# Sturm’s Theorem: determining the number of zeroes of polynomials in an open interval.

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### Sturm’s Theorem: determining the number of zeroes of polynomials in an open interval.

Bachelor’s thesis

Eric Spreen

University of Groningen e.spreen@student.rug.nl

July 12, 2014

Supervisors:

Prof. Dr. J. Top

University of Groningen

Dr. R. Dyer

University of Groningen

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Abstract

A review of the theory of polynomial rings and extension fields is presented, followed by an introduction on ordered, formally real, and real closed fields.

This theory is then used to prove Sturm’s Theorem, a classical result that enables one to find the number of roots of a polynomial that are contained within an open interval, simply by counting the number of sign changes in two sequences. This result can be extended to decide the existence of a root of a family of polynomials, by evaluating a set of polynomial equations, inequations and inequalities with integer coefficients.

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## Contents

1 Introduction 2

2 Polynomials and Extensions 4

2.1 Polynomial rings . . . 4

2.2 Degree arithmetic . . . 6

2.3 Euclidean division algorithm . . . 6

2.3.1 Polynomial factors . . . 8

2.4 Field extensions . . . 9

2.4.1 Simple Field Extensions . . . 10

2.4.2 Dimensionality of an Extension . . . 12

2.4.3 Splitting Fields . . . 13

2.4.4 Galois Theory . . . 15

3 Real Closed Fields 17 3.1 Ordered and Formally Real Fields . . . 17

3.2 Real Closed Fields . . . 22

3.3 The Intermediate Value Theorem . . . 26

4 Sturm’s Theorem 27 4.1 Variations in sign . . . 27

4.2 Systems of equations, inequations and inequalities . . . 32

4.3 Sturm’s Theorem Parametrized . . . 33

4.3.1 Tarski’s Principle . . . 38

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## Introduction

In many of the natural sciences, polynomials and polynomial systems occur as useful approximations to real-world phenomena. As an example of this we give the harmonic oscillator, which is used in physics to approximate dynamical systems that are very close to an equilibrium point. The potential energy of such a (one-dimensional) system takes the form

Vpxq  1 2kx2,

which is a second-degree polynomial in one variable. A similar case occurs in higher-dimensional systems (in 3D, and with multiple bodies). Another example is the hydrogen atom in quantum mechanics, where the radial wave- functions takes on the form: 

Rnlprq  1

reρl 1vpρq, ρ  r an,

where vpρq is a polynomial, and n P N, n ¡ 0 and l P N.1 It is a crucial problem to find the zeroes of these functions in order to determine the electronic structure of the atom, which can be done by finding the zeroes of the polynomial ρl 1vpρq.

It is clear from these examples that finding solutions of polynomial equa- tions is a fundamental problem in applied mathematics. A classical result that enables us to do this numerically is Sturm’s Theorem, named after Jacques Charles Fran¸cois Sturm. This theorem gives the number of zeroes of a polynomial that are contained within a certain open interval, enabling us to determine the zeroes (by partitioning the number line appropriately, up to machine precision) of a polynomial numerically.

We will discuss Sturm’s Theorem in the context of real closed fields, an abstraction of the real number system that has significantly different realizations. The key to the success of Sturm’s Theorem in real closed fields

1We use the convention that 0P N.

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is the analog of the intermediate value theorem for polynomials. It can be shown that this result, and various other key theorems from real analysis hold for polynomials in real closed fields.

After the discussion of Sturm’s Theorem, we will discuss an extension of Sturm’s Theorem that allows us to simplify the problem of the existence of a zero in a certain interval for a whole family of polynomials. The result will be a finite set of systems of polynomial equations, inequations and inequalities with integer coefficients, any one of which may be satisfied by the parameters of the family for the resulting polynomial to have zero in the interval. From this we can quickly establish criteria for the existence of a zero of a whole family of polynomials. A secondary result is that if a polynomial with rational coefficients has a zero in one real closed field, it will have a zero in every other real closed field.

As a high school student I have often wondered whether it would be possible to form an equation of which the solvability is undecidable, in par- ticular when I was unable to solve a particular problem. At the very end we will touch on this question.

A significant portion of this report follows  and . If no citations have been provided, these are the sources. We will assume that the reader has a basic understanding of algebraic structures, such as monoids, groups and rings.

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## Polynomials and Extensions

Before we can begin our study of real closed fields, we will develop the theory of polynomial rings over a field to some extent. This chapter will give basic results on arbitrary polynomial rings, and applications on extension rings and fields. Most of this chapter will follow .

We will say that a subring R of a ring S is generated by a set A  S, if R is the smallest subring that contains A. Also, if u1, . . . , un P S (and

@r P R : uir  rui, 1¤ i ¤ n), then we denote the ring that is generated by RY t u1, . . . , unu by Rru1, . . . , uns. We can readily note that Rru1, u2s  pRru1sqru2s by definition. The existance of such a subring follows from the observation that any arbitrary intersection of subrings of S is again a subring of S.

### 2.1 Polynomial rings

Definition 2.1.1. Given a ring R, its polynomial ring Rrxs is the ring of functions f : N Ñ R such that there exists a k P N with @n P N : n ¥ k ùñ fpnq  0. Addition and multiplication in Rrxs are defined as follows; for fpxq, gpxq P Rrxs and any n P N:

pf gqpnq  fpnq gpnq, pfgqpnq 

¸n i0

fpiqgpn  iq.

and 0 and 1 as obvious. The elements of Rrxs are called polynomials with coefficients in R, and the function values of a polynomial are called the coefficients of a polynomial.

We will establish the basic properties of polynomial rings. The proofs will be ommitted and can be found in [4, Sec.2.10].

Proposition 2.1.1 (Properties of polynomial rings). Let R be a ring and Rrxs its polynomial ring. Then:

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1. There exists an injective homomorphism RÑ Rrxs, so that R may be regarded as a subring of Rrxs.

2. Let x P Rrxs be the polynomial with xp1q  1 and xpnq  0 if n P Nz t 1 u. Then Rrxs is generated by RYt x u. Furthermore, all elements of R commute with x.

3. For any fpxq P N, there exists an n P N and unique a0, . . . , an P R such that fpxq °n

i0aixi.

4. If R is commutative, then so is Rrxs.

Proposition 2.1.2 (Evaluation homomorphism). If R and S are rings, φ : R Ñ S is a homomorphism, u P S and @r P R : φprqu  uφprq, then there exists a unique homomorphism ψ : Rrxs Ñ S such that ψ|R φ and ψpxq  u. Also, the kernel of ψ is an ideal I  Rrxs such that IXR  kerpφq.

This homomorphism is called the evaluation homomorphism in u.

