Cyclotomy: from Euler Through Vandermonde to Gauss Olaf Neumann

Leonhard Euler: Life, Work and Legacy 323 Robert E. Bradley and C. Edward Sandifer (Editors) © 2007 Elsevier B.V. All rights reserved Cyclotomy: From Euler through Vandermonde to Gauss Olaf Neumann Mathematisches Institut Friedrich-Schiller-UniversitätJena D-07737 Jena Germany The word “cyclotomy” is of Greek origin and means “division of the circle.” As a mathematical term it denotes the subdivision of a full circle line into a given number of equal parts. Consider the unit circle x2 + y2 = 1 in the Euclidean plane with Cartesian coordinates (x, y). If this circle is divided into n equal parts beginning with the point (1, 0) then the other 2π·k 2π·k division points will have coordinates (cos n , sin n ) where k runs from 1 to (n − 1). All those points form the edges of a regular n-sided√ polygon. It is well-known that by means of the imaginary quantity i := −1 one can prove the formula (cos α + i · sin α)n = cos nα + i · sin nα (1) which is usually called de Moivre’s formula. But in the form (1) it is due to Leonhard Euler (1707-1783), see [Euler 1748], cap. VIII. In particular, 2π·k the n arguments α = n with 0 ≤ k ≤ n − 1 provide us with the n powers 2 n−1 2π 2π 1, ζn, ζn, . , ζn of the complex number ζn := cos n + i · sin n : 2π · k 2π · k ζk = cos + i · sin (0 ≤ k ≤ n − 1) (2) n n n satisfying the equation xn − 1 = 0. (3) LOL-Ch16-P1 of 40 324 Olaf Neumann This means that Eqn. (3) has exactly n roots which are given in the tran- scendental form (2) and which are the powers of one of them, namely ζn. For these powers we shall adopt the name nth roots of unity common today i among mathematicians. If the exponent i is prime to n then ζn is called a i 2i 3i primitive nth root of unity since its powers 1, ζn, ζn , ζn ,... run through all nth roots of unity. 1 This way, the geometric problem of cyclotomy was entirely reduced to the algebraic problem of solving the equations xn − 1 = 0 in complex numbers. On the other hand, in the 18th century the dominating problem in the theory of equations was still the problem of solving equations “algebraically” or “by radicals” which meant by the extraction of roots. Euler also devoted√ some papers to this topic and he was very well aware that roots like n A are only determined up to factors which are nth roots of unity. These ambiguities led him to investigate the “binomial” equation xn −1 = 0, [Eu- ler 1751], §§ 38-48. It was Euler who succeeded in solving the equations xn − 1 = 0 for n ≤ 10 in terms of radicals with indices < n. In his treatise [Vandermonde 1774] Alexandre-Théophile Vandermonde (1735-1796) over- came the difficulties occuring in the case n = 11 in a truly pioneering way. Eventually, Carl Friedrich Gauss (1777-1855) was the brilliant architect of a fully-fledged cyclotomy theory which among other topics solved the equations (3) by radicals at least for prime number exponents n, [Gauss 1801], Sectio septima. In the present paper we intend to display the evolution of ideas from Euler to Gauss. Our first section is devoted entirely to Euler, in particular to his solution of x7 − 1 = 0 by square and cubic roots. The second section elucidates Vandermonde’sVandermonde innovations in the theory of equations which parallel the work of Joseph-Louis Lagrange (1736-1813) in many aspects. Vandermonde however far excelled Euler as well as Lagrange by his solution of x11 − 1 = 0. We shall expose a version of “pre-Gaussian” cyclotomy theory based on the ideas and tools of Euler, Vandermonde and Lagrange. 2 Our third section mainly discusses Gauss’s relationship to Van- dermonde. On the example of x17 − 1 = 0 we are going to make visible the difference between the cyclotomy theories of these two mathematicians. It is very remarkable that Vandermonde’s ideas also allow us to construct 1 The primitive nth roots of unity should not be confused with the primitive roots a 6≡ 0 (mod n) for prime numbers n which bear their name since the powers 1, a, a2, a3,... run through all residue classes 6≡ 0 (mod n), in other words, a mod n should be a generator of the prime residue class group mod n. 2 In the conceptual framework of our recent set-theoretic mathematics this version −1 amounts to studying the maximal totally real subfields Q(ζn + ζn ) of Q(ζn) first and −1 after that considering the fields Q(ζn) as quadratic extensions of Q(ζn + ζn ). LOL-Ch16-P2 of 40 Cyclotomy: From Euler through Vandermonde to Gauss 325 the regular 17-gon by ruler and compass. 