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

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 2 CONTENTS 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 functions. 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. 4 1 INTRODUCTION 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 + c x + c = 0, c R, 1 0 i ∈ using the famous formula 2 c1 c1 x = c0. − 2 § r 4 − 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 6 2 QUADRATIC, CUBIC, AND QUARTIC EQUATIONS beginning of the fifteenth centuary in Italy, when Scipione del Ferro (1465–1526) succeeded in solving the equation 3 x + 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 + c x2 + c x + c = 0, c C. (1) 2 1 0 i ∈ We will use cube roots of unity. Therefore let

2π 2π w = cos 3 + i sin 3 .

If we make the substitution c x = y 2 , − 3 (in the next section we will show how we find substitutions like that), then we get an equation in y : 3 y + a1y + a0 = 0. 2.2 Solving third and fourth degree equations 7

If a = 0, then this equation has roots y = 0 and y = √ a . Consequently, 0 § − 1 c c x = 2 or x = √ a 2 ; − 3 § − 1 − 3 if a1 = 0, then we get the roots c c c x = √3 a 2 , x = w√3 a 2 and x = w2 √3 a 2 . − 0 − 3 − 0 − 3 − 0 − 3 If a a = 0, then we make the substitution 0 1 6 a y = z 1 . − 3z This substitution gives us an equation in z of the form

a3 a3 z3 + a 1 z6 + a z3 1 = 0, 0 − 27z3 ⇒ 0 − 27 which can be factorized as follows:

a a2 a3 a a2 a3 z3 + 0 + 0 + 1 z3 + 0 0 + 1 = 0. Ã Ã 2 r 4 27!! Ã Ã 2 − r 4 27!!

Theorem 2.1 Both parentheses in the expression

a a2 a3 a a2 a3 z3 + 0 + 0 + 1 z3 + 0 0 + 1 = 0, Ã Ã 2 r 4 27!! Ã Ã 2 − r 4 27!! 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

a1 a1 a1 z1 z2 a1 z1 = z2 z1 z2 = − z1z2 = . (2) − 3z1 − 3z2 ⇔ − − 3 z1z2 ⇔ − 3 The other roots in the parentheses are multiples of w. Therefore a wz w2z = z z = 1 , 1 2 1 2 − 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 a c a c a c x = A 1 2 , x = wA w2 1 2 and x = w2A w 1 2 . − 3A − 2 − 3A − 2 − 3A − 2 8 2 QUADRATIC, CUBIC, AND QUARTIC EQUATIONS

Example 2.2 Solve 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 2 y = z + . z This leads to 8 (z3 2)(z3 4) z3 6 + = 0 − − = 0. − z3 ⇒ z3 We choose to solve z3 2 = 0. The roots to the original equations then are − √3 2 + √3 4 1, w√3 2 + w2 √3 4 1 and w2 √3 2 + w√3 4 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 + c x3 + c x2 + c x + c = 0, c C. (3) 3 2 1 0 i ∈ We begin by making the substitution c x = y 3 , − 4 and obtain 4 2 y + a2y + 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) = a , − 1 r + p 4q2 = a . − 2 2 If a1 = 0, we make the substitution t = y , and obtain a quadratic equation in t. Otherwise, with q = 0, we have 6 a 2p = a + 4q2 1 , 2 − 2q a 2r = a + 4q2 + 1 . 2 2q

Because 2p2r = 4a0, we get a sixth degree equation in q:

64q6 + 32a q4 + 4(a2 4a )q2 a2 = 0. 2 2 − 0 − 1 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 2.3 Solve 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 5 x + 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:

n n−1 x + cn−1x + . . . + 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

n n−1 xk + cn−1xk + . . . + c1xk + c0,

n−1 where xk is a root of (5) as coefficient of y , then the coefficient for this term in our new polynomial will disappear. Therefore, Tschirnhaus transformations will have the general form n−1 n−2 yk = an−1xk + an−2x + . . . + a1xk + a0, k = 1, 2, . . . , n, where xk is a root of (5). In order to make clever substitutions, we need to understand some more theory. 10 2 QUADRATIC, CUBIC, AND QUARTIC EQUATIONS

Definition 2.4 The polynomials

k k k Pk(x1, . . . , xn) = x1 + x2 + . . . + xn, 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) k k k Let Pk(x1, . . . , xn) = x1 + x2 + . . . + xn, where x1, . . . , xn are roots of n n f(x) = x + cn−1x + . . . + c1x + c0 = 0. Then

P1 + cn−1 = 0,

P2 + cn−1P1 + 2cn−2 = 0,

P3 + cn−1P2 + cn−2P1 + 3cn−3 = 0, . .

P − + . . . + c P + c P + c P + (n 1)c = 0, n 1 4 3 3 2 2 1 − 1 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 x )(x x ) . . . (x x ). − 1 − 2 − n Now differentiate this equality with the help of logarithmic differentiation, to obtain f(x) f(x) f(x) f 0(x) = + + . . . + , x x x x x x − 1 − 2 − n where f(x) = xn−1 + a xn−2 + . . . + a x + a . x x n−2 1 0 − i 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, 2 an−3 = xi + cn−1xi + cn−2, . . k−1 k−2 an−k = xi + cn−1xi + . . . + cn−k+1. 2.3 Tschirnhaus transformations 11

It follows that

0 n−1 n−2 f (x) = nx + (P1 + ncn−1)x + . . . + (Pn−1 + cn−1Pn−2 + . . . + c2P1 + nc1).

Differentiating the polynomial in the ordinary way gives

0 n−1 n−2 f (x) = nx + (n 1)c − x + . . . + 2c x + c . − n 1 2 1 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:

n+k n+k−1 k+1 k xi + cn−1xi + . . . + c1xi + c0x = 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 • j polynomial, we can express every y in a polynomial of degree n 1 . k − 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 2.6 Take the equation

n n−1 x + cn−1x + . . . + 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

n n−1 y + bn−1y + . . . + 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 c − + na = 0 a = . − n 1 0 ⇒ 0 n 12 2 QUADRATIC, CUBIC, AND QUARTIC EQUATIONS

To get rid of the xn−1-term we shall thus make the substitution c y = x + n−1 . k k n If one has a second order equation and makes the substitution c y = x + 1 , k k 2 then the equation in y is

c 2 c c2 y 1 + c y 1 + c = 0 y2 1 + c = 0. − 2 1 − 2 0 ⇒ − 4 0 ³ ´ ³ ´ The roots to this equation are

2 2 c1 c1 y1 = + c0 and y2 = + c0. r 4 −r 4 The roots of the equation in x will therefore be

2 2 c1 c1 c1 c1 x1 = + + c0 and x2 = + c0. − 2 r 4 − 2 − r 4 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−3 vanish 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 + c x + c = 0, c , c = 0, (6) 1 0 1 2 6 then the transformation c0 yk = xk, c1 transforms this equation to the form

y5 + ax + a = 0,

5 4 where a = c1/c0. 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 5 x +c0 = 0 for example. Still, the technique with Tschirnhaus transformations is a powerful tool when we are trying to solve equations. 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 n−1 r(w) = x1 + x2w + . . . + xnw , (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 = 0 modulo n, the sum equals 0, otherwise it equals n. We then get the 6 roots to the equation as

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

Example 3.1 For a third degree equation, the method of Lagrange resolvents means by (7) that we put 2 z = x1 + wx2 + w x3, 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 z )(x z ) . . . (x z ) = 0, − 1 − 2 − 6 3 where zi are the six permutations of z. This equation will be quadratic in x , so it is solvable. We now show that the equation above is a quadratic in x3. We list the six different permutations:

z = x + wx + w2x • 1 1 2 3 z = x + wx + w2x = wz • 2 3 1 2 1 z = w2z • 3 1 14 3 SOLVABILITY OF ALGEBRAIC EQUATIONS

z = x + wx + w2x • 4 1 3 2 z = wz • 5 4 z = w2z • 6 4 Therefore we have

(x z )(x z )(x z ) = x3 z3 and (x z )(x z )(x z ) = x3 z3. − 1 − 2 − 3 − 1 − 4 − 5 − 6 − 4 From this we get

(x z )(x z ) . . . (x z ) = (x3 z3)(x3 z3) = x6 (z3 + z3)x3 + z3z3. − 1 − 2 − 6 − 1 − 4 − 1 4 1 4 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: 1 x = (x + x + x + 3 (x + wx + w2x )3 + 3 (x + wx + w2x )3). 1 3 1 2 3 1 2 3 1 3 2 p p 3 3 3 3 2 2 Observe the quantities z1z4 and z1+z4, where z1 = x1+wx2+w x3 and z4 = x1+wx3+w x2, are symmetric in the roots. But the coefficients in a polynomial are symmetric polynomials 3 3 3 3 of its roots. Hence z1z4 and z1 + z4 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 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.

y 6

 T  T  T  T  T - x T  T  T  T  T 

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 σ1σ2 is distance preserving. Also, to each element σ1, there must be an element that reverses −1 the permutation. We call it the inverse and denote it by σ1 . 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 = a x + a x + . . . + a x , a , . . . , a R, 1 1 2 1 n n 1 n ∈ 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 v )(v v ) . . . (v v ) = 0, − 1 − 2 − n 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 . 16 3 SOLVABILITY OF ALGEBRAIC EQUATIONS

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 ) = G G . . . G = Gal(F/F ) = e . 1 ⊇ 2 ⊇ ⊇ k { } 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 3.2 An 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 3.3 Let 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). 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 min(F, u) . The factor ring F [x]/ min(F, u) is a field and is isomorphic to F (u). We can h i h i identify F (u) with the vector space of all elements

n−1 c0 + c1u + . . . + cn−1u , where c F and n is the degree of min F (u). The degree of the extension is denoted [E : F ] i ∈ and equals the degree of the minimal polynomial. We are also interested in finding the smallest field that contains F and some elements u , . . . , u E. We can build such an extension by simple ones, and then get a field which 1 n ∈ 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 3.4 If a polynomial f F [x] splits in an extension field E, i.e., ∈ f(x) = c(x x ) . . . (x x ), − 1 − n where x , . . . , x E, then E is called a splitting field for f. 1 n ∈ Assume that we have a polynomial

f(x) = c + c x + . . . + xn F [x]. 0 1 ∈ If the zeros of f(x) are x , . . . , x E, then the coefficients can be expressed by the 1 n ∈ elementary symmetric polynomials. More precisely, if we define

σk = xi1 xi2 . . . xik , k = 1, 2, . . . , n, i1

Definition 3.5 Let 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. → We will now prove a theorem about Gal(E/F ). 18 3 SOLVABILITY OF ALGEBRAIC EQUATIONS

Theorem 3.6 Assume that we have a field extension E = F (u1, . . . , un). The automor- phisms θ : E E that leaves F fixed only depend on how they permute the set u , . . . , u . → { 1 n}

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

n1−1 c + c u + . . . + c − u , c F , (9) 0 1 2 n 1 2 i ∈ 1 where n1 is the degree of the extension. But F1 can also be identified with the vector space of sums 0 0 0 n2−1 0 c + c u + . . . + c 0 u , c F, 0 1 1 n −1 1 i ∈ where n2 is the degree of the extension. If we substitute this expression for ci in (9), we see 0 that we get a binomial in u1 and u2 with coefficients ci F . The automorphisms θ : E E 0 ∈ → that leaves F fixed, will not affect the ci. 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 u , . . . , u with coefficients c0 F . ¤ 1 n i ∈ Corollary 3.7 If [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 field of the extension E/F . Conversely, if we have a subgroup S of the group of all automorphism of E, then the set

(S) = a E : σ(a) = a for all σ S , F { ∈ ∈ } 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 = (S) for some subgroup S aut(E), then L = (Gal(E/L)). F ⊆ F 2. If H = Gal(E/L) for some subfield L of E, then H = Gal(E/ (H)). F Proof Suppose that L = (S). Then S Gal(E/L) and this fact implies that F ⊆ (Gal(E/L)) (S) = L. But L (Gal(E/L)), and it follows that L = (Gal(E/L)). F ⊆ F ⊆ F F If H = Gal(E/L) for some subfield L of E, then L (Gal(E/L)) and ⊆ F Gal(E/ (Gal(E/L)) Gal(E/L) = H. But H Gal(E/ (H)), so it follows that F ⊆ ⊆ F H = Gal(E/ (H)). ¤ F 3.2 Galois theory 19

Definition 3.9 If F E, G = Gal(E/F ), and (G) = F , then the extension is called ⊆ F 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 If E 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 3.11 If L is an intermediate field and H is a subgroup of G = Gal(E/F ) such that L = (H), then H is normal in G if and only if L is Galois over F , and if this is the F 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−1τσ H, by the definition of a normal group. The field L is fixed by the ∈ automorphisms in H. This implies that σ−1τσ = a, and hence τ(σ(a)) = σ(a) = b. Thus b (H) = L. This means that min(F, a) splits over L and is separable. This shows that ∈ F the extension is Galois. Suppose that L/F is Galois and let θ : G Gal(L/F ) be given by θ(σ) = σ . It is a → |L 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 3.12 Consider the equation, x4 6x2 + 7 = 0. Its roots are 3 √2. If we − § § make the extension p Q Q(√2) Q(√2)( 3 + √2) Q(√2, 3 + √2)( 3 √2) = E, ⊆ ⊆ ⊆ − q q q 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 3.13 We call a field extension F (u1, . . . , ur) radical if there exist integers n1 ni n , . . . , n such that u F and u F (u , . . . , u − ), i = 2, 3, . . . , r. 1 r 1 ∈ i ∈ 1 i 1 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]. 20 3 SOLVABILITY OF ALGEBRAIC EQUATIONS

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 3.15 Consider 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

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

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 3.16 A group G is solvable if there is a chain of subgroups

e = H H . . . H = G, { } 0 ⊆ 1 ⊆ ⊆ n where each Hi is a subgroup of G, Hi / Hi+1, and Hi+1/Hi is abelian. 3.3 Solvable groups 21

An abelian group is always solvable because e = H G satisfies the conditions in { } 0 ⊆ 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 k ab = k ba for every k , k K, 1 2 1 2 ∈ which implies that bab−1a−1 = k−1k K. 2 1 ∈ We need another definition.

Definition 3.17 If 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 group G0 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].

0 Any element h of G is a product c1 . . . cn of commutators, and this gives

−1 −1 −1 −1 −1 ghg = g(c1 . . . cn)g = gc1g gc2g . . . gcng = d1 . . . dk, where d = gc g−1. But d is a commutator, hence ghg−1 G0. ¤ i i i ∈

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.

Theorem 3.20 G is solvable if and only if Gn = e for some n. { } 22 3 SOLVABILITY OF ALGEBRAIC EQUATIONS

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 = H H . . . H = G, { } 0 ⊆ 1 ⊆ ⊆ n where Hi is a subgroup of G such that Hi / Hi+1 and Hi+1/Hi is abelian. 1 0 0 The group G/H − is abelian, so G = G H − because G is the smallest normal n 1 ⊆ n 1 subgroup of G whose factor group is abelian . Similarly, since the factor group Hn−1/Hn−2 0 2 1 0 0 is abelian, H H − , and thus G = (G ) H H − . Continuing this way, it n−1 ⊆ n 2 ⊆ n−1 ⊆ n 2 follows that Gn H = e . ¤ ⊆ 0 { } 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 = H 0 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 = F F = F (w ) . . . F = F − (w ), 0 ⊆ 1 1 ⊆ ⊆ k k 1 k ni where w F − for i = 1, . . . , k. We may assume that F contains all nth roots of unity i ∈ i 1 (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(F /F ) Gal(F /F ) . . . Gal(F /F ) = e . k ⊇ k 1 ⊇ ⊇ k k { }

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(F /F ) is solvable. Since E F and the extension is Galois, Gal(F /E) / Gal(F /F ) k ⊆ k k k 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 = H H . . . H = Gal(E/F ), { } 0 ⊆ 1 ⊆ ⊆ r 3.4 Solvability by radicals 23 such that H / H and H /H is abelian. Let K = (H ). Then K will be Galois i+1 i i i+1 i F i i+1 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 L . . . L . ⊆ 0 ⊆ ⊆ r One can show that this is a radical extension; see Morandi [16] for the details. ¤

Lemma 3.23 If n 5, then S is not a solvable group. ≥ n Proof The group A is simple and non-abelian for n 5, so the derived group of A n ≥ n equals A i.e. A0 = A . This means that Gn = e in Theorem 3.20, because, by Theorem n n n 6 { } 3.19, the only simple groups that are solvable are abelian. Thus An is not solvable. The al- ternating group A is a subgroup of S , so S is not solvable for n 5 by Corollary 3.21. ¤ n n n ≥

Example 3.24 If 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 group. 24 3 SOLVABILITY OF ALGEBRAIC EQUATIONS 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 = √n a , which we can write as x = eln a/n. In analysis, the function ln x is defined by a transcendental integral: x dt ln x = . √ 2 Z1 t Similarly, x dt arcsin x = , 2 0 √1 t Z − 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 4.1 A function f in C is said to be meromorphic if it is analytic except for isolated poles. Definition 4.2 If 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 4.3 A sequence of periods ωn n∈Z+ of a meromorphic function f, which is as- sumed to be non-constant, cannot have{ zer} o 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 ω Z+ of periods of f, then the limit point ω is a { n}n∈ period. In fact, by continuity,

f(z + ω) = lim f(z + ωn) = f(z), n→∞ so ω is a period of f. From the lemma we get the next theorem. 26 4 ELLIPTIC FUNCTIONS

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

Proof If there were a finite accumulation point, then by the Cauchy criterion, ω ω n − m tend to 0, as m, n . But ω ω is a period of f. By Lemma 4.3 this is impossible. ¤ → ∞ n − m

