sturm’s theorem: determining the number of zeroes of...
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Sturm’s Theorem: determining the number ofzeroes of polynomials in an open interval.
Bachelor’s thesis
Eric Spreen
University of Groningen
July 12, 2014
Supervisors:
Prof. Dr. J. TopUniversity of Groningen
Dr. R. DyerUniversity of Groningen
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 thatenables one to find the number of roots of a polynomial that are containedwithin an open interval, simply by counting the number of sign changes intwo sequences. This result can be extended to decide the existence of aroot of a family of polynomials, by evaluating a set of polynomial equations,inequations and inequalities with integer coefficients.
Contents
1 Introduction 2
2 Polynomials and Extensions 42.1 Polynomial rings . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Degree arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Euclidean division algorithm . . . . . . . . . . . . . . . . . . 6
2.3.1 Polynomial factors . . . . . . . . . . . . . . . . . . . . 82.4 Field extensions . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.4.1 Simple Field Extensions . . . . . . . . . . . . . . . . . 102.4.2 Dimensionality of an Extension . . . . . . . . . . . . . 122.4.3 Splitting Fields . . . . . . . . . . . . . . . . . . . . . . 132.4.4 Galois Theory . . . . . . . . . . . . . . . . . . . . . . . 15
3 Real Closed Fields 173.1 Ordered and Formally Real Fields . . . . . . . . . . . . . . . 173.2 Real Closed Fields . . . . . . . . . . . . . . . . . . . . . . . . 223.3 The Intermediate Value Theorem . . . . . . . . . . . . . . . . 26
4 Sturm’s Theorem 274.1 Variations in sign . . . . . . . . . . . . . . . . . . . . . . . . . 274.2 Systems of equations, inequations and inequalities . . . . . . 324.3 Sturm’s Theorem Parametrized . . . . . . . . . . . . . . . . . 33
4.3.1 Tarski’s Principle . . . . . . . . . . . . . . . . . . . . . 38
1
Chapter 1
Introduction
In many of the natural sciences, polynomials and polynomial systems occuras useful approximations to real-world phenomena. As an example of thiswe give the harmonic oscillator, which is used in physics to approximatedynamical systems that are very close to an equilibrium point. The potentialenergy of such a (one-dimensional) system takes the form
V pxq 1
2kx2,
which is a second-degree polynomial in one variable. A similar case occursin higher-dimensional systems (in 3D, and with multiple bodies). Anotherexample is the hydrogen atom in quantum mechanics, where the radial wave-functions takes on the form: [3]
Rnlprq 1
reρρl1vpρq, ρ r
an,
where vpρq is a polynomial, and n P N, n ¡ 0 and l P N.1 It is a crucialproblem to find the zeroes of these functions in order to determine theelectronic structure of the atom, which can be done by finding the zeroes ofthe polynomial ρl1vpρq.
It is clear from these examples that finding solutions of polynomial equa-tions is a fundamental problem in applied mathematics. A classical resultthat enables us to do this numerically is Sturm’s Theorem, named afterJacques Charles Francois Sturm. This theorem gives the number of zeroesof a polynomial that are contained within a certain open interval, enablingus 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 differentrealizations. The key to the success of Sturm’s Theorem in real closed fields
1We use the convention that 0 P N.
2
is the analog of the intermediate value theorem for polynomials. It can beshown that this result, and various other key theorems from real analysishold for polynomials in real closed fields.
After the discussion of Sturm’s Theorem, we will discuss an extensionof Sturm’s Theorem that allows us to simplify the problem of the existenceof a zero in a certain interval for a whole family of polynomials. The resultwill be a finite set of systems of polynomial equations, inequations andinequalities with integer coefficients, any one of which may be satisfied bythe parameters of the family for the resulting polynomial to have zero inthe interval. From this we can quickly establish criteria for the existenceof a zero of a whole family of polynomials. A secondary result is that if apolynomial with rational coefficients has a zero in one real closed field, itwill have a zero in every other real closed field.
As a high school student I have often wondered whether it would bepossible 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 wewill touch on this question.
A significant portion of this report follows [4] and [5]. If no citationshave been provided, these are the sources. We will assume that the readerhas a basic understanding of algebraic structures, such as monoids, groupsand rings.
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Chapter 2
Polynomials and Extensions
Before we can begin our study of real closed fields, we will develop the theoryof polynomial rings over a field to some extent. This chapter will give basicresults on arbitrary polynomial rings, and applications on extension ringsand fields. Most of this chapter will follow [4].
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 byR Y tu1, . . . , un u by Rru1, . . . , uns. We can readily note that Rru1, u2s pRru1sqru2s by definition. The existance of such a subring follows from theobservation that any arbitrary intersection of subrings of S is again a subringof S.
2.1 Polynomial rings
Definition 2.1.1. Given a ring R, its polynomial ring Rrxs is the ring offunctions f : NÑ R such that there exists a k P N with @n P N : n ¥ k ùñfpnq 0. Addition and multiplication in Rrxs are defined as follows; forfpxq, gpxq P Rrxs and any n P N:
pf gqpnq fpnq gpnq, pfgqpnq n
i0
fpiqgpn iq.
and 0 and 1 as obvious. The elements of Rrxs are called polynomials withcoefficients in R, and the function values of a polynomial are called thecoefficients of a polynomial.
We will establish the basic properties of polynomial rings. The proofswill be ommitted and can be found in [4, Sec.2.10].
Proposition 2.1.1 (Properties of polynomial rings). Let R be a ring andRrxs its polynomial ring. Then:
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1. There exists an injective homomorphism RÑ Rrxs, so that R may beregarded as a subring of Rrxs.
2. Let x P Rrxs be the polynomial with xp1q 1 and xpnq 0 if n PNz t 1 u. Then Rrxs is generated by RYtx u. Furthermore, all elementsof R commute with x.
3. For any fpxq P N, there exists an n P N and unique a0, . . . , an P Rsuch that fpxq °n
i0 aixi.
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, thenthere exists a unique homomorphism ψ : Rrxs Ñ S such that ψ|R φ andψpxq 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 asubring 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 withI XR 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 thisproperty are isomorphic): if S is any other ring, and φ : R Ñ S is ahomomorphism, u P S and @r P R : φprqu uφprq, then there exists anx P Rrxs and a unique homomorphism ψ : Rrxs Ñ S so that ψ|R φ,φpxq u and Rrxs is generated by RY tx u. [4, p.124]
Note: Notation of polynomials
From now on, we will denote a polynomial with coefficients in a ring R asfpxq 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 ringRrx1, . . . , xns : Rrx1s . . . rxns. By induction we can then get an evaluationhomomorphism in multiple variables.
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2.2 Degree arithmetic
We have seen that for any polynomial 0 fpxq °ni0 aix
i there is somek P N such that ak 0, but ai 0 if i ¡ k. This observation is a strongtool that we will use often in further arguments. We therefore define
Definition 2.2.1. If R is a ring and 0 fpxq °ni0 aix
i P Rrxs, then thedegree of fpxq is the largest k P N such that ak 0. If fpxq 0, then thedegree 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 0if fpxq 0. This will be denoted by lcpfq or lcpfpxqq. A polynomial fpxqwill be called monic if lcpfq 1.
The following two lemmas can be proven quickly by the definition of thedegree and considering the leading coefficients of fpxq gpxq and fpxqgpxqrespectively.
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 allfpxq, gpxq P Drxs we have degpfgq degpfq degpgq.1 Also, the units ofDrxs will be the units of D.
2.3 Euclidean division algorithm
The proof of the following proposition will be given when we prove algorithm1.
Proposition 2.3.1. Let R be a commutative ring, and fpxq, gpxq P Rrxswith gpxq 0, m degpgq and bm the leading coefficient of gpxq. Thenthere exist k P 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 qpxqgpxqrpxq and degprq degpgq, where bm lcpgq. Now since gpxq 0,we have bm 0 and thus fpxq pbkm qpxqqgpxq pbkm rpxqq. Also, since Fis a domain: degpbkm rpxqq degpbkm qdegprpxqq degprpxqq degpgpxqq.
1It is to be understood here that 8 a 8 for any a P NY t8u.
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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 andthus 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 degpf0qa0 lcpf0q
fi1pxq #bfipxq aix
nimgpxq ni ¥ m
0 ni m
ni1 degpfi1qai1 lcpfi1q
Then there exists a k P N such that fkpxq 0, fk1pxq 0 and nk m.Also2:
bkmfpxq k1
l0
albkl1xnlm
gpxq fkpxq. (2.3)
Proof. Let i P N. We then see that degpfi1q degpbfipxq aixnimgpxq degpfipxqq, since the leading coefficients of bfipxq and aix
nimgpxq are bothaib. This shows that the degree strictly decreases each step, and sincef0pxq 0, there exists some k P N such that fkpxq 0, degpfkq nk mand thus fk1pxq 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 i P N such that i ¤ k we have
bifpxq fipxq °i1
l0 albil1xnlm
gpxq. For i 0 this is clear, so pick
i P N with 0 i ¤ k and assume this holds for i 1. Then:
bifpxq bbi1fpxq bfi1pxq b
i2
l0
albil2xnlm
gpxq
fipxq ai1xni1mgpxq
i2
l0
albil1xnlm
gpxq
fipxq i1
l0
albil1xnlm
gpxq.
This proves our claim. We can then set i k to obtain our final formulafor fpxq, which concludes the proof and also proves proposition 2.3.1, sincedegpfkq 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 usefulfacts. The first of these will concern factors of polynomials. Since we willalmost 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 thatfpxq gpxqhpxq.
