Důkaz (k dopracování):
Řešitelnost
algebraických rovnic
The following proof is based on Galois theory. Historically, Ruffini and Abel's proofs precede Galois theory. One of the fundamental theorems of Galois theory states that an equation is solvable in radicals if and only if it has a solvable Galois group, so the proof of the Abel–Ruffini theorem comes down to computing the Galois group of the general polynomial of the fifth degree. Let y_1 be a real number transcendental over the field of rational numbers Q, and let y_2 be a real number transcendental over Q(y_1), and so on to y_5 which is transcendental over Q(y_1, y_2, y_3, y_4). These numbers are called independent transcendental elements over Q. Let E = Q(y_1, y_2, y_3, y_4, y_5) and let f(x) = (x  y_1)(x  y_2)(x  y_3)(x  y_4)(x  y_5) \in E[x]. Multiplying f(x) out yields the elementary symmetric functions of the y_n: s_1 = y_1 + y_2 + y_3 + y_4 + y_5
The coefficient of x^n in f(x) is thus (1)^{5n} s_{5n}. Because our independent transcendentals y_n act as indeterminates over Q, every permutation \sigma in the symmetric group on 5 letters S_5 induces an automorphism \sigma' on E that leaves Q fixed and permutes the elements y_n. Since an arbitrary rearrangement of the roots of the product form still produces the same polynomial, e.g.: (y  y_3)(y  y_1)(y  y_2)(y  y_5)(y  y_4) is still the same polynomial as (y  y_1)(y  y_2)(y  y_3)(y  y_4)(y  y_5) the automorphisms \sigma' also leave E fixed, so they are elements of the Galois group G(E/Q). Now, since S_5 = 5! it must be that G(E/Q) \ge 5!, as there could possibly be automorphisms there that are not in S_5. However, since the relative automorphisms Q for splitting field of a quintic polynomial has at most 5! elements, G(E/Q) = 5!, and so G(E/Q) must be isomorphic to S_5. Generalizing this argument shows that the Galois group of every general polynomial of degree n is isomorphic to S_n. And what of S_5? The only composition series of S_5 is S_5 \ge A_5 \ge \{e\} (where A_5 is the alternating group on five letters, also known as the icosahedral group). However, the quotient group A_5/\{e\} (isomorphic to A_5 itself) is not an abelian group, and so S_5 is not solvable, so it must be that the general polynomial of the fifth degree has no solution in radicals. Since the first nontrivial normal subgroup of the symmetric group on n letters is always the alternating group on n letters, and since the alternating groups on n letters for n \ge 5 are always simple and nonabelian, and hence not solvable, it also says that the general polynomials of all degrees higher than the fifth also have no solution in radicals. Note that the above construction of the Galois group for a fifth degree polynomial only applies to the general polynomial, specific polynomials of the fifth degree may have different Galois groups with quite different properties, e.g. x^5  1 has a splitting field generated by a primitive 5th root of unity, and hence its Galois group is abelian and the equation itself solvable by radicals. However, since the result is on the general polynomial, it does say that a general "quintic formula" for the roots of a quintic using only a finite combination of the arithmetic operations and radicals in terms of the coefficients is impossible. Q.E.D. History Objev důkazu nebyl snadný:
