are equiv-alent:(1) xTAx ≥ 0 for all x ∈ Rn.(2) All eigenvaues of A are ≥ 0.(3) A = UTU for some n× n matrix U.(4) A is a non-negative linear combination of matrices of the form xxT , x ∈Rn.

(Here we view x as column vector in doing matrix computations. For example

xTx = (x1, ..., xn)

x1

.

.

.xn

= x21 + ...+ x2

n = ||x||2,

whereas xxT =

x1

.

.

.xn

(x1, ..., xn) =

x1x1 · · · x1xn...

......

xnx1 · · · xnxn

the n× n matrix with ij entry xixj.)Proof. (1) ⇒ (2). Since A is symmetric, we know the eigenvalues of Aare real. Let d be an eigenvalue of A with associated eigenvector x. ThenAx = dx so xTAx = xTdx = dxTx = d||x||2. Since xTAx ≥ 0 and x 6= 0,this forces d ≥ 0.(2) ⇒ (3). Since A is symmetric, the Spectral theorem applies. (We recallthat Spectral Theorem says that if A is a real symmetric n×n matrix, thenthere exists an orthogonal n × n matrix Q and a real diagonal matrix ∧

5

such that QTAQ = ∧, and the n eigenvalues of A are the diagonal entriesof ∧.) This yields A = C−1DC, with C orthogonal, entries of D are theeigenvalues of A (counted with multiplicities), say D = diag(d1, ..., dn). By

(2), di ≥ 0 for each i, so D =√DT√

D where√D := diag(

√d1, ...,

√dn).

Thus A = C−1DC = CTDC = CT√DT√

DC = UTU , where U :=√DC.

(3)⇒ (4). A = v1vT1 + ...+ vnv

Tn where v1, ..., vn are the columns of UT .

(4)⇒ (1). Suppose A = r1v1vT1 + ...+ rmvmv

Tm with ri ≥ 0.

Then xTAx =∑m

i=1 rixTviv

Ti x =

∑mi=1 ri(v

Ti x)2 ≥ 0.

We say a square matrix A is positive semidefinite (PSD) if A is realsymmetric and the equivalent conditions of proposition above hold. We sayA is positive definite (PD) if A is real symmetric and xTAx > 0 holds for allnon-zero x ∈ Rn. We note that if A is positive definite, then the eigenvaluesof A are strictly positive and the matrix U in (3) is invertible.

Proposition 2 Let R[X1, ..., Xn] be a polynomial ring. If f ∈ R[X1, ..., Xn],f 6= 0, then there exists a point x ∈ Rn such that f(x) 6= 0.

In fact, one can do much better.

Proposition 3 If f ∈ R[X1, ..., Xn], f 6= 0, then the set Rn \ Z(f) = {x ∈Rn : f(x) 6= 0} is dense in Rn.

The degree of the monomial cxd11 ...xdnn (c ∈ R, d1, ..., dn ≥ 0) is defined to

be∑n

i=1 di. Each f ∈ R[X1, ..., Xn] decomposes (uniquely) as a finite sum ofmonomials. The degree of f is defined to be the maximum of the degrees ofthe various monomials appearing in this decomposition. By convention, thedegree of the zero polynomial is −∞.If deg(f) ≤ d then, collecting together monomials of the same degree, fdecompose (uniquely) as

f = f0 + f1 + ...+ fd

where each fi ∈ R[X1, ..., Xn] is homogeneous of degree i (i.e., a sum ofmonomials of degree i or the zero polynomial).

Corollary 1 Suppose f = f 21 + ... + f 2

k , f1, ..., fk ∈ R[X1, ..., Xn], f1 6= 0.Then(1) f 6= 0.(2) deg(f) = 2max{deg(fi) : i = 1, ..., k}.

Proof. (1) Since f1 6= 0 there exists x ∈ Rn such that f1(x) 6= 0. Then

f(x) = f1(x)2 + ...+ fk(x)2 > 0

so f 6= 0.(2) Decompose fi as fi = fi0 + ...+ fid, where fij homogeneous of degree

j, d := max{deg(fi)|i = 1, ..., k}. Clearly deg(f) ≤ 2d and the homogeneouspart of degree 2d of f is f 2

1d + ... + f 2kd. Since fid 6= 0 for some i, this is not

zero, by(1).Let f ∈ R[x1, x2, ..., xn], the xi different indeterminates.

6

If f is a sum of squares, say f = f 21 + f 2

2 + ...+ f 2r , then f(x) = f1(x)2 +

f2(x)2 + ...+ fr(x)2 ≥ 0, for all x ∈ Rn. It is natural to ask the following:QUESTION: Is the converse true,i.e., is it true that f ≥ 0 on Rn implies

that f is a sum of squares in R[x1, x2, ..., xn]?Hilbert showed that every non-negative polynomial is a sum of squares ofpolynomials only in the following three cases:- univariate polynomials,- quadratic polynomials, and- bivariate polynomial of degree four.

Proposition 4 Suppose f is a non-zero polynomial in the single variable x,and let

f = d∏i

(x− ai)ki∏

((x− bj)2 + c2j)lj

be the factorization of f into irreducibles in R[x]. Then the following areequivalent:

(1) f ≥ 0 on R.(2) d > 0 and each ki is even.(3) f = g2 + h2 for some g, h ∈ R[x].

Proof. For (2)⇒ (3), we use the ”two squares identity” (a2 + b2)(c2 + d2) =(ac− bd)2 + (ad+ bc)2. The other cases are obvious.

The answer is ’no’ if n ≥ 2. A concrete example was given by Motzkinin 1967. The Motzkin example is

M(x, y) = 1− 3x2y2 + x2y4 + x4y2

Proposition 5 For M as above:(1) M ≥ 0 on R2.(2) M is not a sum of squares in R[x, y].

Proof. (1) follows from the standard inequality a+b+c3≥ 3√abc (if a, b, c ≥ 0)

relating the arithmetic mean and the geometric mean, taking a = 1, b = x2y4,and c = x4y2.For (2) suppose, to the contrary, that M =

∑f 2i for some polynomials

fi ∈ R[x, y]. Then each fi can have degree at most 3 (Corollary 1), so issome real combination of

1, x, y, x2, xy, y2, x3, x2y, xy2, y3.

If x3 appears in some fi, then x6 would appear in M with positive coefficient.Thus x3 does not appear. Similarly, y3 does not appear. Arguing in the sameway, we see that x2 and y2 do not appear, and finally that x and y do notappear. Thus fi has the form

fi = ai + bixy + cix2y + dixy

2.

But then∑b2i = −3, a contradiction.

Remarks.(1) The minimum value of M on R2 is zero. This occurs at each of the fourpoints (±1,±1).

7

(2) In fact, one can show that N +M is not a sum of squares in R[x, y], forany real constant N ≥ 0. The argument is exactly the same.(3) In addition to the Motzkin example, many other examples have beenconsidered. These include examples of Robinson in 1969:

f(x, y, z) = x2(x− 1)2 + y2(y − 1)2 + z2(z − 1)2 + 2xyz(x+ y + z − 2).

The Choi-Lam example q(x, y, z) = 1 + x2y2 + y2z2 + z2x2 − 4xyz in 1977;the Schmudgen example in 1979, r(x, y) = 200[(x3 − 4x)2 + (y3 − 4y)2] +(y2 − x2)x(x + 2)[x(x − 2) + 2(y2 − 4)] (produced without prior knowledgeof earlier explicit examples); and the modified Motzkin example p(x, y) =1− x2y2 + x4y2 + x2y4 given by Berg, Christensen and Jensen in 1979. Notethat

p(x, y) =1

27(26 +M(

√3x,√

3y)).

(4) Although M is not a sum of squares of polynomials it is a sum of 4squares of rational functions, for example:

M =x2y2(x2 + y2 + 1)(x2 + y2 − 2)2 + (x2 − y2)2

(x2 + y2)2

=

(x2y(x2 + y2 − 2)

x2 + y2

)2

+

(xy2(x2 + y2 − 2)

x2 + y2

)2

+

(xy(x2 + y2 − 2)

x2 + y2

)2

+

(x2 − y2

x2 + y2

)2

.

(The first term in the numerator gives rise to the first three squares. Thelast term gives the final square.)

Hilbert worked with homogeneous polynomials. Homogeneous polyno-mials are also called forms. Why did Hilbert restrict to this case? If fis any polynomial in R[x1, x2, ..., xn] of degree ≤ d, then f(x0, ..., xn) =xdof(x1

x0, ..., xn

x0) (called the homogenization of f) is homogeneous of degree d

in n+ 1 variables x0, ..., xn.If f(x1, x2, ..., xn) =

∑cxd11 ...x

dnn , then

f(x0, ..., xn) = xd0∑

c

(x1

x0

)d1...

(xnx0

)dn=∑

cxd−

∑di

0 xd11 ...xdnn

=∑

cxd00 xd11 ...x

dnn

where d0 := d−∑di.

Proposition 6 Let Vd,n = the vector space of all polynomials of degree ≤ din n variables with coefficients in R, Fd,n = the vector space of forms ofdegree d in n variables with coefficients in R. f 7→ f defines a vector spaceisomorphism from Vd,n onto Fd,n+1. If d is even, then f ≥ 0 on Rn if andonly if f ≥ 0 on Rn+1, and f is a sum of squares of polynomials if and onlyif f is a sum of squares of forms of degree d

2(if and only if f is a sum of

squares of polynomials).

8

Proof. One checks easily that the map f → f is linear. Since it sends thebasis xd11 ...x

dnn ,∑di ≤ d of Vd,n to the basis xd00 x

d11 ...x

dnn ,∑di = d of Fd,n+1,

it is a vector space isomorphism. Suppose d is even, deg(f) ≤ d. To provef ≥ 0 on Rn+1 =⇒ f ≥ 0 on Rn, use f(x1, ..., xn) = f(1, x1, ..., xn). Toprove f ≥ 0 on Rn =⇒ f ≥ 0 on Rn+1, use f(x0, ..., xn) = xd0f(x1

x0, ..., xn

x0),

if x0 6= 0, and f(0, x1, ..., xn) = limε→0 f(ε, x1, ..., xn), if x0 = 0. If f =∑ki=1 f

2i , then deg(fi) ≤ d

2, and f =

∑ki=1[x

d20 fi(

x1x0, ..., xn

x0)]2, which is a sum

of squares of forms of degree d2. If f =

∑ki=1 g

2i , then f = f(1, x1, ..., xn)

=∑k

i=1 gi(1, x1, ..., xn)2.Remark. Counting the number of monomials xd11 ...x

dnn ,∑di ≤ d, one

sees that

dim(Vd,n) = dim(Fd,n+1) =

(d+ nn

)=

(d+ nd

)We say f ∈ R[x1, ..., xn] is positive semidefinite (on Rn) if f ≥ 0 on Rn.

For d, n ≥ 1, denote by Pd,n the subset of the vector space Fd,n consisting offorms of degree d in n variables which are positive semidefinite, and by

∑d,n

the subset of Pd,n consisting of sums of squares. The case where d is odd isnot interesting.(Or we assume d is even, so that positive polynomials exist.)In his 1888 paper, Hilbert proved the following:

Theorem 1 For d even, Pd,n =∑

d,n if and only if n ≤ 2 or d = 2 or (n =3 and d = 4).

Proof. Applying proposition above, we see that the homogenized Motzkinpolynomial

x6M(y

x,z

x) = x6 + y4z2 + y2z4 − 3x2y2z2

is in P6,3 \∑

6,3. Similarly, the homogenized Choi-Lam Polynomial

w4q(x

w,y

w,z

w) = w4 + x2y2 + y2z2 + z2x2 − 4wxyz

is in P4,4 \∑

4,4. More generally, if d ≥ 6 and n ≥ 3, then xd1M(x2x1, x3x1

) is

in Pd,n \∑

d,n and if d ≥ 4 and n ≥ 4, then xd1q(x2x1, x3x1, x4x1

) is in Pd,n \∑

d,n.Pd,1 =

∑d,1 is trivial. Pd,2 =

∑d,2 is immediate from proposition 4 (using

proposition 6). P2,n =∑

2,n follows from the fact that any quadratic form isexpressible as

f(x1, ..., xn) =n∑

i,j=1

aijxixj

where A = (aij) is a symmetric matrix.If f ≥ 0 on Rn, then the matrix A is PSD, so A factors as A = UTU and

f(x1, ..., xn) = (x1, ..., xn)A

x1...xn

=|| U

x1...xn

||2,which is a sum of squares of linear forms. The case P4,3 =

∑4,3 is highly

non-trivial and first proved by Hilbert.

9

The sets Pd,n and∑

d,nare closed under addition and multiplication bypositive reals, i.e., they are cones in the vector space Fd,n.(see [28].)Hilbert then showed that every bivariate non-negative polynomial is a sumof squares of rational functions and Hilbert’s 17th problem asked whetherthis is true in general. In 1900, at the International Congress of Mathe-maticians in Paris, Hilbert gave a lecture in which he proposed 23 openproblems. Most of these have since been solved, and the solutions haveled to fundamental discoveries in mathematics. As one of his famous set ofproblems, Hilbert posed the following as his 17th problem:HILBERT’S 17th PROBLEM. For any f ∈ R[x], is it true that f ≥ 0 on Rn

=⇒ f is a sum of squares of rational functions?- This is trivial when n = 1.- Hilbert proved it, already in 1893, in the case n = 2 .- Artin furnished an affirmative answer to Hilbert’s 17th problem, joining the”honors class” of mathematicians who solved a Hilbert problem. He provedit in the general case (and with R replaced by an arbitrary real closed field)in 1927. Artin’s work represented a major breakthough. His proof combinedtwo new ingredients. The first ingredient - a description of elements of afield positive at every ordering - has since developed into the larger subjectknown as real algebra. The second ingredient - certain ‘specialization lem-mas’ which exploit the theorem of Sturm for real closed fields - has evolvedover time into what is referred to now as Tarski’s Transfer Principle, whichis an important result in the model theory of real closed fields.

The starting point of the history of the 17th problem of Hilbert wasthe oral defense of the doctoral dissertation of Hermann Minkowski at theUniversity of Konigsberg in 1885. The 21 year old Minkowski expressedhis opinion that there exist real polynomials which are non-negative on thewhole of Rn and cannot be written as a finite sum of squares of real polyno-mials. David Hilbert was an official opponent in this defense. Minkowski hadconvinced Hilbert about the truth of this statement. In 1888 Hilbert provedthe existence of a real polynomial in two variables of degree six which isnon-negative on R2 but not a sum of squares of real polynomials.(The proofof Hilbert used some basic results from the theory of algebraic curves.)

1.2 Ordered Fields, Real Closed Fields

Given any field the following are pertinent questions to ask:- can the field be ordered, and if so, in how many ways?- What sort of interplay is there between the notion of an ordering and otheralgebraic notions definable for fields?- Can we regard the set of orderings of a field as some kind of space?- What happens to orderings under field extensions, and how do the variousorderings on a field reflect the properties of the field itself?The study of these questions (and their applications) constitutes essentiallythe theory of ordered fields.

Definition 1 An ordering on a field F is given by the assignment of a setP ⊆ F which satisfies the following axioms:

10

(1) P + P ⊆ P ,(2) PP ⊆ P ,(3) P ∪ (−P ) = F ,(4) P ∩ (−P ) = {0}.

The first three axioms above imply that all sums of squares in F belongto P . If P defines an ordering on F , we shall speak of the pair (F, P ) as anordered field.

As usual, we can define x ≤P y (resp., x <P y) to mean that y − x ∈ P(resp., y − x ∈ P ∗ = P \ {0} ); this ”inequality” relation results in a totalordering of the elements of F.An ordered field (F, P ) has necessarily characteristic equal to zero, hence wewill always assume F ⊇ Q where Q denotes the field of rational numbers.

The following two ways of ”operating” with orderings are useful. First, ifF0 is a subfield of a field F , then any ordering on F induces an ordering onF0 (by (restriction) an ordering on F0). Thus, one way of finding orderingson a field F0 is to try to imbed F0 into other ordered fields. Secondly, if(F, P ) is an ordered field and σ is any automorphism of F , then F can alsobe ordered by σP . In this way, a given ordering can sometimes be used togenerate many new orderings.Remark: The notion of an ordered field can be alternatively defined as fol-lows:Given a relation ≤ on F, we say that the pair (F,≤) is an ordered field if(i) ≤ is a total order.(ii) x ≤ y implies x+ z ≤ y + z.(iii) x ≤ y and 0 ≤ z imply xz ≤ yz.

