AAbel–Ruffini’s Theorem: Complex but Not Complicated !
A proof, using loops and roots, of the unsolvability of the quintic
Paul Ramond ∗ Laboratoire Univers et Théories, Observatoire de Paris, CNRS,Université PSL, Université de Paris, 92190 Meudon, France
In this article, using only elementary knowledge of complex numbers, we sketch a proof of the celebratedAbel–Ruffini theorem, which states that the general solution to an algebraic equation of degree five or morecannot be written using radicals, that is, using its coefficients and arithmetic operations + , − , × , ÷ , and √ . Thepresent article is written purposely with concise and pedagogical terms and dedicated to students and researchersnot familiar with Galois theory, or even group theory in general, which are the usual tools used to prove thisremarkable theorem. In particular, the proof is self-contained and gives some insight as to why formulae existfor equations of degree four or less (and how they are constructed), and why they do not for degree five or more. INTRODUCTION
Historical background.—
Finding a general expressionfor the solutions of an algebraic equation has been one ofthe oldest and most fruitful problems in mathematics. Thehistory behind what was once called the “theory of equations”[2], is almost as rich and old as the history of mathematicsitself. For example, methods for solving linear and quadraticequations have been known for at least four millennia [1],in independent places in the world. The quadratic formulataught today in school, with modern notation, was firstwritten down by R. Descartes in 1637 [9]. The introductionof the definitive √ notation (with the horizontal overbarcalled the “vinculum”) was only introduced in 1525 [7].Regarding cubic and quartic equations, they too had to waituntil the sixteenth century to be finally solved. By then, agroup of rivaling Italian mathematicians, including S. delFerro, N. Tartaglia, G. Cardano, and L. Ferrari, made theserendipitous discovery of complex numbers while solvingthe general cubic equation. In 1545, a few years beforetheir quarrels settled in a public mathematical contest [2],L. Ferrari solved the quartic equation by reducing it to a cubicone. The quintic equation, however, would still keep thesemathematicians (and all others) in check, while the idea of itbeing unsolvable slowly started to emerge. Unsolvable equations.—
The idea of examining permuta-tions of the solutions to study the (un-)solvability of algebraicequations dates back to the pioneering works of J.-L. Lagrangein 1771 [12]. Lagrange’s ideas matured, and were finallyextended to the quintic equation by P. Ruffini in early 1800[13]. For twenty years, Ruffini tried to convince the mathe-matical community of the importance of his results, withoutsuccess. It is only in 1821, with the help of L.-A. Cauchy,that Ruffini’s work was recognized as a stepping stone in thetheory of algebraic equations. Although it turned out thatRuffini did not prove the theorem that now bears his name perse , his results were strong enough to place serious doubt aboutthe possibility of finding a solution to the general quinticequation. The wait was finally over in 1824 when N.-H. Abel wrote the first complete proof of the theorem (a short proofpublished in 1824 [11] at his own expense, and a longer, moredetailed version two years later [10]). His work still remainedunworthy of interest to the eyes of most mathematicians,including Gauss and Cauchy themselves. Abel died aged 26 in1829, just before his work on the unsolvability of the quinticfinally received all the appreciation it deserved. He receivedposthumously the
Grand Prix de l’Académie des Sciences deParis in 1830, in recognition of his work. The same year alsomarks the publication of É. Galois’s first paper on these topics[14], in which he gives the premises of (now) Galois theory,a novel and elegant extension of all previous results. He toodied young (aged 20 in 1832) and his work also took severaldecades to be fully published and recognized as revolutionary.This short historical account lacks many interesting storiesabout these mathematicians, such as conflict of interests,encrypted communications, fatal duels, long lost and recov-ered memoirs, etc. The interested reader could start withJ. Sesiano’s [1] and J. Stillwell’s [2] books and referencestherein for well-written and thorough presentations of thesefascinating pieces of history.
Aim and content.—
The aim of this article is to sketch an ac-cessible and self-contained proof of the
Abel–Ruffini theorem : No formula exists for the solution to the generalequation of degree five or more, using onlythe operations + , − , × , ÷ , and √ . The word general is important: it emphasizes that a formulathat holds for any coefficients cannot be found. However,the theorem does not prevent some equations to have asolution that can be written in terms of + , − , × , ÷ , √ . In mosttextbooks, the proof of this remarkable theorem relies on apowerful subbranch of mathematics called Galois theory ,developed quasi-exclusively by the French mathematicianÉ. Galois at the beginning of the nineteenth century. Galoistheory solved all “unsolvability problems” once and for all, aswell as other millennia-long problems [2]. However, it is alsorather advanced, usually taught in the second/third years of a r X i v : . [ m a t h . HO ] N ov specialized, university-level mathematics. The first completeproof (by Abel) of the Abel–Ruffini theorem is a few yearsolder than the birth of Galois theory. Moreover, the worksof Galois took several decades to be broadly known to othermathematicians. In other words, neither Ruffini nor Abel usedthe methods developed by Galois to prove that some equationswere unsolvable. Because it usually relies on advancedmathematics, few people in the scientific community areaware of this theorem and its underlying principles. Butbecause Abel did not prove it this way, there must be another,perhaps simpler, way of understanding the reason why thegeneral quintic equation does not have a solution in termsof radicals. In particular, Galois’s, Abel’s, and Ruffini’sideas all rely on a unique, fundamental, common point: thesymmetry of an algebraic equation under the permutation ofits solutions . Based solely on this fundamental symmetry,we propose to sketch a proof of Abel–Ruffini’s theoremusing only elementary knowledge about complex numbers.Familiarity with complex numbers and a pen (and paper!) todraw appropriate figures are the only prerequisites to get agrasp of how the proof works. Everything else is elementarymathematics and useful notations that help present the ideasmore clearly. Motivation.—
The proof given here cannot be considerednew. It is the result of several adaptations and simplificationsof ideas that we feel compelled to attribute to the theoreticalphysicist B. Katz. His ideas are presented concisely inan online video [8], which can be used as complementarymaterial with dynamic illustrations. Katz’s inspiration formaking this video comes from a series of lectures given byphysicist and mathematician V. Arnold, which were nicelycrystallized in a problems-and-solutions book published byV. B. Alekseev, who was Arnold’s student at the time of theselectures. This book, although very well written and complete,is, however, not elementary in any sense. While Katz’s videodoes a very good job at explaining the general idea of theproof, we found that some gaps could be filled, and somearguments could be made much simpler, especially when weget to the end of the proof. Other references dealing with thepresent ideas are rather scarce in the literature (academic ornot). A nonexhaustive selection is located at the conclusionof the article, and can serve as complementary material todeepen one’s understanding of the proof.