Also, a polynomial fpxq P Rrxs of positive degree will be called reducibleif 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. Theywill 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 PRrxs such that fpxq px aqqpxq fpaq.Proof. By the Euclidean division algorithm we may pick qpxq, rpxq P Rrxswith degprq degpx aq 1 and fpxq px aqqpxq rpxq. We thenimmediately see that fpaq pa aqqpaq rpaq rpaq, and since degprq 1we must have rpxq fpaq. Also, since rpxq is fixed in this way, if q1pxq alsosatisfies 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 getqpxq q1pxq.Corollary 2.3.4. If fpxq P Rrxs and a P R. Then a is a zero of fpxq if andonly if px 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, ifpx 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 agreatest common factor (or divisor) of a pair of polynomials pfpxq, gpxqqwe mean a polynomial hpxq such that hpxq fpxq, hpxq gpxq and ifdpxq P F rxs such that dpxq fpxq and dpxq gpxq, then dpxq hpxq. Degreeconsiderations quickly show that two greatest common factors differ by aunit factor in F . Now, for any two polynomials fpxq, gpxq P F rxs we thendefine 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, oneof them must be a unit. However, in polynomial rings over a field, this leads to the sameconcept. 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, andqpxq, 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 ifhpxq gpxq and hpxq rpxq.Proof. Let hpxq P F rxs. If hpxq fpxq and hpxq gpxq, then thereexists 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 PF rxs so that gpxq γpxqhpxq and rpxq ρpxqhpxq, so that fpxq pqpxqγpxqρpxqqhpxq and thus hpxq fpxq and hpxq gpxq.
Algorithm 2:The GCD and Euclidean sequence of twopolynomials over a field
Let fpxq, gpxq P F rxs where F is a field and gpxq 0, and define thefollowing sequence:
h0pxq fpxq h1pxq gpxq
hi2pxq #qi1pxqhi1pxq hipxq, if hi1pxq 0
0, if hi1pxq 0
degphi2q degphi1q, i P N.
(It is understood here that 8 8). Then there exists a 1 ¤ s P N suchthat hspxq 0, but hs1pxq 0. Furthermore, hspxq is a greatest commonfactor of fpxq and gpxq. The finite sequence (terminating at hspxq) thuslydefined is called the Euclidean sequence of fpxq and gpxq.Proof. Let dpxq P F rxs be a common divisor of fpxq and gpxq. Then byrepeated use of lemma 2.3.5 see that this is the case if and only if dpxq isa common divisor of hs1pxq and hspxq. Therefore, every common divisorof fpxq and gpxq will be a factor of hspxq. Also, since hspxq qspxqhspxq hs1pxq and hspxq is a factor of itself, hspxq is a common factor of fpxq andgpxq, 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 particularlyimportant for the study of polynomial equations. We will discuss field ex-tension that are generated by finitely many elements, which can be seen asfields that are obtained by adjoining some elements.
Definition 2.4.1. Let F be a field. An field extension over F is then a fieldE 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 thesmallest subfield containing S.
If E is a field extension over F , and S E, we denote the subfieldgenerated by S Y F by F pSq. If S tu1, . . . , un u is finite, we denote it byF pu1, . . . , unq.
2.4.1 Simple Field Extensions
We will first consider the structure of simple field extensions. That is, fieldextensions over F of the form F puq. The following proposition will be crucialin our discussion. In this discussion we take a slightly different route thanin [4].
Proposition 2.4.1. Let F be a field. Then F rxs 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 alsoclearly pick a non-zero gpxq P Rrxs of minimal degree. Then there existqpxq, 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, wemust have that rpxq 0. Therefore fpxq qpxqgpxq for some qpxq P F rxsand we conclude that I pgpxqq is principal. We have also already seenthat, since F is a domain, F rxs 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 thatfpuq 0. Otherwise, we will call u transcendental over R.
A field extension E over F will be called algebraic if and only if everyelement of E is algebraic over F .
Lemma 2.4.2. If R is a subring of the commutative ring S and u P S istranscendental over R, then F rxs F rus.Proof. If u is transcendental over R, then the kernel of the evaluation homo-morphism ρ : F rxs 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 thereexists some fpxq °n
i1 aixi P F rxs such that x fpuq °n
i1 aiui 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 andirreducible. Then F rxspfpxqq is a field. Furthermore, F rxspfpxqq F puq,where u x mod pfpxqq is a zero of fpxq, when regarded as a polynomialwith coefficients in F rxspfpxqq.Proof. Let J 1 F rxsI be an ideal, where I pfpxqq. Then there existsan ideal J F rxs such that I J and J 1 JI. Since F rxs is a principalideal 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 isirreducible, we have either hpxq P F or gpxq P F . In the first case, we getJ I, so that J 1 t 0 mod I u. In the second case we get J F rxs, sothat J 1 F rxsI. This shows that F rxsI has only the trivial ideals andthus it is a field.
Now set K F rxspfpxqq and let u x mod pfpxqq P K. Then clearlyfpuq fpxq mod pfpxqq 0 mod pfpxqq. We also have F puq K. Nowlet a P 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 °mi0 rix
i. Then a rpuq °mi0 riu
i P F puq. We therefore conclude thatK F puq.Proposition 2.4.4. Let E be a field extension over F and u P F . If u istranscendental over F , then F puq F pxq, the field of fractions of F rxs. Ifu is algebraic over F , there exists some irreducible gpxq P F rxs such thatF puq F rxspgpxqq. Moreover, this gpxq is unique up to a unit multiplier.
Proof. If u is transcendental, then F rxs F rus. Now, since F rus F puq,and the field of fractions of F rus is the smallest field containing F rus (andthe fields of fractions of two isomorphic integral domains are isomorphic),we get F puq F puq.
Now suppose that u is algebraic over F . Then, since F rxs is a prin-cipal ideal domain, there exists some gpxq P F rxs such that the kernel ofthe evaluation homomorphism ρ : F rxs Ñ E in u is I pgpxqq and thusρpF rxsq F rxsI, by the first isomorphism theorem of rings. We claim thatF rxsI 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 kpxqgpxqand hence gpxq hpxqkpxqgpxq. This would imply that degphq 0, andhence hpxq P F . Similarly, if hpxq P I, then fpxq P F . Now let bothfpxq, hpxq R I. Then fpuq 0 hpuq, but fpuqhpuq gpuq 0, so thenE would contain non-zero zero-divisors. From this we conclude that gpxq isirreducible, and by lemma 2.4.3 we conclude that F rxsI is a field.
We now observe that F ρpF rxsq (since I XF t 0 u) and u P ρpF rxsq,so that F puq ρpF rxsq. But if x P ρpF rxsq, then there exists a fpxq °ni1 aix
i P F rxs with x fpuq °ni1 aiu
i P F puq. Therefore, F puq ρpF rxsq F rxspgpxqq.
Now let t 0 u I F rxs be an ideal and fpxq, gpxq P F rxs suchthat I pfpxqq pgpxqq. Then, there exist hpxq, kpxq P F rxs such thatfpxq hpxqgpxq and gpxq kpxqfpxq, so that fpxq hpxqkpxqfpxq. Degreeconsiderations then show that 0 hpxq, kpxq P F . Therefore, fpxq agpxqfor some unit a P 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 F rxspgpxqq 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 spaceover F . In this vector space, the addition is the normal addition of thefield, and the scalar multiplication is the normal multiplication in F , wherethe scalars lie in F . In particular, in vector spaces we have the notion of adimension. This dimension turns out to be of critical importance.
Definition 2.4.4. If E is a field extension over F , the dimensionality (ordegree) of E over F is the dimensionality of E regarded as a vector spaceover F , which shall be denoted as rE : F s.Proposition 2.4.5. Let E be a field extension over F and u P F . Then u isalgebraic over F if and only if rF puq : F s 8. Moreover, if u is algebraic,then rF 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 a P F puq be arbitrary. Then there exists a gpxq °n1
i0 aixi P F rxs
such that a gpuq °n1i0 aiu
i where n degpfq and ai P F for 0 ¤ i ¤ n1. Therefore, p1, u, . . . , un1q spans the vector space F puq over F . Now letb0, . . . , bn1 P F such that
°i0 biu
i 0. Then hpxq °n1i0 bix
i P pfpxqq.But since degphq degpfq we get hpxq 0, so that b0 bn1 0.This shows that p1, u, . . . , un1q is a base for the vector space F puq over F ,and hence rF puq : F s n 8.
Now let u be transcendental over F . Then let n P N and a0, . . . , an P F .We recall that F rxs F pxq F puq. Therefore, 0 °n
i0 aiui °n
i0 aixi
implies that a0 an 0, which shows that there exists no finite basefor F rxs as a vector space over F . Now, since F rxs is a subspace of F puq,we then certainly have that rF 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 extensionover E, which is in turn a field extension over F . Then K is a field extensionover F and rK : 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 tu1, . . . , un u K be a base for K. Then for every a P K there exist
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a1, . . . , an P F E such that°ni1 aiui a. Therefore, tu1, . . . , un u 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
i1 aivi. Also, for 1 ¤i ¤ n there exist bi1, . . . , bim P F such that ai °m
j1 bijuj . This givesus a °n
i1
°mj1 bijujvi, so that the set tujvi | 1 ¤ i ¤ n^ 1 ¤ j ¤ m u
spans K as a vector space over F . Now let c11, . . . , cnm P F such that°ni0
°mj0 cijujvi 0. Then clearly for 1 ¤ i ¤ n: di °m
j0 cijuj 0.This implies that for 1 ¤ i ¤ n and 1 ¤ j ¤ m we have cij 0. Thereforewe have obtained a base for K over F and rK : F s nm rK : EsrE :F s 8.
The following corollary is immediate and will be used later on.
