Why a polynomial is solvable by radicals iff its Galois group is solvable, when $char(F)=0$?

Greets

Recently I studied the fact that if a polynomial ina field of characteristic zero is solvable by radicals, then its Galois group is a solvable group, from the book "A course in abstract algebra" by Jhon Fraleigh, where the author also stated that the other direction of this property is also true, so I want to know where I can find a prove of this, or if the proof is not too hard, I would like to see it here.

Thanks

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See my answer in math.stackexchange.com/questions/152818/…. – lhf Dec 7 '12 at 19:43
I recommend Jacobson's Basic Algebra I – Jyrki Lahtonen Dec 7 '12 at 22:32

Definition 1 Let $K$ be a field of characteristic $0$. Let $p$ be a prime number. Let $\alpha$ be an element of an algebraic closure of $K$ such that $\alpha^p \in K$ and $X^p - \alpha^p$ is irreducible over $K$. Then we call $K(\alpha)/K$ a simple prime radical extension.

Definition 2 Let $K = K_0 \subset K_1 \subset \cdots \subset K_n = L$ be a tower of fields. If $K_i/K_{i-1}$ is a simple prime radical extension for $i =1, \dots, n$, we call $L/K$ is a prime radical extension.

Definition 3 Let $L/K$ be a field extension. If there exists a prime radical extension $E/K$ such that $L$ is a subfield of $E$, we call $L/K$ a prime radically solvable extension.

Lemma 1 Let $K \subset M \subset L$ be a tower of fields. Suppose $M/K$ and $L/M$ are prime radically solvable extensions. Then $L/K$ is also a prime radically solvable extension.

Proof: This is proved in Lemma 10 of my answer to this question.

Lemma 2 Let $L/K$ be a finite Galois extension of prime degree $p$. Suppose $char(K) = 0$ and $K$ has a $p$-th primitive root of unity. Then $L/K$ is a simple prime radical extension.

Proof: By this question, there is an element $\alpha$ of $L$ such that $L = K(\alpha)$ and $\alpha^p$ is an element of $K$. Since $p = [L\colon K]$, $X^p - \alpha^p$ is irreducible over $K$. Hence $L/K$ is a simple prime radical extension. QED

Lemma 3 Let $L/K$ be a finite Galois extension of prime degree $p$. Suppose $char(K) = 0$. Then $L/K$ is a prime radically solvable extension.

Proof: Let $\Omega$ be an algebraic closure of $L$. Let $\zeta$ be a primitive $p$-th root of unity in $\Omega$. Then $L(\zeta)/K(\zeta)$ is a Galois extension and $Gal(L(\zeta)/K(\zeta))$ is isomorphic to a subgroup of $Gal(L/K)$. Hence it is a cyclic group of order $p$ or $1$. By Lemma 2, $L(\zeta)/K(\zeta)$ is a prime radically solvable extension. On the other hand, by Lemma 6 of my answer to this question, $K(\zeta)/K$ is a prime radically solvable extension. Hence, by Lemma 1, $L(\zeta)/K$ is a prime radically solvable extension. Hence $L/K$ is a prime radically solvable extension. QED

Theorem Let $K$ be a field of characteristic $0$. Let $L/K$ be a finite Galois extension. Suppose $G = Gal(L/K)$ is sovable. Then $L/K$ is a prime radically solvable extension.

Proof: There exists a tower of subgroups of $G$: $G = G_n \supset G_{n-1} \supset \cdots \supset G_1 \supset G_0 = 1$, where each $G_i/G_{i-1}$ is a cyclic group of prime degree. Hence there exists a tower of fields: $K_0 = K \subset K_1 \subset \cdots \subset K_n = L$, where each $K_i/K_{i-1}$ is a cyclic Galois extension of prime degree. By Lemma 3, $K_i/K_{i-1}$ is a prime radically solvable extension. Hence, by Lemma 1, $L/K$ is a prime radically solvable extension. QED

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