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 ω nω = tω. But tω is a period of f which is closer to zero than ω. This is impossible 1 − 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 ) = 0. ω2 6 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: ω = r ω + r ω , r , r R. 3 1 1 2 2 1 2 ∈

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

ω n ω n ω = t ω + t ω . 3 − 1 1 − 2 2 1 1 2 2

But ω n ω n ω is a period and it lies inside the parallelogram, a contradiction. ¤ 3 − 1 1 − 2 2

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. 4.1 General theory 27

  £  £  £ £  £ £ £ £ £ £ £ £ : £ £ £ £  £ £ £ £ £ £ Ω  £ :£ £ £ £ £  £ £ £ £c £ £ £ £ £ £  £ £  A lattice induced by the point c and two periods

Figure 2

Definition 4.6 A 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 4.7 A 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 4.8 The 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 4.9 The sum of the residues of an elliptic function f inside a period-parallelogram is zero. Proof Integrating along the boundary ∂Ω of Ω, we have

c+ω1 c+ω1+ω2 c+ω2 c f(z) dz = f(z) dz + f(z) dz + f(z) dz + f(z) dz. Z∂Ω Zc Zc+ω1 Zc+ω1+ω2 Zc+ω2 Make the substitution w = z + ω2 in the third integral. Then

c+ω2 c f(z) dz = f(w + ω2) dw. Zc+ω1+ω2 Zc+ω1 But f(w + ω2) = f(w), and thus the first and third integral cancel. Similarly the second and fourth cancel. Hence f(z) dz = 0. Z∂Ω By the residue theorem, n f(z) dz = 2πi ri, ∂Ω Z Xi=1 where ri are the residues of f, so the sum of residues is zero. ¤ It follows from the above theorem that there cannot be an elliptic function of order one. 28 4 ELLIPTIC FUNCTIONS

Theorem 4.10 The number of zeros a1, . . . , ap of an elliptic function f, where a zero is counted according to its multiplicity, is equal to the order N of the function.

Proof We recall from complex analysis that

p q 1 f 0(z) dz = ord(ai) ord(pj), 2πi ∂Ω f(z) − Z Xi=1 Xj=1 where ord( ) is the order of a zero a or of a pole p of f. The function f 0/f is elliptic, so · i j the integral is zero, i.e.,

p q ord(ai) = ord(pj) = N. ¤ Xi=1 Xj=1

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 4.11 For 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 q ord(a ) ord(q ) = mω + nω i − j 1 2 Xi=1 Xj=1 for some integers m and n.

Proof We again have from complex analysis that

p q 1 f 0(z) ord(ai) ord(qj) = z dz. − 2πi ∂Ω f(z) Xi=1 Xj=1 Z If we make the same substitution as in the proof of Theorem 4.9, we get

f 0(z) c+ω1 f 0(z) f 0(z + ω ) z dz = z (z + ω ) 2 dz f(z) f(z) − 2 f(z + ω ) ZΩ Zc 2 c+ω2 f 0(z) f 0(z + ω ) z (z + ω ) 1 dz − f(z) − 1 f(z + ω ) Zc 1 f(c + ω ) f(c + ω ) = ω log 2 ω log 1 1 f(c) − 2 f(c) = ω log 1 ω log 1 1 − 2 = ω 2πin ω 2πin , 1 1 − 2 2 where n1 and n2 are integers. ¤ 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 ω Im( 1 ) > 0, ω2 otherwise, replace ω1 by ω2 and ω2 by ω1. Define a lattice Λ by

Λ = kω + lω ; k, l Z , { 1 2 ∈ } and let Λ0 = Λ 0 . − { } Theorem 4.13 The sum 1 p 0 ω ωX∈Λ | | converges for p > 2.

Proof For k = 1, 2, . . . we define P to be the parallelogram with corners kω and k § 1 kω , where k is an integer. Such a parallelogram has 8k ω-points on its boundary. Let δ § 2 be the shortest distance from the center of P1 to the boundary. We then have

∞ ∞ 1 8k 8 1 p < p = p p−1 . 0 ω (kδ) δ k ωX∈Λ | | Xk=1 Xk=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 1 G(z) = 2 3 , − 0 (z ω) ωX∈Λ − 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 1 1 G(z + ω1) = 2 3 = 2 3 = G(z), − 0 (z + ω1 ω) − 0 (z ω) ωX∈Λ − ωX∈Λ − 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, 1 1 G( z) = 2 3 = 2 3 = G(z). − 0 (z + ω) 0 (z ω) − ωX∈Λ ωX∈Λ − Hence G is odd. We now introduce the Weierstrass function ℘ as 1 1 1 ℘(z) = 2 + 2 2 . z 0 (z ω) − ω ωX∈Λ − 30 4 ELLIPTIC FUNCTIONS

Observe that ℘0(z) = G(z). It follows that ℘ is an even function. From the above we have 0 0 1 that ℘ (z) = ℘ (z + ω1) for all z. Notice that the number 2 ω is not a pole of ℘. Integrating the periodicity identity for ℘0, we obtain

℘(z) = ℘(z + ω1) + c.

In particular, ℘( 1 ω ) = ℘( 1 ω ) + c. The function ℘ is even, hence c = 0. The Weierstrass − 2 1 2 1 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 g ℘(z) g , (10) − 2 − 3 where 1 1 and g2 = 60 4 g3 = 140 6 . 0 ω 0 ω ωX∈Λ ωX∈Λ The numbers g2 and g3 are called invariants. Proof Make a Laurent expansion of ℘ at z = 0. Since odd terms in the sum cancel, we obtain

1 2 1 4 1 1 g2 2 g3 4 ℘(z) = 2 + 3z 4 + 5z 6 + . . . = 2 + z + z + . . . . z 0 ω 0 ω z 20 28 ωX∈Λ ωX∈Λ This gives 2 g g ℘0(z) = + 2 z + 3 z3 + . . . , −z3 10 7 4 g g ℘0(z)2 = 1 2 z4 3 z6 + . . . , z6 − 10 − 7 1 ³ 3g 3g ´ ℘(z)3 = 1 + 2 z4 + 3 z6 + . . . , z6 20 28 ³ ´ and hence that

℘0(z)2 4℘(z)3 g ℘(z) = g + Az2 + Bz4 + . . . , − − 2 − 3 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 g ℘0 − 2 After division with 2℘0, we get ℘00 = 6℘2 1 g . (11) − 2 2 The right hand side of the differential equation (10) is a third degree polynomial in ℘(z), so it can be factorized as

4℘(z)3 + g ℘(z) g = 4(℘(z) z )(℘(z) z )(℘(z) z ). 2 − 3 − 1 − 2 − 3 4.2 The Weierstrass ℘ function 31

0 To determine z1, z2, and z3, we have to find the zeros of ℘ (z) in a period parallelogram. 1 0 0 Putting z = 2 ω1 in the identity ℘ (z) = ℘ (z + ω1) and remembering that the derivative is 0 −1 1 1 odd gives ℘ ( 2 ω1) = 0. In a similar manner one can show that 2 ω2 and 2 (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

1 1 ℘( 2 ω1) = ℘( 2 (ω1 + ω2)), then the second order elliptic function ℘(z) ℘( 1 ω ) would have two second order zeros − 2 1 counted according to multiplicity, which is impossible. Therefore we have

1 1 1 z1 = ℘( 2 ω1), z2 = ℘( 2 ω2), and z3 = ℘( 2 (ω1 + ω2)). Definition 4.15 The discriminant, D, of a general nth degree polynomial

n n−1 P (x) = cnx + cn−1x + . . . + c1x + c0 is defined to be D = c2n−2 (x x )2, n i − j ≤ ≤ 1 jY

If x1, x2, and x3 are the roots of a general third degree equation, then the discriminant is D = c4(x x )2(x x )2(x x )2. It is obvious that if the discriminant = 0, then the 3 1 − 2 2 − 3 3 − 1 6 equation has three distinct roots. If we take our third degree equation, then we have c = 4, c = 0, 3c = g , and 3 2 1 − 2 c = g . Using the relationship between roots and coefficients we see that the discriminant 0 − 3 for this equation is g2 g3/27. From this fact it follows that if g3 27g2 = 0, g and g 3 − 2 2 − 3 6 2 3 determine a Weierstrass function such that the right hand side of (10) has three distinct zeroes. Example 4.16 Consider 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 g x3 + (2 g )x2 1 = 0 − − 2 − 3 − can be written as x2(4x3 g x g ) x4 + 2x2 1 = 0. − 2 − 3 − − 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

℘(z )2℘0(z )2 (℘(z )2 1)2 = ℘(z )2℘0(z )2 ℘(z )2℘0(z )2 = 0. j j − j − j j − j j 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. 32 4 ELLIPTIC FUNCTIONS

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

1 z 1 σ(z) z 1 ζ(z) = ℘(t) dt and ln = ζ(t) dt. z − − t2 z − t Z0 Z0 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

g (1, τ)3 J(τ) = 2 , g (1, τ)3 27g (1, τ)2 2 − 3 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 4.17 An integral dx , P (x) Z where P (x) is polynomial of degree 4, ispcalled elliptic. ≤ If we have an integral ∞ dt u = , (12) 3 x √4t at b Z − − where a and b are constants, then it is, by definition, an elliptic integral. If the constants a and b satisfy a3 27b2 = 0, we can take them as invariants and construct a Weierstrass − 6 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

v u = dv Z0 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 27g2 = 0, then it is possible to construct a Weierstrass function, see Akhiezer [1] for 2 − 3 6 the details. In this section, we choose g2 = 0 and g3 = c = 0. To get g2 = 0, choose for 2πi/3 6 example ω2/ω1 = e . In Chapter 2, we claimed that the general fifth degree equation can be transformed to the form

P (x) = x5 + cx + c = 0. (13) 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 P 0, we get 5 4 1 4 x + cx + c = (5x + c) 5 x + ( 5 cx + c). If we have a double zero, it follows that

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

We now define the elliptic function fc by

0 fc(z) = ℘(z)℘ (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.

