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. 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
th root. So, an algebraic solution produces a sequence :F_0\subseteq F_1\subseteq \cdots \subseteq F_k of fields, and elements x_i\in F_i such that F_i=F_{i-1}(x_i) for i=1,\ldots, k, with x_i^{n_i}\in F_{i-1} for some integer n_i>1. An algebraic solution of the initial polynomial equation exists if and only if there exists such a sequence of fields such that F_k contains a solution. For having
normal extensions, which are fundamental for the theory, one must refine the sequence of fields as follows. If F_{i-1} does not contain all n_i-th
roots of unity, one introduces the field K_i that extends F_{i-1} by a
primitive root of unity, and one redefines F_i as K_i(x_i). 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. 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 extensions.
Galois correspondence The
Galois correspondence establishes a
one to one correspondence between the
subextensions of a normal field extension F/E and the subgroups of the Galois group of the extension. This correspondence maps a field such E\subseteq K \subseteq F to the
Galois group \operatorname{Gal}(F/K) of the
automorphisms of that leave fixed, and, conversely, maps a subgroup of \operatorname{Gal}(F/E) to the field of the elements of that are fixed by . 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, that is, 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). So, for proving the Abel–Ruffini theorem, it remains to show that the
symmetric group S_5 is not solvable, and that there are polynomials with symmetric Galois groups.
Solvable symmetric groups For , the
symmetric group \mathcal S_n of degree has only the
alternating group \mathcal A_n as a nontrivial
normal subgroup (see ). For , the alternating group \mathcal A_n is
simple (that is, it does not have any nontrivial normal subgroup) and not
abelian. This implies that both \mathcal A_n and \mathcal S_n are not
solvable for . 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 , the symmetric group and all its subgroups are solvable. 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 is the equation :x^n+a_1x^{n-1}+ \cdots+ a_{n-1}x+a_n=0, where a_1,\ldots, a_n are distinct
indeterminates. This is an equation defined over the
field F=\Q(a_1,\ldots,a_n) of the
rational fractions in a_1,\ldots, a_n with
rational number coefficients. The original Abel–Ruffini theorem asserts that, for , this equation is not solvable in radicals. In view of the preceding sections, this results from the fact that the
Galois group over of the equation is the
symmetric group \mathcal S_n (this Galois group is the group of the
field automorphisms of the
splitting field of the equation that fix the elements of , where the splitting field is the smallest field containing all the roots of the equation). For proving that the Galois group is \mathcal S_n, it is simpler to start from the roots. Let x_1, \ldots, x_n be new indeterminates, aimed to be the roots, and consider the polynomial :P(x)=x^n+b_1x^{n-1}+ \cdots+ b_{n-1}x+b_n= (x-x_1)\cdots (x-x_n). Let H=\Q(x_1,\ldots,x_n) be the field of the rational fractions in x_1, \ldots, x_n, and K=\Q(b_1,\ldots, b_n) be its subfield generated by the coefficients of P(x). The
permutations of the x_i induce automorphisms of .
Vieta's formulas imply that every element of is a
symmetric function of the x_i, and is thus fixed by all these automorphisms. It follows that the Galois group \operatorname{Gal}(H/K) is the symmetric group \mathcal S_n. The
fundamental theorem of symmetric polynomials implies that the b_i are
algebraic independent, and thus that the map that sends each a_i to the corresponding b_i is a field isomorphism from to . This means that one may consider 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 cannot be solved in radicals for .
Explicit example The equation x^5-x-1=0 is not solvable in radicals, as will be explained below. Let be x^5-x-1. Let be its Galois group, which acts faithfully on the set of complex roots of . Numbering the roots lets one identify with a subgroup of the symmetric group \mathcal S_5. Since q \bmod 2 factors as (x^2 + x + 1)(x^3 + x^2 + 1) in \mathbb{F}_2[x], the group contains a permutation g 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 also contains g^3, which is a
transposition. Since q \bmod 3 is irreducible in \mathbb{F}_3[x], the same principle shows that contains a
5-cycle. Because 5 is prime, any transposition and 5-cycle in \mathcal S_5 generate the whole group; see . Thus G = \mathcal S_5. Since the group \mathcal S_5 is not solvable, the equation x^5-x-1=0 is not solvable in radicals. ==Cayley's resolvent==