# Abel–Ruffini theorem

Abel–Ruffini theorem Not to be confused with Abel's theorem.

En mathématiques, the Abel–Ruffini theorem (also known as Abel's impossibility theorem) states that there is no solution in radicals to general polynomial equations of degree five or higher with arbitrary coefficients. Ici, general means that the coefficients of the equation are viewed and manipulated as indeterminates.

The theorem is named after Paolo Ruffini, who made an incomplete proof in 1799,[1] (which was refined and completed in 1813[2] and accepted by Cauchy) and Niels Henrik Abel, who provided a proof in 1824.[3][4] Abel–Ruffini theorem refers also to the slightly stronger result that there are equations of degree five and higher that cannot be solved by radicals. This does not follow from Abel's statement of the theorem, but is a corollary of his proof, as his proof is based on the fact that some polynomials in the coefficients of the equation are not the zero polynomial. This improved statement follows directly from Galois theory § A non-solvable quintic example. Galois theory implies also that {style d'affichage x^{5}-x-1=0} is the simplest equation that cannot be solved in radicals, and that almost all polynomials of degree five or higher cannot be solved in radicals.

The impossibility of solving in degree five or higher contrasts with the case of lower degree: one has the quadratic formula, the cubic formula, and the quartic formula for degrees two, three, and four, respectivement.

Contenu 1 Context 2 Preuve 2.1 Algebraic solutions and field theory 2.2 Galois correspondence 2.3 Solvable symmetric groups 2.4 Polynomials with symmetric Galois groups 2.4.1 General equation 2.4.2 Explicit example 3 Cayley's resolvent 4 Histoire 5 References Context Polynomial equations of degree two can be solved with the quadratic formula, which has been known since antiquity. Similarly the cubic formula for degree three, and the quartic formula for degree four, were found during the 16th century. At that time a fundamental problem was whether equations of higher degree could be solved in a similar way.

The fact that every polynomial equation of positive degree has solutions, possibly non-real, was asserted during the 17th century, but completely proved only at the beginning of the 19th century. This is the fundamental theorem of algebra, which does not provide any tool for computing exactly the solutions, although Newton's method allows approximating the solutions to any desired accuracy.

From the 16th century to beginning of the 19th century, the main problem of algebra was to search for a formula for the solutions of polynomial equations of degree five and higher, hence the name the "fundamental theorem of algebra". This meant a solution in radicals, C'est, an expression involving only the coefficients of the equation, and the operations of addition, subtraction, multiplication, division, and nth root extraction.

The Abel–Ruffini theorem proves that this is impossible. Cependant, this impossibility does not imply that a specific equation of any degree cannot be solved in radicals. Au contraire, there are equations of any degree that can be solved in radicals. This is the case of the equation {style d'affichage x^{n}-1=0} for any n, and the equations defined by cyclotomic polynomials, all of whose solutions can be expressed in radicals.

Abel's proof of the theorem does not explicitly contain the assertion that there are specific equations that cannot be solved by radicals. Such an assertion is not a consequence of Abel's statement of the theorem, as the statement does not exclude the possibility that "every particular quintic equation might be soluble, with a special formula for each equation."[5] Cependant, the existence of specific equations that cannot be solved in radicals seems to be a consequence of Abel's proof, as the proof uses the fact that some polynomials in the coefficients are not the zero polynomial, et, given a finite number of polynomials, there are values of the variables at which none of the polynomials takes the value zero.

Soon after Abel's publication of its proof, Évariste Galois introduced a theory, now called Galois theory that allows deciding, for any given equation, whether it is solvable in radicals (this is theoretical, comme, en pratique, this decision may need huge computation which can be difficult, even with powerful computers). This decision is done by introducing auxiliary polynomials, called resolvents, whose coefficients depend polynomially upon those of the original polynomial. The polynomial is solvable in radicals if and only if some resolvent has a rational root.

Proof The proof of the Abel–Ruffini theorem predates Galois theory. Cependant, Galois theory allows a better understanding of the subject, and modern proofs are generally based on it, while the original proofs of the Abel–Ruffini theorem are still presented for historical purposes.[1][6][7][8] The proofs based on Galois theory comprise four main steps: the characterization of solvable equations in terms of field theory; the use of the Galois correspondence between subfields of a given field and the subgroups of its Galois group for expressing this characterization in terms of solvable groups; the proof that the symmetric group is not solvable if its degree is five or higher; and the existence of polynomials with a symmetric Galois group.