Outline.—
The remainder of this article is organized as fol-lows. After some prerequisites and reminders regarding com-plex numbers are introduced, we spend some time on thequadratic equation, explaining why a quadratic formula can-not be built out of only the four basic arithmetic operations(our first impossibility result). Similar ideas are then extendedsuccessively to the cubic and quartic equation, giving strongerimpossibility results at each step. By the time we get to thequintic equation, the reader should be comfortable enoughwith the strategy (hopefully) to see how the quadratic, cubic,and quartic cases foreshadow the proof of the the Abel–Ruffini theorem. Along the way, we also derive the cubic and quarticformulae, scarcely presented in the nonspecialized literature.Although the derivation of these formulae is interesting enoughto justify their presence, they will be especially useful in lightof our temporary results, and will naturally guide us step bystep to Abel–Ruffini’s theorem. Finally, we note that animatedversions of Figures 2, 4, 5 and 6 are available as supplementarymaterial for a better understanding.
PREREQUISITES
In this article, we are dealing with algebraic equations ofdegree 𝑛 ≥
2. These equations are always of the form 𝑧 𝑛 + 𝑐 𝑛 − 𝑧 𝑛 − + · · · + 𝑐 𝑧 + 𝑐 = , (1)where 𝑧 ∈ C is the unknown and the 𝑛 complex numbers ( 𝑐 , ..., 𝑐 𝑛 − ) are the coefficients . It is a remarkable fact, oftencited as the fundamental theorem of algebra, that equation (1)always has exactly 𝑛 complex solutions. (We use solutions ,instead of roots of polynomials, to avoid confusion with the“root” operation √ later on.) These solutions will always bedenoted ( 𝑠 , . . . , 𝑠 𝑛 ) , and we use 𝑠 as a placeholder for any ofthe solutions. Permutations.—
Our strategy will be based on picturing thesolutions ( 𝑠 , . . . , 𝑠 𝑛 ) in the complex plane and make themmove around so as to exchange their positions, i.e., permute them. We will need two kinds of permutation: • transpositions , denoted ( 𝑖 𝑗 ) , exchanging the position of two solutions, i.e., 𝑠 𝑖 ↔ 𝑠 𝑗 . The transposition ( ) isdepicted on the left in Fig 1, • cycles , denoted ( 𝑖 𝑗 𝑘 ) , exchanging the position of three solutions cyclically, i.e., 𝑠 𝑖 → 𝑠 𝑗 , 𝑠 𝑗 → 𝑠 𝑘 , and 𝑠 𝑘 → 𝑠 𝑖 .The cycle ( ) is depicted on the right in Fig 1. Figure 1. The paths-induced transposition ( ) and cycle ( ) onthe solutions ( 𝑠 , 𝑠 , 𝑠 ) of some algebraic equation of degree 𝑛 ≥ See supplementary material for animated version.