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

Proof Since f (z ) = 0, we have ℘(z )℘0(z ) = i√c℘(z ) 2i√c. If we square both c j j j − j − sides, we can use the differential equation for ℘. This manipulation gives

℘(z )2(4℘(z )3 c) = c℘(z )2 4c℘(z ) 4c. j j − − j − j − The ℘2 term cancel, whence 5 ℘(zj) + c℘(zj) + c = 0, so ℘(zj) solves the equation as asserted. ¤

5 If we can show that f (z ) has five simple zeros for c = 5 , then we are very close to c j 6 − 44 solving (13).

Lemma 4.19 If c = 55/44, then the function f has five simple zeros. 6 − c Proof Assume on the contrary that c = 55/44 and that f has a double root, that is 6 − c 0 fc(zj) = ℘(zj)℘ (zj) + i√c℘(zj) + 2i√c = 0, 0 0 2 00 0 fc(zj) = ℘ (zj) + ℘(zj)℘ (zj) + i√c℘ (zj) = 0.

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

f 0(z ) = 10℘(z )3 + i√c℘0(z ) c = 0. (14) c j j j − From the first equation we have

0 i√c℘(zj) + 2i√c ℘ (zj) = . − ℘(zj) 34 4 ELLIPTIC FUNCTIONS

Substituting this expression into (14), we get ℘(z )4 = c/5, that is j − 0 4 P (℘(zj) = 5℘(zj) + c = 0, 5 where P is as above. But by Lemma 4.18, ℘(zj) is a root of x + cx + c = 0. By the discussion above c = 55/44, which is a contradiction to what we assumed. ¤ −

We must show that if z and z are distinct zeros of f , then ℘(z ) = ℘(z ). If we can 1 2 c 1 6 2 show this, then we have solved (13)! We need another three lemmas. Lemma 4.20 If z = z and ℘(z ) = ℘(z ), then ℘0(z ) = ℘0(z ). 1 6 2 1 2 1 − 2 Proof By Theorem 4.12 and the fact that the ℘ has only one pole of order two at zero, we have z + z = mω + nω , where m and n are integers. Thus z = mω + nω z . The 1 2 1 2 2 1 2 − 1 derivative ℘0 is periodic and odd. Thus ℘0(z ) = ℘0(mω + nω z ) = ℘0( z ) = ℘0(z ). ¤ 2 1 2 − 1 − 1 − 1 Lemma 4.21 If z , z are distinct zeros of f , then ℘(z ) = ℘(z ). 1 2 c 1 6 2 Proof Suppose that ℘(z ) = ℘(z ) with z = z . Then 1 2 1 6 2 0 fc(z1) = ℘(z1)℘ (z1) + i√c℘(z1) + 2i√c = 0.

By the previous lemma, we can write the equality fc(z2) = 0 as f (z ) = ℘(z )℘0(z ) + i√c℘(z ) + 2i√c = 0. c 2 − 1 1 1 Subtracting this from the first equality yields 0 2℘(z1)℘ (z1) = 0. But ℘(z ) = 0, because c = 0, whence ℘0(z ) = 0. This means that the function ℘(z) ℘(z ) 1 6 6 1 − 1 assumes the value 0 three times counted according to multiplicity, since it has a double zero at z and a simple zero at z . But the order of ℘(z) ℘(z ) is two, so we have obtained a 1 2 − 1 contradiction. ¤ 5 4 Lemma 4.22 If c = 5 /4 , 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 0 16 √ ℘(z1) = 5/4. Using the differential equation for ℘, we have ℘ (z1) = 15 5. Using the − 0 § numerical values of c, ℘(z1), and ℘ (z1), we get 0 fc(z1) = fc(z1) = 0 and f 00(z ) = 0. The other zeroes x , i = 1, 2, 3, to the quintic are simple, and there are c 1 6 i 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 4.23 The numbers ℘(zj), i = 1, . . . , 5, where fc(zj) = 0, solve the quintic equa- tion x5 + 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. 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 ∞ (m2τ+2mz)πi θ3(z, τ) = e , Im(τ) > 0. m=−∞ X 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 ∞ m2 2πimz θ3(z, τ) = q e . m=−∞ X 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 ∞ (m2τ+2mz)πi+2iπmτ θ3(z + τ, τ) = e . m=−∞ X After some algebraic manipulations, we can write ∞ −πτ−2πiz ((m+1)2τ+2(m+1)z)πi −πτ−2πiz θ3(z + τ, τ) = e e = e θ3(z, τ). m=−∞ X This shows that θ3 is an entire and a quasi-periodic function. There are three other theta-functions connected to θ3 : ∞ m (m− 1 )2 (2m−1)πiz θ (z, τ) = i ( 1) q 2 e , 1 − m=−∞ ∞X m (m− 1 )2 (2m−1)πiz θ (z, τ) = ( 1) q 2 e , 2 − m=−∞ X∞ 2 θ (z, τ) = ( 1)mqm e2mπiz. 4 − m=−∞ X The zeros of these functions are:

θ1 θ2 θ3 θ4 1 1 1 1 m + nτ m + nτ + 2 m + (n + 2 )τ + 2 m + (n + 2 )τ where m and n are integers.

Theorem 4.24 The function θ4 can be expanded in an infinite product as ∞ θ (z, τ) = c (1 q2m−1e2πiτ )(1 q2m−1e−2πiτ ). 4 − − m=1 Y 36 4 ELLIPTIC FUNCTIONS

Proof Consider the function ∞ f(x) = (1 k2m−1x)(1 k2m−1x−1), x C, − − ∈ m=1 Y were k is a constant with k < 1. It converges absolutely for x = 0, as can be seen by | | 6 comparison with the series

∞ log 1 k2m−1x 1 k2m−1x . | − || − | m=1 X 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) = a xj = a k2j+1xj+1, j − j −∞ −∞ j=X j=X 2 from which it follows that a = a ( 1)jkj , and hence that j 0 − ∞ 2 f(x) = a ( 1)jkj xj. 0 − −∞ j=X Then we have ∞ 2 f(e2πiz) = a ( 1)jkj e2πijz. 0 − −∞ j=X 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

∞ θ (z, τ) = 2cq1/4 sin(πz) (1 q2me2πiz)(1 q2me−2πiz), 1 − − m=1 Y∞ 1/4 2m 2πiz 2m −2πiz θ2(z, τ) = 2cq cos(πz) (1 + q e )(1 + q e ), m=1 ∞ Y 2m−1 2πiz 2m−1 −2πiz θ3(z, τ) = c (1 + q e )(1 + q e ), m=1 Y∞ θ (z, τ) = c (1 q2m−1e2πiz)(1 q2m−1e−2πiz), 4 − − m=1 Y where ∞ c = (1 q2m), − m=1 Y 4.5 History of elliptic functions 37 see Akhiezer [1] 2 Because of the factor qm , theta series converges rapidly. They are well suited for calculating elliptic functions such as Weierstrass ℘ function. In fact, we can use theta functions to construct Weierstrass function ℘. Briefly, we can proceed as follows. First choose the invariants g2 and g3. By the differential equation (10), they determine a function ℘(z, ω1, ω2). The right hand side of the differential equation is an ordinary third degree polynomial with three roots zi, i = 1, 2, 3. By using properties of theta functions, we can derive the equality z z θ (0, τ) 1 − 2 = 4 ; z z θ (0, τ) 1 − 3 3 see Wang and Guo [26]. This equation determines τ. By again considering the differential equation, we get the following expression

2 θ2(z, τ) θ3(0, τ)θ4(0, τ) ℘(z, ω1, ω2) = 2 + z1. θ1(z, τ)

For more details, see Wang and Guo [26]. These formulas will be used in Chapter 6, where we study an algorithm for solving the quintic.