Because of (ii) we have in an ordered field x ≤ y ⇒ y − x ≥ 0.Thus the order ≤ of an ordered field F is entirely determined by the set

P = {a ∈ F |a ≥ 0},

called the positive set (or set of positive elements) of ≤.We have P + P ⊆ P and PP ⊆ P ; P ∩ −P = {0};P ∪ −P = F .Conversely, if P is a subset of a field F satisfying properties above, therelation ≤ defined by

x ≤ y ⇔ y − x ∈ P

makes F in to an ordered field (with positive set P ). For this reason we alsocall such a subset P of F an order of F. This allows us to switch freely fromP to ≤ and to call a pair (F, P ) or (F,≤) an ordered field.Let F be an ordered field. Then x2 > 0 for every x ∈ F ∗. Thus all sumsof squares of elements in F ∗ are also strictly positive : x2

1 + ... + x2n > 0 for

every x1, ...xn ∈ F ∗.

Definition 2 The ordering ≤ on F is called Archimedean if for each x ∈ Fthere is an n ∈ N such that x ≤ n. If ≤ is Archimedean we call (F,≤) anArchimedean ordered field.

Lemma 1 If (F,≤) is Archimedean, then Q is (isomorphic to) a dense sub-field of F.

11

Proof. Suppose ≤ is Archimedean, x, y ∈ F and x < y.Choose m ∈ N such that (y − x)−1 < m. Multiplication by the positiveelement y − x gives 1 < m(y − x), where mx < my − 1.Choose n ∈ Z minimal such that my ≤ n+ 1.

Then mx < my − 1 ≤ n < my implies x <n

m< y.

Examples :(1) The fields Q and R are the simplest examples of ordered fields and bothare Archimedean with respect to their ”usual” orderings. For either of thesefields, the usual ordering is the only ordering possible.(2) As another example, consider the field F = Q(θ) where θ2 = 2. Thereare two orderings on F , constructed from the two imbeddings of F into R,one sending θ to

√2, the other sending θ to −

√2. These orderings are ”con-

jugate” under the action of σ ∈ Aut(F ) which sends θ to −θ. In a similarfashion, we can construct four different orderings on Q(

√2,√

3), eight or-derings on Q(

√2,√

3,√

5), ... and uncountably many orderings on Q(√p : p

prime). (see Corollary 3)(3) R(x)(x a single indeterminate)First, for ai ∈ R and ak 6= 0 we define

0 < akxk + ak+1x

k+1 + ...+ anxn :⇔ 0 < ak

Second, for p, q ∈ R[x] \ {0}, 0 < p

q:⇔ 0 < pq.

Finally, for r, s ∈ R(x), r < s :⇔ 0 < s− r.We define f(x)

g(x)≥ 0⇔ f(x)g(x) ≥ 0

This is a total order on R(x) = {f(x)

g(x): f(x), g(x) ∈ R[x], g(x) 6= 0}

which makes (R(x),≤) an ordered field.With this ordering the field R(x) is not Archimedean. It contains in-

finitely small elements (i.e. positive and smaller than 1n

for every n ∈ N,such as x, or f(x) = x − r < 0,∀r ∈ R, r > 0) and also infinitely large

elements (i.e. bigger than n, for any n ∈ N such as1

x) i.e 1

x> n for all

n ∈ N.To get an algebraic criterion for fields which admit some ordering, we gen-eralize the notion of positive cones or orderings to pre-positive cones orpre-orderings of fields.

Definition 3 Suppose T is any subset of F. We call T a pre-positive coneof F or pre-ordering of F if T + T ⊆ T, TT ⊆ T, F 2 ⊆ T , and −1 /∈ T .

If in addition, T ∪−T = F then T is an ordering of F. Any ordering of F isalso a pre-ordering, since for any x ∈ F we have x ∈ P and hence x2 ∈ P or−x ∈ P and hence x2 = (−x)(−x) ∈ P . In particular we have 12 = 1 ∈ P ,which implies −1 /∈ P .

Lemma 2 Let T be a pre-ordering of F and let x ∈ F \T . Then T−xT =: T ′

is a pre-ordering of F with T ∪ {−x} ⊆ T ′.

12

Proof. (T − xT )(T − xT ) ⊆ T − xT + x2T ⊆ T − xT . If −1 were equal tot1 − xt2 for some t1, t2 ∈ T , then x = (1 + t1)( t2

t22) ∈ T contradiction. (t2 6= 0

otherwise t1 = −1 ∈ T contradiction.)

Theorem 2 Every pre-ordering T of F is contained in an ordering P. Fur-thermore,

T =⋂T⊆P

P

Proof. Let P be a maximal pre-ordering of F containing T (such P existsby Zorn’s lemma). To show that P is an ordering, suppose x ∈ F \ P ; thenP − xP is a pre-ordering.−x ∈ P − xP and P ⊆ P − xP , then P = P − xP , since P is maximal; thus−x ∈ P as required.

The inclusion T ⊆⋂T⊆P P is trivial. For the reverse, suppose x ∈ F \T .

Then T − xT is a pre-ordering. By the first half of this theorem, T − xT iscontained in an ordering P which, of course, cannot contain x.

For a field F let∑F 2 denotes the set of finite sums of squares of elements

of F . That is∑F 2 = {x2

1 + ...+ x2r|r ∈ N, xi ∈ F}

If F has characteristic 2, then∑F 2 = F 2 ( as x2

1 +x22 = (x1 +x2)2) and if

charF 6= 2 and −1 ∈∑F 2, it follows from the formula x = (x+1

2)2 − (x−1

2)2

that∑F 2 = F or if −1 =

∑a2i , then x = (1+x

2)2 +

∑(ai(

1−x2

))2.

F is called formally real if −1 /∈∑F 2. This condition is equivalent to

whenever x21 + x2

2 + ...+ x2r = 0 in F , all xi must be zero.∑

F 2 is pre-ordering ⇔ −1 /∈∑F 2 ⇔ F has an ordering; note that if

−1 /∈∑F 2, then F has an ordering containing

∑F 2. Hence, we can quote

Artin’s basic discovery (see [2]) that a field

F is formally real if and only if it admits an ordering.

Remark. If P is an ordering of F , the elements P ∗ = P \ {0} are calledpositive, the elements of −P ∗ are called negative (with respect to P ). Theelements of

⋂P ∗ are totally positive, since they are positive for every or-

dering. Thus the totally positive elements are exactly the non-zero sums ofsquares.

Let K/F be a field extension. Any ordering ≤ on K induces (by restric-tion) an ordering on F . If ≤ is an ordering on F , we seek an extension of ≤to K. Equivalently, given an ordering P ⊆ F , we seek an ordering P ′ ⊆ Ksuch that P ′ ∩ F = P . In that case we shall call (K,P ′) or (K,≤) an order- extension of (F, P ) or (F,≤); and we shall call (F, P ) or (F,≤) an orderedsubfield of (K,P ′) or (K,≤).

Lemma 3 An ordering P of F can be extended to K if and only if

TK(P ) := {n∑i=1

piβ2i : n ∈ N, pi ∈ P, βi ∈ K}

is a pre-ordering of K.

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Proof. (⇒) Let P′be an ordering of K with P

′ ∩F = P . Then TK(P ) ⊆ P′,

whence −1 /∈ TK(P ).(⇐) Suppose TK(P ) is a pre-ordering of K. Let P

′ ⊇ TK(P ) be an order ofK (as every pre-order is contained in an ordering). Then P

′ ∩F := P1 is anorder of F with P ⊆ P1; then P = P1.Remark: If P ⊆ P1 are orderings of F, then P = P1.(For 0 6= x ∈ P1 ⇒ −x /∈ P1 ⇒ −x /∈ P ⇒ x ∈ P .)

Theorem 3 Suppose K = F (√a), a ∈ F \ F 2. Let P be an ordering of F.

Then P extends to an ordering P′

of K if and only if a ∈ P .

Proof. (⇒)P′ ∩ F = P implies a = (

√a)2 ∈ P ′ ∩ F = P .

(⇐) Let a ∈ P . If −1 were of the form∑ai(xi + yi

√a)2 with ai ∈ P and

xi, yi ∈ F , then we would have −1 =∑aix

2i + aiay

2i ∈ P , a contradiction.

Now we apply lemma above.

Theorem 4 Suppose [K : F] is finite and odd. Then every ordering P of Fextends to K.

Proof : Suppose that [K : F ] is finite, > 1, odd, and minimal with respectto the property that P does not extend to K. We write K = F (α) =F [X]/(f), f = Irr(α, F )(the monic irreducible polynomial of α over F ), anddegf = 2n + 1(n ≥ 1), using the primitive element theorem. Since P doesnot extend to K, −1 =

∑mi=1 aiγ

2i , with ai ∈ P, γi ∈ K (Lemma above).

Then there exist f1, ..., fm ∈ F [X] such that

0 = 1 +m∑i=1

aifi(X)2(mod f(X)).

Therefore we get 1 +∑m

i=1 aifi(X)2 = f(X)h(X), for some h ∈ F [X]. We(may) choose these fi to have degrees ≤ 2n; then

deg(1 +m∑i=1

aifi(X)2) ≤ 4n.

Moreover, this degree is even, since all ai ∈ P . Therefore degh is odd and≤ 2n− 1. Let h1 be an odd-degree irreducible factor of h in F [X]. Let β bea zero of h1, and let K

′= F (β). Then

(i) [K′: F ] is odd and ≤ 2n− 1, and

(ii) 1 +∑m

i=1 aifi(β)2 = 0, i.e., −1 =∑m

i=1 aiδ2i with δi ∈ K

′.

Thus P does not extend to K′

and [K′: F ] < [K : F ], a contradiction.

Theorem 5 Every ordering P of F extends to F (x) (where x is a singleindeterminate).

Proof. Otherwise, −1 =∑m

i=1 aifi(X)2 for some ai ∈ P \ {0} and fi ∈F (X)−say, fi = gi/h with gi, h ∈ F [X] and no factor of h divides all giin F [X]. Then −h(0)2 =

∑mi=1 aigi(0)2. If h(0) 6= 0, then −1 ∈ P , contra-

diction. And if h(0) = 0, then 0 =∑aigi(0)2, X|h, what implies that all

gi(0) = 0, hence X|gi for all i, another contradiction.

14

Definition 4 An ordered field (F,≤) is called maximal ordered if ≤ doesnot extend to any proper algebraic field extension K of F.

Lemma 4 Suppose (F,≤) is maximal ordered. Then every non-negativeelement of F is a square; thus ≤ is the only ordering on K.

Proof. First, to prove P≤ ⊆ F 2, suppose, on the contrary, that 0 ≤ a ∈F \ F 2. Then ≤ would extend to K := F (

√a) 6= F , contradicting the

maximality of (F,≤).Second, let P be any ordering of F . Then P≤ ⊆ F 2 ⊆ P implies P≤ = P .

Definition 5 An ordered field is called real closed if it has no proper, real,algebraic extension.

Lemma 5 A field F is real closed if and only if F has a unique ordering ≤,and (F,≤) is maximal ordered.

Proof. (⇒) Suppose P is an ordering of F , and a ∈ P . Then a ∈ F 2 (Lemmaabove). Thus P = F 2, and (F, F 2) is maximal ordered, (using that F hasunique ordering if F 2 is an order and the fact that F is real closed.)(⇐) If F had a formally real, proper, algebraic extension K, then K wouldhave an ordering, which would restrict to the unique ordering ≤ of F , con-tradicting the maximality of (F,≤).

The facts above finally lead to following fundamental characterization ofreal closed fields.

Theorem 6 The following are equivalent:(a) F is real closed;(b) F 2 is an ordering of F, and every odd-degree f ∈ F [x] has a root in F;and(c) F 6= F (

√−1), and F (

√−1) is algebraically closed.

Examples.(1) The field R is real closed.(2) The real algebraic numbers (the numbers algebraic over Q) form a realclosed field denoted by Ralg.(3) Let R(x)∧ (resp. C(x)∧) be the field of puiseux series with real (resp.complex) coefficients, i.e., the set of expressions

∑+∞i=k aix

i/q with k ∈ Z, q ∈N, ai ∈ R (resp. C). Then C(x)∧ is algebraically closed. Since C(x)∧ =R(x)∧[i],R(x)∧ is real closed. A positive element of R(x)∧ is a puiseux seriesof the form

∑+∞i=k aix

i/q with ak > 0.

Theorem 7 Suppose F is real closed; let ≤ be the unique ordering on F.Then for any f ∈ F [x],(a) f factors into F-irreducible polynomials of the form x − a, (a ∈ F ) and(x− a)2 + b2, (a, b ∈ F, b 6= 0);(b) if a < b ∈ F and f(a) < 0 < f(b), then there is some c ∈ F witha < c < b such that f(c) = 0.

15

Proof.(a) F (√−1) is algebraically closed. So any irreducible polynomial in

F [X] must have degree ≤ 2; so if it is monic, it must be of the form X − aor X2− 2aX + c, for some a, c ∈ F . The latter equals (X − a)2 + (c− a2); inorder for it to be irreducible, we must have c−a2 /∈ −F 2, whence c−a2 = b2,for some b ∈ F \ {0}.(b) Factor f as in (a), and note that a sign-change of f(x) for x between aand b can come only from a linear factor.

Definition 6 Let (F,≤) be an ordered field, and (K,≤) an order extensionof (F,≤).(K,≤) (or simply K) is called a real closure of (F,≤) if K is real closed andK/F is algebraic.

Theorem 8 Every ordered field has a real closure.

Theorem 9 Every two real closures of an ordered field (F,≤) are F−isomorphic.

Corollary 2 Let (R,R2) be a real closure of (F, P ). Let F1, F2 be interme-diate fields of R/F , and let σ : F1 → F2 be an isomorphism that fixes Fand respects the orderings induced by R2. Then F1 = F2 and σ = id. Inparticular, Aut(R/F ) = {id}.

Consequently, the F−isomorphism between two real closures R1 and R2 of(F, P ) given by theorem above is unique. On the other hand, there can beother isomorphisms between R1 and R2 that do not fix F .Remark. The hypothesis in (Corollary above) that σ fixes F can not bedropped. For example, let F = F1 = F2 = R({Xr|r ∈ Q and r > 0}),where the Xr are indeterminates. We define a lexicographic ordering < onF by declaring that for each r ∈ Q with r > 0, R({Xs|s ∈ Q and 0 < s <r}) < Xr. Let R be the real closure of (F,<). Then the order preservingpermutation Xr 7→ X2r induces an order-preserving R−automorphism σ :F1 → F2, but σ 6= id. Moreover, one can show that σ extends to a non-trivialautomorphism of R.

Corollary 3 Let (R,R2) be a real closure of (F, P ). Suppose α ∈ R. Thenthe number of extensions of P to an ordering of F (α) is equal to the numberof F-embedding of F (α) in R, i.e., the number of zeros of Irr(α, F ) in R.

The following result is the key to Artin’s solution of Hilbert’s 17th problem.

Theorem 10 (Artin-Lang Homomorphism Theorem). Let R be a real-closedfield and A an R-algebra of finite type. If there exists an R-algebra homo-morphism ϕ : A→ R1 into a real closed extension R1 of R, then there existsan R-algebra homomorphism ψ : A→ R.

Theorem 11 Let R be real closed field and f ∈ R[x1, ..., xn]. If f is non-negative on Rn, then f is a sum of squares in the field of rational functionsR(x1, ..., xn).

16

Proof. If f is not a sum of squares in R(x1, ..., xn), then, by corollary 1,there exists an ordering on R(x1, ..., xn) for which f is negative. Let K bea real closure of R(x1, ..., xn) equipped with this ordering. Then −f has anon-zero square root in K, and hence there is an R-algebra homomorphismR[x1, ..., xn][T ]/(fT 2+1)→ K. By the Artin-Lang homomorphism theorem,there is an R-algebra homomorphism R[x1, ..., xn][T ]/(fT 2 + 1) → R. Thisimplies the existence of a point x in Rn such that f(x) < 0.

1.3 The Space of Orderings of a Field

(This section is from [31]). We saw that for the characterization of sums ofsquares we considered all orderings of a field. So it is important to investigatethe set of all orderings of a field F:

X(F ) := {P : P ordering of F}.

Our aim is to define a canonical topology on X(F ). Every ordering P definesthe sign-function

sP : F ∗ → {±1}, sP (x) = 1⇔ x ∈ P ∗.

The set of all mappings F ∗ → {±1} denoted by {±1}F ∗ or ΠF ∗{±1}. SinceP is completely determined by sP we get an injection

X(F )→ ΠF ∗{±1}.

We give {±1} the discrete topology and Π{±1} the product topology. By theTychnoff theorem Π{±1} is quasi-compact. We consider X(F ) as a subsetof Π{±1} and endow it with the relative topology. X(F ) is called the spaceof ordering of F .

The topology in Π{±1} is defined by taking the following sets as a sub-basis

π−1a (ε) = H(a, ε) = {f : F ∗ → {±1}|f(a) = ε}

where a ∈ F ∗, ε = ±1, and πa : Π{±1} → {±1} is the projection on thea− th factor. Since the following union is disjoint

Π{±1} = H(a, ε) ∪H(a,−ε)

the sets H(a, ε) are open and closed. Such subsets of a topological space arecalled clopen. A topological space is called totally disconnected if any twopoints can be separated by clopen sets. In our case this is possible: For,if f, g ∈ Π{±1} and f(a) 6= g(a), then f and g are separated by H(a, ε)and H(a,−ε). Thus Π{±1} is a boolean space, that is , it is hausdorff,quasi-compact, and totally disconnected.