Note: Evaluation in an overring

From the proposition above, it follows immediately that if we let R be a subring of S and φ the inclusion homomorphism, the kernel of every eval- uation homomorphism in an element of S will be an ideal I of Rrxs with IX R  t 0 u.

We can note that if R is a ring, then the following property is “universal”

for the polynomial ring Rrxs (in the sense that any rings that have this property are isomorphic): if S is any other ring, and φ : R Ñ S is a homomorphism, u P S and @r P R : φprqu  uφprq, then there exists an x P Rrxs and a unique homomorphism ψ : Rrxs Ñ S so that ψ|R  φ, φpxq  u and Rrxs is generated by R Y t x u. [4, p.124]

Note: Notation of polynomials

From now on, we will denote a polynomial with coefficients in a ring R as fpxq and its value under a evaluation homomorphism in some u as fpuq.

Also, if fpuq  0, then u is called a zero of fpxq in S.

Corollary 2.1.3. If R and S are rings, and φ : RÑ S is a homomorphism, then there exists a unique homomorphism ψ : Rrxs Ñ Srx1s such that ψ|R φ and ψpxq  x1.

We will also use the notion of a polynomial in multiple indeterminates.

We can formalize this notion by (for any n P N, n ¡ 1) defining the ring Rrx1, . . . , xns : Rrx1s . . . rxns. By induction we can then get an evaluation homomorphism in multiple variables.

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### 2.2 Degree arithmetic

We have seen that for any polynomial 0  fpxq °n

i0aixi there is some k P N such that ak  0, but ai  0 if i ¡ k. This observation is a strong tool that we will use often in further arguments. We therefore define Definition 2.2.1. If R is a ring and 0 fpxq °n

i0aixi P Rrxs, then the degree of fpxq is the largest k P N such that ak  0. If fpxq  0, then the degree is8. We will denote the degree of fpxq by degpfq or degpfpxqq.

Furthermore, the leading coefficient of fpxq is adegpfq if fpxq  0 and 0 if fpxq  0. This will be denoted by lcpfq or lcpfpxqq. A polynomial fpxq will be called monic if lcpfq  1.

The following two lemmas can be proven quickly by the definition of the degree and considering the leading coefficients of fpxq gpxq and fpxqgpxq respectively.

Lemma 2.2.1. If R is a ring, then for any fpxq, gpxq P Rrxs : degpfpxq gpxqq ¤ maxpdegpfq, degpgqq.

Lemma 2.2.2. If D is a domain, then Drxs is also a domain, and for all fpxq, gpxq P Drxs we have degpfgq  degpfq degpgq.1 Also, the units of Drxs will be the units of D.

### 2.3 Euclidean division algorithm

The proof of the following proposition will be given when we prove algorithm 1.

Proposition 2.3.1. Let R be a commutative ring, and fpxq, gpxq P Rrxs with gpxq  0, m  degpgq and bm the leading coefficient of gpxq. Then there exist kP N, qpxq, rpxq P Rrxs such that:

bkfpxq  qpxqgpxq rpxq ^ degprq   degpgq. (2.1) Corollary 2.3.2. Let F be a field and fpxq, gpxq P F rxs with gpxq  0.

Then there exist unique qpxq, rpxq P F rxs such that:

fpxq  qpxqgpxq rpxq ^ degprq   degpgq. (2.2) Proof. We can find some k P N and qpxq, rpxq P F rxs such that bkmfpxq  qpxqgpxq rpxq and degprq   degpgq, where bm lcpgq. Now since gpxq  0, we have bm  0 and thus fpxq  pbkm qpxqqgpxq pbkm rpxqq. Also, since F is a domain: degpbkm rpxqq  degpbkm q degprpxqq  degprpxqq   degpgpxqq.

1It is to be understood here that8 a  8 for any a P N Y t 8 u.

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Now let q1pxq, r1pxq P F rxs also satisfy (2.2). Then pqpxq  q1pxqqgpxq  r1pxq  rpxq. Without loss of generality we may assume that degprq ¥ degpr1q. It then follows that degpgq ¡ degprq ¥ degpr1 rq  degpq  q1q degpgq and this is only possible if degpq  q1q  8. Then qpxq  q1pxq and thus r1pxq  rpxq.

Algorithm 1: Euclidean Division Algorithm

Let R be a commutative ring and fpxq, gpxq P Rrxs with fpxq, gpxq  0.

Also let m  degpgq P N and 0  b P R the leading coefficient of gpxq.

Define the following three coupled sequences:

f0pxq  fpxq n0  degpf0q

a0  lcpf0q fi 1pxq 

#

bfipxq  aixnimgpxq ni¥ m

0 ni  m

ni 1 degpfi 1q ai 1 lcpfi 1q

Then there exists a k P N such that fkpxq  0, fk 1pxq  0 and nk   m.

Also2:

bkmfpxq 

k1

¸

l0

albkl1xnlm

gpxq fkpxq. (2.3)

Proof. Let iP N. We then see that degpfi 1q  degpbfipxq  aixnimgpxq   degpfipxqq, since the leading coefficients of bfipxq and aixnimgpxq are both aib. This shows that the degree strictly decreases each step, and since f0pxq  0, there exists some k P N such that fkpxq  0, degpfkq  nk  m and thus fk 1pxq  0. We may see this k as the terminal step of the algo- rithm, since from this point on only zero polynomials will be produced.

We will now prove the following: for any iP N such that i ¤ k we have bifpxq  fipxq °i1

l0albil1xnlm

gpxq. For i  0 this is clear, so pick iP N with 0   i ¤ k and assume this holds for i  1. Then:

bifpxq  bbi1fpxq  bfi1pxq b

i2

¸

l0

albil2xnlm

gpxq

 fipxq ai1xni1mgpxq

i2

¸

l0

albil1xnlm

gpxq

 fipxq

i1

¸

l0

albil1xnlm

gpxq.

This proves our claim. We can then set i  k to obtain our final formula for fpxq, which concludes the proof and also proves proposition 2.3.1, since degpfkq  nk  m  degpgq.

2We understand here, that if k 0, the sum evaluates to 0.

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2.3.1 Polynomial factors

The Euclidean division algorithm can be used to prove an array of useful facts. The first of these will concern factors of polynomials. Since we will almost exclusively be concerned with commutative rings from this point on, R will denote a commutative ring in the rest of this chapter.

Definition 2.3.1. If fpxq, gpxq P Rrxs, then gpxq is a factor of fpxq – denoted as gpxq  fpxq – if and only if there exists an hpxq P Rrxs such that fpxq  gpxqhpxq.