Two permutations next to each other are to be performedsuccessively, from left to right . For example, ( )( ) consists in exchanging 𝑠 and 𝑠 , then 𝑠 with 𝑠 . Notice thatthe result is equivalent to the cycle ( ) , hence there is nounique way of writing permutations. However, permutations do not commute in general. Indeed, ( )( ) = ( ) and ( )( ) = ( ) ; therefore ( )( ) ≠ ( )( ) . Loops.—
One way of visualizing permutations of ( 𝑠 , . . . , 𝑠 𝑛 ) is to locate them in the complex plane and make them travelalong some paths . Paths in the complex plane are justcontinuous curves than connect two points (we assume thatthey do not self-intersect, otherwise things get unnecessarilycomplicated). A path that closes , i.e., connects a point toitself, is called a loop and denoted 𝛾 , whereas a path thatconnects two distinct points is simply called an unclosedpath , denoted 𝜔 . These paths will be represented by arrowsin our figures, and will be used to induce permutationson ( 𝑠 , . . . , 𝑠 𝑛 ) . For example, in Figure 1 are depicted thetransposition ( ) on the left, and the cycle ( ) on the right.Notice that to induce ( ) , 𝑠 follows a loop 𝛾 so that only 𝑠 and 𝑠 swap places by following the unclosed paths 𝜔 , 𝜔 .When speaking of permutations of solutions, we will always imagine them traveling on these paths. Roots.—
Now let us examine how roots of complex numbersmove around the complex plane. Fixing some complex number 𝑧 , a root of 𝑧 is some number 𝜁 ∈ C such that 𝜁 𝑘 = 𝑧 for some 𝑘 ∈ N . Such a 𝜁 is then called a 𝑘 th root of 𝑧 ; and 𝑧 admitsexactly 𝑘 such 𝑘 th roots (this follows from the fundamentaltheorem of algebra). We will deliberately use the ambiguousnotation 𝑘 √ as a multivariable notation, i.e., for a given 𝑘 , 𝑘 √ 𝑧 means any of the 𝑘 th roots of 𝑧 . Fixing 𝑘 ∈ N and assumingthat 𝑧 itself follows a loop 𝛾 , let us examine what kind of path 𝑘 √ 𝑧 follows. To this end, we use the exponential form of 𝑧 , i.e, 𝑧 = 𝑟 e i 𝜃 with 𝑟 = | 𝑧 | and 𝜃 = arg 𝑧 , from which we find that all 𝑘 th roots ( 𝜁 , . . . , 𝜁 𝑘 ) can be written explicitly as 𝜁 ℓ = 𝑟 / 𝑘 e i ( 𝜃 + ℓ π )/ 𝑘 , ℓ ∈ { , . . . , 𝑘 } . (2)From equation (2), one can already tell that all 𝑘 th roots of 𝑧 have the same modulus : 𝑟 / 𝑘 . Geometrically, this meansthat they lie on the same circle (of radius 𝑟 / 𝑘 ) in the complexplane. Moreover, we readily see from equation (2) thatarg 𝜁 ℓ = 𝜃𝑘 + ℓ π 𝑘 , (3)which means that all roots are equally spaced on this circle,at angle 2 π / 𝑘 apart. Now suppose that 𝑧 goes on a journeyexploring the complex plane, by traveling on a loop 𝛾 windingonce (say) around the origin, in the counterclockwise direction(in red, on the left in Figure 2). As 𝑧 travels along 𝛾 , its 𝑘 throots also move around, and their position can be trackedfrom equation (2) (see the red paths on the right in Figure2). Since 𝛾 is a loop, the radius 𝑟 comes back to its original(i.e., pre-loop) value, and so does 𝑟 / 𝑘 . In other words: theroots remain on their circle after the path 𝛾 (see the grey,dashed-circle on the right of Figure 2). However, arg 𝑧 wentfrom 𝜃 to 𝜃 + π (one counterclockwise turn). Therefore,from equation (3), each 𝑘 th root 𝜁 ℓ has moved to its closest,counterclockwise neighbor, 𝜁 ℓ + . In particular, the roots havefollowed an unclosed path. Had 𝑧 not wound around theorigin (in blue, on the left in Figure 2), its argument 𝜃 wouldhave seen no net change after the loop 𝛾 , and the roots would Figure 2. When 𝑧 follows a loop 𝛾 that does not wind around theorigin, its roots 𝜁 i follow loops as well (right). However, for the loop 𝛾 that winds once, the roots then follow the red, unclosed paths. Seesupplementary material for animated version. have followed their own loops (in blue on the right in Figure 2).We have seen two example of loops followed by 𝑧 and theresult is not the same for its roots 𝑘 √ 𝑧 : in one case the rootsfollow a loop (blue part of Figure 2), in the other they do not (red part of Figure 2). Consequently, we conclude that when 𝑧 follows a loop, 𝑘 √ 𝑧 does not always follow a loop . Thisconclusion holds for any type of root (i.e., any 𝑘 in 𝑘 √ 𝑧 ). Sincewe will not need to differentiate between all these roots, wewill denote by √ 𝑧 any root of 𝑧 (that is, any 𝑘 th root, whateverthe value 𝑘 ∈ N ). With this notation, the takeaway result ofthis paragraph is simply: When 𝑧 follows a loop, √ 𝑧 doesnot always follow a loop.Formula ingredients.— In this article we question the ex-istence of a general formula for the solutions of the generalalgebraic equation of degree 𝑛 , equation (1). By formula , wemean some equality 𝑠 = Φ ( 𝑐 , . . . , 𝑐 𝑛 − ) , (4)where 𝑠 is a solution of equation (1) and Φ is some functionof its coefficients ( 𝑐 , . . . , 𝑐 𝑛 − ) . The Abel–Ruffini theoremstates that for 𝑛 ≥