4.5 History of elliptic functions Karl Friedrich Gauss (1777–1855) and Adrien-Marie Legendre (1772–1833) investigated elliptic integrals around 1800. Later Niels Henrik Abel (1802–1829) and Karl Jacobi (1804– 1851) observed that if one inverts elliptic integrals, then one gets interesting functions. The name “elliptic” integral has its origin in the fact that the arclength of an ellipse is given by an elliptic integral. Both Jacobi and Abel observed that these inverse functions are doubly periodic and that this was a really characteristic feature. Abel presented his investigation in 1827 in Crelle’s journal. In 1829 Jacobi presented the first book of elliptic functions, Fundamenta Nova Theoriae Functionum Ellipticaum. Gauss had 20 years earlier found similar results, but he did not publish them. Karl Weierstrass (1815–1897) studied the relationship between complex analysis and elliptic functions. Weierstrass also developed more general theory and invented his famous elliptic function ℘. We just mention that Leonhard Euler (1707-1783) and Jacobi Bernoulli (1654–1705) investigated the theta functions. The first mathematician to use elliptic functions in the theory of the quintic were Charles Hermite (1822–1901), Leopold Kronecker (1823–1891), Francesco Brioschi (1824–1897), and Paul Gordan (1837–1912). Hermite, Kronecker, and Brioschi used the same techniques and had similar ideas. Hermite was the first to publish these ideas. See Chapter 5 for more details about these different solution methods to the quintic. These methods for solving the quintic were examined by the German mathematician Felix Klein (1849-1925) in his famous book Vorlesungen ub¨ er das Icosahedron from 1884. In this book Klein investigated Hermite’s and Gordan’s methods and made their results more transparant. Klein examined the different ways for partitioning the icosahedron and how they can be used to solve the quintic. One can compare this book with Lagrange’s R´eflexions sur la r´esolution alg`ebrique des ´equations, mentioned in Chapter 3. Lagrange examined all previous attempts to solve the quintic, and at that time one was not even sure if it was at all possible to solve the equation. About a hundred years later, Klein examined 38 4 ELLIPTIC FUNCTIONS all the previous methods for solving the quintic. For more information about Klein’s book, see [12]. In Chapter 6, we study an algorithm due to Ludwig Kiepert (1846–1934). Kiepert discovered many formulas in the theory of elliptic functions. His algorithm for solving the quintic equation was presented in 1883. During the last century, a lot has happened both in Galois theory and in the theory of special functions. Group theory was developed and was given its modern form. More ways of solving the quintic using special functions were developed. The development of com- puters led to more and better algorithms for solving polynomial equations in an iterative way. Bruce King and E. Rodney Canfield at Georgia University implemented Kiepert’s al- gorithm on a microcomputer in the 1980’s. The commercial program package Mathematica implemented Hermite’s algorithm for solving the quintic in 1994. Kronecker believed that there would be a general theorem for expressing the roots of any polynomial with elliptic functions, but he did not find any theorem of this kind. But in 1983, Hiroshi Umemura proved a general formula for the nth degree equation. We just state his theorem here.

Theorem 4.25 Let

n n−1 P (x) = x + cn−1x + . . . + c1x + c0 = 0.

Then a root of the polynomial is given by the following expression.

1 0 . . . 0 1 1 . . . 0 θ 2 (Ω4) θ 2 2 (Ω4) 0 0 . . . 0 0 0 . . . 0 · µ ¶¸ · µ ¶¸ 0 0 . . . 0 0 1 . . . 0 + θ (Ω4) θ 2 (Ω4) 0 0 . . . 0 0 0 . . . 0 · µ ¶¸ · µ ¶¸ 0 0 . . . 0 0 1 . . . 0 θ (Ω4) θ 2 (Ω4) 1 0 . . . 0 1 0 . . . 0 · µ 2 ¶¸ · µ 2 ¶¸ − 1 0 . . . 0 1 1 . . . 0 2 θ 2 (Ω4) θ 2 2 (Ω4) 0 0 . . . 0 0 0 . . . 0 · µ ¶¸ · µ ¶¸ where Ω is the period matrix of the hyperelliptic curve y2 = F (x), and F (x) = x(x 1)P (x) − if the degree of P (x) is odd, and F (x) = x(x 1)(x 2)P (x), otherwise. − − For a proof, see Mumford [17]. These formulas are complicated and need a lot of memory in a computer, so they are not practically useful. Furthermore, it may be difficult to find the period matrix Ω. More useful algorithms for solving the quintic are examined in the next two chapters. 39

5 Hermite’s and Gordan’s solutions to the quintic

In this chapter we will briefly examine two different methods for solving the quintic. These methods are due to C. Hermite and P. Gordan.

5.1 Hermite’s solution to the quintic Hermite realized that to transform a quintic equation to the Bring–Jerrard form (6), one only has to solve cubics to make the substitution. There already existed a formula for solving a third degree equation

x3 3x + 2c = 0, c R, − 0 0 ∈ where c0 where expressed by trigonometric functions. By observing the similar form of this equation and the Bring–Jerrard quintic, he thought that a corresponding formula would exist for the quintic. We can summarize his technique as follows. Define the numbers k and k0 by z z z z k2 = 2 − 3 and (k0)2 = 1 k2 = 1 − 2 , z z − z z 1 − 2 1 − 3 where the numbers zi are the roots to (10) for a Weierstrass function with periods ω1 and 2 ω2. The number k can also be expressed by theta functions in the following way where we put τ = ω1/ω2: 4 2 θ2(0, τ) k = 4 ; θ3(0, τ) see Wang and Guo [26]. We define the following functions:

4 θ (0, τ) ϕ(τ) = √k = 2 , sθ3(0, τ)

4 θ (0, τ) ψ(τ) = √k0 = 4 . sθ3(0, τ) Observe that ϕ(τ)8 + ψ(τ)8 = 1, because k2 + (k0)2 = 1. We then define the parameters

u = ϕ(nτ) and v = ϕ(τ), where n is an integer. These parameters are linked by an equation of degree n + 1 known as the modular equation. The roots to this equation are

τ + 16m u = ²ϕ(nτ) and u = ϕ , m = 2, 3, . . . , n 1, 1 m n − µ ¶ where ² is 1 if 2 is a quadratic residue to n and ² = 1 otherwise. For the case n = 5, the − modular equation can be expressed in u and v as

v6 u6 + 5v2u2(v2 u2) + 4uv(1 v4u4) = 0. − − − 40 5 HERMITE’S AND GORDAN’S SOLUTIONS TO THE QUINTIC

To relate this sixth degree equation with the Bring–Jerrard form, Hermite defined the function 5 τ + 16 τ + 64 τ + 32 τ + 48 Φ(τ) = ϕ (5τ) + ϕ ϕ ϕ ϕ ϕ . τ 5 − 5 5 − 5 µ µ ¶¶ µ µ ¶ µ ¶¶ µ µ ¶ µ ¶¶ Because of the definition of ϕ, it is built up by theta functions. Therefore, if we, for example, replace τ with τ/5, ϕ will have a regular behaviour. Therefore,

5 τ + 16 τ + 32 τ + 48 Φ (5τ) , Φ , Φ , Φ , and Φ , τ 5 5 5 µ ¶ µ ¶ µ ¶ µ ¶ are the roots of the quintic

Φ5 200 ϕ(τ)4ψ(τ)16Φ 1600√5 ϕ(τ)3ψ(τ)16(1 + ϕ(τ)8) = 0. − − By making the substitution Φ = √4 2253ϕ(τ)ψ(τ)4x, we get the quintic

2(1 + ϕ(τ)8) x5 x = 0. − − √4 55ϕ(τ)2ψ(τ)4

Therefore, if we have transformed a general quintic to the form

x5 x a = 0, − − 0 we have to determine τ so that 2(1 + ϕ(τ)8) a0 = . √4 55ϕ(τ)2ψ(τ)4

This can be done by taking k as one root of the roots to

k4 + z2k3 + 2k2 z2k + 1 = 0, − 4 5 where z = a0 √5 /2. When we have found this value for k, we can calculate the value of τ from the definition of ϕ. The roots of the Bring–Jerrard quintic can then be expressed as

ϕ(τ + 16k) xk = , k = 0, 1, . . . , 4. √4 2453ϕ(τ)2ψ(τ)4

By using the series expansions of the functions involved above, one finds that this way of calculating the roots gives series that converge very rapidly. For more details and a derivation of the last expression, see King [11] and Prasolov and Solovyev [20].

5.2 Gordan’s solutions to the quintic Gordan used a different technique which did not depend on the difficult Tschirnhaus trans- formation. Gordan considered a regular icosahedron inscribed in the Riemann sphere, with its nodes on the sphere. An icosahedron has 20 triangular faces, 12 vertices, and 30 edges (5 edges meeting at each vertex); see Figure 3. It is also one of the five platonic solids. If we take a solid and join the centres of each pair of adjacent faces with an edge, we get a new solid denoted the dual. If we do this for a cube we get an octahedron inscribed in the 5.2 Gordan’s solutions to the quintic 41 cube. Doing the same thing with an octahedron gives a cube inscribed in it. Any symmetry of a solid is clearly a symmetry of the dual. The dual of the icosahedron is the dodecahe- dron which has 12 pentagonal faces, 30 edges, and 20 vertices (three edges meeting at each vertex). Using distance preserving maps, which we dealt with in Chapter 3, shows that the symmetry group of the icosahedron has order 60. The order of A5 is also 60, and one can show that they are isomorphic; see Shurman [22]. Therefore we can view the icosahedron as concrete picture of the group A5. In Chapter 3, we saw that the unsolvability of the quintic was connected to the group A5. Because of this fact, when we study the icosahedral group, we are also studying the quintic. We now consider polyhedral polynomials. They are polynomials with roots that lie on the vertices, • the midpoints of the edges, • the midpoints of the faces • of a polyhedron inscribed in the Riemann sphere.