Also, a polynomial fpxq P Rrxs of positive degree will be called reducible if there exist gpxq, hpxq P Rrxs of positive degree such that fpxq  gpxqhpxq.

Otherwise, fpxq will be called irreducible.3

The following two results characterize the zeroes of a polynomial. They will be used a couple of times in the next chapters.

Lemma 2.3.3. If fpxq P Rrxs and a P R, then there exists a unique qpxq P Rrxs such that fpxq  px  aqqpxq fpaq.

Proof. By the Euclidean division algorithm we may pick qpxq, rpxq P Rrxs with degprq   degpx  aq  1 and fpxq  px  aqqpxq rpxq. We then immediately see that fpaq  pa  aqqpaq rpaq  rpaq, and since degprq   1 we must have rpxq  fpaq. Also, since rpxq is fixed in this way, if q1pxq also satisfies fpxq  px  aqq1pxq rpxq, then px  aqpqpxq  q1pxqq  0. Now, since the leading coefficient of x is 1, which is not a zero divisor, we get qpxq  q1pxq.

Corollary 2.3.4. If fpxq P Rrxs and a P R. Then a is a zero of fpxq if and only ifpx  aq  fpxq.

Proof. By the previous lemma there is a qpxq P Rrxs such that fpxq  px  aqqpxq fpaq. So, if fpaq  0, then px  aq|fpxq. Conversely, if px  aq|fpxq, there exists some hpxq P Rrxs such that fpxq  px  aqhpxq.

But then fpaq  pa  aqhpaq  0.

We can also apply the Euclidean division algorithm to determine a great- est common factor of two polynomials with coefficients in a field F . By a greatest common factor (or divisor) of a pair of polynomials pfpxq, gpxqq we mean a polynomial hpxq such that hpxq  fpxq, hpxq  gpxq and if dpxq P F rxs such that dpxq  fpxq and dpxq  gpxq, then dpxq  hpxq. Degree considerations quickly show that two greatest common factors differ by a unit factor in F . Now, for any two polynomials fpxq, gpxq P F rxs we then define gcdpf, gq P F rxs to be the unique monic greatest common divisor.

3Several other definitions are possible. For example, a polynomial may be called irre- ducible if it is not a unit, and if it can be written as a product of two polynomials, one of them must be a unit. However, in polynomial rings over a field, this leads to the same concept. Hence we adopt this definition.

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Lemma 2.3.5. Let F be a field and fpxq, gpxq P F rxs with gpxq  0, and qpxq, rpxq P F rxs such that degprq   degpgq and fpxq  qpxqgpxq rpxq.

Then for every hpxq P F rxs, hpxq  fpxq and hpxq  gpxq if and only if hpxq  gpxq and hpxq  rpxq.

Proof. Let hpxq P F rxs. If hpxq  fpxq and hpxq  gpxq, then there exists some αpxq, βpxq P F rxs such that fpxq  αpxqhpxq and gpxq  βpxqhpxq. Then rpxq  fpxq  qpxqgpxq  αpxqhpxq  qpxqβpxqhpxq  pαpxq  qpxqβpxqqhpxq, so that hpxq  rpxq and hpxq  gpxq.

Conversely, let hpxq  gpxq and hpxq  rpxq. Then there exist γpxq, ρpxq P Frxs so that gpxq  γpxqhpxq and rpxq  ρpxqhpxq, so that fpxq  pqpxqγpxq ρpxqqhpxq and thus hpxq  fpxq and hpxq  gpxq.

Algorithm 2: The GCD and Euclidean sequence of two polynomials over a field

Let fpxq, gpxq P F rxs where F is a field and gpxq  0, and define the following sequence:

h0pxq  fpxq h1pxq  gpxq hi 2pxq 

#

qi 1pxqhi 1pxq  hipxq, if hi 1pxq  0

0, if hi 1pxq  0

degphi 2q   degphi 1q, iP N.

(It is understood here that8   8). Then there exists a 1 ¤ s P N such that hspxq  0, but hs 1pxq  0. Furthermore, hspxq is a greatest common factor of fpxq and gpxq. The finite sequence (terminating at hspxq) thusly defined is called the Euclidean sequence of fpxq and gpxq.

Proof. Let dpxq P F rxs be a common divisor of fpxq and gpxq. Then by repeated use of lemma 2.3.5 see that this is the case if and only if dpxq is a common divisor of hs1pxq and hspxq. Therefore, every common divisor of fpxq and gpxq will be a factor of hspxq. Also, since hspxq  qspxqhspxq  hs1pxq and hspxq is a factor of itself, hspxq is a common factor of fpxq and gpxq, and thus a greatest common factor.

### 2.4 Field extensions

In the theory of fields, the main subject of study is a field extension. Infor- mally, this is a field that contains some other field. This notion is particularly important for the study of polynomial equations. We will discuss field ex- tension that are generated by finitely many elements, which can be seen as fields that are obtained by adjoining some elements.

Definition 2.4.1. Let F be a field. An field extension over F is then a field E such that F is a subfield of E.

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If S  F , then a subfield K of F is said to be generated by S if it is the smallest subfield containing S.

If E is a field extension over F , and S  E, we denote the subfield generated by SY F by F pSq. If S  t u1, . . . , unu is finite, we denote it by Fpu1, . . . , unq.

2.4.1 Simple Field Extensions

We will first consider the structure of simple field extensions. That is, field extensions over F of the form Fpuq. The following proposition will be crucial in our discussion. In this discussion we take a slightly different route than in .

Proposition 2.4.1. Let F be a field. Then Frxs is a principal ideal domain.

Proof. It is clear that the trivial ideal t 0 u is a principal ideal. Therefore, let t 0 u  I  F rxs be an ideal of F rxs and fpxq P I. We may also clearly pick a non-zero gpxq P Rrxs of minimal degree. Then there exist qpxq, rpxq P F rxs such that fpxq  qpxqgpxq rpxq with degprq   degpgq.

But then rpxq  fpxq  qpxqgpxq P I. Since gpxq was of minimal degree, we must have that rpxq  0. Therefore fpxq  qpxqgpxq for some qpxq P F rxs and we conclude that I  pgpxqq is principal. We have also already seen that, since F is a domain, Frxs is a domain. This concludes the proof.

Definition 2.4.2. If R is a subring of a commutative ring S, and u P S, we will call u algebraic over R if there exists a 0  fpxq P Rrxs such that fpuq  0. Otherwise, we will call u transcendental over R.

A field extension E over F will be called algebraic if and only if every element of E is algebraic over F .

Lemma 2.4.2. If R is a subring of the commutative ring S and u P S is transcendental over R, then Frxs  F rus.

Proof. If u is transcendental over R, then the kernel of the evaluation homo- morphism ρ : Frxs Ñ S in u is t 0 u. This shows that F rxs  A  ρpF rxsq.