Corollary 2.4.7. Let K be a field extension of F with rK : F s 8. Thenfor any field extension E K of F , rE : F s rK : F s. Also, if rK : F s isprime, 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 somemonic polynomial fpxq P F rxs, the u P E is a zero of fpxq if and onlyif px uq fpxq, where we regard fpxq as a polynomial in Erxs (by theinduced inclusion homomorphism). It would of course be great if we couldfind a field extension E over F where we could write fpxq ±n
i1px riqfor r1, . . . , rn P E. Also, taking into account the results we obtained for thedimensionalities of field extensions, we would like that E F pr1, . . . , rnq,since then rE : F s ±n
i1rF pu1, . . . , uiq : F pu1, . . . , ui1qs (where the firstterm in the product is understood to be rF pu1q : F s. Also, if r P E is thena zero of fpxq, then 0 fprq ±n
i1pr riq such that r ri for some1 ¤ i ¤ n. We will call such a field extension a splitting field of fpxq overF .
Definition 2.4.5. Let F be a field and fpxq P F rxs monic. Then a splittingfield of fpxq over F is a field extension E over F , such that in Erxs we canwrite fpxq ±n
i1px riq for r1, . . . , rn P E and E F pr1, . . . , rnq.Proposition 2.4.8. If F is a field and fpxq P F rxs is monic, then thereexists a splitting field E of fpxq over F .
Proof. Let fpxq ±ki1 fipxq where for 1 ¤ i ¤ k, fkpxq P F rxs is monic
and irreducible. Then k ¤ n degpfq. If n k, F itself is a splittingfield of fpxq. Now let n k ¡ 0. Then for some j P t 1, . . . , k u we havedegpfjq ¡ 1. Set K F rxspf1pxqq, which is a field that contains F , andr x mod pfipxqq, so that K F prq and f1prq 0. Then in Krxs
13
we have fpxq ±li1 gipxq, where for 1 ¤ i ¤ l, gipxq P Krxs are the
irreducible factors of fpxq in Krxs. Since these factors can be obtainedby taking the irreducible factors of the fipxq and px rq P Krxs is anirreducible 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 thatfpxq ±n
iipx riq, where r ri for some 1 ¤ i ¤ n. This shows thatE 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[4].
Proposition 2.4.9. Let F be a field, fpxq P F rxs monic and of positivedegree, 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 infactors of degree one is unique. Therefore, the zeroes are unique and thefollowing definition is consistent over every possible splitting field.
Definition 2.4.6. Let F be a field, fpxq P F rxs monic and of positivedegree, and E a splitting field of fpxq over F . Write fpxq ±m
i1px riqki ,where ri rj if i j. We then call ki the multiplicity of ri. Also, a zerori is called a simple zero if and only if it has multiplicity 1. Otherwise, it iscalled a multiple zero.
We lastly make the connection between the derivative of a polynomialand the character of its zeroes. Informally, we will see that a zero (in asplitting field) has multiplicity greater than 1 if and only if the polynomialand its derivative have a common factor of positive degree. For this wedefine the following map on a polynomial ring of a field.
Definition 2.4.7. Let F be a field. We then define the standard derivationin F rxs as the unique function F rxs Ñ F rxs : fpxq ÞÑ f 1pxq so that for anyfpxq, gpxq P F rxs:
1. pf gq1pxq f 1pxq g1pxq2. pfgq1pxq f 1pxqgpxq fpxqg1pxq3. 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 positivedegree, and E any splitting field of fpxq over F . Then all zeroes of fpxq inE are simple if and only if gcdpf, f 1q 1.
14
2.4.4 Galois Theory
Galois theory is one of the pearls of modern mathematics. It allows oneto study solutions of algebraic equations in a purely algebraic way. At theheart of the theory is the connection between the solutions of such equationsand group theory. We will state the fundamental results without proof forlater 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 irreduciblefactors have distinct zeroes in any splitting field.
An algebraic field extension E over F is called seperable over F if andonly 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 polynomialin F rxs 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 GalpEF q of automorphismsof E that reduce to the identity when restricted to F .
Also, if G is any subgroup of the group of automorphisms of E, thenInvG 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 seperablefpxq P F rxs.Lemma 2.4.12. If E is a splitting field of fpxq over F for some monicseperable fpxq P F rxs, then |GalpEF q| rE : F s.Proposition 2.4.13. Let E be a field extension over F . Then the followingstatements are equivalent:
• E is a Galois field extension over F .
• F InvG for some finite subgroup of AutE.
• rE : F s 8, and E is normal and seperable over F .
Theorem 2.4.14 (Fundamental Theorem of Galois Theory). Let E be aGalois field extension over F and define:
• Γ is the set of subgroups of GalpEF q.4It follows that G GalpEF q is the subgroup of AutE such that InvG F .
15
• Σ is the set of subfields K E such that F K.
• γ : Γ Ñ Σ : H ÞÑ InvpHq.• σ : Σ Ñ Γ : K ÞÑ GalpEKq.
Then γ and σ are inverse bijections, and we have the following properties:
1. @H1, H2 P Γ : H1 H2 ðñ InvH1 H2,
2. @H P Γ : |H| rE : InvHs ^ rG : Hs rInvH : F s,3. @H P Γ: H is a normal subgroup of GalpEF q if and only if InvH is a
normal field extension over F . In this case GalppInvHqF q pGHq.
16
Chapter 3
Real Closed Fields
In this chapter we will develop the framework in which we prove Sturm’sTheorem. We will begin with a discussion of ordered fields, showing thata 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 discussreal closed fields and some of their key properties, going on to investigateseveral equivalent characterizations of real closed fields. This serves to il-lustrate the importance of real closed fields in applications. Again we willfollow [4] in our discourse.
3.1 Ordered and Formally Real Fields
Definition 3.1.1. An ordered field is a pair pF, P q where F is a field, andP F such that
1. 0 R 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 that pF, 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 tx P F | Dp P P : x p u. Then P , N and t 0 u are disjointand F P Y t 0 u YN .
Proof. We first note that 0 R 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 andlet a P P X N . Then a P P , so by property 3: 0 a a P P . Thiscontradiction with property 1 shows that P XN H.
Now let a P F z t 0 u. Then by property 2, a P P or a P N , so a P P YNand F P Y t 0 u YN .
17
Definition 3.1.1 of an ordered field is not so intuitive at first glance, but itbecomes more transparent when we recall that P were the positive elementsand 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, thenP tx 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 firstshow 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 , soa ¡ c, which shows that ¡ is transitive. Then, if we take a, b P F , we seeby lemma 3.1.1 that either a b, a b P P , or b a P P . Therefore, eithera b, a ¡ b or b ¡ a, so ¡ is trichotomous and a strict total order.
To prove property 1, let a P F . If a ¡ 0, then @b P F : pabqb a P P ,so @b P F : a b ¡ b. Conversely, if @b P F : a b ¡ b, then this is inparticular the case for b 0: a ¡ 0.
To prove property 2, let a, b P F with a ¡ 0 and b ¡ 0. Then a, b P Pand thus ab P 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 tx 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. Thismeans: a a 0 P P , a 0, or a 0 a P P , so P also satisfies thesecond property. Now let a, b P P . Then by property 1 and the transitivityof ¡: 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 isan ordered field.
Note: Notation of ordered fields
From now on, if we speak of an ordered field pF, P q, and we use the symbol¡, this will denote the induced strict total order. Also, the symbol ¥ willdenote the total order induced by ¡ (defined by a ¥ b ðñ a ¡ 0_ a b).Similarly, for a P 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 12 P P .
18
Proof. Let a P F . Then either a P P or a P P , so that a2 paq2 P P ,since P is closed under multiplication.
We state the following lemma without proof, as it is simply proven byconsidering 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 forany a, b P F :
|a b| ¤ |a| |b|. (3.2)
We will now go on to prove a nice characterization of an ordered fieldin terms of sums of squares. The following definition is due to Artin andSchreier [1]1.
Definition 3.1.2. A formally real field is a field F that satisfies the followingproperty:
@n P N@a1, . . . , an P Fn
i0
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 vanishingof 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 if Ea1, . . . , an P F suchthat
°ni1 a
2i 1.
Proof. Let F be formally real. Now suppose that there exist a1, . . . , an P Fsuch that
°ni1 a
2i 1. Then
°ni1 a
2i 12 1 12 0, which is
forbidden, so no such t ai u exist.Conversely, let there exist no a1, . . . , an P F such that
°ni1 a
2i 1 and
take b0, . . . , bm P F such that°mi0 b
2i 0, and suppose that b0 0 (i.e. one
of the bi is non-zero). Then b20 0 and°mi1 b
2i b20 ùñ °m
i1
bib0
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 a P F z t 0 u. Then either a ¡ 0 or a ¡ 0 and thus a2 paq2 ¡0. We will now show that any sum of non-zero squares is strictly greaterthan zero by induction.
The induction basis was the first step of the proof. Therefore, let k P N,k ¡ 0, a1, . . . , ak1 P F z t 0 u and
°ki1 a
2i ¡ 0. Then a2
k1 ¡ 0 and thus°k1i1 a
2i ¡
°ki1 a
2i ¡ 0.
Therefore, if a1, . . . , an P F and°ni1 a
2i 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.
19
The converse of the foregoing lemma is a theorem that was proved byArtin 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 proofof Jean-Pierre Serre [6]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, andif a P F such that a R P0, then
P1 P0 P0a t p P F | Dx, y P P0 : p x ya uis a subgroup of F that is closed under addition.
Proof. Let p1 x1 y1a, p2 x2 y2a P P1, where x1, y1, x2, y2 P P0. Thenp1p2 px1x2qpy1y2qa P P1, since P0 is closed under addition. Also,p1p2 px1 y1aqpx2 y2aq px1x2 y1y2a
2q px1y2 x2y1qa P P1, sincea2 P P0 and P0 is closed under addition and multiplication.