¨H ¨ @H ¨¨ @ H ¨ XX @ H ¨ T XXHH ¨¨ @ HXXH C@C  T @ ££ @ @. £T  T C CC £ T  C ££ £ T T C £ T  C £ CC T  T££ C £ T  C £ T  C X£ C HXXX T  ¨ H XX  ¨ H XXT.  ¨ HH ¨¨ H ¨ HH¨¨ An icosahedron

Figure 3

Often one expresses equations involving polyhedral polynomials in homogeneous form, that is

n n−1 n n−1 n−1 n x + c − x + . . . + c x + c = 0 u + c − u v + . . . + c uv + c v = 0. n 1 1 0 ⇔ n 1 1 0 This corresponds to the substitution x = u/v. A homogeneous polynomial is often de- noted a form, and the degree is the order of the form. The polyhedral polynomials for the icosahedron are Vertices f = uv(u10 + 11u5v5 v10), − Edges T = u30 + 522u25v5 10005u20v10 10005u10v20 522u5v25 + v30, − − − Faces H = u20 + 228u15v5 494u10v10 228u5v15 v20; − − − − see [11] for a derivation. Observe that the order of each form is equal to the number of faces, edges, and vertices, respectively. 42 5 HERMITE’S AND GORDAN’S SOLUTIONS TO THE QUINTIC

An important relation between these polyhedral polynomials is the equality

1728f 5 H3 T 2 = 0. (15) − − Gordan expressed the different permutations of the icosahedron with the help of a complex 2π 2π fifth root, ω = cos 5 + i sin 5 , and the variables u and v. He then expressed the same rotations of the icosahedron with ω2 and variables denoted w and z. This gives another set of permutations of the icosahedron. Gordan then constructed following function:

ξ = u3w2z + u2vz3 + uv2w3 v3wz2, − which is unchanged by the permutations of the icosahedron. From this function Gordan found two other functions, φ and ψ, with the help of techniques from invariant theory. He then could derive a principal quintic in these three functions:

x5 + 5ξx2 5φx ψ = 0. − − The roots to this equations can now be expressed in the variables u, v, w, and z. Using these roots he was able to derive another equation: H y5 10y3 + 45y = 0, − − f 5 where H and f are parameters. By making the substitutionp y = t√f one gets a Brioschi quintic (see Chapter 6): t5 10ωt3 + 45ω2t ω2 = 0, − − where ω is a parameter. It is possible to solve this equation with the help of elliptic functions and by a partition of the icosahedron. This is done in Kiepert’s algorithm which we study in Chapter 6. For more details, see King [11]. 43

6 Kiepert’s algorithm for the quintic

In this chapter we will study L. Kiepert’s algorithm for solving the quintic equation. It gives good insight in how to use elliptic functions and Tschirnhaus transformation in practice.

6.1 An algorithm for solving the quintic In Chapter 4 we saw that we could use the Weierstrass function for solving the quintic. There are problems with using this technique: we need a difficult Tschirnhaus transformation and we need to find the location of the five points that solves the equation in the period parallelogram. Kiepert used another method. Briefly there are five steps:

Transform the general quintic into a principal quintic. • Transform the principal quintic to a one parameter Brioschi quintic. • Transform the Brioschi quintic into a Jacobi sextic. • Solve the Jacobi sextic with Weierstrass elliptic functions. • Reverse the transformations. • The different types of equations will be explained below.

The principal quintic: Consider an arbitrary fifth degree equation

5 4 3 2 x + c4x + c3x + c2x + c1x + c0 = 0. (16)

First we transform (16) into a principal quintic by the transformation

z = x2 ux + v. (17) − We then get 5 2 z + 5a2z + 5a1z + a0 = 0. (18) The formulas for the transformation are

5a = c (u3 + c u2 + c u + c ) + c (4u2 + 3c u + 2c ) c (5u + 2c ) 10v3, 2 − 2 4 3 2 1 4 3 − 0 4 − 5a = c (u4 + c u3 + c u2 + c u + c ) c (5u3 + 4c u2 + 3c u + 2c ) 5v4 10a v, 1 1 4 3 2 1 − 0 4 3 2 − − 2 a = c (u5 + c u4 + c u3 + c u2 + c u + c ) v5 5a v2 5a v, 0 − 0 4 3 2 1 0 − − 2 − 1 where u is a root of the second order equation

(2c2 5c )u2 + (4c3 13c c + 15c )u + 2c4 8c2c + 10c c + 3c2 10c = 0, 4 − 3 4 − 4 3 2 4 − 4 3 4 2 3 − 1 and v is defined by c u c2 + 2c v = − 4 − 4 3 . 5

The Brioschi quintic: A Brioschi equation is an equation of the form

x5 10w x3 + 45w2x w2 = 0, − 1 1 − 2 44 6 KIEPERT’S ALGORITHM FOR THE QUINTIC

where w1 and w2 are parameters. We now transform the equation (18) to the Brioschi form

y5 10wy3 + 45w2y w2 = 0, (19) − − where w is a parameter. We obtain this by putting λ + µy z = k , k = 1, 2, . . . , 5. (20) k y2/w 3 k − The parameters are obtained from

(a2 + a a a a3)λ2 + ( 11a3a + a a2 2a2a )λ + 64a2a2 27a3a a a2 = 0, 2 2 1 0 − 1 − 2 1 2 0 − 1 0 2 1 − 2 0 − 1 0 and (a λ2 3a λ 3a )3 V = 2 − 1 − 0 , a2(λa a λa2 a a ) 2 2 0 − 1 − 1 0 1 w = , 1728 V 2 − 3 2 V a2 8λ a2 72λ a1 72λa0 µ = − 2 − − . λ a2 + λa1 + a0 In the equation for λ it suffices to solve the equation and take one of the two roots. To understand and find the transformation (20) one studies the polyhedral polynomials mentioned in Section 4.5. Briefly we do the following. We use the midpoints of the edges of the icosahedron to divide it into five octahedrons. We then find the vertex and face polynomials for these octahedrons. The expressions for them are

t = ²3ku6 + 2²2ku5v 5²ku4v2 5²4ku2v4 2²3kuv5 + ²2kv6, k − − − W = ²4ku8 + ²3ku7v 7²2ku6v2 7²ku5v3 + 7²4ku3v5 7²3ku2v6 ²2kuv7 ²kv8, k − − − − − − 2πi/5 where ² = e ; see King [11]. The vertex polynomials tk satisfy the Brioschi equation

t5 10ft3 + 45f 2t T = 0, (21) − − where f and T are the vertex and edge polynomials for the icosahedron. By using this fact and making the substitution λf µf 3 z = W + t W , k H k HT k k µ ¶ µ ¶ where f and T are as above and H is the face polynomial for the icosahedron, we can obtain the coefficients of the principal quintic (18). By reverting these steps we get the transformation (20). For details, see King [11].

The Jacobi equation: Consider the equation

y6 10fy3 + Hy + 5f 2 = 0, (22) − where f and H are parameters. It is denoted the Jacobi sextic equation. One can show that if the parameters satisfy

1728f 5 H3 T 2 = 0, (23) − − 6.1 An algorithm for solving the quintic 45

(compare with (15)) and the roots to the Jacobi sextic are denoted y∞ and yi, i = 0, 1, . . . , 4, then the roots to (21) (observe that f, T , and H are parameters now) can be expressed as

1 tk = (y∞ yk)(yk+2 yk+3)(yk+4 yk+1), k = 0, 1, . . . , 4, (24) s√5 − − − where the subscript are calculated modulo five. For a proof of this result, see Perron [19]. By noticing that (19) is a special case of (21), where f 2 = T and t = yT/f 2, we only have to solve (22) to get the solution to the quintic.