3 Our considerations will be self- contained and entirely based on Vandermonde’s theory of quartic equations and straightforward calculations. This possibility seems to have gone unnoticed until now. This very much begs the question whether Gauss became acquainted with Vandermonde’sVandermonde Mémoire before or after publishing his Disquisitiones Arithmeticae and whether Vandermonde could have exerted some effect on Gauss. Our main thesis is that Gauss was not esssentially influenced by Vandermonde’s algebraic work in con- trast to conjectures formulated by Henri Lebesgue (1875-1941) and adopted or reproduced by other authors without further examination, see [Lebesgue 1940], [Jones 1991], [Waerden], p. 79. In the end some concluding remarks should show the reader how to link our considerations to Gauss’s theory of cyclotomy. Comments on mathematical facts in modern terms and notations will, as a rule, not be given in the text but in footnotes. 1. Euler Here we are going to explain in detail how Euler solved the equations xn − 1 = 0 by radicals for n ≤ 10. We skip the rather easy cases n ≤ 4 and pass at once to the equation x5 − 1 = 0. Division by x − 1 gives the new equation x4 + x3 + x2 + x + 1 = 0 which is mirror-symmetric with regard to the middle term x2. 1.1. Reciprocal Equations Equations of this type were already considered by Euler in his paper [Euler 1738], § 10 seq. There he is dealing with mirror-symmetric equations of the general form y2n + a · y2n−1 + b · y2n−2 + ... + p · yn + ... + b · y2 + a · y + 1 = 0, (4) especially for n = 2, 3, 4, 5, and is calling them reciprocal equations since 1 4 they do not change their form when y is replaced by y . Now Euler observes that the left-hand side of Eqn. (4) is a product of n quadratic factors y2 + α · y + 1, y2 + β · y + 1, y2 + γ · y + 1, y2 + δ · y + 1, etc. 3 It should be emphasized that our construction does not explicitly use a primitive root mod17 as a generator of all residue classes 6≡ 0 (mod 17), contrary to Gauss. 4 1 Aequationes huiusmodi, quae posito y loco y formam non mutant, voco reciprocas. [Euler 1738], § 11. LOL-Ch16-P3 of 40 326 Olaf Neumann where the coefficients α, β, γ, δ, etc. satisfy an equation of degree n the coefficients of which are nothing but linear combinations of the coefficients of Eqn. (4) with some explicitly known rational integers. The idea behind the factorization of Eqn. (4) can be exposed as follows. Divide Eqn. (4) by yn and write the resulting equation in the form (yn + y−n) + a · (yn−1 + y−n+1) + b · (yn−2 + y−n+2) + ... + p = 0. The kth power sum yk + y−k (0 ≤ k ≤ n) is symmetric in y and y−1, hence a polynomial in the elementary symmetric polynomials z := y + y−1 and y · y−1 = 1 according to the recursive relation 1 1 1 yk+1 + = yk + · z − yk−1 + yk+1 yk yk−1 (or by the so-called Girard-Newton formulas for power sums). Hence the auxiliary quantity z := y + y−1 satisfies an equation zn + a0 · zn−1 + b0 · zn−2 + ... + p0 = 0 (5) where the coefficients a0, b0, . , p0 are linear combinations of a, b, c, . , p with rational integer coefficients. Denote by −α, −β, −γ, −δ, etc. the roots of Eqn. (5), in other words zn + a0 · zn−1 + b0 · zn−2 + ... + p0 = (z + α) · (z + β) · (z + γ) · (z + δ) ··· . In view of z = y + y−1 it is obvious that every root of Eqn. (4) satisfies one of the quadratic equations y2 + α · y + 1 = 0, y2 + β · y + 1 = 0, y2 + γ · y + 1 = 0, y2 + δ · y + 1 = 0, etc. whence one gets the desired factorization of Eqn. (4). Below we shall en- counter the reciprocal equations once more in Vandermonde’s work. 1.2. Roots of Unity After these preparations it is rather easy to solve the reciprocal equation x4 + x3 + x2 + x + 1 = 0 by radicals. We put u := −(x + x−1) and obtain √ x2 +x−2 = u2 −2. In summary u2 −u−1 = 0 with the roots p := 1+ 5 and √ 2 1− 5 2 q := 2 . Then one has to solve the two quadratic equations x +p·x+1 = 0 and x2 + q · x + 1 = 0 but that does not cause any problems, [Euler 1751], § 40. It turns out that all 5th roots of unity are rational functions (with √ p √ rational coefficients) of 5 and the square root −10 + 2 5: 1 √ q √ x = −1 − δ 5 + δ −10 − δ 2 5 (6) 4 1 2 1 LOL-Ch16-P4 of 40 Cyclotomy: From Euler through Vandermonde to Gauss 327 5 with δ1, δ2 = ±1, [Euler 1751], § 40.

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