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Analytic Solutions to

Algebraic Equations

Mathematics

Tomas Johansson

LiTH–MAT–Ex–98–13

Supervisor: Bengt Ove Turesson Examiner: Bengt Ove Turesson Link¨oping 1998-06-12

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CONTENTS 1

Contents

1 Introduction 3

2 Quadratic, cubic, and quartic equations 5

2.1 History . . . 5

2.2 Solving third and fourth degree equations . . . 6

2.3 Tschirnhaus transformations . . . 9

3 Solvability of algebraic equations 13 3.1 History of the quintic . . . 13

3.2 Galois theory . . . 16

3.3 Solvable groups . . . 20

3.4 Solvability by radicals . . . 22

4 Elliptic functions 25 4.1 General theory . . . 25

4.2 The Weierstrass ℘ function . . . 29

4.3 Solving the quintic . . . 32

4.4 Theta functions . . . 35

4.5 History of elliptic functions . . . 37

5 Hermite’s and Gordan’s solutions to the quintic 39 5.1 Hermite’s solution to the quintic . . . 39

5.2 Gordan’s solutions to the quintic . . . 40

6 Kiepert’s algorithm for the quintic 43 6.1 An algorithm for solving the quintic . . . 43

6.2 Commentary about Kiepert’s algorithm . . . 49

7 Conclusion 51 7.1 Conclusion . . . 51 A Groups 53 B Permutation groups 54 C Rings 54 D Polynomials 55 References 57

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3

1

Introduction

In this report, we study how one solves general polynomial equations using only the coef-ficients of the polynomial. Almost everyone who has taken some course in algebra knows that Galois proved that it is impossible to solve algebraic equations of degree five or higher, using only radicals. What is not so generally known is that it is possible to solve such equa-tions in another way, namely using techniques from the theory of elliptic funcequa-tions. Galois theory and the theory of elliptic functions are both extensive and deep. But to understand and find a way of solving polynomial equations we do not need all this theory. The aim here is to pick out only the part of the theory which is necessary to understand why not all equations are solvable by radicals, and the alternative procedures by which we solve them. We will concentrate on finding an algorithm for the quintic equation.

In Chapter 2 we examine the history of polynomial equations. We also present the formulas for solutions to equations of degree less then five. A basic tool, the Tschirnhaus transformation, is introduced. These transformations are used to simplify the solution process by transforming a polynomial into another where some coefficients are zero.

In Chapter 3 we introduce some Galois theory. In Galois theory one examines the rela-tionship between the permutation group of the roots and a field containing the coefficients of the polynomial. Here we present only as much theory as is necessary to understand why not all polynomial equations are solvable by radicals. The history of the quintic equation is also presented.

In Chapter 4 we introduce the elliptic functions and examine their basic properties. We concentrate on one specific elliptic function, namely the Weierstrass ℘ function. We show that with the help of this function one can solve the quintic equation. However, this is just a theoretical possibility. To be able to get a practical algorithm, we need to know some facts about theta functions, with which we also deal in this chapter. The history of elliptic functions is included at the end of this chapter. Here we also present the general formula for solving any polynomial equation due to H. Umemura. This formula is also only of theoretical value, so we will look at other ways of solving the quintic. We present three such methods in the last chapter. The one that we will concentrate on is Kiepert’s, because it uses both elliptic functions and Tschirnhaus transformations in a clear and straightforward way.

In Chapter 5 we present two algorithms for solving the quintic, which are due to Hermite and Gordan. Hermite uses a more analytic method compared with Gordan’s solution, which involves the use of the icosahedron.

In Chapter 6 we present Kiepert’s algorithm for the quintic. This lends itself to imple-mentation on a computer. This has in fact been done by B. King and R. Canfield, and at the end of the chapter we discuss their results.

We conclude with an Appendix. Here one can find the most basic facts about groups, rings, and polynomials.

Acknowledgment

I would like to thank the following people: my supervisor Bengt Ove Turesson for helping me with this report and for teaching me LATEX 2ε in which this report is written, Peter Hackman

for helpfull hints about Galois theory, and Gunnar Fogelberg for historical background. I also thank Thomas Karlsson who told me that this examination work was available, and Bruce King for telling me about his work on the quintic. I also express my gratitude to John H. Conway, Keith Dennis, and William C. Waterhouse for giving me information about L. Kiepert.

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5

2

Quadratic, cubic, and quartic equations

In this chapter we shall take a brief look at the history of equations of degree less than five. We will also study the techniques for solving these equations. One such technique, the Tschirnhaus transformation, will be examined in Section 2.3.

2.1 History

Polynomial equations arise quite naturally in mathematics, when dealing with basic math-ematical problems, and so their study has a long history. The attempts to solve certain of these equations have lead to new and exciting theories in mathematics, with the result that polynomial equations has been an important cornerstone in the development to modern mathematics.

It is known that in ancient Babylonia (2000 – 600 B.C.), one was able to solve the second order equation

x2+ c1x + c0 = 0, ci ∈ R,

using the famous formula

x =−c1 2 ±

r c21

4 − c0.

They could, of course, not deal with all of the roots, since they did not have access to the complex numbers. Archaeologists have found tablets with cuneiform signs, where problems concerning second order equations are dealt with. Starting with geometric problems, they were led to these equations by the Pythagorean theorem. The Babylonians also studied some special equations of third degree.

The Greeks (600 B.C. – 300 A.D.) had a more geometric viewpoint of mathematics compared to the Babylonians, but they also considered second order equations. They made geometric solutions to the quadratic and constructed the segment x from known segments c0 and c1.

Problems like trisecting an angle and doubling the volume of the cube led the Greeks to third degree equations which they tried to solve; but they had no general method to for solving such equations.

The Hindus (600 B.C. – 1000 A.D.) invented the number zero, negative numbers, and the position system. Brahmagupta dealt with quadratic equations. The solutions were more algebraic than those of the Babylonians.

The Arabs (500 B.C. – 1000 A.D.) enjoyed algebra. The famous al-Khwarizmi dealt with algebra in his book Al-jabr wa’l muqabala around 800 A.D. In this book, he exam-ined certain types of second order equations (and none of higher degree). Another Arab mathematician, Omar Khayyam (1048–1131) classified third degree equations and exam-ined their roots in some special cases. He also investigated equations of fourth degree. Fore more details about mathematics in these old cultures, see Boyer [2] and van der Waer-den [25].

Europe began to participate in the development of mathematics soon after the Arabs. The Italian Leonardo Fibonacci (1170–1250) wrote Liber Abacci, where he tried to solve quadratic equations in one or more variables.

In Europe in the Middle Ages, competitions in mathematics were popular. This en-couraged the development of the art of solving equations. A new era began around the

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beginning of the fifteenth centuary in Italy, when Scipione del Ferro (1465–1526) succeeded in solving the equation

x3+ c1x + c0 = 0.

In Section 2.3, we will show that every third degree equation can be transformed to the above form, which means that del Ferro had solved the general cubic equation. Before his death, del Ferro told his student Fior of his method for solving cubics. Fior challenged another Italian, Niccolo Tartaglia (1500–1557), to a mathematical duel. This forced Tartaglia, who knew that Fior could solve such equations, to find a general method for solving the cubic. Tartaglia succeded with this and thus won the competition.

Tartaglia told Girolamo Cardano (1501–1576) about his method. Cardano published the method in his book Ars Magna (around 1540). Here he observed that, in certain cases, the method yields roots of negative numbers, and that these could be real numbers. This means that he was touching upon the complex numbers. Cardano made his student Lodovico Ferrari (1522–1565) solve the fourth degree equation, i.e., the quartic. Ferrari succeeded with this, and Cardano also published this method in Ars Magna.

The case when Cardanos formulas lead to roots of negative numbers was studied by Rafael Bombelli (1526–1573) in his book Algebra in 1560.

After the Italians, many famous mathematicians examined these equations, for example Francois Vi`ete (1540–1603) and Ehrenfried Walter von Tschirnhaus (1651–1708).

It was finally the German mathematician Gottfried von Leibniz (1646–1716) who, with-out any geometry, verified these formulas.

More information about these Italians and the solutions to third and fourth degree equations can be found in Stillwell [24].

2.2 Solving third and fourth degree equations

Below we shall present the formulas for the solution of cubic and quartic equations. When solving the second order equation, we just complete the square and get the formula immedi-ately. The formulas for the third and fourth degree equations are a little bit more difficult. One learns the second order formula in high school, and one might wonder why we do not also learn these other formulas. They are not too difficult to remember, but the problem with them is that, in some cases, one has to take roots of negative numbers, even though the equation has real solutions.