Since A contains R and u, we obtain Rrus  A. Also, if x P A, then there exists some fpxq °n

i1aixi P F rxs such that x  fpuq °n

i1aiui P Rrus.

Therefore, A Rrus and we conclude that F rxs  A  Rrus.

Lemma 2.4.3. Let F be a field and fpxq P F rxs of positive degree and irreducible. Then Frxs{pfpxqq is a field. Furthermore, F rxs{pfpxqq  F puq, where u x mod pfpxqq is a zero of fpxq, when regarded as a polynomial with coefficients in Frxs{pfpxqq.

Proof. Let J1  F rxs{I be an ideal, where I  pfpxqq. Then there exists an ideal J  F rxs such that I  J and J1  J{I. Since F rxs is a principal ideal domain, we can find a gpxq P F rxs such that J  pgpxqq. Therefore,

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there exists a hpxq P F rxs such that fpxq  hpxqgpxq. Now, since fpxq is irreducible, we have either hpxq P F or gpxq P F . In the first case, we get J  I, so that J1  t 0 mod I u. In the second case we get J  F rxs, so that J1  F rxs{I. This shows that F rxs{I has only the trivial ideals and thus it is a field.

Now set K  F rxs{pfpxqq and let u  x mod pfpxqq P K. Then clearly fpuq  fpxq mod pfpxqq  0 mod pfpxqq. We also have F puq  K. Now let aP K. Then there exists some bpxq P F rxs so that a  bpxq mod pfpxqq.

But, there also exist qpxq, rpxq P F rxs with degprq   degpfq and bpxq  qpxqfpxq rpxq. Therefore a  rpxq mod pfpxqq  rpuq. Now write rpxq 

°m

i0rixi. Then a rpuq  °m

i0riui P F puq. We therefore conclude that K F puq.

Proposition 2.4.4. Let E be a field extension over F and u P F . If u is transcendental over F , then Fpuq  F pxq, the field of fractions of F rxs. If u is algebraic over F , there exists some irreducible gpxq P F rxs such that Fpuq  F rxs{pgpxqq. Moreover, this gpxq is unique up to a unit multiplier.

Proof. If u is transcendental, then Frxs  F rus. Now, since F rus  F puq, and the field of fractions of Frus is the smallest field containing F rus (and the fields of fractions of two isomorphic integral domains are isomorphic), we get Fpuq  F puq.

Now suppose that u is algebraic over F . Then, since Frxs is a prin- cipal ideal domain, there exists some gpxq P F rxs such that the kernel of the evaluation homomorphism ρ : Frxs Ñ E in u is I  pgpxqq and thus ρpF rxsq  F rxs{I, by the first isomorphism theorem of rings. We claim that Frxs{I is a field.

Suppose that there exist fpxq, hpxq P F rxs such that gpxq  fpxqhpxq.

Now, if fpxq P I, then there exists a kpxq P F rxs such that fpxq  kpxqgpxq and hence gpxq  hpxqkpxqgpxq. This would imply that degphq  0, and hence hpxq P F . Similarly, if hpxq P I, then fpxq P F . Now let both fpxq, hpxq R I. Then fpuq  0  hpuq, but fpuqhpuq  gpuq  0, so then E would contain non-zero zero-divisors. From this we conclude that gpxq is irreducible, and by lemma 2.4.3 we conclude that Frxs{I is a field.

We now observe that F  ρpF rxsq (since I X F  t 0 u) and u P ρpF rxsq, so that Fpuq  ρpF rxsq. But if x P ρpF rxsq, then there exists a fpxq 

°n

i1aixi P F rxs with x  fpuq  °n

i1aiui P F puq. Therefore, F puq  ρpF rxsq  F rxs{pgpxqq.

Now let t 0 u  I  F rxs be an ideal and fpxq, gpxq P F rxs such that I  pfpxqq  pgpxqq. Then, there exist hpxq, kpxq P F rxs such that fpxq  hpxqgpxq and gpxq  kpxqfpxq, so that fpxq  hpxqkpxqfpxq. Degree considerations then show that 0 hpxq, kpxq P F . Therefore, fpxq  agpxq for some unit aP F .

Definition 2.4.3. If E is a field extension over F , and u P E is algebraic

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over F , we call the unique monic polynomial gpxq P F rxs such that F puq  Frxs{pgpxqq the minimum polynomial of u over F .

Also, if E  F puq, we call E a simple field extension over F with gener- ator u, and u is called a primitive element of E.

2.4.2 Dimensionality of an Extension

If we have a field extension E over F , we may regard E as a vector space over F . In this vector space, the addition is the normal addition of the field, and the scalar multiplication is the normal multiplication in F , where the scalars lie in F . In particular, in vector spaces we have the notion of a dimension. This dimension turns out to be of critical importance.

Definition 2.4.4. If E is a field extension over F , the dimensionality (or degree) of E over F is the dimensionality of E regarded as a vector space over F , which shall be denoted asrE : F s.

Proposition 2.4.5. Let E be a field extension over F and uP F . Then u is algebraic over F if and only ifrF puq : F s   8. Moreover, if u is algebraic, thenrF puq : F s is the degree of the minimum polynomial of u.

Proof. Let u be algebraic over F and fpxq P F rxs its minimum polynomial.

Now let aP F puq be arbitrary. Then there exists a gpxq °n1

i0 aixiP F rxs such that a gpuq °n1

i0 aiuiwhere n degpfq and aiP F for 0 ¤ i ¤ n

1. Therefore, p1, u, . . . , un1q spans the vector space F puq over F . Now let b0, . . . , bn1 P F such that °

i0biui  0. Then hpxq °n1

i0 bixi P pfpxqq.

But since degphq   degpfq we get hpxq  0, so that b0      bn1  0.

This shows thatp1, u, . . . , un1q is a base for the vector space F puq over F , and hencerF puq : F s  n   8.

Now let u be transcendental over F . Then let nP N and a0, . . . , anP F . We recall that Frxs  F pxq  F puq. Therefore, 0 °n

i0aiuin

i0aixi implies that a0      an 0, which shows that there exists no finite base for Frxs as a vector space over F . Now, since F rxs is a subspace of F puq, we then certainly have thatrF puq : F s  8. By negation of this argument, we get that if rF puq : F s   8, then u must be algebraic over F .

Proposition 2.4.6 (Dimensionality formula). Let K be a field extension over E, which is in turn a field extension over F . Then K is a field extension over F andrK : F s   8 if and only if rK : Es, rE : F s   8. If rK : F s   8, then:

rK : F s  rK : EsrE : F s (2.4) Proof. It is trivial that K is a field extension over F .