Algebraic solutions and field theory An algebraic solution of a polynomial equation is an expression involving the four basic arithmetic operations (addition, subtraction, multiplication, and division), and root extractions. Such an expression may be viewed as the description of a computation that starts from the coefficients of the equation to be solved and proceeds by computing some numbers, one after the other.

At each step of the computation, one may consider the smallest field that contains all numbers that have been computed so far. This field is changed only for the steps involving the computation of an nth root.

Alors, an algebraic solution produces a sequence {style d'affichage F_{0}subseteq F_{1}subseteq cdots subseteq F_{k}} of fields, and elements {style d'affichage x_{je}in F_{je}} tel que {style d'affichage F_{je}=F_{i-1}(X_{je})} pour {style d'affichage i=1,ldots ,k,} avec {style d'affichage x_{je}^{n_{je}}in F_{i-1}} for some integer {displaystyle n_{je}>1.} An algebraic solution of the initial polynomial equation exists if and only if there exists such a sequence of fields such that {style d'affichage F_{k}} contains a solution.

For having normal extensions, which are fundamental for the theory, one must refine the sequence of fields as follows. Si {style d'affichage F_{i-1}} does not contain all {displaystyle n_{je}} -th roots of unity, one introduces the field {style d'affichage K_{je}} that extends {style d'affichage F_{i-1}} by a primitive root of unity, and one redefines {style d'affichage F_{je}} comme {style d'affichage K_{je}(X_{je}).} Alors, if one starts from a solution in terms of radicals, one gets an increasing sequence of fields such that the last one contains the solution, and each is a normal extension of the preceding one with a Galois group that is cyclic.

inversement, if one has such a sequence of fields, the equation is solvable in terms of radicals. For proving this, it suffices to prove that a normal extension with a cyclic Galois group can be built from a succession of radical extensions.

Galois correspondence The Galois correspondence establishes a one to one correspondence between the subextensions of a normal field extension {displaystyle F/E} and the subgroups of the Galois group of the extension. This correspondence maps a field K such {displaystyle Esubseteq Ksubseteq F} to the Galois group {nom de l'opérateur de style d'affichage {Fille} (F/K)} of the automorphisms of F that leave K fixed, et, conversely, maps a subgroup H of {nom de l'opérateur de style d'affichage {Fille} (F/E)} to the field of the elements of F that are fixed by H.

The preceding section shows that an equation is solvable in terms of radicals if and only if the Galois group of its splitting field (the smallest field that contains all the roots) est résoluble, C'est, it contains a sequence of subgroups such that each is normal in the preceding one, with a quotient group that is cyclic. (Solvable groups are commonly defined with abelian instead of cyclic quotient groups, but the fundamental theorem of finite abelian groups shows that the two definitions are equivalent).

Alors, for proving Abel–Ruffini theorem, it remains to prove that the symmetric group {style d'affichage S_{5}} is not solvable, and that there are polynomials with symmetric Galois group.

Solvable symmetric groups For n > 4, the symmetric group {style d'affichage {mathématique {S}}_{n}} of degree n has only the alternating group {style d'affichage {mathématique {UN}}_{n}} as a nontrivial normal subgroup (see Symmetric group § Normal subgroups). For n > 4, the alternating group {style d'affichage {mathématique {UN}}_{n}} is not abelian and simple (C'est, it does not have any nontrivial normal subgroup). This implies that both {style d'affichage {mathématique {UN}}_{n}} et {style d'affichage {mathématique {S}}_{n}} are not solvable for n > 4. Ainsi, the Abel–Ruffini theorem results from the existence of polynomials with a symmetric Galois group; this will be shown in the next section.

D'autre part, for n ≤ 4, the symmetric group and all its subgroups are solvable. Somehow, this explains the existence of the quadratic, cubic, and quartic formulas, since a major result of Galois theory is that a polynomial equation has a solution in radicals if and only if its Galois group is solvable (the term "solvable group" takes its origin from this theorem).

