Editing Abel–Ruffini theorem (section)

==Proof==
The proof of the Abel–Ruffini theorem predates [[Galois theory]]. However, 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.<ref name="Ayoub"/><ref>{{Citation|last=Rosen|first=Michael I.|author-link=Michael Rosen (mathematician)|title=Niels Hendrik Abel and Equations of the Fifth Degree|journal=[[American Mathematical Monthly]]|year=1995|volume=102|issue=6|pages=495–505|doi=10.2307/2974763|jstor=2974763|zbl=0836.01015|mr=1336636}}</ref><ref name="Tignol_Abel">{{Citation|last=Tignol|first=Jean-Pierre| author-link=Jean-Pierre Tignol| title=Galois' Theory of Algebraic Equations|edition=2nd|publisher=[[World Scientific]]|year=2016|isbn=978-981-4704-69-4|zbl=1333.12001|chapter=Ruffini and Abel on General Equations}}</ref><ref name="Pesic">{{Citation|last=Pesic|first=Peter|title=Abel's Proof: An Essay on the Sources and Meaning of Mathematical Unsolvability|location=Cambridge|publisher=[[MIT Press]]|year=2004|isbn=0-262-66182-9|zbl=1166.01010}}</ref>

The proofs based on Galois theory comprise four main steps: the characterization of solvable equations in terms of [[field theory (mathematics)|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 group]]s; 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 (mathematics)|expression]] involving the four basic [[arithmetic operations]] (addition, subtraction, multiplication, and division), and [[root extraction]]s. 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 (mathematics)|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|{{mvar|n}}th]] root.

So, an algebraic solution produces a sequence
:<math>F_0\subseteq F_1\subseteq \cdots \subseteq F_k</math>
of fields, and elements <math>x_i\in F_i</math> such that 
<math>F_i=F_{i-1}(x_i)</math> for <math>i=1,\ldots, k,</math> with <math>x_i^{n_i}\in F_{i-1}</math> for some integer <math>n_i>1.</math> An algebraic solution of the initial polynomial equation exists if and only if there exists such a sequence of fields such that <math>F_k</math> contains a solution.

For having [[normal extension]]s, which are fundamental for the theory, one must refine the sequence of fields as follows. If <math>F_{i-1}</math> does not contain all <math>n_i</math>-th [[roots of unity]], one introduces the field <math>K_i</math> that extends <math>F_{i-1}</math> by a [[primitive root of unity]], and one redefines <math>F_i</math> as <math>K_i(x_i).</math>

So, 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 group|cyclic]].

Conversely, 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 extension]]s.

===Galois correspondence===
The [[Galois correspondence]] establishes a [[one to one correspondence]] between the [[subextension]]s of a normal field extension <math>F/E</math> and the subgroups of the Galois group of the extension. This correspondence maps a field {{mvar|K}} such <math>E\subseteq K \subseteq F</math> to the [[Galois group]] <math>\operatorname{Gal}(F/K)</math> of the [[field automorphism|automorphisms]] of {{mvar|F}} that leave {{mvar|K}} fixed, and, conversely, maps a subgroup {{mvar|H}} of <math>\operatorname{Gal}(F/E)</math> to the field of the elements of {{mvar|F}} that are fixed by {{mvar|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) is [[solvable group|solvable]], that is, it contains a sequence of subgroups such that each is [[normal subgroup|normal]] in the preceding one, with a [[quotient group]] that is [[cyclic group|cyclic]]. (Solvable groups are commonly defined with [[abelian group|abelian]] instead of cyclic quotient groups, but the [[fundamental theorem of finite abelian groups]] shows that the two definitions are equivalent).

So, for proving the Abel–Ruffini theorem, it remains to show that the [[symmetric group]] <math>S_5</math> is not solvable, and that there are polynomials with symmetric Galois groups.

===Solvable symmetric groups===
For {{math|''n'' > 4}}, the [[symmetric group]] <math>\mathcal S_n</math> of degree {{mvar|n}} has only the [[alternating group]] <math>\mathcal A_n</math> as a nontrivial [[normal subgroup]] (see {{slink|Symmetric group|Normal subgroups}}). For {{math|''n'' > 4}}, the alternating group <math>\mathcal A_n</math> is [[simple group|simple]] (that is, it does not have any nontrivial normal subgroup) and not [[abelian group|abelian]]. This implies that both <math>\mathcal A_n</math> and <math>\mathcal S_n</math> are not [[solvable group|solvable]] for {{math|''n'' > 4}}. Thus, the Abel–Ruffini theorem results from the existence of polynomials with a symmetric Galois group; this will be shown in the next section.