Theorem 12 X(F ) is a closed subset of Π{±1}. Thus X(F ) is a booleanspace with respect to the induced topology.

17

Proof. We show that the complement of X(F ) is open. Let s : F ∗ → {±1}be a function which is not defined by an ordering. One of the three definingconditions of an ordering must be violated. Therefore

s(a) = s(b) = 1 but s(a+ b) = −1

ors(a) = s(b) = 1 but s(ab) = −1

ors(a) = s(−a)(= ε)

for suitable a, b ∈ F ∗. In each of the three cases we can find an openneighborhood of s in the complement of X(F ), namely

H(a, 1) ∩H(b, 1) ∩H(a+ b,−1)

orH(a, 1) ∩H(b, 1) ∩H(ab,−1)

orH(a, ε) ∩H(−a, ε)

respectively.A subbasis for the topology of X(F ) is given by the following sets

HF (a) = H(a) = H(a, 1) ∩X(F ) = {P ∈ X(F )|a ∈ P}.

(This follows from H(a,−1)∩X(F ) = H(−a).) This is the so called Harrisonsubbasis.

1.4 Valuations

Definition 7 Let F be a field.A surjective mapping υ : F → Γ ∪∞ where Γ is an ordered abelian group,amended by an element ∞, being considered larger than every element in Γ,is called a valuation if it satisfies the following axioms:(i) υ(x) =∞⇔ x = 0(ii) υ(xy) = υ(x) + υ(y)(iii) υ(x+ y) ≥ min{υ(x), υ(y)}

Proposition 7 Let F be a field and let υ be a valuation over F. Then(i) υ(1) = 0,(ii) υ(a−1) = −υ(a) for all a 6= 0,(iii) υ(a) = υ(−a),(iv) if υ(a) 6= υ(b), then υ(a+ b) = min{υ(a), υ(b)},(v) υ(

∑ni=1 ai) ≥ min1≤i≤n{υ(ai)} and equality holds if υ(ai) 6= υ(aj) for all

i 6= j,(vi) if

∑ni=1 ai = 0, n ≥ 2, then there exist i 6= j such that υ(ai) = υ(aj).

18

Consider an arbitrary field F and a valuation of F with values in an orderedgroup Γ.

Let Oυ = {x ∈ F : υ(x) ≥ 0}.Then since υ(x) = υ(−x), υ(xy) = υ(x) + υ(y) and since Γ is an orderedgroup it follows that Oυ is a ring.Furthermore, for x ∈ F , if x /∈ Oυ, we have υ(x) < 0.Thus υ(x−1) = −υ(x) > 0, that is x−1 ∈ Oυ.

Hence given x ∈ F , we have x ∈ Oυ or x−1 ∈ Oυ.Furthermore, for x ∈ F , if x ∈ Oυ, then x =

x

1∈ Oυ and if x /∈ Oυ, then

x−1 ∈ Oυ and therefore x =1

x−1∈ quotOυ where quotOυ denotes the field

of quotients of Oυ, which proves that F = quotOυ.Now, x ∈ Oυ is a unit if and only if x−1 ∈ Oυ, that is, υ(x) ≥ 0 and

υ(x−1) = −υ(x) ≥ 0.Therefore, O∗υ = {x ∈ F : υ(x) = 0}.

Let mυ = {x ∈ F : υ(x) > 0} consists of the non-units of Oυ. In fact mυ

is an ideal.Therefore, Oυ is a local ring with maximal ideal mυ.Finally, υ : (F ∗, ·) → (Γ,+) is a group epimorphism with kerυ = O∗υ.

Thus (Γ,+) ∼= (F ∗/O∗υ, ·).The above discussion can be summed up as follows

Proposition 8 If F is a field and υ a valuation over F, then Oυ = {x ∈ F :υ(x) ≥ 0} is a subring of F such that for all x ∈ F , x ∈ Oυ or x−1 ∈ Oυ.In particular, Oυ is a local ring with maximal ideal mυ = {x ∈ F : υ(x) >0} = Oυ \ O∗υ and O∗υ = {x ∈ F : υ(x) = 0}.Furthermore, we have quot Oυ = F and the value group of υ is isomorphicto F ∗/O∗υ.

In our study of rational and elliptic function fields we will be dealing withthe nice class of discrete valuations defined as valuations the value groupof which are (isomorphic to) the additive group Z = (Z,+) of the integers.

Examples.(1) The discrete p-adic valuations(a) Let F = Q,Γ = Z and υ = υp the p−adic valuation, for p ∈ Z a rationalprime. That is , for x ∈ Q∗, we write x = pn a

b, n ∈ Z, p - ab and υp(x) = n.

Similarly, we define the p−adic valuation for every irreducible polynomialp ∈ K[x], K a field. For h ∈ K(x) \ {0} there are f, g ∈ K[x] \ {0} such thath = pr f

g, r ∈ Z where f, g are not divisible by p and we define υp(p

r fg) = r.

We note that in the second case, υp restricted to K is trivial.

(2)There is one more interesting discrete valuation on K(x) trivial on K.This is the socalled degree valuation or infinite valuation υ∞ definedbyυ∞(0) =∞ and , for non-zero polynomials f, g ∈ K[x], byυ∞(f/g) = deg g−degf . We observe that f/g is a unit in the valuation ringV∞ if and only if deg g = degf .

(3) Let F be a field, and let K = {∑

n∈Z anxn : an ∈ F for all n, the set

{n ∈ Z : an 6= 0} is bounded below } be the set of all formal Laurent series

19

over F in one variable x. For a non-zero element f ∈ K there is a minimumexponent of x occurring in f , which implies that we can write f uniquely asf = xng for some n ∈ Z and g ∈ F [[x]] with non-zero constant coefficient.Now,

υ : K∗ → Z,∑n∈Z

anxn 7→ min{n ∈ Z : an 6= 0}

is a valuation on K.

(4) The idea of (3) can be generalized to construct a valuation with valuegroup Q: for a field F we now consider the set K = {

∑q∈Q aqx

q : aq ∈ Ffor all q, the set {q ∈ Q : aq 6= 0} is bounded below and has boundeddenominators } of all formal power series with rational exponents, and thesame map

υ : K∗ → Q,∑q∈Q

aqxq 7→ min{q ∈ Q : aq 6= 0}

as above. υ is a valuation with value group Q. The field K is called the fieldof (formal) puiseux series over F .

20

Chapter 2

Real Holomorphy Rings ofRational Function Fields

2.1 Algebraic Function Fields of One Vari-

able

Definition 8 An algebraic function field F/K of one variable over K is anextension field F ⊇ K such that F is a finite algebraic extension of K(x) forsome element x ∈ F which is transcendental over K.

In other words, there exists x ∈ F , transcendental over K, and there existy1, ..., yn ∈ F , all algebraic over K(x), such that F = K(x, y1, ..., yn).Examples.

1. Rational Function FieldsThe simplest example of an algebraic function field is the rational functionfield; F/K is called rational if F = K(x) for some x ∈ F which is transcen-dental over K.We have a similar definition in any number of variables: a finitely generatedextension F/K is called rational if there exist x1, ..., xr ∈ F algebraicallyindependent over K such that F = K(x1, ..., xr).

In several ways, the rational function fields in one variable are analogousto Q ( and in the same way the function fields are analogous to number fields,i.e. finite extensions of Q). For instance, once a generator x of F/K is given,we can see F as field of quotients of the ring K[x] which is an euclidean ring,analogous to Z.

2. Elliptic Function Fields over RDefinition 9 Let f(x) ∈ R[x], degf = 3 , f(x) /∈ R(x)2 and f withoutmultiple roots in C. Then R(x)(

√f) is called an elliptic function field over

R.

2.2 Real Valuation Rings and Description of

The Real Holomorphy Ring

In this section we want to write down the full list of real valuation rings andto give the description of real holomorphy ring , as well as of its group of

21

units and its group of totally positive units.

2.2.1 Real Valuation Rings of R(x)

We start with the general definitions.

Definition 10 Let V be a subring of a field K. We say that V is a valuationring of K if x ∈ V or x−1 ∈ V for any x ∈ K.

Examples of Valuation rings(The field K is trivial valuation ring.)1. Let K = Q, with a fixed prime p.Let Vp contain 0 and all rationals of the form pr m

n, where r ≥ 0 and p divides

neither m nor n.It follows from the definition that Vp is a valuation ring referred to as thep-adic valuation ring of Q.2. Let K = F (x), where F is a field.Let Vp contain 0 and all rational functions pr f

g, where r ≥ 0, p irreducible

over F [x] and f , g are arbitrary polynomials in F [x] not divisible by p. Vpis called the p-adic valuation ring of K.3. Let K = F (x) , and let V∞ contain 0 and all rational function f

g∈ F (x)

such that deg f ≤ deg g. V∞ is called the degree valuation ring or thevaluation ring at infinity of K.4. Let K be the field of formal Laurent series over F .Thus a non-zero element of K looks like f =

∑∞i=r aix

i with ai ∈ F, r ∈ Zand ar 6= 0. We may write f = arx

rg where g belongs to the ring V = F [[x]]of formal power series over F .Moreover, the constant term of g is 1, and therefore g, hence can be invertedin V (by long division).The ring Oυ in proposition 8 (chap.1) is a valuation ring. We saw therethat every valuation υ on a field K determines a valuation ring V in K.(V = Oυ = {x ∈ K : υ(x) ≥ 0}.)Now we show that conversely, every valuation ring V in K determines avaluation υ on K. The basic sample for this relationship is provided by thep-adic valuations and the p-adic valuation rings.

Proposition 9 If V is a valuation ring and K = quot V, then K∗/V ∗ is anordered group and the natural projection is a valuation with valuation ringV and the value group K∗/V ∗.

Proof. The multiplicative quotient group K∗/V ∗ is an abelian group. Werewrite it additively by setting for the cosets xV ∗ and yV ∗:

xV ∗ + yV ∗ := xyV ∗.

Furthermore we define a binary relation ≤ on K∗/V ∗ by

xV ∗ ≤ yV ∗ ⇔ y

x∈ V

22

It is easily checked that this turns K∗/V ∗ = Γ in to an ordered abeliangroup. The desired valuation is now defined by

υ(x) = xV ∗ ∈ Γ for x ∈ K∗, and υ(0) =∞.

Properties of Valuation RingsLet V be a valuation ring of the field K.

1. The field of quotients of V is K.2. The ideals of a valuation ring V are ordered by inclusion.3. Every overring of a valuation ring is a valuation ring; if {Pλ} is the set ofproper prime ideals of V , then {VPλ} is the set of overrings of V .4. Every valuation ring is a local ring, i.e., a ring with a unique maximalideal.5. Every valuation ring is integrally closed.6. The valuation rings between V and K form a chain.7. If P is a prime ideal of a valuation ring V , then VP and V/P are also avaluation ring.8. If V is a Noetherian valuation ring, then V is a PID. Moreover, forsome prime p ∈ V every ideal is of the form (pm),m > 0. For any such p,∩∞m=1(pm) = 0.9. Let R be a subring of the field K. The integral closure R of R in K isthe intersection of all valuation rings V of K such that V ⊇ R.10. Let V be a valuation ring and let a be an ideal of V .

(1) rad(a) is prime ideal of V .(2) If b = ∩∞n=1a

n, then b is a prime ideal of V which contains everyprime ideal of V which is properly contained in a.Let mV be the maximal ideal of a local ring V . Since the ring is local, weconclude that V −mV is the multiplicative group of invertible elements(units)of the ring V .The quotient field V/mV is called the residue field of V .Let mV be the maximal ideal of a valuation ring V of a field K. Then forany a ∈ K − {0} we have a /∈ V ⇐⇒ a−1 ∈ mV .

A valuation or a valuation ring is called real if its residue field is formallyreal, the valuation rings attached to discrete valuations are called discretevaluations rings.

Let K be a field and υ : K∗ → Z a discrete valuation with valuation ringV and maximal ideal p. Let π ∈ p such that υ(π) = 1 (π is called a primeelement or uniformizing element of the valuation.) Then given x ∈ K∗such that υ(x) = n, we have υ(π−nx) = 0, that is, π−nx ∈ V ∗, so that x canbe written as x = aπn with a ∈ V ∗ and n ∈ Z. This representation is uniquesince if x = bπm with b ∈ V ∗ and m ∈ Z, we have υ(x) = υ(bπm) = m = n.Thus a = b. In particular, if x ∈ p then x = aπn with n ≥ 1, and a ∈ V ∗ sop = (π). Therefore p is principal.

Let b be any ideal of V such that b 6= 0 and b ⊆ p. Let n = min{υ(x) :x ∈ b}. Then n ≥ 1. Then there exists x ∈ b such that υ(x) = n, that is,x = aπn ∈ b with a ∈ V ∗. This implies that πn ∈ b and (πn) ⊆ b. If y is anarbitrary non-zero element of b, we have υ(y) ≥ n. Therefore y ∈ (πn), thatis, b = (πn) = pn.Hence, every non-zero ideal of V is a power of p. We have the followingtheorem:

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Theorem 13 If υ is a discrete valuation over a field K with valuation ringV then the maximal ideal p of V is principal and is generated by any primeelement. Every non-zero ideal of V is a power of p and the groups K∗ andV ∗ × Z are isomorphic.

Proof. To prove the last part, let x ∈ K∗, we can write x = aπn ina unique way, and therefore the function ϕ : K∗ → V ∗ × Z, defined byϕ(x) = (a, n), is the isomorphism needed.

Corollary 4 Let R be a discrete valuation ring with maximal ideal m. Ift ∈ m \m2, then t is a uniformizer.

Proof. m 6= m2. We choose t ∈ m \ m2. We have (t) = mn for some n ≥ 0.We can not have n = 0 because (t) ⊆ m ⊆ R, and we can not have n ≥ 2 bychoice of t. The only possibility is n = 1, hence m = (t).

Corollary 5 (Ideals in a discrete valuation ring). Let V be a discrete val-uation ring with unique maximal ideal p. Then the non-zero ideals of V areexactly the powers pn for some n ∈ N. They form strictly decreasing chain

V = p0 ) p1 ) p2 ) ....

The following theorem is central for what follows.

Theorem 14 Every non-trivial valuation on K(x) which is trivial on Kis either the degree valuation or a p − adic valuation for some irreduciblepolynomial p ∈ K[x].

Proof. We are going to verify that given any non-trivial valuationυ : K(x)→ Γ such that Γ is an ordered group and υ(a) = 0 for all a ∈ K∗,then υ is equivalent to υ∞ or to some υp, where p(x) ∈ K[x] is monic andirreducible.Let V be the valuation ring of υ and let m be its maximal ideal. Now, ifx ∈ V , then K[x] ⊆ V . Let P = m ∩K[x]. We have that P is a prime idealof K[x], K ∩ P = {0} and 1 /∈ P .It follows that P = (p), where p is monic and irreducible polynomial orp = 0.If p = 0, then υ(K[x])∗ = {0}, so υ(K(x)∗) = 0, then we have K(x) = V ,which contradicts the hypothesis that υ is non-trivial. Therefore P = (p)with p 6= 0.Let g, h ∈ K[x] with p - h and h /∈ P , that is h is a unit in V . We haveυ( g

h) ≥ 0, which implies Vp ⊆ V .

Now assume that u(x) ∈ K(x) \ Vp. Then u = gh

with gcd(g, h) = 1 and p|h.If u ∈ V , since g ∈ V ∗p ⊆ V , it follows that h−1 = g−1u ∈ V .However, we have h ∈ P ⊆ V , and this implies that h is a non-unit which isabsurd.Hence u /∈ V and we have V ⊆ Vp, so V = Vp. Therefore v and vp haveisomorphic value groups, and are essentially the same.

If x /∈ V , then y = 1x

= x−1 ∈ V . From the above discussion, we concludethat P ∩K[y] = (l(y)), where l(y) is a monic and irreducible polynomial.Now x = y−1 /∈ V , which is equivalent to saying that y is not a unit. Thus

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y ∈ P ∩K[y] = (l(y)) and l(y)|y, which proves that l(y) = y.Therefore v∞ and v have isomorphic value groups, and are essentially thesame.

As a consequence we obtain the complete list of the valuation rings ofR(x) which are trivial on R. We note that a real valuation ring of anyfield K which contains R must be trivial on R, otherwise we would get anon-archimedean ordering on the field of real numbers which does not exist.

Now, the monic irreducible polynomials p over R are known and we get

(a) deg p = 1. Then p = x− α, Vα := Vx−α, Vα/(x− α) = R,(b) deg p = 2. In this case R[x]/(p) = C,(c) Finally V∞/m∞ = R.

Therefore, the complete list of real valuation rings for R(x) reads asfollows:

Vα, α ∈ P1 := R ∪∞.