If 0 P P1, then Dx, y P P0 : x ya 0, so a xy1 P P0, which leadsto a contradiction. This shows that 0 R P1 and thus P1 F .
Lastly, let p x ya P P1, x, y P P0. Then p1 px yaq1 px yaqpx yaq2 rxppx yaq1q2s ryppx yaq1q2sa P P1, sincex ya 0 and P0 contains all non-zero squares. Therefore, P1 F is asubgroup 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 fieldpK,P q, then L can be ordered as pL,PLq with P PL (i.e. the orderingon L extends that on K) if and only if for all p1, . . . , pn P P K andx1, . . . , xn P L:
°ni1 pix
2i 0 ùñ x1 xn 0.
Proof. Let pL,PLq be an ordered extension field of an ordered field pK,P qwith P PL. Now take p1, . . . , pn P P and x1, . . . , xn P L. We first seethat for every xi either xi 0, in which case x2
i 0, or xi ¡ 0 or xi ¡ 0so that x2
i pxiq2 ¡ 0. Therefore, since each pi ¡ 0 and the positiveelements are closed under addition and multiplication, if one of the xi 0:°ni1 pix
2i ¡ 0. So,
°ni1 pix
2i 0 implies that all xi 0.
Now suppose that the converse is true. Define T as the set of subgroupsof L that are closed under addition and contain all elements of the formpx2 where p P P and x P L.
Clearly the set P0 t°ni1 pix
2i | p P P ^ xi P L u is closed under addi-
tion. Now let x °ni1 pix
2i , y
°mj1 qjy
2j P P0, where all pi, qj P P and
xi, yj P L. Then xy °ni1
°mj1 piqjpxiyjq2 P P0, so P0 is closed under
multiplication. Also, if x P P0, then x1 xx2 P P0, because x 0 (bythe hypothesis) and thus x2 px1q2 P P0. This shows that P0 P T , andthus T is non-empty.
3It is entertaining to note that although this argument was thought up by Serre, it waspresented on a seminar by Elie Cartan.
20
By Zorn’s Lemma we may now pick a maximal element PL P T . Weclaim 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, defineP 1 tx ya | x, y P PL u. Since PL certainly contains all non-zero squares(1 P P ), we can conclude by lemma 3.1.7 P 1 also is a subgroup of L thatis closed under addition. Furthermore, take p P P and x P L. Thenpx2 px2p1 aqp1 aq1 ppx2 px2aqp1 aq1 P P 1, so P 1 P T . Also, ifx P PL, then x xp1 aqp1 aq1 px xaqp1 aq1 P P 1, so PL P 1.Because we took PL to be maximal in T we can now conclude that PL P 1.Lastly, a ap1 aqp1 aq1 pa2 aqp1 aq1 P P 1 PL, since PLcontains all non-zero squares. We can now conclude that either a P PL ora P PL exclusively.
The above showed that pL,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 formallyreal. Conversely, let F be a formally real field. Then its characteristic is 0(for if it has characteristic n ¥ 1, then
°ni1 12 0, which is not the case),
and thus it contains Q as a subfield.Let 0 p1
q1, . . . , pnqn P Q where all pi P Z and qi P Zz t 0 u, and x1, . . . , xn P
F with°ni1
piqix2i 0. Let us multiply with q1 . . . qn:
0 q1 . . . qn
n
i1
piqix2
n
i1
n¹
j1 ji
qj
pix
2i .
This is now a sum of integer multiples of squares, and thus simply a sumof squares. Since F is formally real, we can conclude that all xi 0. Byproposition 3.1.8 we can therefore conclude that there exists an order on Fthat extends the standard order on Q.
As our last result on formally real/ordered fields we will give the followinglemma, which provides us with bounds on the zeroes of a monic polynomial.
Lemma 3.1.10. Let F be an ordered field, fpxq xn °n1i0 aix
i P F rxsmonic and of positive degree, and c P F . Define M maxp1,°n1
i0 |ai|q.Then |c| ¡ M implies that |fpcq| ¡ 0. Conversely, if fpcq 0, then M ¤c ¤M .
Proof. Let c P F with |c| ¡M . We first note c 0, so that: 1 unfpuq °n1i1 aiu
in. Also, |un| 1, and for i P 0, . . . , n 1 we have |uin|
21
|u1|. From this, and the triangle inequality, it follows that:
1 |unfpuq n1
i1
aiuin|
¤ |un||fpuq| n1
i1
|ai||uin|
|fpuq| |u1|n1
i1
|ai|
¤ |fpuq| M1M |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 attemptto capture the characteristic algebraic properties of the real numbers. Thereare several useful examples of formally real fields, which include the realnumbers, the real numbers that are algebraic over Q, the hyperreal numbersand 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 formallyreal and no proper algebraic extension field of F is formally real.
This definition and the foregoing discussion of formally real fields showsthat a real closed field F is closed in the sense that it can be ordered, butno extension of it can be ordered. We will go on to find some more usefulcharacterizations. We first observe the following very useful facts, where wefollow the proof in [1].
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 isirreducible, so F p?γq F rxspx2γq is a proper field extension of F , henceit is not formally real. This shows that there exist α1, . . . , αn, β1, . . . , βn P Fsuch that
γn
ν1
α2ν
n
ν1
β2ν 2
?γ
n
ν1
ανβν n
ν1
pαν?γ βνq2 1.
22
If°nν1 ανβν 0, then
?γ P F , which leads to a contradiction, so that this
sum vanishes. Also, if°nν1 α
2ν 0, then 1 would be a sum of squares in
F , which is also a contradiction, so that sum does not vanish. We can thenconclude that γ is not a sum of squares in F , since otherwise 1 would bea sum of squares in F . Negating this statement leads to the first property.
By the first property we may now pick α, β P F such that:
α2 n
ν1
α2ν , β2 1
n
ν1
β2ν
(observe that 1 12) and thus:
γ 1°nν1 β
2ν°n
ν1 α2ν
β2
α2β
α
2
.
From this we can conclude that either γ is a square, or γ is a square, whichshows the second property.
Now let us pick any polynomial fpxq P F rxs with degpfq 2n1, wheren P N. Without loss of generality we may assume f to be monic, since F isa field. The third statement can then be proven by induction with respectto n.
If n 0, then the polynomial is of first degree and thus of the formfpxq 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 asfpxq gpxqhpxq, where gpxq, hpxq P F rxs are monic and of positive degreestrictly smaller than 2n 1, and one of them (say gpxq)must be of odddegree, since degpfq degpgq degphq. By the induction hypothesis, gpxqthen has a zero in F , and hence so does fpxq.
If fpxq is irreducible, we can form the proper field extension F pαq F rxspfpxqq, where α P F pαq is a zero of fpxq. We then know that F pαq isnot formally real, and thus there exist q1pxq, . . . , qrpxq P F rxs with degreesmaller than 2n 1 such that:
r
ν1
pqνpαqq2 1 P F.
This then shows that there exists some gpxq P F rxs such that:
r
ν1
pqνpxqq2 fpxqgpxq 1.
Now, the degree of the qνpxq2 must be even, and therefore the degree of thesum must be even and positive and strictly less than 4n 2. We therefore
23
conclude that gpxq has odd degree less than or equal to 2n 1. Thereforegpxq has a zero ρ P F . However, then:
1 r
ν1
pqνpρqq2 fpρqgpρq r
ν1
pqνpρqq2.
I.e. 1 is a sum of squares in F , leading to a contradiction. Therefore fpxqmust 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 oneP F such that pF, P q is an ordered field. I.e. a real closed field can beuniquely ordered.
Proof. If F is formally real, then we know that it can be ordered. LetP F be the positive numbers of such an ordering. Then we know thatany non-zero square x2, x P 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, orx y2, in which case x must be positive, in any ordering. Since thiscovers all non-zero elements of F , there exists only one ordering, namely theone where exactly all the non-zero squares are positive.
Note
From now on, when we speak of a real closed field, we will implicitly assumethat it is equipped with this unique order.
The following result is the analog of the classical Fundamental Theoremof Algebra, and shows that real closed fields capture the important propertythat we may obtain an algebraically closed field by adjoining a single squareroot. In particular, this shows that the real numbers R form a real closedfield, 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 algebraicallyclosed, and C F p?1q F rxspx2 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 squaresin F . We can then define the field C F rxspx2 1q. We first define theautomorphism 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 fpxqof Crxs. We then see that if fpxq P Crxs, then fpxqfpxq P F rxs. Also, iffpxqfpxq has a zero r in C, then fprqfprq 0, and hence fpxq has a zeroin C.
We now show that every element of C can be written as a square. Tothis end, let z a bi P C. Then zz a2 b2 P F and non-negative,
24
so that Dα P F : a2 b2 α2. Also, α2 ¥ a2 so that |α| ¥ |a| and henceDc1, c2 P F , where we can pick c1c2 with the same sign as b, such that
c21 a |α|
2, c22 a |α|
2.
Also:
p2c1c2q2 4a |α|
2
a |α|2
a2 pa2 b2q b2.
We can therefore conclude that pc1 c2iq2 c21 c2
2 2c1c2i a bi. Thisshows that there exists no algebraic extension field E of C with rE : Cs 2(since any quadratic equation is reducible).