5, no formula in terms of radicals exists. “Interms of radicals” simply mean that the function Φ in equation(4) can be constructed solely in terms of • the coefficients ( 𝑐 , . . . , 𝑐 𝑛 − ) , • the elementary operations + , − , × , ÷ and √ .Leaving √ aside, if we constrain ourselves to a formula com-bining the coefficients ( 𝑐 , . . . , 𝑐 𝑛 ) and the four operations + , − , × , ÷ , we obtain what we will call an 𝐹 -formula, or simplyan 𝐹 -function . Examples of such 𝐹 -functions are 𝐹 = , 𝐹 = − 𝑐 , 𝐹 = 𝑐 − 𝑐 . (5)They are the elementary building blocks for constructing for-mulae. In particular, they encompass integers, the coefficientsthemselves, as well as polynomials and rational functions ofthe coefficients. Clearly, if two coefficients each follow a loopsimultaneously, then their sum, difference, product, and quo-tient also follow a loop. As they are built with only thesefour operations, 𝐹 -functions enjoy the same property. In otherwords: When ( 𝑐 , . . . , 𝑐 𝑛 − ) follow a loop, 𝐹 -functions also follow a loop. This property of 𝐹 -functions is not shared by √ 𝐹 -functions,i.e., expressions that are roots of 𝐹 -functions, e.g., √ 𝑐 or √ − 𝑐 . (recall the notation in the subsection “ Loops ”). Inparticular, if we denote by 𝐺 -function a combination of 𝐹 -and √ 𝐹 -functions together with + , − , × , ÷ , then we have thefollowing: When ( 𝑐 , . . . , 𝑐 𝑛 − ) follow a loop, 𝐺 -functions do not always follow a loop. A 𝐺 -function is a new type of ingredient as it may includeexpressions with one level of roots, such as 𝐺 = − 𝑐 + √︃ 𝑐 − 𝑐 . (6)We can keep going like this to construct formulae with highernumber of nested roots , i.e., roots in roots. For example, wecan combine 𝐺 -functions and √ 𝐺 -functions with + , − , × , ÷ tomake 𝐻 -functions. These may contain up to two levels ofnested roots, such as 𝐻 = 𝑐 − √︁ 𝑐 + √︂ − 𝑐 + √︃ 𝑐 − 𝑐 , (7)and so on, as summarized in Figure 3. With this nomencla-ture, we can make arbitrarily complex expressions involving + , − , × , ÷ and √ , and at the same time keep track of the num-ber of nested roots appearing in the formula. Conversely, anyformula constructed with + , − , × , ÷ , √ can be built using thisprocedure, provided that we look high enough in the “ . . . ” ofthe list of ingredients ( 𝐹, 𝐺, 𝐻, . . . ) . Figure 3. Ingredients used to build a formula. Combining coefficients ( 𝑐 , ..., 𝑐 𝑛 − ) with + , − , × , ÷ defines an 𝐹 -function. Combining 𝐹 -functions and their roots √ 𝐹 with + , − , × , ÷ defines a 𝐺 -function, etc. See supplementary material for animated version.
We have now covered all the tools necessary: permutationsof ( 𝑠 , . . . , 𝑠 𝑛 ) , loops, and 𝐹, 𝐺, 𝐻 -functions. Let us now applyall these concepts to the degree 𝑛 equation, starting with 𝑛 = 𝑛 = THE QUADRATIC EQUATION
Our journey toward the Abel–Ruffini theorem starts withconsiderations of the much more familiar quadratic equation.In particular, considering only the case 𝑛 =
2, we will provea first impossibility result, actually valid for 𝑛 ≥
2. The ideasdeveloped here are rather simple but also at the heart of theproof of the Abel–Ruffini theorem.
Vieta’s formulae.—
Let us consider the general quadraticequation 𝑧 + 𝑐 𝑧 + 𝑐 = . (8)As mentioned previously, the fundamental theorem of alge-bra informs us that this equation admits exactly two complexsolutions 𝑠 and 𝑠 . Let us then write it in the factored form ( 𝑧 − 𝑠 )( 𝑧 − 𝑠 ) = 𝑧 . We find a new expansion that can be comparedto equation (8). By identification, we obtain the so-calledVieta’s formulae: 𝑐 = −( 𝑠 + 𝑠 ) and 𝑐 = 𝑠 𝑠 . (9)This kind of relation between the coefficients and the solutionscan be established for any degree 𝑛 ≥
2. For example, equation(9) generalizes nicely to 𝑐 𝑛 − = − (cid:205) i 𝑠 i and 𝑐 = (− ) 𝑛 (cid:206) i 𝑠 i ,for any 𝑛 ≥
2. In any case, these formulae always reveal thesame fundamental property:
Coefficients ( 𝑐 , . . . , 𝑐 𝑛 − ) are symmetricfunctions of the solutions ( 𝑠 , . . . , 𝑠 𝑛 ) . In particular, for the 𝑛 = 𝑠 and 𝑠 by moving them continuously in the complex plane(using, for example, the transposition (12) depicted in Figure1), then the coefficients ( 𝑐 , 𝑐 ) will each move on some path,but eventually they must come back to their original locationas they are symmetric in ( 𝑠 , 𝑠 ) . In other words, they willfollow a loop , as depicted in Figure 4. Figure 4. The transposition ( ) on the solutions ( 𝑠 , 𝑠 ) inducesa loop on the coefficients ( 𝑐 , 𝑐 ) . See supplementary material foranimated version.
A first impossibility result.—
Because it is the central ideaat play, let us rephrase the symmetry in Vieta’s formulae geo-metrically:
When ( 𝑠 , . . . , 𝑠 𝑛 ) undergo a permutation ( 𝑐 , . . . , 𝑐 𝑛 − ) each follow a loop. This remarkable fact can be used to obtain a first impossibilityresult, as follows. Suppose that the solutions 𝑠 and 𝑠 of thequadratic equation are given by two formulae of the type 𝑠 = 𝐹 ( 𝑐 , 𝑐 ) and 𝑠 = 𝐹 ( 𝑐 , 𝑐 ) , (10)with 𝐹 , 𝐹 two 𝐹 -functions (i.e., expressions involving ( 𝑐 , 𝑐 ) and the symbols + , − , × , ÷ ). Now, picture ( 𝑠 , 𝑠 ) and ( 𝑐 , 𝑐 ) in the complex plane, and study the following process: • connect 𝑠 and 𝑠 with paths inducing the transposition ( ) , and make them move along these paths (see Fig. 4); • as 𝑠 and 𝑠 move around, 𝑐 and 𝑐 each travel on aloop, as seen previously (see Fig. 4), • while 𝑐 and 𝑐 follow their own loop, the two 𝐹 -functions 𝐹 and 𝐹 will also follow a loop, as arguedearlier (see Fig. 5). Figure 5. A loop followed by ( 𝑐 , 𝑐 ) also induced a loop on the 𝐹 -functions 𝐹 ( 𝑐 , 𝑐 ) and 𝐹 ( 𝑐 , 𝑐 ) . See supplementary materialfor animated version.
At the end of this process, 𝑠 and 𝑠 have permuted, yetboth 𝐹 and 𝐹 have followed a loop. Consequently, 𝐹 and 𝐹 cannot equal 𝑠 and 𝑠 respectively, and no formula such asin (10) exists. This impossibility result actually holds for anequation of any degree 𝑛 ≥
2. Indeed, it suffices to pick two ofthe 𝑛 solutions to the degree 𝑛 equation, name them 𝑠 and 𝑠 ,and apply the above recipe. The conclusion is thus: Using only 𝐹 -functions, no formula solvingthe general equation (1) can be found for 𝑛 ≥ . This is our first impossibility result . In particular, it means thatwe have no chance of finding a formula for the cubic equationusing only 𝐹 -functions either. To see which extra ingredientsare needed, let us examine closely the well-known quadraticformula. Discussion: the quadratic formula.—
The quadratic formulais derived most easily by “completing the square” in equation(8) to get ( 𝑧 + 𝑐 ) = 𝑐 − 𝑐 . Using our notation √ for any ofthe two square roots, we easily obtain a formula for the generalsolution 𝑠 of equation (8) as 𝑠 = − 𝑐 + √︃ 𝑐 − 𝑐 . (11)This formula alone corresponds to two solutions, one for eachof the two square roots on the right-hand side. Moreover,notice how this root indeed points to the same direction asour impossibility result: we need to add √ 𝐹 -function to thelist of ingredients. One last note: just as the Abel–Ruffinitheorem, the impossibility result just derived tells somethingabout the general quadratic equation. However, there exists some quadratic equations with given, explicit coefficient thatadmit a formula in terms of + , − , × , ÷ . THE CUBIC EQUATION
Let us now try to construct a formula for the solutions of thegeneral cubic equation. The equation reads 𝑧 + 𝑐 𝑧 + 𝑐 𝑧 + 𝑐 = . (12)Let ( 𝑠 , 𝑠 , 𝑠 ) be its three complex solutions. Learning fromour previous findings, we now add √ 𝐹 -functions to the listof ingredients. Therefore, we assume that there exists someformulae of the type 𝑠 i = 𝐺 i ( 𝑐 , 𝑐 , 𝑐 ) , for i ∈ { , . . . , } , (13)with 𝐺 , 𝐺 , 𝐺 three 𝐺 -functions (combinations of 𝐹 and √ 𝐹 with + , − , × , ÷ ). Our second impossibility result will consistin showing that such a formula cannot exist. Our previousmethod is not guaranteed to work: yes , the coefficients stillfollow loops as solutions permute, but no , 𝐺 -functions donot generally follow loops in these circumstances, as we havealready seen. We need to change our plan. Introducing commutators.—
Consider the transposition ( ) that induces a loop 𝛾 on 𝐹 and thus an unclosed path 𝜔 on √ 𝐹 . Consider also ( ) , inducing a loop 𝛾 on 𝐹 and a path 𝜔 on √ 𝐹 . Now perform the following sequence of transpositions,called the commutator of ( ) and ( ) , and denoted [( ) , ( )] = ( )( )( ) − ( ) − . (14)Since ( ) − is simply ( ) , and ( ) − = ( ) , it turns outthat [( ) , ( )] is simply the cycle ( ) . In fact, this is truewith any pair of transposition, i.e., [( 𝑖 𝑗 ) , ( 𝑗 𝑘 )] = ( 𝑖 𝑗 𝑘 ) . (15)Therefore, [( ) , ( )] does permute the three solutions ( 𝑠 , 𝑠 , 𝑠 ) . But what is its effect on numbers like 𝐹 and √ 𝐹 ?Clearly, 𝐹 follows a sequence of loops 𝛾 𝛾 𝛾 − 𝛾 − , which isitself a loop. The number √ 𝐹 , however, follows a sequence ofunclosed paths 𝜔 𝜔 𝜔 − 𝜔 − (visiting other roots) but closeson itself by construction; see Figure 6. Figure 6. Effect of the commutator [( ) , ( )] on a coefficient 𝑐 (left), on an 𝐹 -function (center) and on √ 𝐹 -function (right). Afterthe process, both 𝐹 and √ 𝐹 have followed a loop. Notice the loopfollowed by √ 𝐹 consisting in four unclosed paths. See supplementarymaterial for animated version.