Take an arbitrary third degree equation:

x3+ c2x2+ c1x + c0= 0, ci∈ C. (1)

We will use cube roots of unity. Therefore let

w = cos2π3 + i sin2π3 . If we make the substitution

x = yc2 3,

(in the next section we will show how we find substitutions like that), then we get an equation in y :

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2.2 Solving third and fourth degree equations 7 If a0= 0, then this equation has roots y = 0 and y =±√−a1. Consequently,

x =c2 3 or x =± √ −a1− c2 3; if a1 = 0, then we get the roots

x =√3 −a0− c2 3, x = w 3 √ −a0− c2 3 and x = w 2√3 −a0− c2 3. If a0a1 6= 0, then we make the substitution

y = z a1 3z. This substitution gives us an equation in z of the form

z3+ a0− a 3 1 27z3 ⇒ z 6+ a 0z3−a 3 1 27 = 0, which can be factorized as follows:

à z3+ à a0 2 + r a2 0 4 + a3 1 27 !! à z3+ à a0 2 − r a2 0 4 + a3 1 27 !! = 0. Theorem 2.1Both parentheses in the expression

à z3+ à a0 2 + r a2 0 4 + a3 1 27 !! à z3+ à a0 2 − r a2 0 4 + a3 1 27 !! = 0, give the same roots to (1).

Proof From The Fundamental Theorem of Algebra, we know that there are three roots to our original cubic equation. Therefore, there must be at least one root from each paren-thesis which gives the same root to (1). Denote them by z1 and z2. If they are equal, there

is nothing to prove, so assume that this is not the case. We then have z1− a1 3z1 = z2− a1 3z2 ⇔ z1− z2=− a1 3 z1− z2 z1z2 ⇔ z1z2 =− a1 3. (2) The other roots in the parentheses are multiples of w. Therefore

wz1w2z2 = z1z2 =−

a1

3 ,

so they also give rise to the same root because of (2). That is, both parenthesis give the same roots. ¤

In view of Theorem (2.1), to find solutions to (1), just choose one of the parenthesis and solve. Denote one of the third roots by A. Then the solutions to (1) are

x = A a1 3A − c2 2, x = wA− w 2a1 3A − c2 2 and x = w 2A − w3Aa1 − c22.

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Example 2.2Solve the third degree equation

x3+ 3x2− 3x − 11 = 0.

We follow the method described above. Put x = y− 1. We then get the equation y3− 6y − 6 = 0. Now substitute y = z + 2 z. This leads to z3− 6 + 8 z3 = 0 ⇒ (z3− 2)(z3− 4) z3 = 0.

We choose to solve z3− 2 = 0. The roots to the original equations then are

3

2 +√34− 1, w√32 + w2√34− 1 and w2√32 + w√34− 1.

If we had chosen the other parenthesis, we would have found exactly the same roots as asserted.

We now investigate an arbitrary fourth degree equation:

x4+ c3x3+ c2x2+ c1x + c0 = 0, ci ∈ C. (3)

We begin by making the substitution

x = yc3 4, and obtain

y4+ a2y2+ a1y + a0= 0.

Our goal is to factorize this equation and get two second order equations, which are easy to solve. Write

(y2+ 2qy + p)(y2− 2qy + r) = 0 y4+ (r + p− 4q2)y2+ 2q(r− p)y + pr = 0 and collect terms. This gives

pr = a0,

2q(r− p) = a1,

r + p− 4q2 = a2.

If a1 = 0, we make the substitution t = y2, and obtain a quadratic equation in t. Otherwise,

with q6= 0, we have 2p = a2+ 4q2− a1 2q, 2r = a2+ 4q2+ a1 2q. Because 2p2r = 4a0, we get a sixth degree equation in q:

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2.3 Tschirnhaus transformations 9 If we substitute t = q2, then the above equation becomes a third degree equation in t, which

can be solved by our third degree formula. We take one of the roots qi, i = 1, 2, . . . , 6, and

use it to make the factorization. Then we solve both second order equations in y, reverse the transformations, and we find the roots of (3).

Example 2.3Solve the equation

x4− 4x3− 16x + 35 = 0. (4) By letting x = y + 1, this equation is transformed into

y4− 6y2− 24y + 16 = 0. We try to factorize this as above, and get

64q6− 192q4− 112q2− 576 = 16(q2− 4)(4q4+ 4q2+ q) = 0. We then have q = 2, p = 8 and r = 2, so the factorization is

(2 + 4y + y2)(2− 4y + y2) = 0. The roots of (4) will therefore be

x =−1 ± 2i and x = 3±√2.

2.3 Tschirnhaus transformations

Tschirnhaus transformations were invented by the German mathematician Ehrenfried von Tschirnhaus (1651–1708). The Swedish algebraist Erland Bring (1736–1798) showed by a Tschirnhaus transformation that the general quintic equation can be transformed to the form

x5+ c1x + c0 = 0.

The English mathematician George Jerrard (1804–1863) generalized this result. He showed that one can always transform a general nth degree equation to a form where the terms xn−1, xn−2, and xn−3 are absent.

Consider the general nth degree equation:

xn+ cn−1xn−1+ . . . + c1x + c0 = 0. (5)

If we make a substitution where we involve the roots of (5), then, if we are clever, coefficients can disappear. For example, if we succeed to obtain

xnk+ cn−1xn−1k + . . . + c1xk+ c0,

where xk is a root of (5) as coefficient of yn−1, then the coefficient for this term in our new

polynomial will disappear. Therefore, Tschirnhaus transformations will have the general form

yk= an−1xn−1k + an−2xn−2+ . . . + a1xk+ a0, k = 1, 2, . . . , n,

where xk is a root of (5).

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Definition 2.4The polynomials

Pk(x1, . . . , xn) = x1k+ xk2 + . . . + xkn, k = 0, 1, 2, . . . ,

are denoted the kth power polynomials.

Assume that we have a general nth degree polynomial as in (5). Then the following theorem holds.

Theorem 2.5(Newton’s formula for power sums)

Let Pk(x1, . . . , xn) = xk1+ xk2+ . . . + xkn, where x1, . . . , xn are roots of

f (x) = xn+ cn−1xn+ . . . + c1x + c0 = 0. Then P1+ cn−1 = 0, P2+ cn−1P1+ 2cn−2 = 0, P3+ cn−1P2+ cn−2P1+ 3cn−3 = 0, .. . Pn−1+ . . . + c4P3+ c3P2+ c2P1+ (n− 1)c1 = 0, and, for k≥ n, Pk+ . . . + c3Pk−n+3+ c2Pk−n+2+ c1Pk−n+1+ c0Pk−n= 0.

Proof We can write f as

f (x) = (x− x1)(x− x2) . . . (x− xn).

Now differentiate this equality with the help of logarithmic differentiation, to obtain f0(x) = f (x) x− x1 + f (x) x− x2 + . . . + f (x) x− xn , where f (x) x− xi = xn−1+ an−2xn−2+ . . . + a1x + a0.

By comparing coefficients, we get

an−2 = cn−1+ xi,

an−3 = cn−2+ an−2xi,

.. .

a0 = c1+ a1xi.

Substituting the expression for an−k in the expression for an−k−1, we have

an−2 = cn−1+ xi,

an−3 = x2i + cn−1xi+ cn−2,

.. .

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2.3 Tschirnhaus transformations 11 It follows that

f0(x) = nxn−1+ (P1+ ncn−1)xn−2+ . . . + (Pn−1+ cn−1Pn−2+ . . . + c2P1+ nc1).

Differentiating the polynomial in the ordinary way gives

f0(x) = nxn−1+ (n− 1)cn−1xn−2+ . . . + 2c2x + c1.

By comparing coefficients of these two expressions one obtains the first part of the theorem. The last part follows from summing the following identity over i = 1, 2, . . . , n:

xn+ki + cn−1xn+k−1i + . . . + c1xk+1i + c0xk= 0.

We have thus proved the theorem. ¤

We use this in Tschirnhaus transformations in the following way: • Carry out the substitution with the roots as we just described above.

• Calculate Pk(y1, . . . , yn) and express it in xk. Because xk is a root of an nth degree

polynomial, we can express every yjk in a polynomial of degree n− 1 . • Decide which coefficients to eliminate in the new polynomial.

• Look at the Newton identities for both polynomials, then substitute these expressions, and solve for the coefficients.

Example 2.6Take the equation

xn+ cn−1xn−1+ . . . + c1x + c0 = 0.

We want to eliminate the xn−1−term. Substitute

yk= xk+ a0, k = 1, 2, . . . , n,

where a0 is an arbitrary constant. We will then get an equation in y of the form

yn+ bn−1yn−1+ . . . + b1y + b0 = 0.

We calculate P1(y1, . . . , yn) and express in xk. Summation over k in the relation

yk= xk+ a0,

gives

P1(y1, . . . , yn) = P1(x1, . . . , xn) + na0.

But by Newton’s formula, we have

P1(y1, . . . , yn) + bn−1= 0 and P1(x1, . . . , xn) + cn−1 = 0.

Let bn−1= 0. Then

−cn−1+ na0= 0 ⇒ a0=

cn−1

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To get rid of the xn−1-term we shall thus make the substitution

yk = xk+

cn−1

n .

If one has a second order equation and makes the substitution yk= xk+

c1

2, then the equation in y is

³ y−c1 2 ´2 + c1 ³ y−c1 2 ´ + c0 = 0 ⇒ y2−c 2 1 4 + c0 = 0. The roots to this equation are

y1 = r c21 4 + c0 and y2 =− r c21 4 + c0. The roots of the equation in x will therefore be

x1 =− c1 2 + r c2 1 4 + c0 and x2 =− c1 2 − r c2 1 4 + c0.

Thus we have obtained the well-known formula for solving quadratics with the help of a Tschirnhaus transformation.

To find a Tschirnhaus transformation where xn−1, xn−2 and xn−3vanish is quite messy; see King [11].

If we have transformed a fifth degree equation to the Bring-Jerrard form, that is to the form

x5+ c1x + c0 = 0, c1, c2 6= 0, (6)

then the transformation

yk=

c0

c1

xk,

transforms this equation to the form

y5+ ax + a = 0, where a = c5

1/c40.

In Chapter 3 we show that it is impossible to solve equations of degree ≥ 5. Therefore, we cannot find a Tschirnhaus transformation that transforms a general quintic to the form x5+ c

0 = 0 for example. Still, the technique with Tschirnhaus transformations is a powerful

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13

3

Solvability of algebraic equations

In this chapter, we present the history of the quintic equation. We also present the necessary Galois theory, and give a criterion for when an equation is solvable by radicals.

3.1 History of the quintic

After the Italians had solved the cubic and quartic, several mathematicians tried to solve the quintic; of course, no one succeeded in finding a formula involving only radicals. Joseph Louis Lagrange (1736–1814) wrote a book called R´eflexions sur la r´esolution alg`ebrique des ´equations (1770–71); see Stillwell [24]. In this book, Lagrange examined all previous attempts to solve the quintic. He observed that using a special technique called Lagrange resolvents, that worked for the cubic and quartic, one gets higher order equations when applied to an equation of degree≥ 5. For an nth degree equation, we define the Lagrange resolvent as

r(w) = x1+ x2w + . . . + xnwn−1, (7)

where xi, i = 1, . . . , n, are the roots of the equation and w is an nth root of unity. To be

able to express the roots of the equation in terms of this resolvent, consider the sum s = wk+ w2k+ . . . + w(n−1)k,

where k is an integer. We have

(1− wk)(wk+ w2k+ . . . + w(n−1)k) = 1− wnk = 0.

Therefore, if k 6= 0 modulo n, the sum equals 0, otherwise it equals n. We then get the roots to the equation as

nxk= n−1

X

j=1

w−k+1r(wj), k = 1, 2 . . . , n. (8) There exists different ways of finding the Lagrange resolvent for an equation.

Example 3.1For a third degree equation, the method of Lagrange resolvents means by (7) that we put

z = x1+ wx2+ w2x3,

where the xi are solutions to a third degree equation and w is a cube root of unity. We can

permute these roots into each other which gives six different z values z1, . . . , z6. Then we

form an equation

(x− z1)(x− z2) . . . (x− z6) = 0,

where zi are the six permutations of z. This equation will be quadratic in x3, so it is

solvable. We now show that the equation above is a quadratic in x3. We list the six different permutations:

• z1 = x1+ wx2+ w2x3

• z2 = x3+ wx1+ w2x2= wz1

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• z4 = x1+ wx3+ w2x2

• z5 = wz4

• z6 = w2z4

Therefore we have

(x− z1)(x− z2)(x− z3) = x3− z13 and (x− z4)(x− z5)(x− z6) = x3− z43.

From this we get

(x− z1)(x− z2) . . . (x− z6) = (x3− z13)(x3− z43) = x6− (z13+ z43)x3+ z31z43.

We see that it is a quadratic equation in x3 as asserted.

One was led to study the resolvents because one had found the following formula for one of the roots to the cubic:

x1= 1

3(x1+ x2+ x3+

3

p(x1+ wx2+ w2x3)3+p(x3 1+ wx3+ w2x2)3).

Observe the quantities z3

1z43and z13+z43, where z1= x1+wx2+w2x3and z4= x1+wx3+w2x2,

are symmetric in the roots. But the coefficients in a polynomial are symmetric polynomials of its roots. Hence z31z43 and z13 + z43 are known quantities. The problem is to find the correct permutation that correspond to the solution formula for x1 above. The method of

Lagrange resolvents is a simplification of this search for the right combinations.

If we make similar constructions for the quartic, we get an equation of degree 24, which is still also possible to solve. But when we do the same for the quintic, we get an equation of degree 120. Observe that the degree of the resolvents equals the order of S4 and S5,

respectively. We see that the degree grows rapidly and the resolvents get more and more complex. This observation convinced Lagrange that quintic and higher order equations were not solvable by radicals.

In 1799, the Italian mathematician Paolo Ruffini (1765–1822) published a proof that equations of degree five or higher are not solvable by radicals. The proof was not complete, and Ruffini tried during the rest of his life to improve the proof.

The Norwegian mathematician Niels Henrik Abel (1802–1829) presented a proof around 1825. There was an error in the proof which Abel corrected in 1826. He presented his proof in the mathematical journal Crelle’s Journal and other mathematician subsequently began studying the proof. Abel’s proof showed that it was impossible to solve the general equation of degree≥ 5.

Ruffini and Abel looked at Sn. The concepts were not at all clear, but they observed

that certain things that worked with S3 and S4 did not work for S5. From this observation,

they concluded that the quintic and higher equations are not solvable by radicals.

Abel’s and Ruffini’s proofs did not answer the question as to which equations actually can be solved by radicals. Neither did they present an equation not solvable by radicals. The answer to these questions were unknown until Evariste Galois (1811–1832) answered them. Galois did not know Ruffini’s work, but he knew Abel’s.

We shall present parts of Galois work below. In the next section we will use modern concepts to prove a criterion for deciding when an equation is solvable by radicals. It may

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3.1 History of the quintic 15 be a good idea to first look a little informally at Galois’ thoughts to understand better the rather abstract concepts of Galois theory.

The concepts of ‘group’ and ‘ring theory’ were not invented at the time of Galois. In fact, it was Galois who invented them, and understood their full power. Galois also invented the term ‘field’. As mentioned, Galois made a lot of contributions to this area; for example, he defined subgroups, normal groups, cosets, and permutation groups.

T T T T T ·· ·· · T T T T T ·· ·· · y -6 x Figure 1

If one is given a figure with nodes in the plane, as in Figure 1, one can permute the nodes so the distance to the origin is preserved. These permutation are called distance preserving. If one has two such permutations σ1 and σ2, then obviously the permutation

σ1σ2is distance preserving. Also, to each element σ1, there must be an element that reverses

the permutation. We call it the inverse and denote it by σ1−1. Compositions of permutations are also associative. All these properties mean that the set of permutations form a group.

The roots to the equation

x5 = 1,

lie symmetrically on a circle. There are other polynomials where the zeros exhibit a much more antisymmetric appearance when plotted in the complex plane. It is tempting to permute the roots onto each other and investigate the symmetry. That is what Galois did. Galois considered a field F that contained the coefficients of a polynomial. One express the coefficients in terms of the symmetric polynomials (see Section 3.2). Hence, permuting the roots into each other will not affect the original field. Galois also constructed the Galois Resolvent. If we have a polynomial with coefficients in F , then the resolvent

v = a1x1+ a2x1+ . . . + anxn, a1, . . . , an∈ R,

is denoted a Galois resolvent if one gets n! different functions v when permuting the roots onto each other. Then we form the equation

P (v) = (v− v1)(v− v2) . . . (v− vn) = 0,

where each vi is a certain symmetric function of the roots of the polynomial. But the

coefficients can be represented symmetrically in terms of the roots. Hence P is a polynomial with coefficients in F .

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Galois also investigated when an arbitrary function of the roots can be expressed ratio-nally by the resolvent v. Such functions include, for example, the roots of the polynomial. In this case, Galois observed that one gets a permutation group that one can reduce to a single element. More clearly we can put it this way. Assume that we have a polynomial with coefficients in a field F , for example Q. We must adjoin certain nth roots to get a field that contain the roots x1, . . . , xn. The coefficients of the polynomial can be expressed by

the symmetric polynomials, so we can look at extensions of the field Q(x1, x2, . . . , xn),

(for the notation, see Section 3.2 and the Appendix). When we adjoin an nth root to this field, then we destroy symmetry. But by adjoining nth roots in all the other variables we regain symmetry. If we permute the roots onto each other, the original field will be unaffected. These permutations form a group called the Galois group. When we adjoined the roots, we got a chain of extensions

F ⊆ F (a) ⊆ . . . ⊆ E.

If F ⊆ L, then Gal(E/F ) ⊇ Gal(E/L). Therefore we must have a corresponding chain Gal(E/F ) = G1 ⊇ G2 ⊇ . . . ⊇ Gk= Gal(F/F ) ={e}.

This is a key observation, because if it is impossible to have such a chain of groups for the permutation group of index≥ 5, then it is also impossible to solve equations of degree ≥ 5. Galois succeeded in proving this. With the help of his theory he was also able to give a criterion for when equations are solvable. Galois also produced an equation that was not solvable by radicals. That is, Galois settled all the remaining question that Ruffini and Abel had left unanswered.

We see that Galois’ theory is not very different from Lagrange’s theory. But because Galois invented the group and field theory, he was able to prove much more than Lagrange. Because of Galois short life (his life ended in a duel), he did not write down all his thoughts in a strict and concise way. Galois’ brother Alfred and his friend August Chevalier sent his posthumous work to several mathematician, but they did not get any response. Joseph Liouville was the first mathematician to study Galois’ results conscientiously. Liou-ville understood their importance and presented them in 1846.

3.2 Galois theory

The aim of this section, and the remaining sections of this chapter, is to produce a criterion for when an algebraic equation is solvable by radicals.

Suppose that F is a field. If F is a subfield of a field E, then E is called an extension field. We denote the extension as E/F . We begin by giving two definitions.

Definition 3.2An element u ∈ E is algebraic if there exists a nonzero polynomial f (x)∈ F [x] such that f(u) = 0; otherwise u is called transcendental.

Definition 3.3Let u ∈ E. A polynomial m(x) ∈ F [x] that has minimal degree, which is monic, and satisfies m(u) = 0 is called a minimal polynomial of u and is denoted min(F, u).

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3.2 Galois theory 17 An extension is called algebraic if every element of E is algebraic over F . We will only deal with algebraic extensions.

We are interested in finding the smallest field that contains an element u∈ E and F . We denote this field by F (u), and call this a simple extension. We can use the minimal polynomial to find out things about F (u). It is clear that min(F, u) must be irreducible over F [x], and if there is another polynomial f (x) ∈ F [x] with f(u) = 0, then min(F, u) divides f (x). If we define

θ : F [x]→ E by θ(f(x)) = f(u),

(we take a polynomial from F [x] and evaluate it at u), θ will be a homomorphism. The kernel of the homomorphism, ker(θ), is the ideal generated by min(F, u), which we denote hmin(F, u)i. The factor ring F [x]/hmin(F, u)i is a field and is isomorphic to F (u). We can identify F (u) with the vector space of all elements

c0+ c1u + . . . + cn−1un−1,

where ci ∈ F and n is the degree of min F (u). The degree of the extension is denoted [E : F ]

and equals the degree of the minimal polynomial.

We are also interested in finding the smallest field that contains F and some elements u1, . . . , un∈ E. We can build such an extension by simple ones, and then get a field which

we denote by F (u1, . . . , un). The extension F (u1, u2) equals F (u1)(u2), see Nicholson [18].

So we can always build extensions as simple ones.

Definition 3.4If a polynomial f ∈ F [x] splits in an extension field E, i.e., f (x) = c(x− x1) . . . (x− xn),

where x1, . . . , xn∈ E, then E is called a splitting field for f.

Assume that we have a polynomial

f (x) = c0+ c1x + . . . + xn∈ F [x].

If the zeros of f (x) are x1, . . . , xn ∈ E, then the coefficients can be expressed by the

elementary symmetric polynomials. More precisely, if we define σk= X i1<i2<...<ik xi1xi2. . . xik, k = 1, 2, . . . , n, then ck = (−1)n−kσn−k, k = 0, 1, 2, . . . , n− 1.

If we consider the automorphisms θ : E→ E on E = F (u1, . . . , un), u1, . . . , un∈ E, for

which θ(F ) = F , we will have some control over the zeros of f . These automorphisms form a group, so we can also make use of all the theory developed for groups. This motivates the following definition.

Definition 3.5Let E be a field extension of F . The Galois group Gal(E/F ) is the set of automorphism θ : E→ E that leaves F fixed.

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Theorem 3.6 Assume that we have a field extension E = F (u1, . . . , un). The

automor-phismsθ : E→ E that leaves F fixed only depend on how they permute the set {u1, . . . , un}.

Proof We will use the fact that we can build extensions as simple ones. Put F (u1) = F1.

The extension F1(u2) = F (u1, u2) consists of all elements

c0+ c1u2+ . . . + cn−1u2n1−1, ci ∈ F1, (9)

where n1 is the degree of the extension. But F1 can also be identified with the vector space

of sums

c00+ c01u1+ . . . + c0n0−1un12−1, c0i∈ F,

where n2 is the degree of the extension. If we substitute this expression for ci in (9), we see

that we get a binomial in u1and u2with coefficients c0i ∈ F . The automorphisms θ : E → E

that leaves F fixed, will not affect the c0i. Because of this fact we only have to know what the automorphisms do with the u1 and u2. Similarly for extensions with more than two

elements, but we will get a multinomial in u1, . . . , unwith coefficients c0i∈ F . ¤

Corollary 3.7If [E : F ] is finite, then Gal(E/F ) is also finite.

Proof We can write E = F (u1, . . . , un) because of the definition of [E : F ]. By the

above theorem, there are only finitely many possibilities for each ui, which means that

Gal(E/F ) is finite. ¤

The idea behind the Galois group is to go back and forth between E and the Galois group as a pair, and just like a function and its derivative in analysis, we can get a lot of information by doing this.

If we have F ⊆ L ⊆ E, then Gal(E/L) is a subgroup of Gal(E/F ), and we call L an intermediate fieldof the extension E/F . Conversely, if we have a subgroup S of the group of all automorphism of E, then the set

F(S) = {a ∈ E : σ(a) = a for all σ ∈ S},

is called the fixed field of S and is a subfield of E. What we really have is a one-to-one inclusion reversing bijection between the set of intermediate fields L and the set of subgroups H of Gal(E/F ).

Theorem 3.8

1. If L =F(S) for some subgroup S ⊆ aut(E), then L = F(Gal(E/L)). 2. If H = Gal(E/L) for some subfield L of E, then H = Gal(E/F(H)).

Proof Suppose that L = F(S). Then S ⊆ Gal(E/L) and this fact implies that F(Gal(E/L)) ⊆ F(S) = L. But L ⊆ F(Gal(E/L)), and it follows that L = F(Gal(E/L)).

If H = Gal(E/L) for some subfield L of E, then L ⊆ F(Gal(E/L)) and Gal(E/F(Gal(E/L)) ⊆ Gal(E/L) = H. But H ⊆ Gal(E/F(H)), so it follows that H = Gal(E/F(H)). ¤

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3.2 Galois theory 19 Definition 3.9If F ⊆ E, G = Gal(E/F ), and F(G) = F , then the extension is called Galois.

We are interested in when an extension is Galois. Assume that L is a fixed field of a group G of automorphisms of an extension E/F . One can show that then [E : L] = |G|; see Nicholson [18]. We call a polynomial separable if it has distinct roots in every splitting field. If an element u∈ E has a minimal polynomial m which is separable, then we have | Gal(E/F )| = [E : F ]. This is easy to see, because if the degree of m is n, then m has n distinct roots in E. But by Theorem 3.6, the automorphisms σ of E only depend on what they do with u, so σ(u) must be a root of m. Consequently, there are n possibilities and therefore | Gal(E/F )| = [E : F ]. From this we see that if E is a splitting field of some separable polynomial, then Gal(E/F ) is Galois. Indeed, let L be the fixed field of Gal(E/F ). Then [E : F ] =| Gal(E/F )| = [E : L], and from the inclusions F ⊆ L ⊆ E, it follows that F = L, so we have almost proved the following theorem

Theorem 3.10 IfE is a splitting field of some separable polynomial in F [x], then Gal(E/F ) is Galois.

We shall now state and prove part of what is often called the main theorem in Galois theory.

Theorem 3.11If L is an intermediate field and H is a subgroup of G = Gal(E/F ) such thatL =F(H), then H is normal in G if and only if L is Galois over F , and if this is the case, Gal(L/F ) ∼= G/H.

Proof Suppose that H is normal in G. Let b∈ E be any zero of a minimal polynomial m(F, a) over L, where a ∈ L. Let σ ∈ G such that σ(a) = b. Such a σ exists because G consists of the automorphisms of E. Take τ ∈ H. Then τ(b) = τ(σ(a)) = σσ−1τ (σ(a)).

Because H / G, σ−1τ σ ∈ H, by the definition of a normal group. The field L is fixed by the automorphisms in H. This implies that σ−1τ σ = a, and hence τ (σ(a)) = σ(a) = b. Thus

b∈ F(H) = L. This means that min(F, a) splits over L and is separable. This shows that the extension is Galois.