With the foregoing in mind, we now let fpxq P F rxs be a monic polyno-mial of even degree. We define E to be a splitting field over F of fpxqpx21q,such that C E. Then E is Galois over F (since F is of characteristic 0and thus any polynomial in F rxs is seperable; hence E is the splitting fieldof a separable polynomial). We write |GalEF | 2em, where m is odd.By Sylow’s theorem, GalEF contains a subgroup H with |H| 2e. LetH be the subfield of E containing F corresponding to H under the Galoispairing. Therefore, 2em rE : F s rE : HsrH : F s 2erH : F s so thatrH : F s m. But since every polynomial of odd degree in F has a zero inF , F has no proper odd-dimensional algebraic extension fields. Therefore,m 1, H GalEF , and H E. We can now conclude that because|GalEF | is even, we can obtain E by repeatedly adjoining square roots.However, since we obtained C by adjoining a square root and C contains allpossible square roots, we must have that C E. Therefore, C is a splittingfield of fpxqpx2 1q and hence contains all zeroes of fpxq4. This shows thatevery polynomial in F rxs has a zero in C. By the reasoning above we canthen conclude that C is algebraically closed.
We will now go on to show the converse. Let F be a field that is notalgebraically closed, but let C F piq F rxspx21q be algebraically closed.We then clearly see that
?1 R F , since otherwise x21 would be reducibleand C would not be a field. Now let a, b P F . We can show in the same wayas before that every element of C can be written as a square, and so we pickz P C such that z2 a bi. Then a2 b2 pa biqpa biq z2z2 pzzq2and zz P F . This shows that every sum of squares in F can be writtenas a square. In particular, 1 is not a square, and hence not a sum ofsquares, so that F is formally real. We can also see that C is an algebraicclosure of F (since C is generated by i ?1 with minimum polynomialx2 1, so that rC : F s 2 8 and C is algebraically closed), so that everyalgebraic extension of F is contained within C. But then, C is the onlyproper algebraic extension field, and C is not formally real (as i2 1), sothat F is real closed.
4We note that x2 1 splits in Crxs as px iqpx iq.
25
3.3 The Intermediate Value Theorem
In this section we will discuss a very important theorem for real continousand differentiable functions that holds in the context of polynomials withcoefficients in a real closed field. This is the familiar intermediate valuetheorem, and it will be the key to our success in the next chapter.
Theorem 3.3.1 (Intermediate Value Theorem). Let F be a real closed field,fpxq P F rxs, a, b P F and a b. Then if fpaqfpbq 0, there exists a c P Fsuch that a c b and fpcq 0.
Proof. From theorem 3.2.3 we already know that the only irreducible poly-nomials in F rxs are going to be those of degree 1 or 2. Furthermore, apolynomial x2 αx β P Rrxs is going to be irreducible if and only ifα2 4β 0. This follows in the same way as for second degree polynomialswith real coefficients.
Now let us pick fpxq P F rxs to be monic and of positive degree. Thegeneral case then follows quickly by dividing out the leading coefficient andby noting that the premise cannot hold for polynomials of degree zero. Wecan write fpxq in terms of its irreducible factors as:
fpxq m¹i1
px riqs¹j1
gjpxq,
where r1, . . . , rm P R and g1pxq, . . . , gspxq P Rrxs with:
gjpxq x2 ajx bj , a2j 4bj 0, 1 ¤ j ¤ s.
For j P t 1, . . . , s u we can, by lemma 3.2.1, find 0 cj P R such thatc2j 1
4p4bj a2j q. We can then write:
gjpxq x aj
2
2 c2j ,
so that for all u P R, gjpuq ¡ 0.We first rule out the case that fpxq has no irreducible factors of first
degree. If this would be the case, then fpaqfpbq ±sj1 gjpaqgjpbq ¡ 0,
contradicting our hypothesis.Now, if @i P t 1, . . . ,m u : a ri ^ b ri, then fpaqfpbq ±m
i1pa riqpbriq±2
j1 gjpaqgjpbq ¡ 0. Similarly, if @i P t 1, . . . ,m u : a ¡ ri^b ¡ ri,then also fpaqfpbq ¡ 0. We conclude that there exists a i P t 1, . . . ,m u suchthat a ri b and fpriq 0, which concludes the proof.
The key property in the proof above was that every positive element ofR can be written as a square, which is a characteristic property of real closedfields. It turns out that analogues of several other important theorems inreal analysis, such as Rolle’s Theorem and the Mean Value Theorem, holdfor polynomials in a real closed field as well.
26
Chapter 4
Sturm’s Theorem
In this chapter we will study the classical method for determining the num-ber of zeroes of a polynomial with real coefficients that are contained withinan open interval, which is based on a theorem by J.C.F. Sturm, published in1829 [7]. In particular, this method allows us to symbolically locate the ze-roes of a polynomial up to an arbitrary precision. We will study this methodin the context of real closed fields, which we have shown to encompass thereal number system.
We will give two versions of the theorem. The first gives a decisionmethod in terms of variations in sign of a sequence of numbers. The secondanswers when a parametrized family of polynomials has zero in a certaininterval, by reducing it to a set of polynomial equations and inequationsfor the parameters of the family, where the equations and inequations haveinteger coefficients. From the last theorem we can then quickly show that ifa polynomial with rational coefficients has a zero in one real closed field, itwill have a zero in any real closed field.
Throughout this chapter, R will denote a real closed field, equipped withthe strict total order ¡. Also, if a, b P R and a b we will use the notationsra, bs tx P R | a ¤ x ¤ b u and sa, br tx P R | a x b u for closed andopen intervals respectively.
Most of this chapter draws from [4], but several definitions and theoremshave been modified to streamline the discussion and to get some more generalresults.
4.1 Variations in sign
Definition 4.1.1. Let pc0, . . . , cnq P Rn1 be a sequence of numbers in R.Then the number of variations in sign of this sequence is defined to be
| t i P t 1, . . . , n1 u | c1i1c1i 0 u |,
27
where pc10, . . . , c1n1q is the subsequence obtained by dropping the zero elementsof the original sequence.
Definition 4.1.2. Let fpxq P Rrxs and a, b P R with a b. Then a Sturmsequence for fpxq on ra, bs is a sequence of polynomials pf0pxq, . . . , fspxqq PRrxss1 such that f0pxq fpxq and:
1. f0paqf0pbq 0,
2. @c P ra, bs : fspcq 0 (i.e. fspxq has no zeroes in ra, bs),3. If c P ra, bs and fjpcq 0 for some j P t 1, . . . , s 1 u, then fj1pcqfj1pcq
0,
4. If c P ra, bs and fpcq 0, there exist open intervals sc1, cr, sc, c2r Rsuch that @u Psc1, cr: f0puqf1puq 0 and @u Psc, c2r: f0puqf1puq ¡ 0.
In the proposition below we will show that a Sturm sequence can be usedto calculate the number of distinct (i.e. not counting multiplicity) zeroes ofthe polynomial that lie in some open interval.
Proposition 4.1.1. Let fpxq P Rrxs be of positive degree, a, b P R witha b, and pf0pxq, . . . , fspxqq a Sturm sequence for fpxq on ra, bs. For anyc P ra, bs, denote the number of variations in sign of pf0pcq, . . . , fspcqq as Vc.Then the number of distinct zeroes of fpxq within sa, br is Va Vb.
Proof. Since the number of zeroes of all the fipxq within ra, bs is finite, wecan write them down as a a0 a1 am b so that no fjpxqhas a zero in any of the open intervals sai1, air, 1 ¤ i ¤ s. Now pick for1 ¤ i ¤ m: ci Psai1, air.
First we see that no fjpxq has a zero in sa0, c1r. Then by the nega-tion of theorem 3.3.1 we have fjpa0qfjpc1q ¡ 0 for j P t 0, . . . , s u. Now letk P t 0, . . . , s u with fkpa0q 0. Then clearly 0 k s, since f0pa0q 0 fspa0q, and so fk1pa0qfk1pa0q 0. Then fk1pa0qfk1pa0qfk1pcqfk1pcq ¡0 implies that fk1pcqfk1pcq 0. Taking into account all such k, we getVa0 Vc1 . In exactly the same way we may prove that Vcm Vam .
We now let i P 1, . . . ,m 1. Then if fpaiq 0, we can carry throughthe same argument to get Vci Vci1 0. If fpaiq 0, we note that (pos-sibly by repicking our ci and ci1 to comply with property 4 of a Sturmsequence) f0pciqf1pciq 0 and f0pci1qf1pci1q ¡ 0. Furthermore, the argu-ment above again shows that if 1 j s, then fj1pciq, fjpciq, fj1pciq andfj1pci1q, fjpci1q, fj1pci1q have the same number of variations in sign.Therefore in this case Vci Vci1 1.
We can now write:
Va Vb pVa Vc1q m1
i1
pVci Vci1q pVcm Vamq m1
i1
δi,
28
where δi 1 if fpaiq 0 and δi 0 if fpaiq 0. Now since all of the zeroesof fpxq that lie within sa, br per definition are one of the ai, we have countedall the zeroes. Therefore, Va Vb is the total number of distinct zeroes offpxq that lie within sa, br.
Now that we have a method of determining how many distinct zeroesa polynomial has in some open interval, given a Sturm sequence, we willneed a method to actually produce a Sturm sequence. If we do this, wehave a full-blown algorithm to determine the zeroes of a polynomial in someinterval. Even better, if we can find a bound on the absolute values of thezeroes of a polynomial and strategically disect the resulting interval, we canlocate the zeroes numerically up to an arbitrary precision! It turns out thatwe can construct a Sturm sequence in a formalized way, using the Euclideandivision algorithm.
Definition 4.1.3. Let fpxq P Rrxs be of positive degree and f 1pxq P Rrxsits formal derivative. Then define the following sequence, terminating whenfs1pxq 0:
f0pxq fpxqf1pxq f 1pxq
fi1pxq qipxqfipxq fi1pxq degpfi1q degpfiq, 1 ¤ i ¤ s
(4.1)
where qipxq P Rrxs. Then pf0pxq, . . . , fspxqq is called the standard sequenceof fpxq.
Note: Existence and uniqueness
The polynomials fi1pxq and qipxq exist and are unique by corollary 2.3.2.Note however that we have picked fi1pxq rpxq. This is the key inproducing a Sturm sequence.
We notice that if pf0pxq, . . . , fspxqq is the standard sequence for somefpxq P Rrxs, then fspxq is a common factor of fpxq and f 1pxq and all fipxq,and any such common factor will be a factor of fspxq. Temporarily pass-ing to the field of fractions of Rrxs, we can then define a derived sequencepg0pxq, . . . , gspxqq by setting gipxq fipxqfspxq1 for 0 ¤ i ¤ s and observ-ing that each gipxq P Rrxs.Lemma 4.1.2. Let fpxq P Rrxs be of positive degree and pf0pxq, . . . , fspxqqbe its standard sequence. Define the derived sequence of fpxq as pg0pxq, . . . , gspxqq,where gipxq fipxqfspxq1 P Rpxq1 for 0 ¤ i ¤ s. Then each gipxq P Rrxs,and the derived sequence is a Sturm sequence for g0pxq on every intervalra, bs such that g0paqg0pbq 0.
Furthermore, @c P R : fpcq 0 ðñ g0pcq 0.
1Rpxq denotes the field of fractions of Rrxs.
29
Proof. We showed above that fspxq is a common factor of all the fipxq.Therefore, for every 0 ¤ i ¤ s we have some hipxq P Rrxs such that fipxq hipxqfspxq and thus gipxq hipxqfspxqfspxq1 hipxq P Rrxs.
We will now show that the derived sequence is a Sturm sequence. Leta, b P R with a b and g0paqg0pbq 0. Then clearly property 1 holds.Furthermore, gspxq 1, so that gspxq has no zeroes in R and hence not inra, bs. We now use the definition of the standard sequence to see that for1 ¤ i ¤ s (where it is understood that gs1pxq 0:
gi1pxq fi1pxqfspxq1
pqipxqfipxq fi1pxqqfspxq1
qipxqgipxq gi1pxq.Suppose that c P ra, bs and gjpcq 0 for 0 j s. Then gj1pcqgj1pcq qjpcqgjpcqgj1pcq pgj1pcqq2 pgi1pcqq2 ¤ 0. Also, gj1pcq gj1pcq,so if gj1pcq 0, then gjpcq 0 gj1pcq and by induction we can thenshow that gspcq 0, which is not the case. Therefore property 3 holds.
Lastly, suppose that c P ra, bs and g0pcq 0. Then fpcq g0pcqfspcq 0,so there exist hpxq P Rrxs and e P N such that fpxq pxcqehpxq, e ¡ 0 andhpcq 0. Also, f 1pxq epxcqe1hpxqpxcqeh1pxq. Therefore, pxcqe1
is a common factor of fpxq and f 1pxq and hence a factor of fspxq. It followsthat there exists a kpxq P Rrxs such that fspxq pxcqe1kpxq and kpcq 0.Then hpxq kpxqlpxq and h1pxq kpxqmpxq for some lpxq,mpxq P Rrxs withlpcq 0 mpcq. Then g0pxq px cqlpxq and g1pxq px cqmpxq elpxqand thus g1pcq elpcq 0. We may then choose2 an interval rc1, c2s suchthat c P rc1, c2s and the interval contains no zeroes of g1pxq nor lpxq. Thenby theorem 3.3.1, g1pxqlpxq ¡ 0, so that for γ P rc1, c2s : g0pγqg1pγq pγ cqg1pγqlpγq which has the same sign as γ c and thus is negative whenγ Psc2, cr and positive when γ Psc, c1r. Hence property 4 holds and thederived sequence is a Sturm sequence for g0pxq in ra, bs.
By combining the foregoing lemma and proposition, we may now provethe main result of this section.
Theorem 4.1.3 (Sturm’s Theorem). Let fpxq P Rrxs be of positive degreeand pf0pxq, . . . , fspxqq its standard sequence. For all c P R, let Vc be thenumber of variations in sign of pf0pcq, . . . , fspcqq. Then, if a, b P R, a band fpaqfpbq 0, the number of distinct zeroes of fpxq in the interval sa, bris Va Vb.
Proof. Let pg0pxq, . . . , gspxqq be the derived sequence of fpxq. We have seenthat fpxq and g0pxq have the same distinct zeroes, so the derived sequenceis a Sturm sequence for g0pxq on ra, bs. Also, since fpaq 0 fpbq, neither
2E.g. by choosing a random such interval and then filtering out the zeroes of g1pxq andlpxq by taking the ones closest to c and averaging with c
30
px aq nor px bq are common factors of fpxq and fpxq. It then followsthat fspaq 0 fspbq and thus the sequences
fipaq gipaqfspaq and fipbq gipbqfspbqhave the same variations in sign as the gipaq and gipbq respectively. Now, bythe foregoing proposition and the observation above, the number of distinctzeroes of fpxq in sa, br is equal to the number of distinct zeroes of gpxq inthe interval, which is Va Vb.
We can use the foregoing result to form a useful algorithm, that runs inpolynomial time with respect to the degree of the polynomial in question.
Algorithm 3:Calculating the total number of zeroes of apolynomial
Let fpxq °ni0 aix
i P Rrxs be monic and of positive degree. Define µ 1maxp1,°n1
i0 |ai|q. Calculate the standard sequence pf0pxq, . . . , fspxqq offpxq by repetitive use of algorithm 1. For c P R, let Vc denote the number ofvariations in sign of the sequence pf0pcq, . . . , fspcqq. Then the total numberof distinct zeroes of fpxq in R is Vµ Vµ.
Proof. We have found in lemma 3.1.10 that all zeroes of fpxq are contained inthe interval rM,M s, where M maxp1,°n1
i0 |ai|q. Therefore, all zeroes offpxq are certainly contained in the open interval sµ, µr, where µ 1M .If we combine this with Sturm’s theorem, we get Vµ Vµ as the totalnumber of distinct zeroes of fpxq.Example. We let fpxq x3 3x 1 P Rrxs. Then f 1pxq 3x2 3 and theEuclidean sequence of fpxq and f 1pxq (and thus the standard sequence of fpxq is:
f0pxq x3 3x 1
f1pxq 3x2 3
f2pxq p2x 1q
f3pxq 15
4.
We observe that all zeroes of fpxq will lie in the interval sM 1,M 1r, whereM maxp1, 4q 4. We therefore evaluate the standard sequence at 5 and 5.
f0p5q 139 0 f0p5q 141 ¡ 0
f1p5q 78 ¡ 0 f1p5q 78 ¡ 0
f2p5q 9 ¡ 0 f2p5q 11 0
f3p5q 15
4 0 f3p5q
15
4 0
From this we see that V5 V5 2 1 1, so fpxq has 1 distinct zero in any realclosed field.
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4.2 Systems of equations, inequations and inequal-ities
This section serves as a preamble to the next section. We will now developthe notion of a system of equations, inequations and inequalities, whichare expressions vpt1, . . . , trq 0, vpt1, . . . , trq 0, and vpt1, . . . , trq ¡ 0respectively, where v P Zrt1, . . . , trs for indeterminates ti, 1 ¤ i ¤ r. Notethat will write vptiq for vpt1, . . . , trq if it is more convenient. We can considerany ordered field F , which will contain Z as a subring. We then havean evaluation homomorphism Zrt1, . . . , trs Ñ F induced by the inclusionhomomorphism, that sends Z to Z and ti to some ci P F . In this way wecan look for solutions of such an expression in the extension field F .
We further note, that if vpc1, . . . , crq 0 and wpc1, . . . , crq 0, thensince the solutions of these two inequations are in a field F , we can rewritethis equivalently as vpc1, . . . , crqwpc1, . . . , crq 0. So, any finite set of in-equations can be replaced by a single inequation. We can now state thefollowing definition.
Definition 4.2.1. An r-system (of equations, inequations and inequalities)is a triple
Γ ppv1, . . . , vsq, v, pv¡1, . . . , v¡uqqPY8i1Zrt1, . . . , trspiq
Zrt1, . . . , trs
Y8i1Zrt1, . . . , trspiq
.
Moreover, if pF, P q is an ordered field, then the solution set of Γ is the setΓpF q of pc1, . . . , crq P F prq such that:
v1pciq vspciq 0,
vpciq 0,
v¡1pciq, . . . , v¡upciq ¡ 0.
If we wish to specify a system without equalities, we can specify thetrivial equality 0 0. Similarly, we can adjoin the trivial inequation 1 0and inequality 1 ¡ 0. In this chapter, we shall not use inequalities much,and when we do not need them, we shall drop the last term in the triple,assuming the trivial inequality is to be adjoined. Also, when no inequation(the second element in the triple) has been specified, we assume that thetrivial inequation must be adjoined.
We can now ask when a set of systems covers all possible cases. Thefollowing definition will make this formal.
Definition 4.2.2. An r-cover is a finite set of r-systems δ t∆1, . . . ,∆s usuch that for any ordered field F :¤
∆Pδ
∆pF q F prq.
32
Also, a refinement of an r-cover γ is an r-cover δ, such that for anyordered field F of K: @∆ P δ DΓ P γ : ∆pF q ΓpF q.Definition 4.2.3. If Γ and ∆ are r-systems, their join is defined to be ther-system Γ [ ∆3 that has as its equalities and inequalities both those of Γand ∆, and as inequality the product of the inequalities of Γ and ∆.
We will give the following lemmas without proof, as they are quitestraightforward if you just write out the definitions.
Lemma 4.2.1. Let Γ and ∆ be r-systems. Then:
pΓ[∆qpF q ΓpF q X∆pF q,for any ordered field F .