Conclusion.—
With the permutation ( ) written as thecommutator [( ) , ( )] , we reach the same conclusion as inthe quadratic case: while ( 𝑠 , 𝑠 , 𝑠 ) undergoes the permuta-tion ( ) , both 𝐹 and √ 𝐹 follow a loop (and thus any 𝐺 -function). Consequently, there cannot be equalities given by(13). Again, this holds for the general equation of degree 𝑛 ≥ 𝑛 ≥ 𝑠 , 𝑠 , 𝑠 , and apply the above recipe. Therefore,we conclude: Using only 𝐺 -functions, no formulasolving the general equation (1) can be found for 𝑛 ≥ . This is our second impossibility result . We must emphasizethat it works only if we apply the cycle ( ) as a commutatorsuch as in equation (14). Had we just applied the cycle ( ) directly (i.e., without writing it as a commutator), there wouldhave been no guarantee that √ 𝐹 followed a loop. It is thecommutator that allows us to discard one level of roots, andthus √ 𝐹 , from the list of ingredients. Let us now put this newimpossibility result to the test, by solving explicitely the cubicequation. Discussion: the cubic formula.—
We follow the classicalmethod found by Italian mathematicians of the sixteenth cen-tury. First, perform the change of variables 𝑍 = 𝑧 + 𝑐 /
3, which“removes” the 𝑧 term in equation (12), transforming it into 𝑍 + 𝑃𝑍 + 𝑄 = , (16)where 𝑃 = 𝑐 − 𝑐 and 𝑄 = 𝑐 + 𝑐 − 𝑐 𝑐 . Notice that both 𝑃 and 𝑄 are 𝐹 -functions of ( 𝑐 , 𝑐 , 𝑐 ) . To solve equation(16), one then writes 𝑍 = 𝑣 + 𝑤 , where 𝑣, 𝑤 are two complexnumbers to be chosen freely later on. Then, equation (16)becomes 𝑣 + 𝑤 + ( 𝑣𝑤 + 𝑃 )( 𝑣 + 𝑤 ) + 𝑄 =
0, from which wecan remove the second term by imposing that 𝑣, 𝑤 satisfy 𝑣𝑤 = − 𝑃 . By cubing the latter, we then obtain two equationsfor two unknowns, namely 𝑣 + 𝑤 = − 𝑄 and 𝑣 𝑤 = − 𝑃 . (17)These equations can be solved simultaneously for 𝑣 and 𝑤 ,since they explicitly give their sum and product, respectively.(These are nothing but Vieta’s formulae for 𝑛 =
2; See equation(9).) Using the quadratic formula, one obtains 𝑣 and 𝑤 interms of 𝑃 and 𝑄 , takes their cube root and adds the resultto obtain 𝑣 + 𝑤 = 𝑍 . Going back to the original unknown 𝑧 = 𝑍 − 𝑐 / 𝑠 = − 𝑐 + √︃ − 𝑄 + √︁ 𝑄 + 𝑃 + √︃ − 𝑄 − √︁ 𝑄 + 𝑃 . (18)This formula gives three solutions ( 𝑠 , 𝑠 , 𝑠 ) , one for eachcube root. It is clear that this expression involves more than 𝐹 and √ 𝐹 functions: indeed, the two cube roots are actually √ 𝐺 -functions. In a sense, the cubic formula above contains“two levels” of roots, whereas 𝐺 -functions can only containone, by definition. This kind of expression is thus called a nested root . Our “commutator trick” was only able to remove one level of roots. Perhaps two levels of commutators willremove two? If so, then it looks like a pattern is emerging. . . THE QUARTIC EQUATION
We now turn to the quartic equation 𝑧 + 𝑐 𝑧 + 𝑐 𝑧 + 𝑐 𝑧 + 𝑐 = . (19)For the cubic, we saw that 𝐺 -functions are not enough to con-struct a formula, as we also needed √ 𝐺 functions. Therefore,we start by assuming the existence of some formula for thefour solutions 𝑠 i = 𝐻 i ( 𝑐 , 𝑐 , 𝑐 , 𝑐 ) , for 𝑖 ∈ { , . . . , } . (20)As before, the four functions 𝐻 i are assumed to be 𝐻 -functions, i.e., 𝐺 - and √ 𝐺 -functions combined with + , − , × , ÷ .As should be clear by now, it turns out that even with the extraingredient √ 𝐺 , no general quartic formula can be constructed.Once again we will prove this by constructing an appropriatepermutation of ( 𝑠 , 𝑠 , 𝑠 , 𝑠 ) . A brief checkpoint.—
Once again, just as the first method didnot work for cubic equations, the method used for the cubiccase is not guaranteed to work for quartic equations either.Indeed, the commutator of transpositions induced a loop on 𝐹 and √ 𝐹 (and thus on 𝐺 ). But a loop on 𝐺 generally doesnot induce a loop on √ 𝐺 , as we have seen many times. Asummary of these previous methods is given on Table I. ingredient 𝐹 -functions 𝐺 -functionsnested roots 0 1discarded by transpositions commutator of transpositionswith the path (12) [(12),(23)] = (123)for degree 𝑛 ≥ 𝑛 ≥ But now a natural solution presents itself: what if wetake the commutator of, say, ( ) and ( ) , written ascommutators themselves, using equation (15)? Let us examinethis in detail. Commutators, yet again.—