Suppose that L/F is Galois and let θ : G→ Gal(L/F ) be given by θ(σ) = σ|L. It is a

homomorphism and the kernel is Gal(L/F ) = H. Hence H is normal. The isomorphism theorem now shows that Gal(L/F ) ∼= G/H. ¤

Example 3.12Consider the equation, x4− 6x2+ 7 = 0. Its roots are ±p3±√2. If we make the extension

Q⊆ Q(√2)⊆ Q(√2)( q 3 +√2)⊆ Q(√2, q 3 +√2)( q 3√2) = E,

all the roots will lie in the extension field E. We adjoined a root of an element in each extension to get the field E. It is clear that if we can solve the equation by radicals, then we can make an extension E like the one above.

Definition 3.13We call a field extension F (u1, . . . , ur) radical if there exist integers

n1, . . . , nr such that un11 ∈ F and unii ∈ F (u1, . . . , ui−1), i = 2, 3, . . . , r.

Equations like xn− 1 = 0 will be of interest in Section 3.4. An element that satisfies

such an equation is called an nth root of unity. If charF does not divide n, an nth-root will exist; see Morandi [16].

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Theorem 3.14 If F is a field that contains an nth root of unity w, then the extension F (u)/F , where u is in some field E, is Galois and Gal(F (u)/F ) is abelian.

Proof Assume that un = a. Then xn− a = 0 has roots u, uw, . . . , uwn−1. The poly-nomial xn− a is separable over F (u). Hence F (u) is Galois by Theorem 3.10. This means

that if σ and τ ∈ Gal(F (u)/F ), then σ(u) and τ(u) are zeros of xn− a. So assume that

σ(u) = uwi and τ (u) = uwj. We then have

σ(τ (u)) = uwj+i= τ (σ(u)), where i and j are integers. Thus Gal(F (u)/F ) is abelian. ¤ We end this section with an example.

Example 3.15Consider the equation

x4− 7x2+ 10 = 0.

We can factorize the left hand side as (x2 − 2)(x2 − 5), so the zeros are ±2 and ±5.

We make the extensions F = Q ⊆ Q(√2) ⊆ Q(√2,√5) = E which are radical. What is Gal(E/F )? Because an irreducible polynomial splits in E, we know that the extension is Galois. This gives±√2→ ±√2 and±√5→ ±√5, because if u is a root of a minimal poly-nomial, then σ(u), where σ∈ Gal(E/F ), should also be a root of this minimal polynomial. The field Q should be fixed so for example √2 → 1 is impossible. So we have |G| = 4, as predicted. The elements of the Galois group are

ε = µ √2 − √ 2√5√5 √ 2√2√5√5 ¶ , σ1 = µ √ 2 −√2 √5 −√5 −√2 √2 √5 −√5 ¶ , σ2 = µ √2 − √ 2 √5 √5 √ 2 −√2 −√5 √5 ¶ , σ3 = µ √ 2√2 √5√5 −√2 √2√5 √5 ¶ .

It is easy to verify that this is a group which is isomorphic to a subgroup of S4.

3.3 Solvable groups

We begin with a definition.

Definition 3.16A group G is solvable if there is a chain of subgroups {e} = H0 ⊆ H1⊆ . . . ⊆ Hn= G,

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3.3 Solvable groups 21 An abelian group is always solvable because {e} = H0 ⊆ G satisfies the conditions in

the definition above.

We are interested in when a factor group G/K is abelian. Take two cosets Ka and Kb. Suppose that KaKb = KbKa, i.e., Kab = Kba. Then k1ab = k2ba for every k1, k2 ∈ K,

which implies that bab−1a−1= k−1

2 k1 ∈ K.

We need another definition.

Definition 3.17If G is a group and x, y ∈ G, then xyx−1y−1 is called the commutator of

x and y and is denoted [x, y]. The subgroup G0, generated by all products of commutators, is the commutator subgroup (or the derived group) of G.

Lemma 3.18 The inverse of a commutator is a commutator.

Proof If z = [x, y] = xyx−1y−1, then z−1= yxy−1x−1= [y, x]. ¤

It is clear that G0 is closed under the operation in G, and because of the lemma above, we see that G0 really is a subgroup of G. But even more can be said.

Theorem 3.19 The groupG0 is a normal subgroup of G.

Proof We have to show that ghg−1 ∈ G0 for every g∈ G and every g ∈ G0. The trick

is to insert g−1g because we have

g[x, y]g−1 = gxg−1gyg−1gx−1g−1gyg−1 = [gxg−1, gyg−1]. Any element h of G0 is a product c

1. . . cn of commutators, and this gives

ghg−1= g(c1. . . cn)g−1 = gc1g−1gc2g−1. . . gcng−1= d1. . . dk,

where di = gcig−1. But di is a commutator, hence ghg−1 ∈ G0. ¤

It follows from the theorem that G/G0 is abelian. We also see from the definition

preceeding Definition 3.17 that a normal subgroup H of G contains every commutator (G0 ⊆ H) if and only if G/H is abelian. Conversely, if G0 ⊆ H, where H is a subgroup of

G, g∈ G, and h ∈ H, then

ghg−1 = (ghg−1h−1)h = [g, h]h∈ G0h⊆ Hh = H,

thus H / G and G/H is abelian. Consequently, G0 is the smallest normal subgroup of G

whose factor group is abelian.

We now look at the following chain:

G⊇ G1⊇ G2⊇ . . . ⊇ Gn⊇ . . . ,

where G1 = G0 and Gi+1= (Gi)0. In this chain Gi+1/ Gi and Gi/Gi+1 is abelian. This is

connected to solvable groups by the following theorem.

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Proof If Gn ={e}, then G is solvable because of the definition of solvability.

Conversely, assume that G is solvable. Then there is a chain {e} = H0 ⊆ H1⊆ . . . ⊆ Hn= G,

where Hi is a subgroup of G such that Hi/ Hi+1 and Hi+1/Hi is abelian.

The group G/Hn−1 is abelian, so G1 = G0 ⊆ Hn−1 because G0 is the smallest normal

subgroup of G whose factor group is abelian . Similarly, since the factor group Hn−1/Hn−2

is abelian, Hn−10 ⊆ Hn−2, and thus G2 = (G1)0 ⊆ Hn−10 ⊆ Hn−2. Continuing this way, it

follows that

Gn⊆ H0= {e}. ¤

Corollary 3.21 If G is a solvable group, then every subgroup of G is solvable.

Proof Let H be a subgroup of G, then H = H0 ⊆ G0= G. Assume that Hk ⊆ Gk for

some k∈ Z+. Then

Hk+1 = (Hk)0 ⊆ (Gk)0 = Gk+1. After n steps we get Hn⊆ Gn⊆ {e}, so H is solvable. ¤

We have now developed all the theory necessary to give a criterion for when an equation is solvable by radicals. This will be the subject of the next section.

3.4 Solvability by radicals

We have defined radical extensions and we have examined solvability. The following theorem by Galois connects these two concepts.

Theorem 3.22 Suppose that char(F ) = 0. Let f (x) ∈ F [x] and let E be a splitting field over F . Then the equation f (x) = 0 is solvable by radicals if and only if Gal(E/F ) is a solvable group.

Proof If f (x) = 0 is solvable by radicals, there exists a sequence of fields F = F0⊆ F1 = F (w1)⊆ . . . ⊆ Fk= Fk−1(wk),

where wni

i ∈ Fi−1 for i = 1, . . . , k. We may assume that F contains all nth roots of unity

(otherwise adjoin them), so every extension will be Galois by Theorem 3.14. By Theorem 3.11 this implies that Gal(Fk/Fi) / Gal(Fk/Fi−1). Consider the chain

Gal(Fk/F )⊇ Gal(Fk/F1)⊇ . . . ⊇ Gal(Fk/Fk) ={e}.

Every subgroup is normal in the one preceding it. The field Fi is a Galois extension of

Fi−1, hence Gal(Fi/Fi−1) is isomorphic to Gal(Fk/Fi−1)/ Gal(Fk/Fi) by Theorem 3.11. But

Gal(Fi/Fi−1) is abelian and so is each quotient group in the chain above. This means that

Gal(Fk/F ) is solvable. Since E⊆ Fk and the extension is Galois, Gal(Fk/E) / Gal(Fk/F )

and Gal(Fk/E) ∼= Gal(Fk/F )/ Gal(Fk/E), by Theorem 3.11. Thus Gal(Fk/E) is a

homo-morphic image of Gal(Fk/F ) which is a solvable group, i.e., Gal(Fk/E) is solvable.

For the converse, suppose that Gal(E/F ) is a solvable group. We then have a chain {e} = H0⊆ H1 ⊆ . . . ⊆ Hr = Gal(E/F ),

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3.4 Solvability by radicals 23 such that Hi+1/ Hi and Hi/Hi+1 is abelian. Let Ki =F(Hi). Then Ki+1 will be Galois

over Li by Theorem 3.11 and Gal(Ki+1/Ki) ∼= Hi/Hi+1. Let w be an nth root of unity,

where n is the least common multiple of the orders of the elements in G. Set Li = Ki(w).