If rK : F s   8, then rE : F s   8, since E is a subspace of K. Now let t u1, . . . , unu  K be a base for K. Then for every a P K there exist

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a1, . . . , anP F  E such that°n

i1aiui  a. Therefore, t u1, . . . , unu spans K as a vector space over E. We conclude that rK : Es ¤ n   8.

Now let rK : Es, rE : F s   8 and pick bases pu1, . . . , umq  E and pv1, . . . , vnq  K for E over F and K over E respectively. Pick any a P K.

Then there exist a1, . . . , an P E such that a  °n

i1aivi. Also, for 1 ¤ i ¤ n there exist bi1, . . . , bim P F such that ai  °m

j1bijuj. This gives us a  °n

i1°m

j1bijujvi, so that the set t ujvi | 1 ¤ i ¤ n ^ 1 ¤ j ¤ m u spans K as a vector space over F . Now let c11, . . . , cnm P F such that

°n

i0

°m

j0cijujvi  0. Then clearly for 1 ¤ i ¤ n: di  °m

j0cijuj  0.

This implies that for 1¤ i ¤ n and 1 ¤ j ¤ m we have cij  0. Therefore we have obtained a base for K over F and rK : F s  nm  rK : EsrE : Fs   8.

The following corollary is immediate and will be used later on.

Corollary 2.4.7. Let K be a field extension of F withrK : F s   8. Then for any field extension E  K of F , rE : F s  rK : F s. Also, if rK : F s is prime, then the only subfields of K that contain F are K and F themselves.

2.4.3 Splitting Fields

We have seen before that if we have a field extension E over F and some monic polynomial fpxq P F rxs, the u P E is a zero of fpxq if and only if px  uq  fpxq, where we regard fpxq as a polynomial in Erxs (by the induced inclusion homomorphism). It would of course be great if we could find a field extension E over F where we could write fpxq  ±n

i1px  riq for r1, . . . , rnP E. Also, taking into account the results we obtained for the dimensionalities of field extensions, we would like that E  F pr1, . . . , rnq, since then rE : F s ±n

i1rF pu1, . . . , uiq : F pu1, . . . , ui1qs (where the first term in the product is understood to be rF pu1q : F s. Also, if r P E is then a zero of fpxq, then 0  fprq  ±n

i1pr  riq such that r  ri for some 1¤ i ¤ n. We will call such a field extension a splitting field of fpxq over F .

Definition 2.4.5. Let F be a field and fpxq P F rxs monic. Then a splitting field of fpxq over F is a field extension E over F , such that in Erxs we can write fpxq ±n

i1px  riq for r1, . . . , rnP E and E  F pr1, . . . , rnq.

Proposition 2.4.8. If F is a field and fpxq P F rxs is monic, then there exists a splitting field E of fpxq over F .

Proof. Let fpxq  ±k

i1fipxq where for 1 ¤ i ¤ k, fkpxq P F rxs is monic and irreducible. Then k ¤ n  degpfq. If n  k, F itself is a splitting field of fpxq. Now let n  k ¡ 0. Then for some j P t 1, . . . , k u we have degpfjq ¡ 1. Set K  F rxs{pf1pxqq, which is a field that contains F , and r  x mod pfipxqq, so that K  F prq and f1prq  0. Then in Krxs

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we have fpxq  ±l

i1gipxq, where for 1 ¤ i ¤ l, gipxq P Krxs are the irreducible factors of fpxq in Krxs. Since these factors can be obtained by taking the irreducible factors of the fipxq and px  rq P Krxs is an irreducible factor of f1pxq, we obtain n ¥ l ¡ k, so that n  l   n  k.

By induction we then obtain an extension field E  Kpr1, . . . , rnq such that fpxq  ±n

iipx  riq, where r  ri for some 1 ¤ i ¤ n. This shows that E F prqpr1, . . . , rnq  F pr1, . . . , rnq is a splitting field of fpxq over F .

We state the following proposition without proof, which can be found in .

Proposition 2.4.9. Let F be a field, fpxq P F rxs monic and of positive degree, and E and E1 splitting fields of fpxq over F . Then E  E1.

We may now quickly see that the splitting of a monic polynomial in factors of degree one is unique. Therefore, the zeroes are unique and the following definition is consistent over every possible splitting field.

Definition 2.4.6. Let F be a field, fpxq P F rxs monic and of positive degree, and E a splitting field of fpxq over F . Write fpxq ±m

i1px  riqki, where ri  rj if i  j. We then call ki the multiplicity of ri. Also, a zero ri is called a simple zero if and only if it has multiplicity 1. Otherwise, it is called a multiple zero.

We lastly make the connection between the derivative of a polynomial and the character of its zeroes. Informally, we will see that a zero (in a splitting field) has multiplicity greater than 1 if and only if the polynomial and its derivative have a common factor of positive degree. For this we define the following map on a polynomial ring of a field.

Definition 2.4.7. Let F be a field. We then define the standard derivation in Frxs as the unique function F rxs Ñ F rxs : fpxq ÞÑ f1pxq so that for any fpxq, gpxq P F rxs:

1. pf gq1pxq  f1pxq g1pxq 2. pfgq1pxq  f1pxqgpxq fpxqg1pxq 3. x1  1.

As in real analysis we may quickly derive all the familiar algebraic prop- erties of polynomial derivatives. We can now state the following proposition, the proof of which can be found in [4, Sec. 4.4].

Proposition 2.4.10. Let F be a field, fpxq P F rxs monic and of positive degree, and E any splitting field of fpxq over F . Then all zeroes of fpxq in E are simple if and only if gcdpf, f1q  1.

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2.4.4 Galois Theory

Galois theory is one of the pearls of modern mathematics. It allows one to study solutions of algebraic equations in a purely algebraic way. At the heart of the theory is the connection between the solutions of such equations and group theory. We will state the fundamental results without proof for later use. An extensive treatment of this subject may (again) be found in [4, Ch. 4]. Throughout this subsection, F denotes a field.

Definition 2.4.8. fpxq P F rxs is called seperable if and only if its irreducible factors have distinct zeroes in any splitting field.

An algebraic field extension E over F is called seperable over F if and only if the minimum polynomial over F of every element of E is seperable.

Also, E is called normal over F if and only if every irreducible polynomial in Frxs that has a zero in E splits into factors of degree 1.

Lemma 2.4.11. Any field extension E over F of characteristic 0 is seper- able.

Definition 2.4.9 (The Galois group). Let E be a field extension over F . The Galois group of E over F is then the group GalpE{F q of automorphisms of E that reduce to the identity when restricted to F .