2.2.2 Real Holomorphy Rings

At the beginning of the extension of Artin-Schreier theory which Beckerstarted in the 1970s valuation rings and higher level orderings were the mostcrucial notions. Later on, it is found that real holomorphy rings have greatinfluence in the study of sums of 2n−th powers. In fact the ideal structure ofholomorphy rings and the structure of its group of units have great influence.

Definition 11 The real holomorphy ring H(K) of a formally real field K isthe intersection of all real valuation rings of K, i.e. valuation rings of K witha formally real residue class field. By E(K) we denote the group of units ofH(K), by E+(K) := H(K)∗ ∩

∑K2 the subgroup of totally positive units.

In the case of R(x) we know all real valuation rings. Consequently, thereal holomorphy ring of R(x) denoted by H(R(x)) and its group of units canbe described as follows.

H(R(x)) =⋂α∈R Vα

⋂V∞ =

⋂α∈R∪∞ Vα =

⋂Vα∈P1

= {fg

: g(α) 6= 0,∀α ∈ R}⋂{fg

: deg f ≤ deg g}

= {fg

: deg f ≤ deg g, g(α) 6= 0,∀α ∈ R}

As to the units we start with the observation that H(R(x))∗ =⋂α∈P1 V ∗α . It

follows that the units of can be described in the following way:

(E(R(x))) = {fg

: f, g 6= 0, gcd(f, g) = 1, f and g without roots in R, deg f = deg g}.

(E+(R(x))) = {fg

: f, g 6= 0, gcd(f, g) = 1, f and g positive definite on R, deg f = deg g}.

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2.3 Real Places

In this section we are summarizing general facts about real places which areneeded to sketch the general theory of the real holomorphy rings of arbitraryformally real fields.

Definition 12 A place of a field K is a mapping ϕ of K in to F ∪ {∞}(where F may be any field) such that(i) ϕ(x+ y) = ϕ(x) + ϕ(y)(ii) ϕ(xy) = ϕ(x)ϕ(y)(iii) ϕ(1) = 1.

Here for all a ∈ F the following operations are defined:

a+∞ =∞+ a =∞, a.∞ =∞.a =∞.∞ =∞

The operations ∞+∞, 0.∞ and ∞.0 are not defined.

Consider the set Vϕ of elements a ∈ K such that ϕ(a) 6= ∞. Then Vϕ is asubring of K, and ϕ is a homomorphism of this ring into F . Since F is afield, the kernel of this homomorphism is a prime ideal p of Vϕ.

Let b be an element in K which is not in Vϕ. We contend that ϕ(1/b) = 0.For if this were not true, we would get 1 = ϕ(1) = ϕ(b)ϕ(1/b) =∞, by (ii).Thus, 1/b ∈ p, and thus p is precisely the set of non-units of Vϕ.

Since any ideal strictly containing p would contain a unit, we see that pis a maximal ideal and hence the image of Vϕ in K is again a field. We shalltherefore assume that F is precisely the image of Vϕ by ϕ is a mapping ontoF ∪ {∞}.OR, Given a place ϕ we define Vϕ = {a ∈ K : ϕ(a) 6= ∞} = ϕ−1(F ). ThenVϕ is an integral subdomain of K.

We observe that ϕ : Vϕ → F is a homomorphism such that

kerϕ = {a ∈ Vϕ : ϕ(a) = 0} = mϕ = ϕ−1(0).

Then Vϕ/mϕ∼= ϕ(Vϕ) ⊆ F and kerϕ = mϕ is a prime ideal of Vϕ. (We know

that if b ∈ K \ Vϕ, ϕ(b) = ∞ and ϕ(1/b) = 0. Thus for any x ∈ K we havex ∈ Vϕ or x−1 ∈ Vϕ i.e. Vϕ is a valuation ring.

We saw above how to obtain a valuation ring from a place. Conversely,consider a valuation ring V,m its maximal ideal and K = quotV .

Let F be the field V/m and F1 = F ∪ {∞}.Let ϕ : K → F1 be given by

ϕ(x) =

{x mod m if x ∈ V∞ if x /∈ V

Then ϕ is a place.Thus we summarize it by re-writing the definition of a place again.

Definition 13 A place λ : K → F ∪∞, where K and F are fields, is a ringhomomorphism λ : V → F of a valuation ring V of K into F with kerλ = mwhereby we set λ(x) = ∞ if x /∈ V. Places λ : K → R ∪ ∞ are called realplaces.

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Examples of Places

(1) The real places of R(x), λα, α ∈ P1

The complete list of real valuation rings of K := R(x) is known: Vα, α ∈ P1.They all have residue field R. Consider any real place λ : K → R ∪∞ andits associated valuation ring Vλ. This latter ring is a real valuation ring, say,Vλ = Vα, some α ∈ P1. Hence, R ⊆ Vλ,mλ = mα and we deduce that theplace λ induces, by restriction, a field homomorphism λ0 : R→ R = Vα/mα.There is just one field homomorphism, namely the identity.We summarize: Vλ = Vα, λ|mα = 0 and λ|R = id. These three informationsalone allow to determine the place uniquely.First consider Vα, α 6=∞ and any r ∈ Vα, i.e. r = f/g, f, g ∈ R[x], g(α) 6= 0.Then r(α) := f(α)/g(α) ∈ R is well-defined and we find that r − r(α) =(f · g(α)− f(α) · g)/g · g(α) ∈ mα = (x− α)Vα. Finally, λ(r) = r(α).This place is denoted by

λα, defined by λα(r) = r(α) ∀r ∈ Vα.

We next turn to r ∈ Vλ = V∞. Following the reasoning just given we have tofind r(∞) with r − r(∞) ∈ m∞. Then we would get λ(r) = r(∞). In viewof the presentation r = f/g, f, g ∈ R[x], g 6= 0, degf ≤ degg we can assumef =

∑k0 aix

i, g =∑k

0 bixi, bk 6= 0. Then r − (ak/bk) ∈ m∞ = 1/xV∞ and

we obtain λ(r) = ak/bk. In particular, as claimed, λ is uniquely determineddenoted by λ∞.There is a more conceptual way of describing this ”place at infinity”. Infact one verifies for any r ∈ V∞ the equation

λ∞(r) = lim|γ|→∞

r(γ).

Therefore, the complete list of real places of R(x) reads as follows:

λα, α ∈ P1.

(2) Let K be a field.For all u ∈ K((t))∗, let us write f(u) = ∞ if u /∈ K[[t]] and define f(u)

to be the constant term of u if u ∈ K[[t]].Then f is a place of K((t)), with residue field K and ring K[[t]].For K[[t]] is a valuation ring of K((t)) and the restriction of f to K[[t]] is

identified with the cannonical homomorphism of K[[t]] onto its residue field.(3) Let S be a connected complex analytic variety of dimension 1 and K thefield of meromorphic functions on S.

For all z0 ∈ S, the mapping f 7→ f(z0) from K to C ∪ {∞} is a place ofK whose ring is the set of f ∈ K which are holomorphic at z0 and whoseideal is the set of f ∈ K which are zero at z0.

Space of real PlacesDetails can be found in [2],[19].Let K be a formally real field and let P

be an ordering of K. Then set

A(P ) = {a ∈ K : r ± a ∈ P for some r ∈ Q, r > 0}.

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A(P ) is shown to be a valuation ring of K. That is P determines a valuationυP whose ring A(P ) is the convex hull of the rationals Q with respect to P .

A(P ) is called the ring of finite elements relative to P.In A(P ) we have the ideal

I(P ) := {a ∈ K : r ± a ∈ P for all r ∈ Q, r > 0}

of the infinitely small elements, that are elements which, with respect toP , are smaller in absolute value than all positive rationals.

Theorem 15 (i) A(P ) is a real valuation ring with maximal ideal I(P ),(ii) P := {a+ I(P ) : a ∈ P ∩A(P )} is an Archimedean order of the residuefield A(P )/I(P ). Thus A(P )/I(P ) is naturally embedded in R, and we canthink about it as a subfield of R,(iii) given any real valuation ring V of K there exists an ordering P suchthat A(P ) ⊆ V .

Above, we had seen how places arise from valuation rings. This can beapplied to the valuation ring A(P) and yields the first statement of thefollowing theorem. The second one uses a method of pulling back orderingsfrom a residue field to the the field in question.

Theorem 16 (i) An order P of K gives rise to a real place λP : K → R∪∞with valuation ring A(P ),(ii) every real place on K is of the type λP for some ordering P of K,(iii) for P ∈ X(K), λP is the only real-valued place ξ : K → R∪∞ with theproperty ξ(P ) ≥ 0.

Corollary 6 A field K is formally real if and only if it admits a real placeλ : K → R ∪∞.

From theorem 15 we draw the conclusion that H(K) can already be de-scribed by the rings A(P) or the real places. To this end we set:

M(K) = {λP : P ∈ X(K)}, where X(K) is the space of orderings of K.

M(K) is called the space of real places of the field K. We now get:

H(K) =⋂

P∈X(K)

A(P )

H(K) = {a ∈ K : ∀ζ∈M(K) : ζ(a) 6=∞}.

The following lemma provides a large supply of elements in H(K).

Lemma 6 Let K be a formally real field, and b1, ..., bn ∈ K. Then(i) 1

1+b21+...+b2n∈ H(K);

(ii) bi1+b21+...+b2n

∈ H(K) for all i;

(iii) K is the field of quotients of H(K).

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As before, for a formally real field K, X(K) denotes the set of all order-ings on K, topologized by taking as subbasis the Harrison setsH(a) = {P ∈ X(K) : a ∈ P} where a ∈ K∗.

From above, we have a map Λ : X(K)→M(K) which sends an orderingP ∈ X(K) to λP , its associated real place. We know that λP has beencharacterized as the only place ξ ∈ X(K) with the property that ξ(P ) ≥ 0.

Proposition 10 The map Λ : X(K)→M(K) is onto.

Using the surjection Λ in proposition above, we can view M(K) as a quotientset of X(K).Therefore, we can make M(K) into a topological space by giving it thequotient topology T induced by the topology on X(K). Since X(K) isquasi-compact, (M(K),T) is also quasi-compact. To prove that it is in facteven a Hausdorff space one has to discusss functions on this space.For any element a ∈ K, we have the following ”evaluation map” at a:

a : M(K)→ R ∪∞,

defined by a(ξ) = ξ(a) for any ξ ∈M(K).We view P1 := R∪∞ as the one-point compactification of R. In other words,we identify R ∪∞ with the unit circle S1 with its usual topology, details inthe subsequent section.

Theorem 17 For any a ∈ K, the evaluation map a : M(K) → R ∪ ∞ iscontinuous on (M(K),T).

Lemma 7 The ring of evaluation maps {a : a ∈ H(K)} from M(K) to Rseparates points on M(K).

That we are dealing with a ring of functions M(K) → R follows from theproperties:

a+ b = a+ b, a · b = a · b ∀a, b ∈ H(K).

Theorem 18 (M(K),T) is a Hausdorff space.

Corollary 7 The continuous map Λ : X(K)→M(K) is a closed mapping.

Using the results obtained above, we can now describe the topology on M(K)in a somewhat concrete way.

Theorem 19 The topology T on M(K) coincides with the coarest topologyT′

for which all the functions {a : a ∈ H(K)} into R are continuous. Asubbasis for this topology is given by the setsH′(a) := {ξ ∈M(K) : ξ(a) > 0} = a−1(0,∞), where a ∈ H(K).

The description of the topology onM(K) given above points to the funda-mental difference between the space M(K) and the space X(K). In X(K),the Harrison sets H(a), (a ∈ K∗) are closed with respect to complements(since X(K) \ H(a) = H(−a)), so each H(a) is a clopen (closed and openset), and so X(K) is a totally disconnected (hence Boolean) space. In M(K),

29

however, the complement of a subbasic set H′(a) = {ξ ∈ M(K) : ξ(a) > 0}

is {ξ ∈M(K) : ξ(a) ≤ 0}, which is no longer of the form H′(b), and may not

be even open. Therefore, we have no reason to expect M(K) to be a Booleanspace. In the special case when all the orderings on K are archimedean, themap Λ is a bijection, so M(K) homeomorphic to X(K), hence a Booleanspace. In general, however, M(K) need not be totally disconnected. In fact,there exist many examples of fields K for which M(K) even turn out to beconnected.

Let C(M(K),R) be the ring of real-valued continuous functions onM(K).Since M(K) is compact, we can define the sup-norm on C(M(K),R), by

||f || = sup{f(ξ) : ξ ∈M(K)} (f ∈ C(M(K),R)).

In this context, the classical Stone-Weierstraß Theorem can be brought intoplay. In fact, consider the family of functions

E := {a : a ∈ H(K)} ⊆ C(M(K),R).

Since E separates points on M(K), and E contains the constant function1(= 1), the Stone-Weierstraß Theorem implies that this set lies dense inC(M(K),R). This is an important fact. In order to formulate it in a mostmeaningful way we introduce the ring homomorphism

Φ : H(K)→ C(M(K),R), a 7→ a.

That Φ is a ring homomorphism was already noted above. With all this weget:

Theorem 20 The representation Φ : H(K) → C(M(K),R) has a denseimage.

For the application of theorem above we consider the problem ”separat-ing” two sets of orderings in X(K). Let A,B be two disjoint closed sets inX(K). We say that we can separate A from B if there exists an elementc ∈ K∗ which is positive with respect to any ordering in A, and negativewith respect to any ordering in B. If this is the case, we shall say that A isseparated from B by c.

Proposition 11 (SEPARATION CRITERION). Let A, B be disjoint closedsets in X(K). The following three statements are equivalent:(1) Λ(A) is disjoint from Λ(B) in M(K);(2) we can separate A from B by an element c ∈ H(K) which is a valuationunit at any place in Λ(A ∪B);(3) we can separate A from B by an element c ∈ K∗ which is a valuationunit at any place in Λ(A).

2.4 Topological Representation of H(R(x))We show that P1 is a compact (= quasi compact + Hausdorff) space, andthat is homeomorphic to S1 via the stereographic projection.

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The one-point compactification of RS1 = {(x1, x2) ∈ R2 : x2

1 + x22 = 1}.

We stereographically project the one-dimensional unit circle S1 from its”north-pole” N = (0, 1) onto R.

Every point x ∈ S1 \ {N} is thus associated to the point ϕ(x) at whichthe line containing N and x intersects R. This gives us a continuous bijection

ϕ : S1 \ {N} → R, ϕ(x1, x2) = x11−x2

and the inverse map ϕ−1 : R→ S1 \ {N}, ϕ−1(x) = 2xx2+1

+ ix2−1x2+1

is also continuous.If we extend ϕ to a bijection ϕ′ : S1 → P 1 = R ∪∞where ϕ′(x) = ϕ(x) for x 6= N,ϕ(N) =∞

Then both ϕ′ and its inverse are continuous.We may thus identify P1 and the circle S1 as topological spaces. We there-fore can state theFact : P1 is compact, connected topological space and carries a metric(inheritedfrom S1).

We next show that the elements of R(x) operates on P1 as a continuousfunctions with values in P1.

Let f = gh

: g, h ∈ R[x], gcd(g, h) = 1 or g = 0, h 6= 0

For α ∈ R,

f(α) =

{∞ if h(α) = 0g(α)h(α)

if h(α) 6= 0

For α =∞,

f(∞) =

∞, if k > l;akbl, if k = l;

0, if k < l

where f(x) = akxk+...+a0

blxl+...+b0

Each f ∈ R(x) gives rise to a continuous function f : P1 → P1,where f(α) = f(α), f(∞) = f(∞) as defined above. For later use, we notethat for all α ∈ P1 we have f(α) = λα(f).

So we get a representation Ψ : R(x)→ C(P1,P1), f 7→ f , the latter beingthe set of all continuous functions between these spaces.Clearly Ψ is injective and one can observe that

f ∈ H := H(R(x))⇔ f ∈ C(P1,R).

Now applying Stone-Weierstraß Theorem we get that the image Ψ(H) isdense in C(P1,R) relative to the supremum norm.From the description of H we deduce the following facts (where we drop thereference to the field R(x) as in the case of H):

f ∈ E⇔ f has no zero on P1, f ∈ E+ ⇔ f > 0 on all points of P1.

At present we are facing two representations. i.e. ring homomorphismsΦ : H → C(M,R), f 7→ f ,Ψ : H → C(P1,R), f 7→ f where we setM = M(R(x)). That they essentially the same will follow below.

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Identification of M with P1

We have to show that M is homeomorphic with P1 and that the represen-tation Ψ is nothing but the concrete geometric version of the representationΦ : H → C(M,R) of the general theory .

We begin the proof by recalling the bijection σ : P1 → M,α 7→ λα andwant to prove that it is a homeomorphism. Since both spaces in questionare compact it is enough to deduce that the bijection σ is continuous, bygeneral topology of compact spaces.Further note the equation f ◦ σ = f what follows from

f(σ(α)) = f(λα) = λα(f) = f(α) = f(α).