On the other hand, for {{math|''n'' ≤ 4}}, the symmetric group and all its subgroups are solvable. This explains the existence of the [[quadratic formula|quadratic]], [[cubic formula|cubic]], and [[quartic formula|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 {{mvar|n}} is the equation
:<math>x^n+a_1x^{n-1}+ \cdots+ a_{n-1}x+a_n=0, </math>
where <math>a_1,\ldots, a_n</math> are distinct [[indeterminate (variable)|indeterminates]]. This is an equation defined over the [[field (mathematics)|field]] <math>F=\Q(a_1,\ldots,a_n)</math> of the [[rational fraction]]s in <math>a_1,\ldots, a_n</math> with [[rational number]] coefficients. The original Abel–Ruffini theorem asserts that, for {{math|''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 {{mvar|F}} of the equation is the [[symmetric group]] <math>\mathcal S_n</math> (this Galois group is the group of the [[field automorphism]]s of the [[splitting field]] of the equation that fix the elements of {{mvar|F}}, where the splitting field is the smallest field containing all the roots of the equation).

For proving that the Galois group is <math>\mathcal S_n,</math> it is simpler to start from the roots. Let <math>x_1, \ldots, x_n</math> be new indeterminates, aimed to be the roots, and consider the polynomial
:<math>P(x)=x^n+b_1x^{n-1}+ \cdots+ b_{n-1}x+b_n= (x-x_1)\cdots (x-x_n).</math>
Let <math>H=\Q(x_1,\ldots,x_n)</math> be the field of the rational fractions in <math>x_1, \ldots, x_n,</math> and <math>K=\Q(b_1,\ldots, b_n)</math> be its subfield generated by the coefficients of <math>P(x).</math> The [[permutation]]s of the <math>x_i</math> induce automorphisms of {{mvar|H}}. [[Vieta's formulas]] imply that every element of {{mvar|K}} is a [[symmetric function]] of the <math>x_i,</math> and is thus fixed by all these automorphisms. It follows that the Galois group <math>\operatorname{Gal}(H/K)</math> is the symmetric group <math>\mathcal S_n.</math>

The [[fundamental theorem of symmetric polynomials]] implies that the <math>b_i</math> are [[algebraic independence|algebraic independent]], and thus that the map that sends each <math>a_i</math> to the corresponding <math>b_i</math> is a field isomorphism from {{mvar|F}} to {{mvar|K}}. This means that one may consider <math>P(x)=0</math> 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 {{mvar|n}} cannot be solved in radicals for {{math|''n'' > 4}}.

====Explicit example====
{{see also|Galois theory#A non-solvable quintic example}}

The equation <math>x^5-x-1=0</math> is not solvable in radicals, as will be explained below.

Let {{mvar|q}} be <math>x^5-x-1</math>.
Let {{mvar|G}} be its Galois group, which acts faithfully on the set of complex roots of {{mvar|q}}.
Numbering the roots lets one identify {{mvar|G}} with a subgroup of the symmetric group <math>\mathcal S_5</math>.
Since <math>q \bmod 2</math> factors as <math>(x^2 + x + 1)(x^3 + x^2 + 1)</math> in <math>\mathbb{F}_2[x]</math>, the group {{mvar|G}} contains a permutation <math>g</math> that is a product of disjoint cycles of lengths 2 and 3 (in general, 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 {{mvar|G}} also contains <math>g^3</math>, which is a [[transposition (mathematics)|transposition]]. Since <math>q \bmod 3</math> is irreducible in <math>\mathbb{F}_3[x]</math>, the same principle shows that {{mvar|G}} contains a [[cyclic permutation|5-cycle]]. Because 5 is prime, any transposition and 5-cycle in <math>\mathcal S_5</math> generate the whole group; see {{slink|Symmetric group|Generators and relations}}. Thus <math>G = \mathcal S_5</math>. Since the group <math>\mathcal S_5</math> is not solvable, the equation <math>x^5-x-1=0</math> is not solvable in radicals.