First we need to check that thecommutator of ( ) and ( ) does indeed permute the foursolution ( 𝑠 , 𝑠 , 𝑠 , 𝑠 ) . Fortunately it does, as a quick checkreveals that [( ) , ( )] = ( )( ) , (21)which is a particular case of the more general formula [( 𝑖 𝑗 𝑘 ) , ( 𝑗 𝑘ℓ )] = ( 𝑖ℓ )( 𝑗 𝑘 ) . Therefore, our commutator [( ) , ( )] does indeed permute ( 𝑠 , 𝑠 , 𝑠 , 𝑠 ) . Now, letus examine how it affects 𝐺 - and √ 𝐺 -functions, one step at atime: • First, we apply the cycles ( ) = [( ) , ( )] then ( ) = [( ) , ( )] . Since they are commutators, 𝐺 -functions will follow two loops 𝛾 , 𝛾 successively, comingback to their original positions. However, quantities like √ 𝐺 will move along two (generally unclosed) paths 𝜔 and 𝜔 .All this is exactly as in the cubic case. • Second, we apply these two paths backwards, in reversei.e., ( ) = [( ) , ( )] and then ( ) = [( ) , ( )] .During these two, 𝐺 -functions will follow 𝛾 − 𝛾 − , i.e. theprevious loops backwards. Similarly, √ 𝐺 -functions will travelalong 𝜔 − 𝜔 − .What just happened is exactly the same as in the cubic case,except with 𝐺 -functions in place of 𝐹 -functions. In partic-ular, 𝐺 -functions follow the loop 𝛾 = 𝛾 𝛾 𝛾 − 𝛾 − ; and √ 𝐺 -functions a sequence of unclosed paths 𝜔 𝜔 𝜔 − 𝜔 − , whichcloses on itself by construction. In other words, both 𝐺 -and √ 𝐺 -functions followed a loop and thus any 𝐻 -functionwill, too. Our conclusion has therefore been reached: while ( 𝑠 , 𝑠 , 𝑠 , 𝑠 ) undergoes the permutation ( )( ) written asa commutator of commutators, any 𝐻 -function follows a loop.Consequently, no formula (20) can exist. This result extendsto any equation of degree 𝑛 ≥
4, as before, and constitutes our third impossibility result:Using only 𝐻 -functions, no formula solvingthe general equation (1) can be found for 𝑛 ≥ . In particular, we can extend Table I with an additional columnfor the new ingredient, 𝐻 -functions. ingredient 𝐻 -functionsnested roots 2discarded by commutator of commutator of transpositionswith the path [[(12),(23)],[(23),(34)]] = (14)(23)for degree 𝑛 ≥ 𝐻 -functions, for degree 𝑛 ≥ Discussion: the quartic formula.—
As for the cubic case,our impossibility result does not imply that there is no quartic(nor quintic) formula. It just means that to construct one, oneneeds at least three levels of nested roots, and 𝐻 -functionscontain only two. It turns out that the quartic equation can besolved as follows and, indeed, it involves three levels of nestedroots. As for the cubic case, we start by removing the 𝑧 termby the change of variables 𝑍 = 𝑧 + 𝑐 /
4. This brings equation(19) into the form 𝑍 + 𝑃𝑍 + 𝑄𝑍 + 𝑅 = , (22)where 𝑃, 𝑄, 𝑅 are three 𝐹 -functions of ( 𝑐 , 𝑐 , 𝑐 , 𝑐 ) , whoseexpressions are long, but easily obtained. The next step isto transform equation (22) into one that is quadratic in 𝑍 .For now, nothing guarantees that 𝑃𝑍 + 𝑄𝑍 + 𝑅 is a perfectsquare, but if it were, then equation (22) could be factoredinto two equations quadratic in 𝑍 . One way is to write 𝑍 in the equivalent form 𝑍 = ( 𝑍 + 𝐴 ) − 𝐴𝑍 − 𝐴 , for somecomplex number 𝐴 to be chosen freely later on. Inserting thisin equation (22) gives ( 𝑍 + 𝐴 ) + ( 𝑃 − 𝐴 ) 𝑍 + 𝑄𝑍 + 𝑅 − 𝐴 = . (23) Now we can choose 𝐴 in equation (23) such that the quadraticpart ( 𝑃 − 𝐴 ) 𝑍 + 𝑄𝑍 + 𝑅 − 𝐴 has the form of a perfect square.This will be the case if its discriminant 𝑄 + ( 𝑃 − 𝐴 ) 𝐴 vanishes. The latter amounts to8 𝐴 − 𝑃 𝐴 − 𝑄 = , (24)which is a cubic equation 𝐴 . It can be solved using thecubic formula, giving a value of 𝐴 in terms of 𝑃 and 𝑄 that is an 𝐻 -function (recall the cubic formula (18)).Once 𝐴 takes this special value, equation (23) becomes ( 𝑍 + 𝐴 ) + ( 𝑃 − 𝐴 )( 𝑍 − 𝐴 ) =
0, which can be factored easilyinto two quadratic polynomials in 𝑍 . The latter equations aresolved easily using the quadratic formula. Since 𝐴 is an 𝐻 -function, the solution for 𝑍 will necessarily involve some √ 𝐻 quantities, something which we did not include in equation(20). This confirms our impossibility result, once again. THE QUINTIC EQUATION
It seems at this point that things are becoming repetitive,and that a clear pattern emerges. For 𝑛 = , ,
4, commutatorscould be used to reject formulae with too few nested roots intheir expressions. However, we were still be able to solve theequation simply by allowing more levels of roots. But at 𝑛 = 𝑛 = any number of roots.Let us pretend that we found a quintic formula, e.g., 𝑠 i = Φ i ( 𝑐 , . . . , 𝑐 ) , for 𝑖 ∈ { , . . . , } , (25)with the five functions Φ i built out of 𝐻 – and √ 𝐻 –functions.If we follow the previous methods, summarized in Tables Iand II, it should be clear that (1) all 𝐻 -functions will follow aloop from a commutator of commutators of the solutions (asin the quartic case), but (2) we will need one more level ofcommutators for the √ 𝐻 terms.As we now have five solutions to play with, let us considerfor example the permutations ( ) and ( ) to construct afirst commutator [( ) , ( )] . An