Also, if G is any subgroup of the group of automorphisms of E, then Inv G E is the subfield of elements that are invariant under all automor- phisms in G.4

Definition 2.4.10. A field extension E over F is called a Galois field ex- tension if and only if E is a splitting field of fpxq over F for some seperable fpxq P F rxs.

Lemma 2.4.12. If E is a splitting field of fpxq over F for some monic seperable fpxq P F rxs, then | GalpE{F q|  rE : F s.

Proposition 2.4.13. Let E be a field extension over F . Then the following statements are equivalent:

• E is a Galois field extension over F .

• F  Inv G for some finite subgroup of Aut E.

• rE : F s   8, and E is normal and seperable over F .

Theorem 2.4.14 (Fundamental Theorem of Galois Theory). Let E be a Galois field extension over F and define:

• Γ is the set of subgroups of GalpE{F q.

4It follows that G GalpE{F q is the subgroup of Aut E such that Inv G  F .

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• Σ is the set of subfields K  E such that F  K.

• γ : ΓÑ Σ : H ÞÑ InvpHq.

• σ : ΣÑ Γ : K ÞÑ GalpE{Kq.

Then γ and σ are inverse bijections, and we have the following properties:

1. @H1, H2 P Γ : H1 H2 ðñ Inv H1  H2,

2. @H P Γ : |H|  rE : Inv Hs ^ rG : Hs  rInv H : F s,

3. @H P Γ: H is a normal subgroup of GalpE{F q if and only if Inv H is a normal field extension over F . In this case GalppInv Hq{F q  pG{Hq.

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## Real Closed Fields

In this chapter we will develop the framework in which we prove Sturm’s Theorem. We will begin with a discussion of ordered fields, showing that a field can be (compatibly) ordered if and only if the field is formally real (meaning that no non-zero element is a sum of squares). We will then discuss real closed fields and some of their key properties, going on to investigate several equivalent characterizations of real closed fields. This serves to il- lustrate the importance of real closed fields in applications. Again we will follow  in our discourse.

### 3.1 Ordered and Formally Real Fields

Definition 3.1.1. An ordered field is a pairpF, P q where F is a field, and P  F such that

1. 0R P ,

2. @a P F : a  0 _ a P P _ a P P , 3. @a, b P P : a b P P ^ ab P P .

We call the elements of P the positive elements of F .

We also say that a field F can be ordered if and only if a P  F exists so thatpF, P q is an ordered field.

Lemma 3.1.1. If pF, P q is an ordered field, define the set of negative el- ements N  t x P F | Dp P P : x  p u. Then P , N and t 0 u are disjoint and F  P Y t 0 u Y N.

Proof. We first note that 0R P by property 1. This implies that 0  0 R N.

Therefore P X t 0 u  H  N X t 0 u. Now suppose that P X N  H and let a P P X N. Then a P P , so by property 3: 0  a  a P P . This contradiction with property 1 shows that P X N  H.

Now let aP F z t 0 u. Then by property 2, a P P or a P N, so a P P Y N and F  P Y t 0 u Y N.

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Definition 3.1.1 of an ordered field is not so intuitive at first glance, but it becomes more transparent when we recall that P were the positive elements and we consider the following:

Proposition 3.1.2. Any ordered field pF, P q induces a strict total order ¡ by:

@a, b P F : a ¡ b ðñ a  b P P, (3.1) with the following properties:

1. @a P F : ra ¡ 0 ðñ @b P F : a b ¡ bs, 2. @a, b P F : a ¡ 0 ^ b ¡ 0 ùñ ab ¡ 0.

Conversely, if ¡ is a strict total order with the properties above, then P  t x P F | x ¡ 0 u defines an ordered field pF, P q.

Proof. Let pF, P q be an ordered field and define ¡ as above. We shall first show that¡ is a strict total order. Let a, b, c P F such that a ¡ b and b ¡ c.

Then a b P P and b  c P P . We then have a  c  a  b b  c P P , so a¡ c, which shows that ¡ is transitive. Then, if we take a, b P F , we see by lemma 3.1.1 that either a b, a  b P P , or b  a P P . Therefore, either a b, a ¡ b or b ¡ a, so ¡ is trichotomous and a strict total order.

To prove property 1, let aP F . If a ¡ 0, then @b P F : pa bqb  a P P , so @b P F : a b ¡ b. Conversely, if @b P F : a b ¡ b, then this is in particular the case for b 0: a ¡ 0.

To prove property 2, let a, bP F with a ¡ 0 and b ¡ 0. Then a, b P P and thus abP P , which shows that ab ¡ 0.

Let us now suppose that we are given a strict total order ¡ on F satis- fying properties 1 and 2. Define P  t x P F | x ¡ 0 u. Then clearly 0 R F , since otherwise 0¡ 0, which is the first property of an ordered field. Also, by the trichotomy of¡, we have for any a P F : a ¡ 0, a  0, or 0 ¡ a. This means: a  a  0 P P , a  0, or a  0  a P P , so P also satisfies the second property. Now let a, bP P . Then by property 1 and the transitivity of ¡: a ¡ 0 ùñ a b ¡ b ¡ 0 ùñ a b P P . Also, by property 2:

a¡ 0 ^ b ¡ 0 ùñ ab ¡ 0 ùñ ab P P , which finally shows that pF, P q is an ordered field.

Note: Notation of ordered fields

From now on, if we speak of an ordered fieldpF, P q, and we use the symbol

¡, this will denote the induced strict total order. Also, the symbol ¥ will denote the total order induced by¡ (defined by a ¥ b ðñ a ¡ 0 _ a  b).

Similarly, for aP F we write |a|  a if a  0 or a ¡ 0 and |a|  a if a   0.

If the set P is not used, we may also just write: “the ordered field F ”.

Lemma 3.1.3. If pF, P q is an ordered field, then for any a P F : a2 P P . In particular we see that 1 12P P .

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Proof. Let aP F. Then either aP P or a P P , so that a2  paq2 P P , since P is closed under multiplication.

We state the following lemma without proof, as it is simply proven by considering the various possibilities of the signs of a and b.

Lemma 3.1.4 (Triangle inequality). If pF, P q is an ordered field, then for any a, bP F :

|a b| ¤ |a| |b|. (3.2)

We will now go on to prove a nice characterization of an ordered field in terms of sums of squares. The following definition is due to Artin and Schreier 1.

Definition 3.1.2. A formally real field is a field F that satisfies the following property:

@n P N@a1, . . . , anP F

 n

¸

i0

a2i  0 ùñ @i P t 1, . . . , n u : ai  0

 , i.e. the zero of the field is not the sum of non-zero squares, or the vanishing of a sum of squares implies the vanishing of all the individual squares.

