Editing Abel–Ruffini theorem (section)

===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.