Polynomials with symmetric Galois groups General equation The general or generic polynomial equation of degree n is the equation {style d'affichage x^{n}+un_{1}x^{n-1}+cdots +a_{n-1}x+a_{n}=0,} où {style d'affichage a_{1},ldots ,un_{n}} are distinct indeterminates. This is an equation defined over the field {displaystyle F=mathbb {Q} (un_{1},ldots ,un_{n})} of the rational fractions in {style d'affichage a_{1},ldots ,un_{n}} with rational number coefficients. The original Abel–Ruffini theorem asserts that, for n > 4, this equation is not solvable in radicals. In view of the preceding sections, this results from the fact that the Galois group over F of the equation is the symmetric group {style d'affichage {mathématique {S}}_{n}} (this Galois group is the group of the field automorphisms of the splitting field of the equation that fix the elements of F, where the splitting field is the smallest field containing all the roots of the equation).

For proving that the Galois group is {style d'affichage {mathématique {S}}_{n},} it is simpler to start from the roots. Laisser {style d'affichage x_{1},ldots ,X_{n}} be new indeterminates, aimed to be the roots, and consider the polynomial {style d'affichage P(X)=x^{n}+b_{1}x^{n-1}+cdots +b_{n-1}x+b_{n}=(x-x_{1})cdots (x-x_{n}).} Laisser {displaystyle H=mathbb {Q} (X_{1},ldots ,X_{n})} be the field of the rational fractions in {style d'affichage x_{1},ldots ,X_{n},} et {displaystyle K=mathbb {Q} (b_{1},ldots ,b_{n})} be its subfield generated by the coefficients of {style d'affichage P(X).} The permutations of the {style d'affichage x_{je}} induce automorphisms of H. Vieta's formulas imply that every element of K is a symmetric function of the {style d'affichage x_{je},} and is thus fixed by all these automorphisms. It follows that the Galois group {nom de l'opérateur de style d'affichage {Fille} (H/K)} is the symmetric group {style d'affichage {mathématique {S}}_{n}.} The fundamental theorem of symmetric polynomials implies that the {displaystyle b_{je}} are algebraic independent, and thus that the map that sends each {style d'affichage a_{je}} to the corresponding {displaystyle b_{je}} is a field isomorphism from F to K. This means that one may consider {style d'affichage P(X)=0} as a generic equation. This finishes the proof that the Galois group of a general equation is the symmetric group, and thus proves the original Abel–Ruffini theorem, which asserts that the general polynomial equation of degree n cannot be solved in radicals for n > 4.

Explicit example See also: Galois theory § A non-solvable quintic example The equation {style d'affichage x^{5}-x-1=0} is not solvable in radicals, as will be explained below.

Let q be {style d'affichage x^{5}-x-1} . Let G be its Galois group, which acts faithfully on the set of complex roots of q. Numbering the roots lets one identify G with a subgroup of the symmetric group {style d'affichage {mathématique {S}}_{5}} . Depuis {style d'affichage q{dans un sens {2}}} factors as {style d'affichage (x^{2}+x+1)(x^{3}+x^{2}+1)} dans {style d'affichage mathbb {F} _{2}[X]} , the group G contains a permutation g that is a product of disjoint cycles of lengths 2 et 3 (en général, when a monic integer polynomial reduces modulo a prime to a product of distinct monic irreducible polynomials, the degrees of the factors give the lengths of the disjoint cycles in some permutation belonging to the Galois group); then G also contains {style d'affichage g^{3}} , which is a transposition. Depuis {style d'affichage q{dans un sens {3}}} est irréductible en {style d'affichage mathbb {F} _{3}[X]} , the same principle shows that G contains a 5-cycle. Car 5 est premier, any transposition and 5-cycle in {style d'affichage {mathématique {S}}_{5}} generate the whole group; see Symmetric group § Generators and relations. Ainsi {style d'affichage G={mathématique {S}}_{5}} . Since the group {style d'affichage {mathématique {S}}_{5}} is not solvable, the equation {style d'affichage x^{5}-x-1=0} is not solvable in radicals.

Cayley's resolvent Testing whether a specific quintic is solvable in radicals can be done by using Cayley's resolvent. This is a univariate polynomial of degree six whose coefficients are polynomials in the coefficients of a generic quintic. A specific irreducible quintic is solvable in radicals if and only, when its coefficients are substituted in Cayley's resolvent, the resulting sextic polynomial has a rational root.