We have seen above that the topology on M is characterized as the small-est(= coarsest) topology such that all maps f : M → R, λ 7→ λ(f), f ∈ Hare continuous. A subbasis is given by the sets Uf := f−1(0,∞). In view of

f ◦ σ = f we obtain σ−1(Uf ) = f−1(0,∞). As f is continuous this latter setis open, and σ is shown to be continuous, finally a homeomorphism betweenP1 and the space of real places M(R(x)).

The relation above between yields φ(f) ◦ σ = Ψ(f) ∀f ∈ H, meaningthat both representations Φ and Ψ are equivalent. In other words, Ψ isnothing but the geometric version of the general type representation Φ.

2.5 H(R(x)) a Dedekind Ring

There are various ways to prove that in this case the real holomorphy ring isa Dedekind ring. We have chosen an approach where we start by achievingexplicit information how the prime ideals can be generated by finitely manyelements. After sketching the proof of the known result that H is a Pruferring anyway we then conclude that H is a Dedekind ring where, as said,we explicitly know generators of the maximal ideals and their squares. Thisinformation allows to derive a somewhat explicit presentation in proposition16 and chapter 4.

Fractional Ideals

Here we introduce the notion and results needed to prove that the real holo-morphy rings of formally real fields are Prufer rings.

Let R be an integral domain with quotient field K.A fractional ideal is any R−submodule a ⊆ K such that there is

f ∈ R, f 6= 0 with fa ⊆ R.Example. Set of fractional ideals in Q = {rZ | r ∈ Q}.Let a and b be fractional ideals of R.

Their sum a+b = {a+b : a ∈ a, b ∈ b} and their product ab = {∑

finite aibi :ai ∈ a, bi ∈ b} as well as their intersection a ∩ b are fractional ideals of R.

If both fractional ideals are finitely generated (as R−modules), say a =(a1, ..., ar), b = (b1, ..., bs), then ab = (a1b1, ..., arbs). Multiplication of frac-tional ideals is compatible with the order relation between fractional ideals:

32

if a ⊆ b, then ac ⊆ bc

The non-zero fractional ideals form a multiplicative semi-group with Ras its unit element.

Definition 14 A fractional ideal a of a ring R is invertible if there exists afractional ideal b of R such that ab = R.

Proposition 12 Let a 6= (0) be an invertible fractional ideal such that ab =R for a fractional ideal b. Then b = [R : a], i.e if a is invertible thena−1 = [R : a] := {r ∈ K : ra ⊆ R}.

We note that the invertible ideals form an abelian group.

Lemma 8 Let a be a fractional R-ideal. If a is invertible, then a is finitelygenerated as R-module.

Definition 15 A ring R is called a Prufer domain if and only if all non-zerofinitely generated fractional ideals are invertible.

Theorem 21 H(K) = H is a Prufer ring.

Proof. Let a = (a1, ..., ar) be a non-zero finitely generated H−module.

a = {r∑i=1

fiai : fi ∈ H}

a2 = a · a is generated by (∑r

i=1 fiai)(∑r

j=1 gjaj), fi, gj ∈ H.

=∑i,j

(figj)(aiaj)

= (..., aiaj, ...) ⊇ (r∑i=1

a2i )

=⇒ a2 = (..., aiaj, ...)

Now asaiaj∑ri=1 a

2k∈ H we have that aiaj =

aiaj∑ri=1 a

2k

∑ri=1 a

2i ∈ H ·

∑ri=1 a

2i

=⇒ a2 ⊆ H · (r∑i=1

a2i ) ⊆ a2

i.e. (a1, ..., ar)2 = (

∑ri=1 a

2i ). If f =

∑ri=1 a

2i , then it follows that

(a1, ..., ar)[(a1, ..., ar).f−1] = H as principal ideal is invertible.

i.e. a is invertible and hence H is Prufer. We now turn to the proof that

H(R(x)) is a Dedekind Ring.

33

Given α ∈ P1 let να, Vα, Iα, λα be the corresponding (real) valuation,valuation ring, maximal ideal and real place. We know that

Iα = (x− α)Vα if α ∈ R, I∞ = (1/x)V∞, να(x− α) = 1, ν∞(1/x) = 1

From H ⊆ Vα we get that mα := H ∩Iα is a maximal ideal of H with residuefield R. We are going to prove the following three facts:

1. mα is generated by elements x−α1+x2

, (x−α)2

1+x2if α ∈ R,

2. m∞ is generated by 11+x2

, x1+x2

,

3. the prime ideals of H are the mα, α ∈ P1 and the zero ideal (0). Thus,the Krull dimension of H is equal to 1.

The proofs start with the observation that each non-zero element of H canbe written in the form

a = ε · Πr1

x− αi(1 + x2)k

, ε ∈ H∗, r ≤ 2k.

So, in dealing with ideals we can ignore the units and focus on elementsf = Πr

1x−αi

(1+x2)k

Now let p be a non-zero prime ideal in H. We look at the elements f 6= 0in this prime ideal with a smallest r ≥ 0.Case 1: r = 0. Then 1

1+x2∈ p since the ideal is prime ideal. From [ x

1+x2]2 =

[ x2

1+x2].[ 1

1+x2] ∈ p we get that x

1+x2∈ p since this ideal is a prime ideal. Define

the H-ideal a := ( 11+x2

, x1+x2

). These two generators also lie in I∞, hencea ⊆ p and a ⊆ m∞. To show equality in both cases it will be proven thatH/a = R, in particular, that a is a maximal ideal, whence the claim. Firstly,from R ⊆ H we get R ⊆ H/a. Secondly, we have to prove that for eachf ∈ H there exists α ∈ R such that f ≡ α mod a, i.e. f + a = α. To obtainthis it is enough to consider the elements g := xi

(1+x2)kwhere i ≤ 2k since an

arbitrary f ∈ H is a R-linear combination of those elements. If

i ≤ 2k − 2 we have g =xi

(1 + x2)k−1· 1

1 + x2∈ a.

If

i = 2k − 1 then g =x2k−1

(1 + x2)k=

x2k−2

(1 + x2)k−1· x

1 + x2∈ a.

Finally, look at

g =x2k

(1 + x2)k.

This special element g does not belong to the ideal a. Instead we get g−1 ∈ aas can be shown as follows. From x2

1+x2≡ 1 mod a we get by taking the k-th

power g ≡ 1 mod a as well. This settles this special subcase and m∞ = a = pis proven.

Case 2: r ≥ 1. Here 11+x2

/∈ p. Consider an element f with smallestnumber r. If r = 2k, minimality of r and the fact that p is a prime leads tor = 2. So, f · [ 1

1+x2] = [ x−α

1+x2].[ x−β

1+x2] ∈ p. Primality of p shows that for some

34

α the element g := x−α1+x2

∈ p. Then [ (x−α)2

1+x2].[ 1

1+x2] = g2 ∈ p. Since 1

1+x2/∈ p

we get (x−α)2

1+x2∈ p. This time we get that the H-ideal b := ( (x−α)2

1+x2, x−α

1+x2) is

contained in mα and p. Similarly if r < 2k. To prove equality of the threeideals we proceed in an analogue manner. We first note that any polynomialcan be presented as linear combination of the powers (x − a)i, i = 0, 1, . . . .

So we have to look at the elements (x−α)i

(1+x2)kfor 0 ≤ i ≤ 2k. If 1 ≤ i ≤ 2k then

(x−α)i

(1+x2)k∈ b readily follows. It remains to look at the elements gk := 1

(1+x2)k.

If k = 1 then

1

1 + x2− 1

1 + a2= −[(x− a)2 + 2a(x− a)]/[(1 + x2)(1 + a2)] ∈ b

implying gk ≡ 1(1+a2)k

mod b. This again proves mα = b = pSo far it has been proven that each prime ideal is finitely generated,

in fact the zero ideal by 0, otherwise it is maximal and generated by twoelements. Since a ring in which each of its prime ideal is finitely generated isNoetherian Ring (See Kaplansky: Commutative rings, Theorem 8. ). So wecould apply this fact and get that H is Noetherian. Since H is Prufer ring,using the Noetherian property we see that each ideal of H is invertible. SoH turns out to be a Dedekind domain.(See Kaplansky: Commutative rings,Theorem 96.)Then each non-zero ideal a of H is a product of finitely many maximal ideals.This representation as a product of maximal ideals is unique up to the orderof the maximal ideals.(Kaplansky: Commutative rings, Theorem 97.)Therefore, we have proven

Theorem 22 H is a Dedekind domain.

A bit more has been proven in fact. The ideals mα, α ∈ P1 form acomplete list of the maximal ideals of H. From mα = H ∩ Iα ⊆ Vα we getHmα ⊆ Vα. Localizations of Dedekind rings are discrete valuations rings,hence maximal subrings of their field of quotients. So, Hmα = Vα.

Invoking the general theory of Dedeking rings we therefore get the for-mula,

if f ∈ K∗, then fH = Πα∈P1mυα(f)α .

As said above, this formula describes the unique factorization of a principalfractional ideal as a product of maximal ideals.

We note that the formula implicitly takes into account that the valuesυα(f) are equal to 0 apart from finitely many α’s.

Lemma 9 1. For each f ∈ R(x)∗ the sum∑

α υα(f) is an even integer,

2. for any two, not necessarily distinct maximal ideals m1 and m2 theproduct m1m2 is a principal ideal.

Proof. (1) Any non-zero element f can be written asf = const ·

∏r1(x − αi)

ei · q where the αi are pairwise distinct and q is anon-zero sum of squares in R(x). Note first that υ(q) ∈ 2Z for any real

35

valuation of R(x). If υ is a real valuation different from the υαi , i = 1, . . . , rand υ∞ then υ(f) = υ(q) ≡ 0 mod 2. Next, we get υαi(f) ≡ ei mod 2 andυ∞(f) ≡ −

∑i ei mod 2. Hence claim (1) is settled.

(2) We will provide explicit generators for the product.

First consider distinct elements α, β ∈ R and set f := (x−α)(x−β)1+x2

. We see thatυ(f) = 0 for any real valuation 6= υα, υβ and υ(f) = 1 if υ = vα or = vβ.This implies fH = mαmβ. In a similar fashion one proves x−α

1+x2H = m∞mα.

Again, by calculating the values υ(f) one verifies 11+x2

H = m2∞ and (x−α)2

1+x2H =

m2α.

Proposition 13 (1) No maximal ideal in H(R(x)) is a principal ideal,(2) given finitely many distinct ideals m1, ...,mk and exponents e1, ..., ek ∈ Z,then Πk

1meii is a principal fractional ideal if and only if

∑k1 ei is even.

Proof. (1) The assumption fH = m would contradict the first statement ofthe previous lemma. (2) If the given product is a principal fractional idealthen, by the previous lemma, the sum

∑k1 ei is even. Conversely, assume

this to hold. If ei is even then, by the lemma above, the fractional idealmeii is a principal fractional ideal. Otherwise, ei = 2fi + 1 and we get

that meii = (aiH)mi. Altogether, our product in question is a product of a

principal fractional ideal with a product∏

1 of an even number of maximalideals. By the second statement of the last lemma

∏1 is a principal fractional

ideal as well. Hence, the proof is complete.To prepare the next statement let H denote an arbitrary real holomorphy

ring of a formally real field K.

Proposition 14 Let a be an H-fractional ideal, finitely generated by a1, ..., ak.Then for each n ∈ N we get

a2n = (k∑i=1

a2ni ).

Proof. b := a2n is generated as an H-module by the finitely many productsai1 .ai2 ...ai2n where for all indices 1 ≤ ij ≤ k. Denoting such a product by band considering c := b∑k

i=1 a2ni

gives that υ(c) ≥ 0 for any real valuation υ of

K. Thus c is contained in H. Finally one gets that b is a principal fractionalideal generated by the element as announced.

We apply this proposition to the case K = R(x) and its maximal idealmα.

Proposition 15 Let m be a maximal ideal of H(R(x)) and l ≥ 1. Then(1) if l is odd then ml is not a principal ideal,(2) if l is even then ml = (al + bl) for some a, b ∈ H.

Proof. The first statement is just a special case of a previous result. Thesecond one results from the previous proposition and the fact that a maximalideal is generated by two elements.

The representation of a principal fractional ideal allows a nice descrip-tion of arbitrary non-zero sums of squares in R(x). To spell it out we usegenerators of the squares of the maximal ideals of the real holomorphy ringsas given above.

36

Proposition 16 Any non-zero sum of squares in R(x) admits a unique pre-sentation of the type

f = ε · [ 1

1 + x2]e∞∏α∈R

[(x− α)2

1 + x2]eα where ε ∈ E+.

Proof. We know that υ(f) is even for any real valuation. Taking this intoaccount the formula for the fractional ideal fH directly translates into theproduct formula for the element f where the unit factor must be sum ofsquares necessarily.

37

Chapter 3

Real Holomorphy Rings ofElliptic Function Fields

3.1 Quadratic Extension of Fields

Definition 16 Let L be a field extension of a field K. We can always regardL as a vector space over K where addition is field addition multiplication byK is simply multiplication.We say that the degree of L as an extension of K is the dimension of thevector space (denoted [L : K]).

Extension of degree 2 (i.e [L : K] = 2)are called quadratic extensions.

Theorem 23 Suppose char(K) 6= 2. Then any extension K ⊂ L of degree2 can be obtained by adjoining a square root: L = K(β), where β2 = b ∈ K.Conversely if β is an element of L and if β2 ∈ K, but β /∈ K then K(β) isa quadratic extension.

Proof. Let K−basis be 1, α. Then L = K1 +Kα. (α ∈ L \K).L 3 α2 = aα + b, (a, b ∈ K)α satisfies quadratic equation x2−ax−b = 0, i.e.α2−aα−b = 0 (a, b ∈ K).Then (α− a

2)2 = b+ (a

2)2. Let β = α− a

2

1, β K−basis, β2 = γ ∈ K, γ 6= 0, γ /∈ K2 and L generated by 1,√γ.

Irr(α,K) = x2− ax− b irreducible over K. If not x2− ax− b = (x− r)(x−s), (r, s ∈ K).Inserting x = α, we get 0 = (α− r)(α− s)⇒ α = r ∈ K or α = s ∈ K⇒ L = K ⇒ [L : K] = 1.i.e. L/K quadratic x2 − ax− b ∈ K[x], (a, b ∈ K) irreducible.Conversely, every quadratic irreducible polynomial is associated with a quadraticfield extension of K. To show this, let f ∈ K[x], f = x2 + ax + b, a, b ∈ Kbe irreducible.1st: (f) = fK[x] is maximal (true for all PIDs).Assume not!Then (f) $ m $ K[x]. Let g ∈ m \ (f)=⇒ gcd(g, f) = h. Then h|f ⇒ f = h.h1.f irreducible ⇒ h = const. or h1 = const.h = const 6= 0⇒ gcd(g, f) = 1. Then 1 = Ag +Bf ∈ m. Contradiction.

38

h1 = const.⇒ p = gcd(g, f)⇒ f |g, thus g = f.f1 ∈ (f). Contradiction.2nd: K[x]/(f) = L is a field. K ↪→ L, α 7→ α + (f) field embedding.Claim: L = K +Kx, x = x+ (f).g + (f) ≡ h+ (f), g ≡ hmodf, f |(g − h).1, x, x2 ≡ −ax− bmodf ⇒ ...⇒ xk ≡ (αk + βk)modf=⇒ 1, x K−basis for L, [L : K] = 2.That is to say if L/K is a quadratic extension there is associated irreduciblequadratic polynomial x2 − βx − α ∈ K[x], β, α ∈ K. Conversely quadraticirreducible polynomial is associated with a quadratic field extension of K.Remark: f irreducible of degf = n⇒ [K[x]/(f) : K] = n.

Proposition 17 Let K be a formally real field, L = K(√a) a quadratic

extension. Then the following are equivalent:(i) L is formally real,(ii) a /∈ −

∑K2,

(iii) there exists an ordering P of K with a ∈ P .

Proof. Assume(i). If (ii) were not true, then −1 ∈∑L2, a contradiction.

Assume (ii). If no ordering as in(iii) exists, then −a ∈ ∩P =∑K2, a

contradiction.Assume (iii). We consider the set T = {

∑n1 x

2i pi|n ∈ N, xi ∈ K, pi ∈ P}.

Then T is a preordering of L, so it is contained in an ordering P′

of L. Thismeans that L is formally real.

Corollary 8 Let L/K be a field extension of finite degree n and let α ∈ L.Then α is algebraic over K and its degree divides n.

Corollary 9 Every irreducible polynomial in R[x] has degree 1 or 2.

From above theorems we get that given any γ /∈ K2, then x2 − γ ∈ K[x] isirreducible.L = K[x]/(x2 − γ) = K1 +Kx, x2 = γ. That is to say quadratic extension(char(K) 6= 2) they come from γ /∈ K2.Let now f ∈ R[T ] and degf odd. Then f /∈ R(T )2 and it follows thatx2 − f ∈ R(T )[x] is irreducible.