easy check shows that thelatter is equal to ( ) , and this commutator therefore permutesthree of our solutions. In general, the following result holds at 𝑛 = [( 𝑖 𝑗 𝑘 ) , ( 𝑘ℓ𝑚 )] = ( 𝑗 𝑘𝑚 ) . (26)But now, contrary to the previous cases, we have somethingrather remarkable with equation (26). It shows that any cycle ( 𝑗 𝑘𝑚 ) can be written as a commutator of two other cycles,namely [( 𝑖 𝑗 𝑘 ) , ( 𝑘ℓ𝑚 )] . But notice that this is true for any cycle ( 𝑗 𝑘𝑚 ) , including ( 𝑖 𝑗 𝑘 ) and ( 𝑘ℓ𝑚 ) on the left-hand sideof equation (26) itself. In other words, this formula can beapplied to itself, again and again, allowing us to write ( 𝑗 𝑘𝑚 ) as a commutator of as many commutators as needed. Since anumber 𝑁 ∈ N of commutators allows us to discard precisely 𝑁 levels of roots in a formula (see Tables I and II), we canactually discard any number of roots in any candidate quinticformula. The Abel–Ruffini theorem follows immediately fromthis remark, but let us give a more detailed explanation.Suppose that, in the quintic formula, equation (25), we usea Φ -function made of + , − , × , ÷ , along with 𝑁 levels of roots,for some 𝑁 ∈ N . To construct this Φ -function, we have atour disposal several ingredients: 𝐹 -functions (no roots), 𝐺 -functions (one level of root), 𝐻 -functions (two levels of root),and so on. As always, we start by choosing a permutation of thesolutions, say ( ) , that discards any 𝐹 -functions (no roots).Next, using equation (26), we write ( ) as a commutator,for example: ( ) = [( ) , ( )] . (27)When applied to ( 𝑠 , . . . , 𝑠 ) , this commutator discards the 𝐺 -functions from the list of ingredients (one level of roots). Nowwe keep going: we write the cycles ( ) , ( ) appearing inequation (27) as commutators themselves, again using equa-tion (26). We obtain ( ) expressed with two commutators: ( ) = [[( ) , ( )] , [( ) , ( )]] , (28)which removes 𝐻 -functions (expression with two levels ofroots). By iterating equation (26) 𝑁 − ( ) as a combination of 𝑁 commutators. Whenthe latter is applied to ( 𝑠 , . . . , 𝑠 ) , the solutions 𝑠 , 𝑠 , 𝑠 willpermute; and yet any expression of the coefficients with 𝑁 orless roots will follow a loop. Since a Φ -function is made upof all these ingredients, Φ , Φ , Φ go back to their originalposition. Clearly this contradicts equation (25). This resultgeneralizes to an equation of any degree 𝑛 ≥ 𝑁 is arbitrary, weconclude that no number of roots will be sufficient to write aformula. Our conclusion is therefore: No formula exists for the solution to the generalequation of degree five or more, using onlythe operations + , − , × , ÷ , and √ ,i.e., the Abel–Ruffini theorem itself. A last remark is in order.Why the fifth degree, and not the fourth or sixth? This allboils down to the possibility of writing a permutation of atleast two solutions as a sequence of commutators. A formulasuch as equation (26) can only be iterated indefinitely whenit involves five or more elements. For four or fewer elements,any sequence of commutators of transposition and/or cycleswill necessarily end, i.e., end up being the trivial permutationthat “does nothing.” The reader familiar with group theorywill here recognize the notions of perfect or solvable group. CONCLUSIONS
To conclude this article, we would like to first make somecomments on the various advantages and disadvantages ofthis proof, compared to the usual proof using Galois theory.First of all, the present proof does not say that no equations ofdegree five or higher can be solved; but only that a generalformula (valid for the general equation) cannot be writtenusing only + , − , × , ÷ , and √ . Indeed, some equations ofdegree 𝑛 ≥ 𝑧 + 𝑎𝑧 + 𝑏 = , sin , . . . ) in the list of ingredients.Indeed, just like + , − , × , ÷ , these functions follow a loopwhen the coefficients do. Galois theory is unable to providefor this, as it only accounts for expressions in terms of radicals.We hope that the present proof will be seen not only as asimplified and elementary demonstration of the Abel–Ruffini’stheorem, but also as a complementary result, as it helps to ex-plain why the 𝑛 = fora free digital copy). Devised as a problems-and-solutionsbook, it discusses many advanced concepts in a very pedagog-ical and extremely well-written manner. ∗ [email protected][1] Sesiano, J. (1999). An introduction to the history of algebra:solving equations from Mesopotamian times to the Renaissance. (Pierrehumbert A. transl.) American Mathematical Society.[2] Stillwell, J. (2010).
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