The following lemma illustrates a different characterization2.

Lemma 3.1.5. A field F is formally real if and only ifEa1, . . . , anP F such that°n

i1a2i  1.

Proof. Let F be formally real. Now suppose that there exist a1, . . . , anP F such that °n

i1a2i  1. Then °n

i1a2i 12  1 12  0, which is forbidden, so no sucht aiu exist.

Conversely, let there exist no a1, . . . , anP F such that°n

i1a2i  1 and take b0, . . . , bm P F such that°m

i0b2i  0, and suppose that b0  0 (i.e. one of the bi is non-zero). Then b20  0 and°m

i1b2i  b20 ùñ °m

i1

bi

b0

2



1, which gives us a contradiction. Therefore, F is formally real.

Lemma 3.1.6. Any ordered field pF, P q is formally real.

Proof. Let aP F z t 0 u. Then either a ¡ 0 or a ¡ 0 and thus a2  paq2 ¡ 0. We will now show that any sum of non-zero squares is strictly greater than zero by induction.

The induction basis was the first step of the proof. Therefore, let kP N, k ¡ 0, a1, . . . , ak 1 P F z t 0 u and °k

i1a2i ¡ 0. Then a2k 1 ¡ 0 and thus

°k 1

i1 a2i ¡°k

i1a2i ¡ 0.

Therefore, if a1, . . . , an P F and °n

i1a2i  0, we must have that a1 

    an 0 and thus F is formally real.

1Artin and Schreier chose this as one of the key properties of the real number system, in an effort to characterize the real numbers in a purely algebraic way.

2This was actually the original definition of Artin and Schreier.

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The converse of the foregoing lemma is a theorem that was proved by Artin and Schreier [1, Satz 1.], and gives the definite answer on the connec- tion between the sums of squares in a field and orderings. We follow a proof of Jean-Pierre Serre 3. We first prove the following

Lemma 3.1.7. If P0 is a subgroup of the multiplicative group F of a field, such that P0 is closed under addition and contains all non-zero squares, and if aP F such thata R P0, then

P1  P0 P0a t p P F | Dx, y P P0: p x ya u is a subgroup of F that is closed under addition.

Proof. Let p1  x1 y1a, p2  x2 y2aP P1, where x1, y1, x2, y2 P P0. Then p1 p2  px1 x2q py1 y2qa P P1, since P0is closed under addition. Also, p1p2  px1 y1aqpx2 y2aq  px1x2 y1y2a2q px1y2 x2y1qa P P1, since a2 P P0 and P0 is closed under addition and multiplication.

If 0P P1, then Dx, y P P0 : x ya  0, so a  xy1 P P0, which leads to a contradiction. This shows that 0R P1 and thus P1  F.

Lastly, let p  x ya P P1, x, y P P0. Then p1  px yaq1  px yaqpx yaq2  rxppx yaq1q2s ryppx yaq1q2sa P P1, since x ya  0 and P0 contains all non-zero squares. Therefore, P1  F is a subgroup of the multiplicative group of F that is closed under addition.

Proposition 3.1.8 (Serre). If L is an extension field of an ordered field pK, P q, then L can be ordered as pL, PLq with P  PL (i.e. the ordering on L extends that on K) if and only if for all p1, . . . , pn P P  K and x1, . . . , xnP L: °n

i1pix2i  0 ùñ x1      xn 0.

Proof. Let pL, PLq be an ordered extension field of an ordered field pK, P q with P  PL. Now take p1, . . . , pn P P and x1, . . . , xn P L. We first see that for every xi either xi  0, in which case x2i  0, or xi ¡ 0 or xi ¡ 0 so that x2i  pxiq2 ¡ 0. Therefore, since each pi ¡ 0 and the positive elements are closed under addition and multiplication, if one of the xi 0:

°n

i1pix2i ¡ 0. So,°n

i1pix2i  0 implies that all xi  0.

Now suppose that the converse is true. Define T as the set of subgroups of L that are closed under addition and contain all elements of the form px2 where pP P and x P L.

Clearly the set P0  t°n

i1pix2i | p P P ^ xi P Lu is closed under addi- tion. Now let x °n

i1pix2i, y °m

j1qjyj2 P P0, where all pi, qj P P and xi, yj P L. Then xy  °n

i1°m

j1piqjpxiyjq2 P P0, so P0 is closed under multiplication. Also, if xP P0, then x1  xx2 P P0, because x  0 (by the hypothesis) and thus x2  px1q2 P P0. This shows that P0 P T , and thus T is non-empty.

3It is entertaining to note that although this argument was thought up by Serre, it was presented on a seminar by ´Elie Cartan.

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By Zorn’s Lemma we may now pick a maximal element PL P T . We claim that this PL makes L an ordered field that extends K. To see this, let a P L. If both a and a P PL, then 0 P PL, which is a contradiction, so a and a cannot be simultaneously in PL. Now, if a R PL, define P1  t x ya | x, y P PLu. Since PL certainly contains all non-zero squares (1P P ), we can conclude by lemma 3.1.7 P1 also is a subgroup of L that is closed under addition. Furthermore, take p P P and x P L. Then px2  px2p1 aqp1 aq1 ppx2 px2aqp1 aq1P P1, so P1 P T . Also, if xP PL, then x xp1 aqp1 aq1  px xaqp1 aq1 P P1, so PL P1. Because we took PLto be maximal in T we can now conclude that PL P1. Lastly, a  ap1 aqp1 aq1  pa2 aqp1 aq1 P P1  PL, since PL

contains all non-zero squares. We can now conclude that eithera P PL or aP PLexclusively.

The above showed thatpL, PLq is an ordered field. Now, if p P P  K, then p p12 P PL, so P  PL and the order extends the order on K.

Now we are ready to prove

Theorem 3.1.9. A field F can be ordered if and only if it is formally real.

Proof. We already saw that if a field F can be ordered, then it is formally real. Conversely, let F be a formally real field. Then its characteristic is 0 (for if it has characteristic n¥ 1, then °n

i112  0, which is not the case), and thus it contains Q as a subfield.

Let 0  pq11, . . . ,pqn

n P Q where all pi P Z and qi P Zz t 0 u, and x1, . . . , xnP F with°n

i1 pi

qix2i  0. Let us multiply with q1. . . qn: 0 q1. . . qn

¸n i1

pi qix2

¸n i1

 n

¹

j1 ji

qj

pix2i.

This is now a sum of integer multiples of squares, and thus simply a sum of squares. Since F is formally real, we can conclude that all xi  0. By proposition 3.1.8 we can therefore conclude that there exists an order on F that extends the standard order on Q.