Lemma 4.2.2. If Γ is an r-system and δ t∆1, . . . ,∆s u is a finite r-cover,and we define Γj Γ[∆j for 1 ¤ j ¤ s, then
sj1 ΓjpF q ΓpF q for every
ordered field F .
Lemma 4.2.3. Let γ tΓ1, . . . ,Γu u and δ t∆1, . . . ,∆s u be r-coversand define Γ1j Γ1 [ ∆j. Then γ1 tΓ11, . . . ,Γ
1s,Γ2, . . . ,Γu u is again an
r-cover, and a refinement of γ.
4.3 Sturm’s Theorem Parametrized
We will now consider a family of polynomials in a formally real field R whosecoefficients are parametrized as multivariate polynomials over its prime ringZ. That is, the family of polynomials is represented by a polynomial inZrt1, . . . , trsrxs. The ti represent parameters, and the x represents a variablewe wish to solve for. Using Sturm’s Theorem we will show that we can,algorithmically, obtain a cover of systems in Z such that a member of thisfamily has a zero in a certain interval if and only if the parameters andboundaries satisfy one of those systems. This method could be extendedto parametrize the systems that the coefficients have to satisfy with respectto the boundaries of the system, but that extension will not be consideredhere.
In order to get to our main result, we first let K Z and R be a realclosed field. We also let r P N, r ¥ 1 and define A Krt1, . . . , trs, where theti, 1 ¤ i ¤ r are indeterminates. Now, if we pick pc1, . . . , crq P Rprq, we havea homomorphism AÑ R that extends the inclusion homomorphism K Ñ Rand sends ti ÞÑ ci. Therefore, we have an extension of this homomorphismArxs Ñ Rrxs that maps each parametrized polynomial to a polynomial withcoefficients in F : F pti;xq ÞÑ F pci;xq.
3This is not standard notation, but it proves intuitive given lemma 4.2.1.
33
Since A is a commutative ring, we can perfectly well perform Euclideanpolynomial division in Arxs. If we now make the connection with the eval-uation in pc1, . . . , crq we can make the following important observation.
Lemma 4.3.1. Let F pti;xq, Gpti;xq P Arxs with Gpti;xq 0 and vmptiq theleading coefficient of G. Then there exists an even e P N and Qpti;xq, Rpti;xq PArxs with degpRq degpGq and:
vmptiqeF pti;xq Qpti;xqGpti;xq Rpti;xq.Also, if pc1, . . . , crq P Rprq and vmpciq 0, then the qpxq, rpxq P Rrxs withF pci;xq qpxqGpci;xq rpxq and degprq degpGpciqq differ from Qpci;xqand Rpci;xq by a common positive multiplier.
We also note that the choice of the Qpti;xq, Rpti;xq and e are indepen-dent of which real closed field we use.
Proof. The existence of an arbitrary e P N and the Qpti;xq, Rpti;xq P Arxsfollows from the Euclidean division algorithm. However, if e is odd, we maymultiply the entire equation by vmptiq and so obtain a new Qpti;xq andRpti;xq and an even e so that the equation still holds.
Now, if pc1, . . . , crq P Rprq such that vmpciq 0, then since e is even wehave vmpciqe ¡ 0. Then evaluating the equation in the ci and dividing byvmpciqe, we obtain:
F pci;xq vmpciqeQpci;xqGpci;xq vmpciqeRpci;xq qpxqGpci;xq rpxq,
where the qpxq, rpxq P Rrxs are as above. And since such qpxq and rpxq areunique in the polynomial ring of a field, we have Qpci;xq vmpciqeqpxq andRpci;xq vmpciqerpxq.
We are now ready to state the following proposition, that allows us touse Sturm’s theorem on the parametrized polynomials.
Proposition 4.3.2. Let F pti;xq, Gpti;xq P Arxs with Gpti;xq °mj0 vjptiqxj
0. Define Gkpti;xq °kj0 vjptiqxj and the r-systems Γk ppvj , j ¡ kq, vkq
for 0 ¤ k ¤ m and Γ8 ppv0, . . . , vmq, 1q4. Then we can obtain, ina finite number of steps, an r-cover δ t∆1, . . . ,∆h u that is a refine-ment of the cover γ tΓ8,Γ0, . . . ,Γm u and h sequences of polynomialspFj0pti;xq, . . . , Fjsj pti;xqq in Arxs such that, if pc1, . . . , crq P ∆jpRq, then theterms of pFj0pci;xq, . . . , Fjsj pci;xqq differ from the terms of the Euclideansequence of F pci;xq and Gpci;xq by a positive multiplier.
Also, if this property holds in one real closed field, then it holds for anyreal closed field.
4The k and 8 correspond to the degree of Gpci;xq if pciq P ΓkpRq.
34
Proof. We consider any k P t 0, . . . ,m u with vkptiq 0 (or equivalentlyΓkpRq H), for else Gkpti;xq Gjpti;xq for some j k and Γk would notbe contributing to the cover γ. We can then just as well omit Γk in ourrefinement. Now find Qkpti;xq, Rkpti;xq P Arxs as in the foregoing lemma.We have to consider two cases.
If Rkpti;xq 0, we can take the sequence pF,G, 0q and Γk as thecorresponding system. This suffices because if pc1, . . . , crq P ΓkpRq, thenGpci;xq Gkpci;xq and thus the Euclidean sequence would be pF pciq, Gkpciqq.Note that we will use this case as an induction basis in the next case.
Now let Rkpti;xq 0. If k m ¡ degpF q, then we see that Rkpti;xq F pti;xq. We may then obtain the result for Gpti;xq and Rkpti;xq, by goingthrough the argument again and seeing that this case is then excluded.Otherwise degpRkq degpGkq degpF q degpGq, so by induction on thesum of the degrees, we may obtain a cover δk t∆k0, . . . ,∆khk u and hksequences pFkl0pti;xq, . . . , Fklsklpti;xqq so that the required property holdsfor Gkpti;xq and Rkpti;xq. We now define Γkl Γk[∆kl for l P t 0, . . . , hk u.Then, if pciq P ΓklpRq ΓkpRq, we have Gkpci;xq Gpci;xq. Also, sinceFkl0pci;xq Gkpci;xq Gpci;xq and Fkl1pci;xq Rkpci;xq, we can takethe sequences pF pci;xq, Fkl0pci;xq, . . . , Fklsklpci;xqq, whose terms differ fromthe Euclidean sequence of F pci;xq and Gpci;xq by a positive multiplier, andpair these with the respective Γkl.
If we now let δ consist of the systems obtained above, and pair these withtheir respective sequences, including Γ8 with pF, 0, 0q, we have obtained arefinement of γ that satisfies our requirements. We also note now that thechoice of the systems and sequences did not depend on the real closed fieldin question, so that the property holds for any real closed field.
Example. Let F pp, q;xq x2 px q and Gpp, q;xq 2x p. We then haveΓ8pRq Γ0pRq H and Γ1pRq Rp2q. We therefore consider only k 2,G2pp, q;xq Gpp, q;xq. We first observe that:
22F pp, q;xq p2x pqGpp, q;xq pp2 4qq.
We therefore set R2pp, q;xq p2 4q. Now, since R2pp, q;xq P A, another step(only possible if p2 4q 0) will give us the 0 polynomial. We therefore have the2-cover and corresponding sequences:
Γ1 : a2 4b 0 Ø pF,Gq
Γ2 : a2 4b 0 Ø pF,G,R2q
If we now recall that the standard sequence of a polynomial fpxq issimply the Euclidean sequence of fpxq and its formal derivative, we canquickly prove the following theorem, which is our second main result. Notethat this version is more general than the one in [4], as the systems we haveto obtain include requirements on the bounds of our interval.
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Theorem 4.3.3 (Parametrized version of Sturm’s Theorem). Let F pti;xq PArxs. Then there exists a finite set of r 2-systems5 ω in K – which wecan obtain in a finite number of steps – such that for every pc1, . . . , cr, a, bq PRpr2q with a b, F pci;xq has a zero in ra, bs if and only if F pci; aqF pci; bq 0 or F pci; aqF pci; bq 0 and there is some Ω P ω such that pc1, . . . , cr, a, bq PΩpRq.
We can restate this theorem as follows: Let F pti;xq be a family of poly-nomials whose coefficients are parametrized by polynomials with integer co-efficients. Then for any interval sa, br we can obtain a finite set of systemsof equations, inequations and inequalities, so that F pci;xq has a zero in thatinterval if and only if the coefficient parameters pciq and the boundaries aand b satisfy one of those systems (provided that F pci; aqF pci; bq 0).
Proof of the parametrized version of Sturm’s Theorem, 4.3.3. Let F pti;xq °nν1 uνptiqxν , where uνptiq P A, and unptiq is the leading coefficient. Now,
if F 1pti;xq 0, then F pti;xq u0ptiq is constant and we can take the solesystem ppu0qq.
We can therefore now assume that 0 F 1pti;xq °nν1 νuνptiqxν1.
Then by proposition 4.3.2 we can obtain a cover δ t∆0, . . . ,∆h u andcorresponding sequences pFj0pti;xq, . . . , Fjsj pti;xqq (0 ¤ j ¤ h) such thatif pc1, . . . , crq P ∆j , then the terms of pFj0pci;xq, . . . , Fjsj pci;xqq differ fromthe terms of the standard sequence of F pci;xq by positive multipliers. Inparticular, we see that at any point, the number of sign changes is the same.