Then we have a chain of fields

F ⊆ L0 ⊆ . . . ⊆ Lr.

One can show that this is a radical extension; see Morandi [16] for the details. ¤ Lemma 3.23 If n≥ 5, then Sn is not a solvable group.

Proof The group An is simple and non-abelian for n ≥ 5, so the derived group of An

equals An i.e. A0n= An. This means that Gn6= {e} in Theorem 3.20, because, by Theorem

3.19, the only simple groups that are solvable are abelian. Thus An is not solvable. The

al-ternating group Anis a subgroup of Sn, so Snis not solvable for n≥ 5 by Corollary 3.21. ¤

Example 3.24If f (x) = x5− 6x + 2 ∈ Q, then it follows from Eisenstein’s criterion that

this polynomial is irreducible over Q. It has three real roots and two complex conjugate roots. Because | Gal(E/Q)| = [E : Q] and [Q(u) : Q] = 5, where u is any root to a minimal polynomial of f over E, its Galois group must contain an element of order 5. Then Gal(E/Q) contain a 5-cycle and a 2-cycle which generate S5; see Nicholson [18]. Therefore

Gal(E/Q) ∼= S5. But S5 is not solvable by radicals by the above lemma, hence the equation

f (x) = 0 is not solvable by radicals.

With this theory one can show that the general polynomial of degree n≥ 5 is not solvable by radicals. However, for example, xn= 1 is solvable, because it has a well-behaved Galois

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25

4

Elliptic functions

We have seen in the previous chapter that, in general, it is impossible to solve a polynomial equation by radicals if the degree is greater than four. However, there are other ways one can solve such equations. In this chapter we shall describe one possible method.

4.1 General theory

Let us first try to motivate why elliptic functions appear in the theory of algebraic equations. The equation xn = a has the solution x = √na , which we can write as x = eln a/n. In

analysis, the function ln x is defined by a transcendental integral: ln x = Z x 1 dt √ t2. Similarly, arcsin x = Z x 0 dt √ 1− t2,

and sin x is the inverse of this integral. If we put some other polynomial in the integral, we might hope to get a function, or the inverse, which solves x5+ ax + a = 0. We can think of

these functions as super-roots. This leads to the elliptic functions, and we shall investigate their properties below.

Definition 4.1A function f in C is said to be meromorphic if it is analytic except for isolated poles.

Definition 4.2If there exists a complex number ω such that f (z) = f (z +ω), for all z∈ C, then ω is called a period of f .

How many periods can a function have? In order to answer this question, we need a couple of results.

Lemma 4.3A sequence of periods n}n∈Z+ of a meromorphic function f , which is

as-sumed to be non-constant, cannot have zero as an accumulation point.

Proof Assume on the contrary that zero is an accumulation point. For a regular point z of f we introduce the function g by

g(w) = f (z)− f(z + w).

Then g is analytic in a neighbourhood of 0. Furthermore, g(ωn) is zero for every integer

n. By the uniqueness theorem for analytic functions, g is zero in a neighbourhood of 0, which means that f (z) = f (z + w) in this neighbourhood. Hence f is constant function, a contradiction. ¤

If there is a convergent sequence n}n∈Z+ of periods of f , then the limit point ω is a

period. In fact, by continuity,

f (z + ω) = lim

n→∞f (z + ωn) = f (z),

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Theorem 4.4A sequence of periods n}n∈Z+ cannot have a finite accumulation point.

Proof If there were a finite accumulation point, then by the Cauchy criterion, ωn− ωm

tend to 0, as m, n→ ∞. But ωn−ωm is a period of f . By Lemma 4.3 this is impossible. ¤

From the theorem we see that if we have a period ω, then among all periods with this direction in the complex plane, there must be a period with least distance to 0. Such a period is denoted a primitive period. If we have a primitive period ω, then every integer multiple of ω also is a period, because

f (z + nω) = f (z + (n− 1)ω + ω) = f(z + (n − 1)ω) = . . . = f(z + ω) = f(z). Could rω be a period even for real numbers r? The answer is no, because if there existed a non-integer multiple of ω, denoted ω1, then ω1/ω = r, where r is a real number. By writing

r = n + t, where n is the integer part and t is the fractional part of r, it would follow that ω1− nω = tω. But tω is a period of f which is closer to zero than ω. This is impossible

because of the discussion above. We say that two periods are different if they are not integer multiples of each other. Obviously two different periods satisfy Im(ω1

ω2)6= 0.

We can now prove the following theorem.

Theorem 4.5 (Jacobi theorem) There does not exist a non-constant meromorphic func-tion with more than two primitive periods.

Proof In view of Theorem 4.4, we can always create a parallelogram which is spanned by two different periods, such that there are no periods in it other than the vertices. Assume that there is a third period ω3. Then it can be represented as a linear combination of ω1

and ω2:

ω3 = r1ω1+ r2ω2, r1, r2 ∈ R.

Let ri= ni+ ti, were ni is the integer part and ti is the fractional part of ri. Then we have

ω3− n1ω1− n2ω2= t1ω1+ t2ω2.

But ω3− n1ω1− n2ω2 is a period and it lies inside the parallelogram, a contradiction. ¤

If a function f has two different periods, then the function is called doubly-periodic. The parallelogram with corners c, c + ω1, c + ω1+ ω2, and c + ω2, where c is an arbitrary point

in the complex plane, is called a period parallelogram and is denoted by Ω. The only node that is assumed to belong to the parallelogram is c, and the edges in it are the ones between c and c + ω1 and c and c + ω2. The complex plane is divided into a lattice induced by the

point c and the two periods. We only have to study a doubly-periodic function inside this parallelogram.

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4.1 General theory 27 ¤¤ ¤¤º ¤¤ ¤¤º »»»»: »»»»¤ ¤¤¤ »»»» ¤¤ ¤¤ ¤¤ ¤¤ »»»» ¤¤ ¤¤»»» » »»»» »»»»: ¤¤ ¤¤

A lattice induced by the point c and two periods

»»»» ¤¤ ¤¤ »»»» ¤ ¤ ¤¤»»»»¤ ¤¤¤ »»»»¤ ¤¤¤ »»»»¤ ¤¤¤ c Ω Figure 2

Definition 4.6A meromorphic, doubly periodic function is called an elliptic function. Since the singularities of a doubly periodic function are isolated, we can assume that no pole lies on the boundary of the period-parallelogram. In fact, if there are poles on the boundary, then just move the parallelogram a little by choosing a new node c.

Theorem 4.7A non-constant elliptic function f cannot be regular inside a period paral-lelogram.

Proof If f did not have a singularity inside a period-parallelogram, then it would have a maximal absolute value M . By periodicity,|f(z)| ≤ M in the whole plane, so by Liouville’s theorem, it would be a constant, which contradicts what we assumed. ¤

Definition 4.8The number of poles of a periodic function f in a period-parallelogram, where a pole of order n is counted n times, is called the order of f .

Theorem 4.9The sum of the residues of an elliptic functionf inside a period-parallelogram is zero.

Proof Integrating along the boundary ∂Ω of Ω, we have Z ∂Ω f (z) dz = Z c+ω1 c f (z) dz + Z c+ω1+ω2 c+ω1 f (z) dz + Z c+ω2 c+ω1+ω2 f (z) dz + Z c c+ω2 f (z) dz. Make the substitution w = z + ω2 in the third integral. Then

Z c+ω2 c+ω1+ω2 f (z) dz = Z c c+ω1 f (w + ω2) dw.

But f (w + ω2) = f (w), and thus the first and third integral cancel. Similarly the second

and fourth cancel. Hence Z

∂Ω

f (z) dz = 0. By the residue theorem,

Z ∂Ω f (z) dz = 2πi n X i=1 ri,

where ri are the residues of f , so the sum of residues is zero. ¤

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Theorem 4.10The number of zeros a1, . . . , ap of an elliptic function f , where a zero is

counted according to its multiplicity, is equal to the orderN of the function. Proof We recall from complex analysis that

1 2πi Z ∂Ω f0(z) f (z) dz = p X i=1 ord(ai)− q X j=1 ord(pj),

where ord(· ) is the order of a zero ai or of a pole pj of f . The function f0/f is elliptic, so

the integral is zero, i.e.,

p X i=1 ord(ai) = q X j=1 ord(pj) = N. ¤

The function F (z) = f (z)− C, where C is a constant, is also an elliptic function, having the same poles as f in a period-parallelogram. We thus have the following corollary. Corollary 4.11For every C ∈ C, the number of points z in Ω where f(z) = C, is equal to the order of f .

Theorem 4.12 Let f be an elliptic function which has p zeros at a1, . . . , ap and q poles at

p1, . . . , pq. Then p X i=1 ord(ai)− q X j=1 ord(qj) = mω1+ nω2

for some integersm and n.

Proof We again have from complex analysis that

p X i=1 ord(ai)− q X j=1 ord(qj) = 1 2πi Z ∂Ω zf 0(z) f (z) dz.