Solving the Jacobi sextic: We shall solve (22) using elliptic functions. We begin by expressing the coefficients of (21) in terms of the elliptic invariants g2 and g3. Put D = g3 27g2. Write the equation (21) in the form 2 − 3 10 45 216 t5 + t3 + t g = 0. (25) D D2 − D3 3 We then get the relationships 1 216g f = and T = 3 . D D3 Substituting this into the identity (23), we get

HD2 3 D = 27g2. − 12 − 3 µ ¶ From this we can calculate H as D 27g2 12 H3 = 123 − 3 H = g . D6 ⇒ − D 2 µ ¶ The Jacobi equation corresponding to (25) is then

10 12 5 y6 + y3 g y + = 0. (26) D − D2 2 D2 By comparing (25) with (19), we see that

1 D = . −w In order to solve (26), the key equation is

℘0 ℘00 ℘(3) ℘(4) σ(5z) 1 ℘00 ℘(3) ℘(4) ℘(5) = ¯ ¯ = 0, (27) σ25(z) 2882 ¯ ℘(3) ℘(4) ℘(5) ℘(6) ¯ ¯ ¯ ¯ ℘(4) ℘(5) ℘(6) ℘(7) ¯ ¯ ¯ ¯ ¯ ¯ ¯ where σ is one of the Weierstrass functions¯ mentioned in Chapter¯ 4. This equation was derived by Kiepert; see [10]. The values of z that are roots to this equation are zkl = (kω1 + lω2)/5, where ω1 and ω2 are the periods that corresponds to the invariants g2 and 46 6 KIEPERT’S ALGORITHM FOR THE QUINTIC g constructed above, and k and l are integers satisfying k + l = 0. There are 24 different 3 | | | | 6 pairs k, l that gives different roots to this equation inside a period parallelogram of ℘. Now expand the determinant (27) using the fact that the derivatives can be expressed by ℘ and ℘0; see Section 4.2. We then arrive at the equation

((℘00)2 12℘(℘0)2)3 16(℘0)4℘00((℘00)2 12℘(℘0)2) 64(℘0)8 = 0, (28) − − − − which is of degree 12 in ℘. These 12 roots correspond to the 24 values of zkl above by ℘(z−k−l) = ℘(zkl) because ℘ is an even function. We shall solve the equation (26) using these roots. Put ℘ = ℘(z ) andχ = ℘ ℘ . We will use the following formula due kl kl kl 2k2l − kl to Weierstrass: ℘(z)0℘(z)(3) (℘(z)00)2 ℘(z) ℘(2z) = − . − 4(℘(z)0)2

Using this equality and the fact that ℘00 and ℘(3) can be expressed by ℘ and ℘0, we get the following formula for χkl: 00 2 0 2 (℘kl) 12℘kl(℘kl) χkl = − 0 2 . (29) 4(℘kl)

We will now derive properties valid for every pair χkl, ℘kl. Therefore we will drop the subscripts for χkl and ℘kl below. By substituting (29) into (28) we arrive at an equation for χ3 ℘00χ (℘0)2 = 0. (30) − − If we solve for (℘00)2 in (29) and use (30), we get

(℘00)2 = (4χ + 12℘)(χ3 ℘00χ). (31) − Using the differential equation for ℘0 and the identity (11) in (30) and in (31) gives two equations:

1 χ3 (6℘2 g )χ (4℘3 g ℘ g ) = 0, (32) − − 2 2 − − 2 − 3 1 1 (6℘2 g )2 (4χ + 12℘)(χ3 6℘2χ + g χ) = 0. (33) − 2 2 − − 2 2 Now introduce the variable ℘ = 1/2(z χ). Use this in the expressions above and solve for − z2. We then get 2 2 2 3z = 5χ + g2 + 2χ 5χ + 3g2. Write equation (32) as p z(z2 3χ2 g ) = 2g . − − 2 3 Square this and use the expression for z2 to get

5χ6 2χ5 5χ2 + 3 + g3 = 27g3. − 2 2 2 p Rearranging, squaring, and using D = g3 27g2, we have 2 − 3 5χ12 12g χ10 + 10Dχ6 + D2 = 0. − 2 6.1 An algorithm for solving the quintic 47

To obtain (26), we introduce the variable χ2 = 1/s and multiply by s6/D2. From this we can derive the following expression for the roots of (26):

2ω + 48kω 4ω + 96ω −1 √s = ℘ 1 2 ℘ 1 2 , (34) k 5 − 5 µ µ ¶ µ ¶¶ −1 2ω1 4ω1 √s∞ = ℘ ℘ ; (35) 5 − 5 µ µ ¶ µ ¶¶ see Kiepert [10]. The thing to do know is to evaluate ℘ at these points, which means that we must evaluate an infinite series. We do this with the help of theta functions. By using the identity z z θ (0, τ)4 1 − 2 = 4 z z θ (0, τ)4 1 − 3 3 (see Wang and Guo [26]), where zi is a root to the left hand side in the differential equation (10) for ℘, and using the same differential equation and the fact that theta function can be expanded as infinite products, one can derive the following formulas:

∞ 1 m j(6m+1)2 (6m+1)2/60 √sj = ( 1) ² q , j = 0, 1, 2, 3, 4, B − m=−∞ X∞ √5 m 5(6m+1)2/12 √s∞ = ( 1) q , B − m=−∞ X and ∞ 2 B = √6 D ( 1)mq(6m+1) /12, − m=−∞ X where ² = e2πi/5 and q = eπiω1/ω2 . To find these expressions, one uses properties of the σ functions. For a complete proof, see Akhiezer [1] and King [11]. We have now an algorithm for solving the general quintic using only its coefficients. But one thing remains: we must find the periods of the elliptic functions since the formula contains the factor q = eπiω1/ω2 . In fact, it is sufficient to determine q. To do this we use the elliptic integral (12) in the following way. Solve the equation

4x3 g x g = 0. − 2 − 3

Denote the roots by z1, z2, and z3. Then calculate z z z z k2 = 2 − 3 and (k0)2 = 1 k2 = 1 − 2 . (36) z z − z z 1 − 2 1 − 3 The number k2 can also be expressed by theta functions in the following way:

4 2 θ2(0, q) k = 4 ; θ3(0, q) see Wang and Guo [26]. Now form the expression

0 4 4 1 √k √z1 z3 √z1 z2 2L = − = 4 − − 4 − . 1 + √k0 √z1 z3 + √z1 z2 − − 48 6 KIEPERT’S ALGORITHM FOR THE QUINTIC

With the help of theta function this can be written

4 θ2(0, q ) 2L = 4 . θ3(0, q )

From this we have q + q9 + q25 + . . . L = , 1 + 2q4 + 2q16 + . . . whence q = L + 2q4L q9 + 2q16L q25 + . . . . − − By substituting L + 2q4L q9 + 2q16L q25 + . . . − − for the value of q in the right hand side, we have

q = L + 2L5 + 15L9 + 150L13 + 1707L17 + O(L21).

Even if L is close to 1, this expression gives a good numerical value for q. The expression in L for q is denoted the Jacobi nome. When we solve (10) we get three roots. It is important how we order the roots, because for half of the orderings we will get q > 1 and then | | the series for the roots does not converge. We must therefore make a check of this if we implement this method of solving the quintic on a computer. To be sure that we have done everything right, we can carry out the multiplication

4 (s s∞) (s s ), − − k kY=0 which should equal (26).

Reversing the transformations: We now only have to reverse the transformations that we have done. Use

2 1 y = (s∞ sk)(sk+2 sk+3)(sk+4 sk+1), k √5 − − − ω2 yk = 2 2 2 , yk + 10/Dyk + 45/D where the first equation comes from (24) and the last is a rewritten version of (21). To reverse the first transformation use the following equation derived by Kiepert

c + (z v)(u3 + c u2 + c u + c ) + (z v)2(2u + c ) x = 0 k − 4 3 2 k − 4 . k −u4 + c u3 + c u2 + c u + c + (z v)(3u2 + 2c u + c ) + (z v)2 4 3 2 1 k − 4 3 k − By writing the transformation (17) as

(x u)2 = (z v) u(x u), − − − − one can derive an equation for xk which gives the equation above. We now have a complete algorithm for solving the quintic. 6.2 Commentary about Kiepert’s algorithm 49

6.2 Commentary about Kiepert’s algorithm As we have mentioned, B. King and R. Canfield rediscovered Kiepert’s work. They suc- ceeded to implement it on a microcomputer in the beginning of the 1980’s. Since a lot has happened in computing since that time, King believes that on a modern computer and with a modern programming language, Kiepert’s algorithm would be fast and also give good numerical values for the roots. 50 6 KIEPERT’S ALGORITHM FOR THE QUINTIC 51

7 Conclusion

7.1 Conclusion We have seen that algebraic equations have interested mathematicians for a long time. A new era began when Galois proved the impossibility of solving equations of degree higher than four, using only radicals. After this attention concentrated on finding other ways of solving these equations. This was done in several ways, mainly using the theory of special functions, such as elliptic and theta functions. This lead to several algorithms for solving the quintic; these are presented in Chapter 5 and 6. These algorithms are not only interesting from a theoretical point of view, but are also practically useful. The program package Mathematica has already implemented a part of Hermite’s approach of solving the quintic, and Kiepert’s algorithm has also been implemented on a computer. But these solution methods only deal with the quintic. A goal for future research might be to find new algorithms, or to simplify Umemuras solution, so that one can solve higher degree equations in a similar way. With this in mind we can finally say that polynomial equations have been a part of mathematics for a long time, but there are still things to do; so they will certainly accom- pany us into the 21th century. 52 7 CONCLUSION 53

Appendix

A Groups

Definition A.1 A nonempty set G, on which a binary operation is defined, is called a ◦ group if, for arbitrary a, b, c G, ∈ (a b) c = a (b c) (associativity); • ◦ ◦ ◦ ◦ there exists an element e G such that e a = a e = a (existence of an identity • ∈ ◦ ◦ element);

for each a G there exists an element a−1 G such that a a−1 = a−1 a = e • ∈ ∈ ◦ ◦ (existence of inverse).

