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This is an extract of Wikipedia's page on Abel-Ruffini theorem

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 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 $y_n$

$$ s_1 = y_1 + y_2 + y_3 + y_4 + y_5 $$ $$ s_2 = y_1y_2 + y_1y_3 + y_1y_4 + y_1y_5 + y_2y_3 + y_2y_4 + y_2y_5 + y_3y_4 + y_3y_5 + y_4y_5 $$ $$ s_3 = y_1y_2y_3 + y_1y_2y_4 + y_1y_2y_5 + y_1y_3y_4 + y_1y_3y_5 + y_1y_4y_5 +y_2y_3y_4 + y_2y_3y_5 + y_2y_4y_5 + y_3y_4y_5 $$ $$ s_4 = y_1y_2y_3y_4 + y_1y_2y_3y_5 + y_1y_2y_4y_5 + y_1y_3y_4y_5 + y_2y_3y_4y_5 $$ $$ s_5 = y_1y_2y_3y_4y_5. $$ The coefficient of $x^n$ in $f(x)$ is thus $(-1)^{5-n} s_{5-n}$. 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 isomorphism|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 group|simple and non-abelian, 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 root of unity|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.

I understand all vocabulary used here, but my doubts are in the connection beetween the splitting field and the Galois group. Why the condition of normal group? Why the condition of abelian group?

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What condition of normal group? What condition of abelian group? – Chris Eagle Oct 29 '12 at 18:24
This uses a result in Galois theory about radical extensions. Maybe read – Cocopuffs Oct 29 '12 at 18:33
@ChrisEagle I think he is asking why we normal group concept is used here and what is the detail behind this. – zinking Nov 3 '12 at 2:41
@zinking: Yes, is that. – dot dot Nov 5 '12 at 13:56
up vote 1 down vote accepted

Tht's the main idea of Galois theory: To solve by radivcals mean sthat you walk your way up from $\mathbb Q$ to $E$ via intermediate fields $F_0=\mathbb Q, F_1, \ldots, F_n=E$, where each is obtained from the previous by adjoining a root of some element. For the intermedite fields $F_i$ in this situation (i.e. with $\mathbb Q$ at the bottom), the groups $G(E/F_i)$ have nice properties: Each is a subgroup of the next and is in fact the kernel of a homomorphism $G(E/\mathbb Q)\to G(F_i/\mathbb Q)$, hence normal. Moreover, the single steps are of the from $F_{i+1}=F_i[\sqrt[ k] \alpha]$ and it is a fact that for such extensions $G(F_{i+1}/F_i)$ is abelian.

Since $S_5$ is not solvable, there is no such sequence of fields because there is no suitable sequnce of Galois groups.

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