Special case: If degf = 3, then f /∈ R(T )2. If in addition f is withoutmultiple roots in C over R we call R(T )(

√f) an Elliptic Function Field

over R.Let γ /∈ K2 and assume L = K(

√γ).

Then σ : L→ L be given by σ(α+ β√γ) := α− β√γ is a K-automorphism

of L.As usual we define the trace and norm:

trace = tr(α + β√γ) = (α + β

√γ) + σ(α + β

√γ) = 2α

norm = N(α + β√γ) = (α + β

√γ)σ(α + β

√γ) = α2 − β2γ

Each z ∈ L satisfies z2 − tr(z)z +N(z) = 0.

39

3.2 Extension of Valuation Rings from R(x)to Elliptic Function Fields

Our aim here is to understand valuation rings of elliptic function fields onthe basis of valuation rings of R(x).By a valued field we mean a pair (K,V ) where V is a valuation ring of K. If(K,V ) and (L,W ) are valued fields, then we say that (L,W ) is an extensionof (K,V ) (we write (K,V ) ⊆ (L,W )) if K is a subfield of L and W ∩K = V .For field extensions K ⊆ L and a valuation ring W of L, V = W ∩K is avaluation ring of K.Then if (L,W ) is an extension of (K,V ), we have that W ∩ K = V andmW ∩K = mW ∩ V = mV .Consequently,the field V/mV is naturally isomorphic to a subfield the fieldW/mW .(V → W/mW , a 7→ a = a+ mW )Then Ker = mW ∩K = mV and thus V/mV ↪→ W/mW .

The relation W ∩ K = V and W ∗ ∩ K = W ∗ ∩ V = V ∗ (which followsfrom the equality W ∩K = V ) imply that the linearly ordered group ΓV =K∗/V ∗ is naturally isomorphic to a subgroup of the linearly ordered groupΓW = L∗/W ∗. We often identify V/mV with the corresponding subfield ofW/mW (V/mV ⊆ W/mW ) and ΓV with the corresponding subgroup of ΓW(ΓV ⊆ ΓW ).We recall some basic facts which can be found in [12].The following theorem is crucial for what follows.

Theorem 24 (Chevalley’s Theorem) Let K be a field, R ⊆ K be a subringand P ⊆ R be a prime ideal of R. Then there exists a valuation ring V of Ksuch that

R ⊆ V and m ∩R = PWhere m is the maximal ideal of V.

Remark 1 We can add the condition that P 6= {0} to the theorem if wewant to avoid the choice that R = K.

Theorem 25 Let L/K be a field extension, and let V be a valuation ring ofK. Then there is an extension W of V in L.

Next we state an application of Chevalley’s Theorem to characterize theintegral closure of a domain D by means of valuation rings containing D.

Theorem 26 Let D be a subring of a field K. Then the integral closure ofD in K is the intersection of the valuation rings of K containing D.

Integral Elements and Integral Extensions

Definition 17 Given rings R ⊆ S and b ∈ S , we say that b is integral overR if b is the root of a monic polynomial f(x) ∈ R[x].

40

S is integral over R if every b ∈ S is integral over R. In this case we call San integral extension of R.E.g.

√2 is integral over Z.

Ring Z[√

2] is an integral extension of Z.12

is not integral over Z (although it is algebraic.)

Lemma 10 (Integral elements and finite modules). Let S be a ring, R ⊆ Sa subring, and s ∈ S. Then the following statements are equivalent:(a) The element s is integral over R.(b) The subalgebra R[s] ⊆ S generated by s is finitely generated as an R-module.(c) There exists an R[s]- module M with Ann(M) = {0} such that M isfinitely generated as an R-module.

The following theorem is in perfect analogy to the result that a finitely gen-erated field extension is finite if and only if it is algebraic. It also impliesthat sums and products of integral elements are again integral.

Theorem 27 (Generated by integral elements implies integral). Let S be aring and R ⊆ S a subring such that S is finitely generated as an R-algebra,i.e. S = R[a1, ..., an]. Then the following statements are equivalent:(a) All ai are integral over R.(b) S is integral over R.(c) S is finitely generated as an R- module.

Corollary 10 (Integral elements form a subalgebra). Let S be a ring andR ⊆ S a subring. Then the set

S ′ := {s ∈ S : s is integral over R} ⊆ S

is an R-subalgebra.

Corollary 11 (Towers of integral extensions). Let T be a ring and R ⊆S ⊆ T subrings. If T is integral over S and S is integral over R, then T isintegral over R.

Theorem 28 Let R be a commutative ring, u an invertible element of a ringcontaining R. Then u−1 is integral over R if and only if u−1 ∈ R[u].

Theorem 29 Let R be an integral domain contained in a field L. If L isintegral over R, then R is a field.

Let (K1, V1) ⊂ (K2, V2) be an arbitrary extension of valued fields . To eachVi, i = 1, 2, corresponds a valuation υi : Ki � Γi ∪ {∞}. We recall thatυi|K∗i : K∗i � Γi is a group homomorphism with kernel V ∗i and K∗i /V

∗i∼= Γi.

Moreover, the composite mapping

K∗1id↪→ K∗2 � K∗2/V

∗2∼= Γ2

41

has kernel V ∗2 ∩ K∗1 = V ∗1 , whence Γ1∼= K∗1/V

∗1 ↪→ K∗2/V

∗2∼= Γ2, by the

homomorphism theorem. Therefore we could replace the valuation υ1 bythe restriction of υ2 and may regard Γ1 as an ordered subgroup of Γ2. Thisprocedure works well in the general situation where we follow this fashion.However,in the context of discrete valuation which are usually assumed tohave the value group Z, i.e. they are normalized, one prefers to keep thisnice fact and accepts that the value groups are not contained in each other.Details will follow below.Let us call e := e(V2|V1) := [Γ2 : Γ1] the ramification index of this extension.

Similarly, denoting the maximal ideals by m1,m2 the composite mapping

V1id↪→ V2 � V2/m2 = K2

has kernel m2 ∩ V1 = m1. Thus , K1 = V1/m1 ↪→ V2/m2 = K2. Therefore wemay regard K1 as a subfield of K2.Now, we define f := f(V2|V1) := [K2 : K1] as the residue degree of thisextension.Remark: The ramification index and the residue degree are multiplicative.If (K1, V1) ⊂ (K2, V2) ⊂ (K3, V3) are valued extension of fields, then

e(V3|V1) = e(V3|V2)e(V2|V1) andf(V3|V1) = f(V3|V2)f(V2|V1).

For the following lemmas, corollary and theorems upto fundamental equalityone can see the proof in [12].

Lemma 11 Suppose (K1, V1) ⊆ (K2, V2), and , for i = 1, 2,

υi : Ki � Γi ∪ {∞}

is the valuation corresponding to Vi.Choose w1, ..., wf ∈ V2 and π1, ..., πe ∈ K∗2 so that:(1) the residues w1, ..., wf ∈ K2 are linearly independent over K1;(2) the values υ2(π1), ..., υ2(πe) are representatives of the distinct cosets ofΓ2|Γ1.Then for all aij ∈ K1,

υ2(

f∑i=1

e∑j=1

aijwiπj) = min{υ2(aijwiπj) : 1 ≤ i ≤ f, 1 ≤ j ≤ e}.

In particular, the product {wiπj : i = 1, ..., f, j = 1, ..., e} are linearly inde-pendent over K1.

The above lemma has the following immediate consequence:

Corollary 12 Suppose (K1, V1) ⊆ (K2, V2), and set n = [K2 : K1], e =e(V2|V1), and f = f(V2|V1). If n <∞, then e, f <∞ and ef ≤ n.

More generally, we can state:

42

Theorem 30 For (K1, V1) ⊆ (K2, V2) with K2 algebraic over K1 the follow-ing statements hold:(1) for every γ ∈ Γ2 there is an n ∈ N such that nγ ∈ Γ1; i.e., Γ2/Γ1 is atorsion group;(2) K2 is algebraic extension of K1.

Theorem 31 (Conjugation Theorem). Suppose L/K is an arbitrary normalextension of fields, V is a valuation ring of K, and V1 and V2 are valuationrings in L extending V. Then there exists σ ∈ Aut(L/K) with σ(V1) = V2.

Lemma 12 Let N/K be a finite Galois extension of degree n. Assume thatV1, ..., Vr are all extensions of the valuation ring V from K to N. Then ref ≤n, where e = e(Vi/V ) and f = f(Vi/V ) for all i with 1 ≤ i ≤ r.

Theorem 32 (Fundamental Inequality). Let L/K be a finite extension ofthe valued field (K, V), and assume that V1, ..., Vr are all extensions of V toL. Then

r∑i=1

e(Vi/V )f(Vi/V ) ≤ [L : K].

The next theorem shows that in case V has value group Z and L/K is sepa-rable, even equality holds in Fundamental Inequality.

Theorem 33 (Fundamental Equality). Let (K, V) be a valued field withvalue group Z, and let V1, ..., Vm be all extensions of V to a finite separableextension L of K. Then

[L : K] =m∑i=1

e(Vi/V )f(Vi/V ).

Now, let L = R(x)(√f) , f is a cubic polynomial over R with no multiple

roots in C.f = k(x − a)(x − b)(x − c), a, b, c are distinct or f = k(x − a)(x2 + bx + c)where x2 + bx+ c is quadratic irreducible.If W is a real valuation ring of L, then W ∩ R(x) is a real valuation ring ofR(x).So W ∩R(x) = Vα, α ∈ P1 since Vα, α ∈ P1 is the collection of real valuationrings of R(x).

First we consider the case α ∈ R.

Let y =√f . Then y2 = f(x) ∈ R[x] ⊆ Vα ⊆ W

Passing to the residue field W/mW = L, y2 = f(x).

Now f(x) − f(α) = (x − α)g(x) ∈ mα = mV (as α is a root of the lefthand side .)

⇒ (f(x)− f(α) = 0 i.e. f(x) = f(α), y2 = f(α) in L.

⇒ f(α) ≥ 0 because y2 ≥ 0.

Thus if Vα has a real extension, then f(α) ≥ 0

43

Therefore, if L is an elliptic function field, which is a quadratic extensionof R(x) and if W is a real valuation ring of L, then W ∩ R(x) = Vα, α ∈ P1

and f(α) ≥ 0 or α = ∞. This latter condition is a necessary condition forthe Vα, to be extendible to a real valuation ring of L. It remains to be shownthat it is also a sufficient condition.

Again Vα, α ∈ P1 denotes the collection of real valuation rings of R(x)and να the list of their associated discrete valuations where the value groupis Z. The valuation rings of L are denoted by W , their normalized valuationby ω, always with subscript added. L = R(x)(

√f), f is a cubic polynomial

without multiple roots.Since we are keeping the feature that both discrete valuations admit thevalue group Z ramification now gets expressed in the form

ω|K∗ = e(ω|υ).υ.

Then Vα extends to a real valuation ring of L if and only if f(α) ≥ 0 orα =∞.It remains to show that the right hand side yields a sufficient condition whatwill be done in three steps.(1) f(α) = 0

From the fundamental equality∑g

i=1 eifi = [L : K], where fi = [L : K],ei = [Γi : Γ], g = number of extensions.Since [L : K] = 2, it follows that g ≤ 2.

Let ω|υα, then ω|K∗ = e(ω|υ)υα.We know that extending pairs W |V with associated discrete valuations ω, υexist and that in each case ω(K∗) = e(W |V )Z.y2 = f(x) = (x− α)g(x), g(α) 6= 0.Then 2ω(y) = ω(x− α) + ω(g(x)) where ω(g(x)) = 0 necessarily.Thus 2ω(y) = ω(x− α) =⇒ 2|ω(x− α) as ω(x− α) = eυα(x− α) it followsthat 2|e since υα(x− α) = 1. From the fundamental equality it follows thate = 2 and hence as

∑g1 eifi = 2, and efg = 2 we have f = g = 1

Then Vα has exactly one extension, it is also real and ramified.Also ω(y) = 1. Then y is a generator of maximal ideal in this case.

(2) f(α) > 0Let W be an extension of Vα. Then in the residue field L of W we have

y2 = f(α) > 0, so y = a ∈ R, a =√f(α). This shows in L, y − a ∈ mW .

(since y − a = y − a = y − a = 0 in L = W/mW ). Passing to the conjugatevaluation ring W ′ we obtain σ(y − a) = −y − a ∈ m

′. This shows W 6= W ′

(Otherwise y − a,−y − a ∈ mW ) that leads to 2a ∈ mW .A contradiction as a is a real number different from zero, hence a unit.Therefore, we have at least two extensions, and from the fundamental equal-ity we conclude that Vα has exactly two extensions W±

α each of which hasresidue degree equal to 1. Hence both extensions are real.

(3) ω|υ∞. Then ω|K = eυ∞.From y2 = f(x), 2ω(y) = ω(y2) = eυ∞(f(x)) = −3e.

=⇒ 2|e. Then e = 2 from the fundamental equality and ω(y) = −3.In particular, there is unique extension, it is ramified, has residue degree 1,therefore a real valuation.

44

Then ω( y1+x2

) = −3− eυ∞(1 + x2) = −3− 2(−2) = 1.Thus y

1+x2is a generator of maximal ideal in this case.

Regarding the extension of the real valuation rings Vα, α ∈ P1 of R(x) toelliptic function field L we can now summarize:

Theorem 34 1. Vα, where f(α) = 0, and V∞ have just one extension W toL, it is ramified with index 2 and real, also σ(W ) = W ,2. if f(α) < 0, then Vα has just one extension to L, it is not real, has theresidue field C, so residue degree f = 2,3. if f(α) > 0, then Vα has two extensions to L, both are real and conjugateunder σ.

3.3 The Valuation Theoretic Approach to El-

liptic Function Fields

Let L be an elliptic function field (or L = R(x)(√f), f - monic, cubic polyno-

mial without multiple roots in C). We will be proceeding by quoting generaltheorems on Dedekind rings, e.g. cf. [16, 38], and then applying those toour special situation.

Theorem 35 The integral closure of a Dedekind ring in a finite-dimensionalfield extension is a Dedekind ring.

Theorem 36 If A ⊆ B ⊆ K, K the quotient field of A and A is a Dedekindring, then B is a Dedekind ring.

Both theorems are needed in the proof of the next proposition. Assume L/Kbe a finite field extension and hence it is an algebraic extension. Further, letboth fields be formally real.

Proposition 18 (1) H(K) ⊆ H(L),(2) if H(K) is a Dedekind domain, then H(L) is as well.

Proof. To show the first statement we note that if W is a real valuationring of L, then its restriction V := W ∩K is a real valuation ring since theresidue field of V is a subfield of the residue field of W and the property ofbeing formally real goes down from larger field to any of its subfield.For the second statement pass to the integral closure R of H(K) in L. Thenthe ring R is a Dedekind domain by the theorem 35 above. Since H(L) isintegrally closed in L, being a prufer domain, we get that H(K) ⊆ R ⊆H(L). So , by theorem 36 above, we get that H(L) is a Dedekind ringprovided it can be shown that L is the quotient field of R.

To prove this let a ∈ L. Since the field extension L/K is algebraic weget an equation an +

∑n1 αia

n−i = 0 where the coefficients αi belong to K.This implies that we have a representation αi = hi/s where hi, s ∈ H(K).Inserting this in the equation for a and multiplying the whole equation by sn

we find that the element sa is integral over H(K), so belongs to R. Hencea = (as)/s ∈ quot(R), and the final claim is proven.

45

Theorem 37 A, K as in preceding theorem. If A = ∩Vα, each Vα discretevaluation ring of K, then the family (Vα) is the family of all localizationsAM , M running through all maximal ideals of A.

Theorem 38 If A is Dedekind, f ∈ K∗, then fA = Πmmυm(f) where m runs

over all maximal ideals of A and υm is the normalized discrete valuationattached to the valuation ring Am.

Let K = R(x) and L = R(x)(√f), where f is a cubic polynomial with

no multiple roots. Invoking proposition 18 we first state

Proposition 19 H(L) is a Dedekind ring.

By definition, H(L) = ∩W , where W runs through the real valuation ringsof L.Since W ∩ R(x) is a real valuation ring of R(x), it follows thatW ∩ R(x) = Vα, some α ∈ P1. We know that the α′ s that occur arecharacterized by f(α) ≥ 0 or α =∞.

Therefore,

H(L) = (⋂

α:f(α)=0

Wα) ∩W∞ ∩⋂

α:f(α)>0

(W+α ∩W−

α ).

As L = R(x)(√f) where f is monic, cubic polynomial without multiple

roots in C. i.e. L = {a + b√f : a, b ∈ R(x)}, we can write out H(L) even

more explicitly.