If we make the same substitution as in the proof of Theorem 4.9, we get Z Ω zf 0(z) f (z) dz = Z c+ω1 c zf 0(z) f (z) − (z + ω2) f0(z + ω 2) f (z + ω2) dz − Z c+ω2 c zf 0(z) f (z) − (z + ω1) f0(z + ω 1) f (z + ω1) dz = ω1log f (c + ω2) f (c) − ω2log f (c + ω1) f (c) = ω1log 1− ω2log 1 = ω12πin1− ω22πin2,

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4.2 The Weierstrass℘ function 29

4.2 The Weierstrass ℘ function

Let f be an elliptic function with periods ω1 and ω2. We may assume that

Im(ω1 ω2

) > 0,

otherwise, replace ω1 by ω2 and ω2 by ω1. Define a lattice Λ by

Λ ={kω1+ lω2; k, l∈ Z},

and let

Λ0 = Λ− {0}. Theorem 4.13 The sum

X

ω∈Λ0

1 |ω|p

converges for p > 2.

Proof For k = 1, 2, . . . we define Pk to be the parallelogram with corners ±kω1 and

±kω2, where k is an integer. Such a parallelogram has 8k ω-points on its boundary. Let δ

be the shortest distance from the center of P1 to the boundary. We then have

X ω∈Λ0 1 |ω|p < ∞ X k=1 8k (kδ)p = 8 δp ∞ X k=1 1 kp−1.

The series on the right hand side converges for p > 2. Hence we have proved the theorem. ¤ In view of Theorem 4.13, the function

G(z) =−2X

ω∈Λ0

1 (z− ω)3,

converges absolutely and uniformly in each bounded region of the z-plane that does not intersect Λ0. Hence the sum defines a meromorphic function.

We have G(z + ω1) =−2 X ω∈Λ0 1 (z + ω1− ω)3 =−2 X ω∈Λ0 1 (z− ω)3 = G(z),

because of the infinite summation range, and similarly G(z + ω2) = G(z). Thus, we see

that G is doubly-periodic with poles kω1+ lω2. By definition, this means that G is elliptic.

Moreover, G(−z) = 2 X ω∈Λ0 1 (z + ω)3 = 2 X ω∈Λ0 1 (z− ω)3 =−G(z). Hence G is odd.

We now introduce the Weierstrass function ℘ as ℘(z) = 1 z2 + X ω∈Λ0 1 (z− ω)2 − 1 ω2.

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Observe that ℘0(z) = G(z). It follows that ℘ is an even function. From the above we have

that ℘0(z) = ℘0(z + ω1) for all z. Notice that the number 12ω is not a pole of ℘. Integrating

the periodicity identity for ℘0, we obtain

℘(z) = ℘(z + ω1) + c.

In particular, ℘(−1

2ω1) = ℘(12ω1) + c. The function ℘ is even, hence c = 0. The Weierstrass

function has a pole of order two for z = 0 and no more poles inside the period-parallelogram with 0 as a node. All this means that ℘ is an elliptic function of order two.

We will now prove an important property of the Weierstrass function. Theorem 4.14 The Weierstrass function℘ satisfies the differential equation

℘0(z)2= 4℘(z)3− g2℘(z)− g3, (10) where g2= 60 X ω∈Λ0 1 ω4 and g3= 140 X ω∈Λ0 1 ω6.

The numbersg2 andg3 are called invariants.

Proof Make a Laurent expansion of ℘ at z = 0. Since odd terms in the sum cancel, we obtain ℘(z) = 1 z2 + 3z 2 X ω∈Λ0 1 ω4 + 5z 4 X ω∈Λ0 1 ω6 + . . . = 1 z2 + g2 20z 2+ g3 28z 4+ . . . . This gives ℘0(z) = −2 z3 + g2 10z + g3 7 z 3+ . . . , ℘0(z)2 = 4 z6 ³ 1− g2 10z 4g3 7 z 6+ . . .´, ℘(z)3 = 1 z6 ³ 1 +3g2 20 z 4+3g3 28 z 6+ . . .´,

and hence that

℘0(z)2− 4℘(z)3− g2℘(z) =−g3+ Az2+ Bz4+ . . . ,

where A and B are constants. The left hand side is an elliptic function, and in view of the right hand side, it is entire, and hence a constant. Thus A = B = 0. ¤

If we differentiate (10), we see that all the derivatives of ℘ can be expressed in terms of ℘ and ℘0. For example,

2℘00℘0 = 12℘0℘2− g2℘0

After division with 2℘0, we get

℘00 = 6℘212g2. (11)

The right hand side of the differential equation (10) is a third degree polynomial in ℘(z), so it can be factorized as

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4.2 The Weierstrass℘ function 31 To determine z1, z2, and z3, we have to find the zeros of ℘0(z) in a period parallelogram.

Putting z =−1

2ω1 in the identity ℘

0(z) = ℘0(z + ω

1) and remembering that the derivative is

odd gives ℘0(1

2ω1) = 0. In a similar manner one can show that 12ω2 and 12(ω1+ ω2) are also

zeros of ℘0. There cannot be any other zeros in a period parallelogram, because the order

of ℘0 is three. The values of ℘(z) at these points are all distinct, because if for example ℘(12ω1) = ℘(12(ω1+ ω2)),

then the second order elliptic function ℘(z)− ℘(12ω1) would have two second order zeros

counted according to multiplicity, which is impossible. Therefore we have z1= ℘(12ω1), z2 = ℘(12ω2), and z3 = ℘(12(ω1+ ω2)).

Definition 4.15The discriminant, D, of a general nth degree polynomial P (x) = cnxn+ cn−1xn−1+ . . . + c1x + c0

is defined to be

D = c2n−2n Y

1≤j<i≤n

(xi− xj)2,

where xi are the roots of the polynomial.

If x1, x2, and x3 are the roots of a general third degree equation, then the discriminant

is D = c43(x1− x2)2(x2− x3)2(x3− x1)2. It is obvious that if the discriminant 6= 0, then the

equation has three distinct roots.

If we take our third degree equation, then we have c3 = 4, c2 = 0, 3c1 = −g2, and

c0=−g3. Using the relationship between roots and coefficients we see that the discriminant

for this equation is g2

3 − g23/27. From this fact it follows that if g32 − 27g23 6= 0, g2 and g3

determine a Weierstrass function such that the right hand side of (10) has three distinct zeroes.

Example 4.16Consider the function

f (z) = ℘(z)℘0(z) + ℘(z)2− 1.

This is an elliptic function of order 5. We know that it takes every value five times inside a period parallelogram, and in particular it takes the value zero five times.

The fifth degree equation

4x5− x4− g2x3+ (2− g3)x2− 1 = 0

can be written as

x2(4x3− g2x− g3)− x4+ 2x2− 1 = 0.

If we put x = ℘(zj), where zj is one of the five zero of f , and use the differential equation

(10) for ℘(z) , we get

℘(zj)2℘0(zj)2− (℘(zj)2− 1)2 = ℘(zj)2℘0(zj)2− ℘(zj)2℘0(zj)2= 0.

Thus the function f has five points that solve a fifth degree equation. This shows that the elliptic functions seem to have the properties we were looking for at the beginning of this chapter. This observation will be extended in the next section.

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There are other functions that are connected to the Weierstrass function. These are the Weierstrass ζ and σ functions which are defined by the expressions

ζ(z) = 1 z − Z z 0 ℘(t) 1 t2 dt and ln σ(z) z = Z z 0 ζ(t) 1 tdt.

The zeta function is an meromorphic function with simple poles. It is not doubly periodic but it is odd. The sigma function is an entire function with simple zeros.

Another important class of functions in the theory of elliptic functions are the elliptic modular functions. One studies a function depending on a parameter τ , with Im(τ ) > 0, under certain transformations, for example a Moebius transformation. We get these different functions when we transform the function and they form a group called the modular group. An example of a elliptic modular function is

J(τ ) = g2(1, τ )

3

g2(1, τ )3− 27g3(1, τ )2

,

where g2 and g3 are the invariants of ℘ and τ = ω2/ω1. With the aid of modular functions

one can prove Theorem 4.25 below.

We end this section by showing a connection between elliptic functions and elliptic integrals.

Definition 4.17An integral

Z dx pP (x), where P (x) is polynomial of degree≤ 4, is called elliptic.

If we have an integral u = Z ∞ x dt √ 4t3− at − b, (12)

where a and b are constants, then it is, by definition, an elliptic integral. If the constants a and b satisfy a3− 27b2 6= 0, we can take them as invariants and construct a Weierstrass function.

If we see x as a function of u, then one might wonder if x is an elliptic function. It is, and to prove this, make the substitution t = ℘(v). Then we get the integral

u = Z v

0

dv

We get u = v, but this implies that ℘(u) = x, so x is an elliptic function of u as asserted. Thus, in this case, an elliptic integral is the inverse of an elliptic function.