Now pick any j P t 0, . . . , h u. Now, if we let γ be the same cover as inproposition 4.3.2, then δ is a refinement of γ. Therefore, if pciq P ∆jpRq,we have either unpciq u1pciq 0 – in which case F pci;xq has azero if and only if u0pciq 0 – or there is some k P t 1, . . . , n u such thatukpciq 0 but ulpciq 0 for l ¡ k. In the first case we set ωj t ppu0qq u:the sole equation u0 0. In the latter case we may construct the followingtwo sequences:
αjlpti;xa, xbq umptiq2nlFjlpti;xaq P Arxa, xbsβjlpti;xa, xbq umptiq2nlFjlpti;xbq P Arxa, xbs,
for 0 ¤ l ¤ sj , and where nl degpFjlq and xa and xb are new indeter-minates. Then all the αjlpci; a, bq and βjlpci; a, bq differ from Fjlpci; aq andF pci; bq respectively – and hence from the standard sequence of F pci;xq atthose points – by a positive multiplier. If F pci; aqF pci; bq 0, we now con-clude by Sturm’s Theorem that F pci;xq has a zero in sa, br if and only if thenumber of variations in sign of the sequences pαj0pci; a, 0q, . . . , αjsj pci; a, 0qqand pβj0pci; 0, bq, . . . , βjsj pci; 0, bqq are not equal. We therefore now take all
5The 2 extra parameters in the systems of ω correspond to the bounds of our interval.They serve to keep the set of systems independent of the real closed field that we chooseto use.
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possible r 2-systems on K that can be formed by the elements of thosesequences (which is finite), and filter out the ones that lead to a differ-ence in the number of variations in sign between the sequences and foreach take the join with ∆j
6, to form the set of systems ωj . Then, if apci; a, bq P ΩpRq for some Ω P ωj , then pciq P ∆jpRq, so that the above ap-plies, and there is a difference between the variation in sign in the sequencespαjlpci, a, 0qq and pβjlpci, 0, bqq, so that F pci;xq has a zero in sa, br, providedthat F pci; aqF pci; bq 0. We also observe that if F pci; aqF pci; bq 0, thenF pci;xq has a zero in ra, bs.
If we now let ω Yhj0ωj , we obtain the set of r 2-systems we require,since δ is a cover of K.
Example. In our last example we obtained a 2-cover and the corresponding se-quences F pp, q;xq x2 px q and F 1pp, q;xq 2x p. We write:
∆1 : p2 4q 0 Ø pF, F 1q
∆2 : p2 4q 0 Ø pF, F 1, p2 4qq.
We can then define the corresponding αjl and βjl as follows:
α10pp, q;xa, xbq x2a pxa q β10pp, q;xa, xbq x2b pxb q
α11pp, q;xa, xbq 2xa p β11pp, q;xa, xbq 2xb p
α20pp, q;xa, xbq x2a pxa q β20pp, q;xa, xbq x2b pxb q
α21pp, q;xa, xbq 2xa p β21pp, q;xa, xbq 2xb p
α22pp, q;xa, xbq p2 4q β22pp, q;xa, xbq p2 4q
For ∆1 we get the sole system Ω11 ppp2 4qq, 1, ppx2a pxa qqp2xa pq, px2b pxb qqp2xb pqqq, that is, α1l will change sign, but β1l will not.
For ∆2, we have to consider three cases: the α2l change sign once, and the β2ldon’t (one zero), the α2l change sign twice, and the β2l don’t (two zeroes), or theα2l change sign twice and the β2l change sign once (one zero). These cases can alloccur in several ways, and so we end up with a whole pile of systems.
Note: Existence of a zero
We have introduced two extra indeterminates in our systems in order toaccount for the zeroes of our polynomial. However, if we choose to inves-tigate the problem of the existence of a zero in the entire field, we candrop those two parameters. This can be done by noting that we do nothave to consider a and b up to the point that we define the sequencespαjlpciqq and pβjlpciqq. In particular, at that point we can see that for anypciq P Rprq, if ρ P R is to be a zero of F pci;xq, then necessarily µ ρ µ,
6Technically, we now have to transform the r-system ∆j to an r 2-system by usingthe inclusion homomorphism A Ñ Arxa, xbs on all the elements of the system.
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where µ pk 1q °m1ν0 uνpciq2ukpciq2. We may from that point on let
aptiq, bptiq P A depend on the parameters and modify the αptiq and βptiqaccordingly, and finish the argument in the same way. We can thereforestate the following corollary.
Corollary 4.3.4. Let F pti;xq P Arxs. Then we can construct a finite set ofr-systems ω in K such that for any real closed field R, and pc1, . . . , crq P Rprq:F pci;xq has a zero in R if and only if pc1, . . . , crq P ΩpRq for some Ω P ω.
Restated: Let F pti;xq P Arxs be a family of polynomials whose coef-ficients are parametrized by polynomials with integer coefficients. Then wecan construct a finite set of systems of polynomial equations, inequations andinequalities with integer coefficients – independent of the real closed field inquestion – so that for some choice pciq of the parameters, F pci;xq P Rrxshas a zero in R if and only if the pciq satisfy one of the constructed systems.
Now let fpxq °ni0 aix
i P Rrxs and let F pti;xq °ni0 tix
i. We thensee that F pai;xq fpxq. Suppose that all the ai P Q R and that fpxq hasa zero in R. Then by corollary 4.3.4 we can construct a set of n-systems inZ such that the paiq satisfy one of those systems. Now let R1 be another realclosed field. Then clearly all ai P R1 (by an isomorphism of the prime fields)and they still satisfy one of those systems. Therefore, the correspondingpolynomial in R1rxs will also have a zero in R1.
Corollary 4.3.5. If a polynomial fpxq with rational coefficients has a zeroin one real closed field, it will have a zero in any real closed field.
This last corollary is i.a. of the utmost importance for computer calcula-tions. E.g. the computable numbers, described by Turing as “the numberswhose expressions as a decimal are calculable by a machine”[9], can be shownto be real closed[2]. This then gives the result, that a polynomial with ra-tional coefficients has a zero in the real numbers, if and only if it has a zeroin the computable numbers. Therefore, for any polynomial with rationalcoefficients, we are in principle able to compute all its real zeroes with acomputer (or any realization of a Turing machine).
4.3.1 Tarski’s Principle
The question now naturally arises whether we can generalize this procedureto families of polynomials in multiple indeterminates. The answer turns outto be positive. The idea we can pursue is to replace an equation in multipleindeterminates to a set of equations in one less indeterminates. We may goon with this procedure to eventually obtain a set of equations that have tobe satisfied for the original equation to be solvable. If we then invoke theparametrized version of Sturm’s Theorem for each of these, we obtain a setof systems that will have to be satisfied by the parameters for our equationto be solvable. [4, Sec. 5.6]
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This method has an important application in the so-called field of meta-mathematics, where the properties of mathematics itself are studied. Inparticular, it implies that every “elementary” sentence in the logic of a realclosed field is decidable. This was shown by Tarski in 1948 for the real num-bers. [8] Note that in the logic of a real closed field, we mean the first-orderlogic that remains when only the axioms of the field itself are assumed. Set-theoretic sentences are not allowed. This does however, to quote Tarski,“gives the mathematician the assurance that he will be able to solve everysuch problem (an elementary problem in a real closed field) by working at itlong enough.” And with that assurance we can continue to make algebraicexercises for high school students.
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Epilogue
Sturm’s Theorem has provided us with a very simple way to determine thezeroes of a polynomial that lie within a certain interval. It is interesting tonote that despite the simplicity of this method, it is not widely taught inundergraduate calculus courses. Perhaps this can be attributed to the inef-ficiency of the algorithm compared to more modern root-finding methods,the amount of algebra involved, or simply its age (almost 200 years!). Ineither case I would like to express my hopes that the tides could change inthis respect.
Nevertheless, the theorem not only provides us with this calculationmethod, it also leads to several important theoretical implications. As ex-amples we have seen the decidability of the theory of real closed fields (inmetamathematics), and the fact that if a polynomial with rational coeffi-cients is going to have a zero in one real closed field, then it is going tohave one in every real closed field. The last result finds an application incomputer science, where we can conclude that we can compute every zero ofa polynomial with rational (even computable!) coefficients with a computerprogram.
I have personally enjoyed this project very much due to the large amountof new algebra I have come to learn, and the discovery of an obscure, but funand useful result. I know that I will definitely have use for Sturm’s Theoremin the future.
Lastly, I would like to acknowledge Prof. Dr. Jaap Top and Dr. RamsayDyer for their support during the course of this project. Prof. Top hasrecommended this project, and they have both provided me with very usefulfeedback on the report, for which I am very grateful.
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Bibliography
[1] E. Artin and O. Schreier. Algebraische konstruktion reele korper. Ab-handlungen aus dem Mathematischen Seminar der Universitat Hamburg,5(1):85–99, December 1927. Conference proceedings from June 1926.
[2] M. Braverman. On the complexity of real functions. In Proceedings ofthe 2005 46th Annual IEEE Symposium on Foundations of ComputerScience. IEEE, 2005.
[3] D.J. Griffiths. Introduction to Quantum Mechanics. Pearson Education,2nd edition, 2005.
[4] N. Jacobson. Basic Algebra, volume 1. Dover, dover edition, 2009.
[5] N. Jacobson. Basic Algebra, volume 2. Dover, dover edition, 2009.
[6] J.P. Serre. Extensions de corps ordonnes. In Comptes rendus des seancesde l’Academie des Sciences, pages 576–577, September 1949.
[7] J.C.F. Sturm. Memoire sur la resolution des equations numeriques. Bul-letin des Sciences de Ferussac, 11:419–425, 1829.
[8] A. Tarski. A Decision Method for Elementary Algebra and Geometry.RAND Corporation, 1948.
[9] A.M. Turing. On computable numbers, with an application to theentscheidungsproblem. Proceedings of the London Mathematical Soci-ety, 42:230–265, 1937.
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