Now, we calculate

H(L) ∩K =⋂

f(α)=0

(Wα ∩K) ∩ (W∞ ∩K) ∩⋂

f(α)>0

[(W+α ∩K) ∩ (W−

α ∩K)]

=⋂

f(α)=0

Vα ∩ V∞ ∩⋂

f(α)>0

Vα =⋂

f(α)≥0

Vα ∩ V∞

.Clearly,

H(K) =⋂β∈R

Vβ ∩ V∞ ⊆ H(L) ∩K

Now, by theorem 26 the integral closure ofH(L)∩K in L can be computedas follows. First of all:

˜H(L) ∩K =⋂

H(L)∩K⊆W

W.

From the presentation of H(L) ∩ K, theorem 37 and the fact that W is areal valuation ring we have that W = Wα,W

±α , f(α) ≥ 0 or α =∞.

=⇒ ˜H(L) ∩K =⋂f(α)≥0Wα ∩ (W±

α ) ∩W∞i.e. ˜H(L) ∩K = H(L). Hence we have

46

Theorem 39 H(L) is the integral closure of H(L) ∩K.

Using trace and the norm we can characterize when an element is integral.Using the above informations, let z ∈ L that is z = a+ b

√f, a, b ∈ R(x) and

z be integral over H(L) ∩K.Then we know that (i) z is integral implies that σ(z) integral,

(ii) Sums and products of integral elements also integral.We can conclude that z integral over H(L) ∩K if and only if tr(z), N(z) inH(L) ∩K. As tr(z) = 2a, N(z) = a2 − b2f , thus z integral over H(L) ∩Kif and only if a ∈ H(L) ∩K, a2 − b2f ∈ H(L) ∩K. From statements in (i)and (ii) above we get that b2f ∈ H(L) ∩K = Vα, f(α) ≥ 0 or α =∞ . Nowυα(b2f) ≥ 0 since b2f ∈ H(L)∩K where υα is a valuation correspond to thevaluation ring Vα in K. Thus, υα(b) ≥ 0 since υα(f) = 1, f = (x− α)g(x)

For α =∞, υα(b) ≥ 2.Therefore if z ∈ L is integral over H(L) ∩ K it is of the form z =

a+ b√f, a, b ∈ H(L) ∩K, υ∞(b) ≥ 2.

In contrast, L = R(x)√f = {a+ b

√f : a, b ∈ R(x)} where f is a monic,

cubic polynomial over R without multiple roots in C.A similar presentation can be obtained for H(L). To this end we choose anyt ∈ K subject to υα(t) = 0 if f(α) ≥ 0, υ∞(t) = 2, e.g. t = 1

1+x2. Then, it

can be deduced from the result above:

Proposition 20 i. H(L) = {z = a+ b√f : a, b ∈ H(L) ∩K, υ∞(b) ≥ 2}

ii. Choose any t ∈ H(L) ∩ K with υ∞(t) = 2, υα(t) = 0 if f(α) ≥ 0 thenH(L) = {z = a+ b · t

√f : a, b ∈ H(L) ∩K}.

The remaining part of this section is devoted to explicit computations inthe elliptic function field L. The first task is to calculate the values of realvaluations of L. This will then be applied to determing generators of themaximal ideals of H(L).

Conjugate valuations and calculation of ω(z)As before K = R(x), L = R(x)

√f where f is monic, cubic polynomial with

out multiple roots in C over R.The quadratic extension L/K admits the conjugation

σ : L→ L, a+ by 7→ a− by. This is a field automorphism, trivial on K.L/K is a Galois extension with Galois group {id, σ}.

Lemma 13 If W is any valuation ring of L with maximal ideal I, residuefield k, then σ(W ) is also a valuation ring of L. Its maximal ideal σ(I), itsresidue field is isomorphic to k.

As a consequence we get that real valuation rings are transferred into realvaluation rings.Using our information on the real valuation rings of L, cf. theorem 34, weobtain:

Proposition 21 (i) If Wα is ramified over Vα,i.e. f(α) = 0 or α =∞,then σ(Wα) = Wα,(ii)if f(α) > 0, then σ(W+

α ) = W−α , σ(W−

α ) = W+α .

47

This fact has nice consequences for the normalized discrete valuations at-tached to the valuation rings. Let W denote any discrete valuation ring andW′

= σ(W ), z = επk, ε a unit, π a uniformizing element. Then I = (π), soI′

= (σ(π)) is the maximal ideal of W′, i.e. σ(π) = ηπ

′, η a unit in W

′. We

get σ(z) = σ(ε)ηkπ′k. Now, units map to units under σ|W : W → W

′. As a

result we get:

Corollary 13 (i) If f(α) = 0, then ωα = ωα ◦ σ,(ii) if f(α) > 0, then ω−α ◦ σ = ω+

α , ω+α ◦ σ = ω−α .

These relationships allow an efficient way to calculate the valuesω(a+ by), a, b ∈ K = R(x), y2 = f(x).(1) If f(α) = 0, then ωα(z) = υα(N(z)),(2) if f(α) > 0, then ω+

α (z) + ω−α (z) = υα(N(z)).Where for z = a+ by,N(z) = a2 − b2f = z.σ(z), tr(z) = 2a = z + σ(z).To prove we use the fact ω(z) + (ω ◦ σ)(z) = ω(N(z)) = e(w|v).υ(N(z))and the corollary. Note that we know how to calculate the valuations υα, soυα(N(z)) is computable using the rules above.

We now turn to the determination of generators of the maximal idealsm = I∩H(L) where I runs through the maximal ideals of the real valuationsrings W of L. υm denotes the normalized discrete valuation belonging toHm = W or to m, for short. The use of indices will indicate which realvaluation ring we are considering. We recall that in our situation ofH = H(L): if a = (a1, ..., an) then a2 = (

∑i a

2i ).

This fact allow us to find generators of maximal ideals of H in a systematicfashion . Let m be a maximal ideal of H. Then clearly there is no ideal strictlybetween m2 and m. Now, we know that m2 = (a). Choose any b ∈ m \ m2.Then we get m = (a, b).

So, it remains to find such elements a, b. Applying theorem (38) we get(1) m2 = (a) iff υm(a) = 2, υm′ (a) = 0 for all maximal ideals m

′ 6= m,(2) b ∈ m \m2 iff b ∈ A, υm(b) = 1.

In the sequel α ∈ R, and we assume f(0) = 0 what can be reached bysubstituting x for x− α where f(α) = 0. The ideals are ideals of H(L).(1) m∞ must contain the elements 1/(1 + x2), x/(1 + x2), y/(1 + x2) sincethey are in H(L) and their values with respect to ω∞ are 4, 2, 1 respectively.In case of ω0 we have the values 0, 2, 1, in case of ωα, α 6= 0, f(α) = 0 (if thiscase occurs at all) the values are 0, 0, 1, in case of ω±α , f(α) > 0 the valuesare 0, 0, 0. So we find that

a := (1/(1 + x2)) + (y/(1 + x2))2 = (1 + x2 + y2)/(1 + x2)2

satisfies the condition (1). In doing the calculation we apply the fact thatω(∑

i qi) = min{ω(qi)} holds for any real valuation ω and sums of squaresqi. Finally we get that m∞ = (a, b := y/(1 + x2)),(2) the ideal m0 must contain the elements x/(1+x2), x2/(1+x2), y/(1+x2).This time a = x2

1+x2+ ( y

1+x2)2, b = y

1+x2,

(3) in the case that α exists with α 6= 0, f(α) = 0 the ideal mα must containthe elements (x − α)/(1 + x2), (x − α)2/(1 + x2), y/(1 + x2). This time

a = (x−α)2(1+x2+y2)(1+x2)2

, b = y1+x2

,

48

(4) finally assume f(α) > 0, and let β2 = f(α). Then the ideal m+α must

contain the elements (x − α)/(1 + x2), (x − α)2/(1 + x2), (y − β)/(1 + x2) .

This time a = (x−α)2(1+x2+(y−β)2)(1+x2)2

, b = x−α1+x2

.

To get m−α we change β into −β.

3.4 The Geometric Approach to Elliptic Func-

tion Fields

The following facts from ring theory are useful in understanding the geomet-ric approach.(1) Let A be a Dedekind domain with field of quotients K, m a maximalideal of A. Then the inclusion ι : A → Am, a 7→ a

1induces an isomorphism

ι : A/m → Am/mAm. Hence, Am is a real valuation ring iff the residue fieldA/m is formally real.(2) Let φ : A → B be an epimorphism of rings. Then the assignmentm 7→ φ−1(m) is a bijection between the set of maximal ideals of B and theset of maximal ideals of A which contain the the kernel ker(φ). This bijec-tion preserves the residue fields, i.e. φ induces an isomorphismφ : A/φ−1(m)→ B/m.

As before, let K = R(x), L = K(√f), y =

√f, and Wα,W

±α where

f(α) ≥ 0,W∞ the collection of the real valuation rings of L. As usual,deg (f) = 3, no multiple roots in CSet A = R[x],then B = R[x, y] = R[x] + R[x]y is the integral closure of Ain L. This can be directly be seen by looking at the trace and norm of theelements in L.Then the Wα,W

±α are overrings of B, and W∞ is not, the latter in view of

w∞(x) = −2. The ring B, being the integral closure of the Dedekind domainA (in fact a PID), is a Dedekind domain with field of quotients equal to L.By the first statement above, the Wα,W

±α are exactly the localizations Am

where m runs through the maximal ideals of A with a formally real residuefield which in this case amounts to having R as residue field.

The kernel of the epimorphism

φ : R[X, Y ]→ B,F (X, Y ) 7→ F (x, y).

equals the principal ideal (Y 2 − f(X)). The maximal ideals of R[X, Y ] withresidue field R are the ideals

ma,b := (X − a, Y − b) = {F (X, Y ) ∈ R[X, Y ] : F (a, b) = 0}, (a, b) ∈ R2

.Finally we get that

(Y 2 − f(X)) ⊆ ma,b ⇔ b2 − f(a) = 0.

Hence, we are led to consider the elliptic curve defined by the equationY 2− f(X) = 0. More precisely, we have to study the set Γ of real points onthis curve:

Γ = {(α, β) ∈ R2 : β2 = f(α)}.

49

What was demonstrated above means, first of all, that the Wα,W±α are in

one-to-one correspondence with the points of the elliptic curve. But muchmore is true: we will use this fact to describe the real valuation rings andthe real places of L in a different, i.e. geometric manner.

We first turn to

The valuation rings W 6= W∞ and their real places

With this curve at hand there is now no need to preserve the notationW±α . Instead we are using uniformily the latter γ = (α, β) to denote the

points on Γ. A point γ gives rise to a maximal ideal

Mγ = {F (X, Y ) : F (γ) = 0} = (X − α, Y − β) ⊆ R[X, Y ].

As said above, these are all the maximal ideals containing the kernel of Φ.Invoking the fact (2) from the beginning of this section we obtain that theimages mγ := Φ(Mγ) constitute the complete list of the maximal ideals ofthe R-algebra admitting a real residue field. Using fact (1) from above weget all real valuation rings containing B as the localizations at these mγ.

We can therefore summarize:

Theorem 40 Let γ = (α, β) ∈ Γ. Then

mγ = {F (x, y) : F (γ) = 0} = (x− α, y − β)

is a maximal ideal of B with a real residue field inducing the real valuationring

Wγ = Bmγ = {F (x, y)

G(x, y): F,G ∈ R[X, Y ], G(γ) 6= 0}.

In this way we obtain all real valuation rings of L, 6= W∞, in a one-to-onecorrespondence.

The result allows us to determine the real places attached to the real valua-tion ring Wγ. The real place attached to Wγ is denoted by λγ.

Corollary 14 Given z ∈ L, thenλ(z) 6= ∞ ⇔ z = F (x, y)/G(x, y) where F,G ∈ R[X, Y ], G(γ) 6= 0. In thatcase λ(z) = F (γ)/G(γ).

50

Finally, we address

The case of W∞

W∞ is the only extension of V∞, ramification index 2, ω∞(y) = −3,ω∞(y/(1+x2) = 1. We set y := y/(1+x2) and find that y ∈ J∞, by definitionthe maximal ideal of W∞. Also, y ∈ H(L) and

W∞ = {a(x) + b(x)y : a(x), b(x) ∈ V∞}.

Similar to the geometric interpretation of the infinite place λ∞ of R(x) weconsider the associated elliptic curve Γ and get completely analogous results.

If γ ∈ R2 we let |γ| denote the Euclidean norm. First note that Γ admitspoints γ with |γ| → ∞.

By F we denote the rational function a(X) + b(X)Y ∈ R(X, Y ), setz = F (x, y) and evaluate z at points of Γ. There are special instances, whichfollow directly:

lim|γ|→∞

y(γ) = 0, lim|γ|→∞

a(γ) exists ⇔ a ∈ V∞.

With the help of this remark and the arguments regarding the infiniteplace λ∞ on R(x) one derives the following

Proposition 22 (1) limγ∈Γ,|γ|→∞F (γ) exists iff z ∈ W∞,(2) if z ∈ W∞, then λ∞(z) = limγ∈Γ,|γ|→∞F (γ) = lim|u|→∞a(u).

3.5 Topological Representation

In this section we briefly present the geometric interpretation of the generalrepresentation

H → C(M,R)

along the line we had followed for the rational functional field in section 2.4.Since the arguments are completely parallel we omit the details.

Suggested by the previous section we are led to consider the one-pointcompactification Γ := Γ∪∞. In geometric terms, this means we are passingfrom the affine curve, given by Y 2 − f(X) = 0, to its projective closuredescribed by Y 2Z−f ∗(X,Z) = 0 where f ∗(Y, Z) = Z3f(X/Z). One realizesthat only one point ”at infinity” gets added, with homogeneous coordinates[0 : 1 : 0].

Topologically, passing from Γ to Γ means to add to the unbounded com-ponent (which always exists) of Γ a point at infinity. In this way, the un-bounded component gets ”compactified” to a homeomorphic image of S1.The bounded component, if it exists, is also homeomorphic to S1. Hence,up to homeomorphisms, Γ is of the topological type of one or two copies ofS1.

In topological terms, the last two propositions of section 3.4 states thateach z ∈ H(L) induces by evaluation a continous function z : Γ → R.

Appealing to the bijection between the space of real places and the Γ andtransfering the arguments from the situation in the rational function fieldcase, cf. section 2.4, we get this time analogously:

51

Theorem 41 1. The map Γ→M(L), γ 7→ λγ is a homeomorphism,

2. the representation H(L)→ C(Γ,R) has a dense image.

52

Chapter 4

Totally Positive Units andSums of Powers

4.1 The Units of H

In this section we confine ourselves to the rational function field case. Thisis done for getting greater clarity regarding results and proofs. It is clearthat statements of this type can be obtained in greater generality but at theprice of applying a more abstract approach. Instead, as explained in theintroduction, we want to make use of all favourable features of the specialfields we are considering. Not all what has been proven directly for R(x) hasfound a similar proof for the elliptic case so far.

As said we consider the field R(x) in this section. Here we will prove thateach totally positive unit is a sum of 2n−th powers for every n, in particularthis is true for the totally positive unit 1+x2

2+x2.

The proof will make use of the density result and the famous Hilbert’sidentities. These results are the basis for the bounds on Pythagoras numbers.

We will make use of the characterizationf ∈ E⇔ f has no zero on P1,f ∈ E+ ⇔ f > 0 on P1.

We note also that∑

i a2ni ∈ H implies that all ai ∈ H and that every

sum of squares is a sum of two squares.

Hilbert’s Identities

(1.1) For k, n ∈ N (k∑i=0

x2i

)n

=l∑

i=1

αiLi(x0, ..., xk)2n

(1.2) (k∑i=0

x2i

)n

+ 2nx20

(k∑i=0

x2i

)n−1

=l′∑

i=1

αiLi(x0, ..., xk)2n

with αi ∈ Q, αi > 0, linear forms Li ∈ Q[x0, ..., xk] and

l ≤(

2n+ kk

), l′ ≤

(2n+ kk

)53

Let a ∈ E+.We look at the associated continuous function a : P1 → R. a is strictly

positive on the compact space. Consider a−1 and 2.a−1. Both are strictlypositive continuous functions. One can find a strictly positive functionf : P1 → R whose 2n− th power strictly lies between these two functions. fcan be approximated as closely as wanted (Stone-Weierstraß ). In particular,this means we can find b ∈ H such that

b2n also lies strictly between the two functions.

We now pass to the element c := ab2n. Then we get c lies strictly between 1

and 2. As b2n is strictly between a−1 and 2.a−1 i.e. to say a−1 < b2n < 2.a−1

multiplying throughout by a we get 1 < ab2n < 2 where ( c = ab2n). Thus

c− 1 is strictly positive on P1, and all its value are strictly smaller than 1.This implies that c−1 = u2 +v2, u, v ∈ H. It follows that the absolute valuesof u, v are strictly smaller than 1. As 1 ≤ n, both elements n−u2, n− v2 aresums of squares, i.e. n− u2 = u2

1 + u22, n− v2 = v2

1 + v22, where all elements

are in H. We now apply the Hilbert identity (1.2) for k = 2, x0 = u, xi = uiand k = 2, x0 = v, xi = vi respectively.