If a b = b a for all elements in G, then the group is called abelian . A subset H of G ◦ ◦ that is closed with respect to the binary operation is a subgroup of G. A subgroup H such that ghg−1 H for every g G and h H is a normal subgroup and this fact is denoted ∈ ∈ ∈ H /G. A group is simple if the only normal subgroups are e and the group itself. A group { } is cyclic if there exist a generator g so every element can be written in the form a = gn. The number of elements in a group is its order and is denoted G . | | Definition A.2 If (G, ) and (G , ) are groups, then a mapping σ : G G is called an ◦ 1 · → 1 homomorphism if σ(a b) = σ(a) σ(b) for every a, b G. ◦ · ∈

If σ also is a bijection, then the mapping is called an isomorphism, and if G = G1, then the mapping is called an automorphism. The kernel of a homomorphism is the set

ker(σ) = k G : σ(k) = e , { ∈ } and one can show that this is a normal subgroup of G.

Definition A.3 Let H be a subgroup of a group G and a an arbitrary element of G. Then the right coset is Ha = h a : h H and similary the left coset is aH = a h : h H . { ◦ ∈ } { ◦ ∈ } Two cosets aH and bH are equal if and only if ab−1 H. We have also that aH = bH or ∈ aH bH = , and the same is true for the left cosets. This means that the cosets will form ∩ ∅ a partition of G. If H / G, then the distinct cosets form a group called the factor group and is denoted G/H. By definition aH bH = a bH. If G is finite, then Ha = aH = H , ◦ ◦ | | | | | | and we have the following famous theorem due to Lagrange.

Theorem A.4 Let H be a subgroup of G, then H divides G . | | | | Next we state the isomorphism theorem.

Theorem A.5 Let σ : G G be a homomorphism and K = ker(σ). Then we have an → 1 isomorphism σ : G/K σ(G) given by σ(Kg) = σ(g) for all Kg G/K. → ∈ 54 C RINGS

B Permutation groups

The set of all permutations of the set 1, 2, 3, . . . , n is called the symmetric group and is { } deoted Sn. The order of Sn is n!. We write a permutation as

1 2 3 . . . n σ = i i i . . . i µ 1 2 3 n ¶ A permutation is called even or odd depending on the sign ( 1) of § σ(k) σ(i) − . k i i

C Rings

A nonempty set R is said to form a ring with respect to two binary operations, often called addition and multiplication and denoted by + and , if the following holds: ∗ R is an abelian group with respect to addition; • (a b) c = a (b c); • ∗ ∗ ∗ ∗ a (b + c) = a b + a c; • ∗ ∗ ∗ (b + c) a = b a + c a. • ∗ ∗ ∗ The identity element with respect to addition is denoted 0. A subset of R is called a subring if it is closed with respect to the operations.

Definition C.1 We define char(R) to be the smallest integer such that n a = 0 for all ∗ elements in R. If no such integer exists, then char(R) = 0.

Definition C.2 An element a = 0 is a divisor of zero if there exists an element b = 0 of R 6 6 such that a b = 0 or b a = 0. ∗ ∗ We list some common types of rings.

Commutative: A ring for which multiplication is commutative. • Ring with unity: A ring with a multiplicative identity element. This element is • often denoted with 1.

Integral domain: A commutative ring with unity having no divisors of zero. • Field: A ring whose nonzero elements form an abelian multiplicative group. • Definition C.3 A subgroup H of a ring R having the property that r x H for all x H ∗ ∈ ∈ and r R is called left ideal. Similary a subgroup H having the property x r H for all ∈ ∗ ∈ x H and r R is called a right ideal. If a subgroup is both a left and right ideal, then it ∈ ∈ is called an ideal. 55

Observe that the additative group is abelian, so the subgroups will be normal. If a subgroup g is an ideal, we can define factor groups R/g = r + g : r R as the set of all { ∈ } different cosets of g. This will in fact be a ring, denoted the factor ring.

Definition C.4 Assume that R and R are two rings, then a mapping σ : R R is a 1 → 1 homomorphism if it preserves addition and multiplication.

The kernel of σ is the set ker(σ) = k R : σ(k) = 0 . { ∈ } There are a similar isomorphism theorems as for groups.

Theorem C.5 Let σ : R R be a homomorphism and K = ker(σ). Then there exists an → 1 isomorphism σ : R/K σ(G) given by σ(K + g) = σ(g) for all K + g R/K. → ∈

D Polynomials

Let R be a ring and let x be any symbol not in R. By a polynomial in x over R is meant an expression n n j P (x) = cnx + . . . + c1x + c0 = cjx . Xj=0 j n Every cjx is called a term and the coefficients of x is denoted the leading coefficient. Two polynomials are equal if every term are equal. If the operation in R are denoted by and , ◦ ∗ then we define addition as

f(x) g(x) = (f g ) + (f g )x + . . . + (f g )xn, ◦ 0 ◦ 0 1 ◦ 1 n ◦ n and multiplication as

n k k f(x) g(x) = (f g ) + (f g f g )x + . . . = f g − x . ∗ 0 ∗ 0 0 ∗ 1 ◦ 1 ∗ 0 j ∗ k j Xk=0µXj=0 ¶ With this operation, the polynomials form a ring denoted R[x].

Theorem D.1 (Division Algorithm) Let f, g R[x], where R is a ring. Suppose that ∈ f = 0 and that the leading term in f is a unit. Then there exist uniquely determined 6 polynomials q and r such that

g = q f + r and r = 0 or deg(r) < deg(f). ∗ Definition D.2 Let f R[x] and a R. Then a is a root of f if f(a) = 0. ∈ ∈ From complex analysis we have the following theorem

Theorem D.3 (Fundamental Theorem of Algebra) If f is a nonconstant polynomial in C[x], then f has a root in C.

When a polynomial has a root, it is reducible in that ring. We end this section with a useful criterion for when an polynomial in Z is irreducible in Q. 56 D POLYNOMIALS

n n−1 Theorem D.4 (Eisenstein’s Criterion) Let P (x) = cnx + cn−1x + . . . + c1x + c0, be a polynomial in Z, i.e., every c Z. If there exists a prime p such that i ∈ p divides each c , i = 0, . . . , n 1, • i − p does not divide c , • n p2 does not divide c , • 0 then P is irreducible in Q[x]. REFERENCES 57

Commentary about the references

In the references I have put together the original articles by the inventors of the different subjects in this field. If one wants to find other articles by these famous authors, then [12] contains a list of them. As I studied this subject, I realized L. Kiepert had done a lot of work in this area. He wrote many papers about the quintic and elliptic functions, and, as we have seen, invented an algorithm for solving the quintic. Despite this, he did not seem to exist; no historical book had anything written about him. After hard effort I found a source about him, namely the article [4]. I can also mention that he has written a famous calculus book, and was a student of Weierstrass.

References

[1] Akhiezer, N., Elements of the Theory of Elliptic Functions, AMS, Rhode Island, 1990.

[2] Boyer, B. and Merzbach, C., A History of Mathematics, John Wiley & Sons, New York, 1989.

[3] Brioschi F., Sur les ´equations du sixi`eme degr´e, Acta math. 12 (1888), 83-101.

[4] Eddy, R. H., The Conics of Ludwig Kiepert – A Comprehensive Lesson in the Ge- ometry of the Triangle, Math. Mag. 67 (1994), 188-205.

[5] Gordan, P., Sur les ´equations du cinqui`eme degr´e, Journal de math. 1 (1885), 445–458.

[6] Gordan, P., Ub¨ er Gleichungen funften¨ grades, Math. Annalen 28 1 (1887), 152–166.

[7] Hausner, A., The Bring-Jerrard Equation and Weierstrass Elliptic functions, Amer. Math. Monthly 69 (1962), 193–196.

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Spr˚ak Rapporttyp ISBN Language Report category ISRN Svenska/Swedish Licentiatavhandling X Engelska/English X Examensarbete C-uppsats Serietitel och serienummerISSN D-uppsats Title of series, numbering 0348–2960 Ovrig¨ rapport LiTH–MAT–Ex–98–13

URL f¨orelektronisk version

Titel Analytic Solutions to Algebraic Equations Title

F¨orfattare Tomas Johansson Author

Sammanfattning Abstract This report studies polynomial equations and how one solves them using only the coefficients of the polynomial. It examines why it is impossible to solve equations of degree greater than four using only radicals and how instead one can solve them using elliptic functions. Although the quintic equation is the main area of our investigation, we also present parts of the history of algebraic equations, Galois theory, and elliptic functions.

Nyckelord Algebraic equation, Elliptic function, Galois theory, Polynomial, Quintic Keyword equation Theta function, Tschirnhaus transformation, Weierstrass func- tion