As our last result on formally real/ordered fields we will give the following lemma, which provides us with bounds on the zeroes of a monic polynomial.

Lemma 3.1.10. Let F be an ordered field, fpxq  xn °n1

i0 aixi P F rxs monic and of positive degree, and c P F . Define M  maxp1,°n1

i0 |ai|q.

Then |c| ¡ M implies that |fpcq| ¡ 0. Conversely, if fpcq  0, then M ¤ c¤ M.

Proof. Let cP F with |c| ¡ M. We first note c  0, so that: 1  unfpuq 

°n1

i1 aiuin. Also, |un|   1, and for i P 0, . . . , n  1 we have |uin|

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|u1|. From this, and the triangle inequality, it follows that:

1 |unfpuq 

n¸1 i1

aiuin|

¤ |un||fpuq|

n¸1 i1

|ai||uin|

|fpuq| |u1|

n¸1 i1

|ai|

¤ |fpuq| M1M  |fpuq| 1, from which we can conclude that|fpuq| ¡ 0.

If we now negate this statement, then fpcq  0 implies that M ¤ c ¤ M .

### 3.2 Real Closed Fields

Artin and Schreier defined a refinement of formally real fields in an attempt to capture the characteristic algebraic properties of the real numbers. There are several useful examples of formally real fields, which include the real numbers, the real numbers that are algebraic over Q, the hyperreal numbers and the computable numbers. Let us state the definition.

Definition 3.2.1. A field F is called real closed if and only if F is formally real and no proper algebraic extension field of F is formally real.

This definition and the foregoing discussion of formally real fields shows that a real closed field F is closed in the sense that it can be ordered, but no extension of it can be ordered. We will go on to find some more useful characterizations. We first observe the following very useful facts, where we follow the proof in .

Lemma 3.2.1. If F is a real closed field, then:

• Every sum of squares in F can also be written as a single square.

• @x P F Dy P F : x  y2_ x  y2.

• Every polynomial fpxq P F rxs of odd degree has a zero in F .

Proof. Let γ P F not be a square. Then the polynomial x2 γ P F rxs is irreducible, so Fp?γq  F rxs{px2γq is a proper field extension of F , hence it is not formally real. This shows that there exist α1, . . . , αn, β1, . . . , βnP F such that

γ

¸n ν1

α2ν

¸n ν1

βν2 2? γ

¸n ν1

ανβν 

¸n ν1

ν?

γ βνq2  1.

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If°n

ν1ανβν  0, then ?γ P F , which leads to a contradiction, so that this sum vanishes. Also, if°n

ν1α2ν  0, then 1 would be a sum of squares in F , which is also a contradiction, so that sum does not vanish. We can then conclude that γ is not a sum of squares in F , since otherwise1 would be a sum of squares in F . Negating this statement leads to the first property.

By the first property we may now pick α, βP F such that:

α2 

¸n ν1

α2ν, β2  1

¸n ν1

βν2

(observe that 1 12) and thus:

γ  1 °n

ν1βν2

°n

ν1α2ν  β2 α2 

β α

2

.

From this we can conclude that either γ is a square, orγ is a square, which shows the second property.

Now let us pick any polynomial fpxq P F rxs with degpfq  2n 1, where nP N. Without loss of generality we may assume f to be monic, since F is a field. The third statement can then be proven by induction with respect to n.

If n  0, then the polynomial is of first degree and thus of the form fpxq  x  a, where a P F is a zero of fpxq.

Now let n¥ 1 and the statement be true for all gpxq P F rxs, degpgq  2k 1, k P N and k   n. If fpxq is reducible, then it can be written as fpxq  gpxqhpxq, where gpxq, hpxq P F rxs are monic and of positive degree strictly smaller than 2n 1, and one of them (say gpxq)must be of odd degree, since degpfq  degpgq degphq. By the induction hypothesis, gpxq then has a zero in F , and hence so does fpxq.

If fpxq is irreducible, we can form the proper field extension F pαq  Frxs{pfpxqq, where α P F pαq is a zero of fpxq. We then know that F pαq is not formally real, and thus there exist q1pxq, . . . , qrpxq P F rxs with degree smaller than 2n 1 such that:

¸r ν1

pqνpαqq2  1 P F.

This then shows that there exists some gpxq P F rxs such that:

¸r ν1

pqνpxqq2 fpxqgpxq  1.

Now, the degree of the qνpxq2 must be even, and therefore the degree of the sum must be even and positive and strictly less than 4n 2. We therefore

(26)

conclude that gpxq has odd degree less than or equal to 2n  1. Therefore gpxq has a zero ρ P F . However, then:

1 

¸r ν1

pqνpρqq2 fpρqgpρq 

¸r ν1

pqνpρqq2.

I.e. 1 is a sum of squares in F , leading to a contradiction. Therefore fpxq must be reducible and the third statement has been proven.

Lemma 3.2.2. If a field F is real closed, there exists one and only one P  F such that pF, P q is an ordered field. I.e. a real closed field can be uniquely ordered.

Proof. If F is formally real, then we know that it can be ordered. Let P  F be the positive numbers of such an ordering. Then we know that any non-zero square x2, xP F must be positive.

Now, if F is real closed, it is formally real and can thus be ordered. Also,

@x P F Dy P F such that x  y2, in which case x must be positive, or

x  y2, in which case x must be positive, in any ordering. Since this covers all non-zero elements of F , there exists only one ordering, namely the one where exactly all the non-zero squares are positive.

Note

From now on, when we speak of a real closed field, we will implicitly assume that it is equipped with this unique order.

The following result is the analog of the classical Fundamental Theorem of Algebra, and shows that real closed fields capture the important property that we may obtain an algebraically closed field by adjoining a single square root. In particular, this shows that the real numbers R form a real closed field, since C is obtained by adjoining?

1 and is algebraically closed.

Theorem 3.2.3. A field F is real closed if and only if it is not algebraically closed, and C  F p?

1q  F rxs{px2 1q is algebraically closed.

Proof. Let F be a real closed field. Then we see that x2 1 is irreducible, and hence has no zeroes in F , since otherwise1 would be a sum of squares in F . We can then define the field C  F rxs{px2 1q. We first define the automorphism z  a bi ÞÑ ¯z  a  bi of C, where i P C denotes a zero (any one of the two) of x2 1. This induces an automorphism fpxq ÞÑ ¯fpxq of Crxs. We then see that if fpxq P Crxs, then fpxq ¯fpxq P F rxs. Also, if fpxq ¯fpxq has a zero r in C, then fprq ¯fprq  0, and hence fpxq has a zero in C.

We now show that every element of C can be written as a square. To this end, let z  a bi P C. Then z¯z  a2 b2 P F and non-negative,

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