4.3 Solving the quintic

If g3

2− 27g326= 0, then it is possible to construct a Weierstrass function, see Akhiezer [1] for

the details. In this section, we choose g2 = 0 and g3 = c 6= 0. To get g2 = 0, choose for

example ω2/ω1 = e2πi/3. In Chapter 2, we claimed that the general fifth degree equation

can be transformed to the form

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4.3 Solving the quintic 33 Recall from algebra that a polynomial has a double zero if and only if the zero is a zero to the derivative of the polynomial. The derivative of P is 5x4+ c. Using long division on P and P0, we get

x5+ cx + c = (5x4+ c)15x + (45cx + c). If we have a double zero, it follows that

c(45x + 1) = 0, and we have x =−5 4 and c =− 55 44.

We now define the elliptic function fc by

fc(z) = ℘(z)℘0(z) + i√c℘(z) + 2i√c.

The order of fc is five, so it assumes every value five times inside a period parallelogram.

Especially, there are five points zj, j = 1, 2, . . . , 5, counted according to multiplicity, were

fc takes the value zero.

Lemma 4.18 The numbers℘(zj) are the roots to the equation x5+ cx + c = 0.

Proof Since fc(zj) = 0, we have ℘(zj)℘0(zj) = −i√c℘(zj)− 2i√c. If we square both

sides, we can use the differential equation for ℘. This manipulation gives ℘(zj)2(4℘(zj)3− c) = −c℘(zj)2− 4c℘(zj)− 4c.

The ℘2 term cancel, whence

℘(zj)5+ c℘(zj) + c = 0,

so ℘(zj) solves the equation as asserted. ¤

If we can show that fc(zj) has five simple zeros for c6= −5

5

44, then we are very close to

solving (13).

Lemma 4.19 If c6= −55/44, then the function fc has five simple zeros.

Proof Assume on the contrary that c6= −55/44 and that f

c has a double root, that is

fc(zj) = ℘(zj)℘0(zj) + i√c℘(zj) + 2i√c = 0,

fc0(zj) = ℘0(zj)2+ ℘(zj)℘00(zj) + i√c℘0(zj) = 0.

Now use the differential equation (10) and the formula (11) for the second derivative of ℘. Then the derivative becomes

fc0(zj) = 10℘(zj)3+ i√c℘0(zj)− c = 0. (14)

From the first equation we have

℘0(zj) =−

i√c℘(zj) + 2i√c

℘(zj)

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Substituting this expression into (14), we get ℘(zj)4 =−c/5, that is

P0(℘(zj) = 5℘(zj)4+ c = 0,

where P is as above. But by Lemma 4.18, ℘(zj) is a root of x5 + cx + c = 0. By the

discussion above c =−55/44, which is a contradiction to what we assumed. ¤

We must show that if z1 and z2 are distinct zeros of fc, then ℘(z1) 6= ℘(z2). If we can

show this, then we have solved (13)! We need another three lemmas. Lemma 4.20 If z1 6= z2 and ℘(z1) = ℘(z2), then ℘0(z1) =−℘0(z2).

Proof By Theorem 4.12 and the fact that the ℘ has only one pole of order two at zero, we have z1+ z2 = mω1+ nω2, where m and n are integers. Thus z2= mω1+ nω2− z1. The

derivative ℘0 is periodic and odd. Thus

℘0(z2) = ℘0(mω1+ nω2− z1) = ℘0(−z1) =−℘0(z1). ¤

Lemma 4.21 If z1, z2 are distinct zeros of fc, then ℘(z1)6= ℘(z2).

Proof Suppose that ℘(z1) = ℘(z2) with z1 6= z2. Then

fc(z1) = ℘(z1)℘0(z1) + i√c℘(z1) + 2i√c = 0.

By the previous lemma, we can write the equality fc(z2) = 0 as

fc(z2) =−℘(z1)℘0(z1) + i√c℘(z1) + 2i√c = 0.

Subtracting this from the first equality yields

2℘(z1)℘0(z1) = 0.

But ℘(z1)6= 0, because c 6= 0, whence ℘0(z1) = 0. This means that the function ℘(z)−℘(z1)

assumes the value 0 three times counted according to multiplicity, since it has a double zero at z1 and a simple zero at z2. But the order of ℘(z)− ℘(z1) is two, so we have obtained a

contradiction. ¤

Lemma 4.22 If c = 55/44, then fc has a double and three simple zeros.

Proof Because ℘ is elliptic it assumes every value, so there is a point z1 such that

℘(z1) =−5/4. Using the differential equation for ℘, we have ℘0(z1) = ±1615

5. Using the numerical values of c, ℘(z1), and ℘0(z1), we get

fc(z1) = fc0(z1) = 0

and fc00(z1) 6= 0. The other zeroes xi, i = 1, 2, 3, to the quintic are simple, and there are

numbers zi, i = 1, 2, 3 such that ℘(zi) = xi. As above one can show that they are zeroes

to fc. From Lemma 4.21 we conclude that they are distinct. Hence fc has one double and

three simples zeroes and we have proved the lemma. ¤ We have now proved the following theorem,

Theorem 4.23The numbers ℘(zj), i = 1, . . . , 5, where fc(zj) = 0, solve the quintic

equa-tionx5+ cx + c = 0.

There are of course problems making this theory work in practice. We will look at this in Chapter 6. There we will need some theory about theta functions, which we deal with in the next section.

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4.4 Theta functions 35

4.4 Theta functions

Theta functions are entire functions with one ordinary period and one quasi period. A function is quasi periodic if it shows a simple behaviour when one adds a multiple of a complex number ω. For example, a function is quasi periodic if

f (z + ω) = z2f (z), for all z ∈ C. We define a function θ3(z, τ ) = ∞ X m=−∞ e(m2τ +2mz)πi, Im(τ ) > 0.

Since Re(iτ ) < 0, it follows that the series converges uniformly in every set|z| ≤ R, where R is a real number. One often puts q = eiπτ. Then we have

θ3(z, τ ) = ∞

X

m=−∞

qm2e2πimz.

If we replace z with z + 1 in the function above, we get the same series. Therefore θ3(z) =

θ3(z + 1), so θ3 is periodic. Now, if we substitute z + τ for z above, we get

θ3(z + τ, τ ) = ∞

X

m=−∞

e(m2τ +2mz)πi+2iπmτ. After some algebraic manipulations, we can write

θ3(z + τ, τ ) = e−πτ −2πiz ∞

X

m=−∞

e((m+1)2τ +2(m+1)z)πi= e−πτ −2πizθ3(z, τ ).

This shows that θ3 is an entire and a quasi-periodic function.

There are three other theta-functions connected to θ3:

θ1(z, τ ) = i ∞ X m=−∞ (−1)mq(m−12)2e(2m−1)πiz, θ2(z, τ ) = ∞ X m=−∞ (−1)mq(m−12)2e(2m−1)πiz, θ4(z, τ ) = ∞ X m=−∞ (−1)mqm2e2mπiz. The zeros of these functions are:

θ1 θ2 θ3 θ4

m + nτ m + nτ +12 m + (n +12)τ +12 m + (n +12)τ where m and n are integers.

Theorem 4.24 The functionθ4 can be expanded in an infinite product as

θ4(z, τ ) = c ∞

Y

m=1

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Proof Consider the function f (x) = ∞ Y m=1 (1− k2m−1x)(1− k2m−1x−1), x∈ C,

were k is a constant with |k| < 1. It converges absolutely for x 6= 0, as can be seen by comparison with the series

X

m=1

log|1 − k2m−1x||1 − k2m−1x|. Notice that f satisfies the functional equation

f (x) =−kxf(k2x). If we expand f in a Laurent series and use this fact, we get

f (x) = ∞ X j=−∞ ajxj =− ∞ X j=−∞ ajk2j+1xj+1,

from which it follows that aj = a0(−1)jkj

2

, and hence that f (x) = a0 ∞ X j=−∞ (−1)jkj2xj. Then we have f (e2πiz) = a0 ∞ X j=−∞ (−1)jkj2e2πijz.

By putting q = k, this equals a0θ4, and we have proved the theorem. ¤

To find the value of the constant c = a0, see Akhiezer [1].

We can in a similar way find product expansions for the other theta functions. If we put k = q, then we have

θ1(z, τ ) = 2cq1/4sin(πz) ∞ Y m=1 (1− q2me2πiz)(1− q2me−2πiz), θ2(z, τ ) = 2cq1/4cos(πz) ∞ Y m=1 (1 + q2me2πiz)(1 + q2me−2πiz), θ3(z, τ ) = c ∞ Y m=1 (1 + q2m−1e2πiz)(1 + q2m−1e−2πiz), θ4(z, τ ) = c ∞ Y m=1 (1− q2m−1e2πiz)(1− q2m−1e−2πiz), where c = ∞ Y m=1 (1− q2m),

References

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