In the first case we get(2∑i=0

x2i

)n

+ 2nx20

(2∑i=0

x2i

)n−1

=l′∑

i=1

αiLi(x0, x1, x2)2n

i.e. (x20 + x2

1 + x22)n + 2nx2

0(x20 + x2

1 + x22)n−1

= (u2 + u21 + u2

2)n + 2nu2(u2 + u21 + u2

2)n−1

= nn + 2n.u2.nn−1 ∈∑H2n

Dividing by 2nn we get

nn + 2n.u2.nn−1

2nn=

1

2+ u2 ∈

∑H2n

The same conclusion holds for v i.e 12

+ v2 ∈∑H2n

Then 12

+u2 + 12

+v2 ∈∑H2n. So we derive that c = 1+u2 +v2 ∈

∑H2n

As c := ab2n, finally that a ∈∑H2n

That means we have proved (for K = R(x)) the following.

Theorem 42 E+ ⊆ ∩n∑K2n

In elliptic function field L also H∗ ∩∑L2 ⊆ ∩

∑L2n, i.e E+ ⊆ ∩n

∑L2n

Remark: J.Schmidt has proved that even more E+ =∑

(E+)2n .(See [4] for the details)Note: This Schmidt’s theorem has interesting application for the bound onPythagoras numbers.

In 1981, at a conference on Real Algebraic Geometry and Ordered Fieldsin Rennes, France, Professor Becker gave a talk in which he proved that

B(t) :=1 + t2

2 + t2∈ Q(t)

54

is a sum 2n− th powers of elements in Q(t) for all n.That is for all k ≥ 1, there exists 0 < λj,k ∈ Q and polynomials fj,k and gj,kin Q[t] such that

B(t) =1 + t2

2 + t2=∑

λj,k

(fj,k(t)

gj,k(t)

)2k

Explicit version for this are known for k = 1, 2 and there has been interestin finding them for all k. (For complete solution on R(t) see ([26], [27]).

4.2 The Valuation Criterion for Sums of Pow-

ers

We begin with the following lemma.

Lemma 14 Every fractional H−ideal is generated by two elements.

Proof. Let a be a fractional ideal. We know that it is finitely generated andwe find a2 = (

∑ki a

2i ). Since every sum of squares is a sum of two squares we

get that a2 = (b2 + c2) = b2 where b = (b, c). Since all non-zero fractionalideals are invertible in this case we obtain that a2b−2 = (ab−1)2 = H. Thatis we get c2 = H. This implies that c = H and finally a = b.

Theorem 43 (Becker’s Valuation Criterion). Given a ∈ R(x) then thefollowing statements are equivalent:(1) a is a sum of 2n-th powers in R(x),(2) (a) a is a sum of squares, i.e. totally positive in R(x),

(b) υ(a) ∈ 2nZ for every (surjective) real valuation υ : R(x)→ Z ∪∞.

Proof. Assume (2) and consider the element a with the given properties.

We take aH = Πα∈P1mυα(a)α . Since all values υ(a) are multiples of 2n we get

that aH = a2n, a a fractional H-ideal. We know that every fractional ideal isgenerated by two elements. This leads to aH = (b2n + c2n). Two generatorsof a principal fractional ideal differ by a unit of H. Hence, a = ε[b2n + c2n], εa unit of H. Now as a is assumed to be a sum of squares, the unit ε is totallypositive as well, i.e. ε ∈ E+. Hence the claim.The above theorem holds for elliptic case as well.

Theorem 44 Let L be an ellptic function field. Thenf ∈

∑L2n ⇔ f ∈

∑L2, 2n|υγ(f), ∀γ ∈ Γ ∪∞

4.3 Sums of Mixed Powers

It was R. Berr who published in [6] an important generalization of the Val-uation Criterion above to include sums of mixed even powers, i.e. elementsof∑

i

∑K2ni where the ni belong to some finite or infinite set of natural

numbers. We quote it for arbitrary formally real fields.

55

Theorem 45 Let K be any formally real field and I a non-empty set ofnatural numbers. Then the following statements are equivalent:(1) a ∈

∑n∈I∑K2n

(2) (i) a ∈∑K2,

(ii) for each real valuation υ we have υ(a) ∈ ∪n∈I2nΓ,Γ the value group ofυ.

From Berr’s theorem we get that a polynomial f ∈ R[x] admits a represen-tation

f ∈∑n∈I

∑R(x)2n

if and only if f satisfies:(1) f is positive semidefinite.(2) degf ∈ ∪n∈I2nZ.(3) ord(α) ∈ ∪n∈I2nZ for every real root α of f .

Before moving on we consider special cases.(1) Let q1, ..., qk ∈ R[x] be irreducible, positive definite quadratic poly-

nomials.Then a polynomial f of the form

f = (x− r1)2n1 ...(x− rt)2ntq1...qk

admits in R(x) a representation as a sum of mixed powers of degree I, i.e.I = {n1, ..., nt,

∑ni + k}.

(2) Let f = (x− 1)4(x− 2)6(x2 + 1).degf = 12. By the valuation criterion f /∈

∑R(x)4 and f /∈

∑R(x)6.

As f is positive semidefinite (f ≥ 0), degf ∈ ∪n∈I2nZ where I = {2, 3, 6}and ord(α) ∈ ∪n∈I2nZ for α = 1, 2 which are real roots of f .

Therefore, f ∈∑

R(x)4 +∑

R(x)6.How can we find such a presentation? The idea to find it in this special

case is already strong enough to provide the basis for a quite general resultwhich will be proven in a moment. In our given case we set

g = (x− 1)(x− 2)2, h = (x− 1)(x− 2) then

g4+h6 = (x−1)4(x−2)6[(x−1)2+(x−2)2] = f ·ε where ε =2x2 − 6x+ 5

x2 + 1∈ E+

This impliesf = ε−1(g4 + h6).

ε−1, as a totally positive unit, is contained in∑

R(x)4 ∩∑

R(x)6. Hence,the claim on f is proven.

To prepare the next result we recall some of the features of H = H(R(x)):H is a Dedekind ring, the square of a maximal ideal is a principal ideal, thetotally positive units are sums of 2nth powers for every n. In fact,

m2∞ = (

1

1 + x2),m2

α = ((x− α)2

1 + x2).

56

To facilitate notation we denote by p’s the above generators of the square ofthe maximal ideals. Note that they are sums of squares.Now consider any non-zero f ∈

∑R(x)2. We know that all values

υα(f), α ∈ P1 are even, only finitely many of them are non-zero. Seteα = (1/2)υα(f). Then

fH =∏α∈P1

m2eαα implying f = η

∏α∈P1

peαα , η ∈ E+.

To simplify the notation even further we denote by p1, . . . , pk the generatorsoccuring in the product formula, i.e. admitting a non-zero exponent ei. Inanalogy to the example above we set

ai = pi ·∏j 6=i

plij where liei ≥ ej if j 6= i.

By summing up aeii , i = 1, . . . , k we calculate

k∑1

aeii = f ·∑i

∏j 6=i

pliei−ejj .

The sum is denoted by S and belongs to the real holomorphy ring H sinceall exponents are non-negative. The individual terms in this sum are sumsof squares, so we can easily compute the value υα(S) by calculating theminimum of the values of the individual terms. Note that (pi) = m2

i . Ifmα 6= mi for all i = 1, . . . , k then all υα(

∏j 6=i) = 0, whence υα(S) = 0. If

mα = mi then υα(∏

j 6=i) = 0 as υα(pj) = 0 for j 6= i. Therefore, υα(S) = 0as well.

Thus we have proven that S ∈ E+. Setting ε = S−1 we can formulate

Theorem 46 Let f ∈∑

R(x)2, f 6= 0, set υα(f) = 2eα for α ∈ P1. Denoteby e1, . . . , ek the list of the non-zero eα’s. Then there are non-zero sums ofnon-squares q1, . . . , qk and a totally positive unit ε such that

f = ε ·k∑1

qeii .

This theorem implies Berr’s characterization of sums of mixed powers.To see this note that a totally positive unit is a sum of 2n-th powers forevery n and that an e-th power of a sum of squares is a sum of 2e-th powersdue to the first Hilbert identity.

Analyzing the ingredients of the proof of this theorem we see that onlyfacts are used which are known likewise for the real holomorphy rings ofelliptic function fields: Dedekind property and sums-of-squares generationof the squares of the maximal ideals. As a consequence, the last theoremcan be proven for such fields by appealing to the family of all discrete realvaluations ωγ, γ ∈ Γ ∪∞.

57

4.4 Topics of Complexity

Let f ∈ R[x] be a polynomial in one variable and let n ≥ 2 be a naturalnumber. In the context of the study of sums of powers the following questionsarise quite natural.

(1) For which f does there exist a representation f =∑N

1 (gih

)2n, someN ∈ N with gi, h ∈ R[x]?(2) What can be said about the length N of such representation?(3) What can be said about the degree of h?

The topic addressed in (1) is of a qualitative nature and finds a very satis-factory answer by the Valuation Criterion of section 4.2.Recall that f ∈ R[x] has a representation (1) if and only if

(a) f ≥ 0 (i.e. f(a) ≥ 0 for all a ∈ R,(b) 2n|degf , and(c) 2n divides the multiplicity of x− a in f for every a ∈ R.In contrast, the questions (2),(3) address topics of a quantitative nature

and looks the complexity of a representation by sums of powers. Complexitytopics are difficult to deal with, only a few results are known. A sampleof them will be presented here, sometimes passing to more general formallyreal fields .

As usual, given any formally real field K, we call the least N ∈ N suchthat

∑K2n =

∑N1 K

2n the 2n − th Pythagoras number p2n(K) of K.By a result of Becker [3], this number is known to be finite for all n ∈ Nprovided p2(K) <∞. So, all Pythagoras numbers of R(x) are finite.

In the special case n = 2, in the paper [10] of Choi, Lam, Prestel andReznick it was shown that 3 ≤ p4(R(x)) ≤ 6 for a real closed field R.Concerning topic (3) very little is known. Gilbert Stengle in [36] showedthat for the simple polynomial y4+ty2+1 , viewed as a one-parameter familyof elements of R[y] if y4 + ty2 + 1 =

∑(fk/f0)4, t ≥ 0 the degree of f0 tends

to infinity with t in any such family and there exists a family and a constantC in which the degree of f0 has the bound C log t for all large t.

The following result is a improved version of an older result by Becker.The main progress came from the observation that Caratheodory’s theorem,cf. [28], about the generation of cones in R- or Q-vectorpaces lead to shorterlength in Hilbert’s identities.

We recall the Hilbert’s identity

( k∑i=0

Xi2

)n+ 2nX0

2

( k∑i=0

Xi2

)n−1

=m∑i=1

βiLi(X0, · · · , Xk)2n

where k, n ∈ N, βi ∈ Q, βi > 0, linear forms Li ∈ Q[X0, · · · , Xk] andm ≤

(2n+k

2n

). Caratheodory’s theorem yields this upper bound for the length

of the sum on the right hand side since this bound is the Q-dimension of thespace Q[X0, . . . , Xk]2n

Proposition 23 Let R ⊆ K, K be formally real, p := p2(K) <∞ thenp2n(K) ≤ p2

(2n+p

2n

).

58

Corollary 15 If p2 ≤ 2 then p2n ≤ 2(2n+ 1)(2n+ 2).

Proof. In Hilbert’s identity the coefficients βi can be merged with the linearforms since they are 2n-powers.1st step:

∑i a

2ni = ε

∑p1 b

2nj where ε is a totally positive unit ε, in particular

ε =∑p

1 x2i . This can be obtained by considering fractional ideals: (

∑i a

2ni ) =

(. . . , a2i , . . . )

n = (∑

i a2i )n = (

∑p1 b

2i )n = (b2

1, . . . , b2p)n = (

∑p1 b

2ni ).

2nd step: we assume ε < 1. Since ε is totally positive we have ε =∑p

1 x2i .

Then for all i we get x2i < 1, hence

1− x2i =

∑pj=1 y

2ij, i.e. 1 = x2

i +∑

j y2ij. Plugging in in Hilbert’s identity we

find for each index i

1 + 2nx2i =

∑N1 x

2nij , N =

(2n+p

2n

).

Summing over all i and dividing by p this means

1 + (2n/p)ε =∑Np

1 y2nj .

3rd step: Given any totally positive unit ε find, by the density of the imageof H in C(M,R), a totally positive unit γ and define another totally positiveunit η such that

1 < ε · γ2n < 1 + 2n/p, define η by: 1 + 2n/p · η = εγ2n.

η satisfies the condition in the second step, and ε turns out to be a sum ofNp 2n-th powers. Multiplying ε with the sum

∑p1 b

2nj we get the claim on

p2n.

J.Schmid’s approachThe bounds for p2n from above might be correct assymptotically but aremuch too large for small exponents and small quadratic Pythagoras numberp2. In such cases the theorem of Schmid

Theorem 47 E+ =∑

En+ for all n.

can be applied to produce smaller bounds for p2n.To outline the idea we introduce the Pythagoras number

p+n := pn(E+) = min{k | E+ =

∑k1 En+} or ∞.

The new idea runs as follows:

s =∑

i a2ni = [

∑i a

2ni /q

n] · qn where q =∑

i a2i .

The element in [] is a totally positive unit ε as all real valuations assigns

the value 0 to it, and it is a sum of squares. Hence, ε =∑p+n

1 εni where allεi ∈ E+. Assuming p2 ≤ 2 we have εi = η2

1 + η22, q = a2 + b2. This leads to

s =∑

i[(η21 + η2

2)(a2 + b2)]n =∑p+n

1 (a2i + b2

i )n

59

since the product of two sums of two squares is a sum of two squares. Inthe book of Reznick ”Sums of even powers of real linear forms” [28] we findidentities

(u2 + v2)n =n+1∑

1

w2ni

[equation 1.50, page 23].This gives

Proposition 24 If p2 ≤ 2 then p2n ≤ (n+ 1)p+n .

The following argument below shows that p2 ≤ 2 implies the statementp+

2 ≤ 3. For function fields in one variable over R it is known that evenp+

2 ≤ 2, cf. [4,10]: so in the general case p4 ≤ 9 or in the special case p4 ≤ 6.Now the general arguments for the case n = 2 .As a function on M(K) ε is strictly positive, hence strictly larger than

some rational number 1/n. Then (ε − 1/n) induces a function on M whichis strictly positive. Hence, this element is a sum of squares. Therefore

ε = 1/n+∑p

1 x2i

To keep the argument simple we restrict to the case p ≤ 2 and R ⊆ K. So,ε = 1/n(1 + u2 + v2). Necessarily, u, v ∈ H. The basic discovery of Schmidis this lemma:

Lemma 15 If u ∈ H then 1 + u2 = η2 + ω2, η, ω ∈ E+.

To prove this we consider the continuous function

f = (1/√

1 + u2)[u− i] : M → S1.

The values of f lie in the lower half-circle. Therefore we can use the stereo-graphic projection to find a, b ∈ H where a2 + b2 = 1 such that the functiong := a + bi is close to f , as close as we want and need in the sequel (usingthe density):f ≈ (a+ bi). Multiplying with the inverse function g−1 = a− bi we find thatfg−1 is close to the constant function 1, as close as needed for an argumentbelow. Calculation shows:

1. fg−1 = (1/√

1 + u2)[v+ wi], v = ua−b, w = −ub−a, v2 +w2 = 1+u2,

2. v ≈√

1 + u2, w ≈ 0, so approximating sufficiently near:

3. v ∈ E+.

Hence, 1 + u2 = v2 +w2 where v is a totally positive unit. However, w neednot be a unit, but w is small as wanted. This means that v ± w are totallypositive units, and we have 1 + u2 = [(1/

√2)(v + w)]2 + [(1/

√2)(v − w)]2,

and the claim is proven.As a consequence we get that ε =

∑31 ε

2i , εi ∈ E+.

Proposition 25 If R ⊆ K, p2 ≤ 2 then p4 ≤ 9

Proof.∑

i a4i = (

∑31 ε

2i )(a

2 + b2)2. Use the identity(X2 + Y 2)2 = (X + 1/

√3Y )4 + (X − 1/

√3Y )4 + 8/9Y 4.

60

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63

PSALM 124A SONG OF ASCENTS. OF DAVID.

1 If the LORD had not been on our side − let Israel say −2 if the LORD had not been on our side when men attacked us,3 when their anger flared against us, they would haveswallowed us alive;4 the flood would have engulfed us, the torrent would haveswept over us,5 the raging waters would have swept us away.6 Praise be to the LORD, who has not let us be torn by theirteeth.7 We have escaped like a bird out of the fowler’s snare; thesnare has been broken, and we have escaped.8 Our help is in the name of the LORD, the Maker of heavenand earth.(NIV)

64