Proving that extension by radicals implies solvable group I'm trying to understand the following excerpt from Fraleigh's A First Course In Abstract Algebra, Seventh Edition, pp. 472-473:

56.4 Theorem Let $F$ be a field of characteristic zero, and let $F\subseteq E\subseteq K\subseteq \overline F$, where $E$ is a normal extension of $F$ and $K$ is an extension of $F$ by radicals. Then $G(E/F)$ is a solvable group.
Proof. Since $K$ is an extension by radicals, $K = F(\alpha_1,\ldots,\alpha_r)$ where $\alpha_1^{n_1}\in F$ and each $\alpha_i^{n_i}\in F(\alpha_1,\ldots,\alpha_{i-1})$ for $1< i\leq r$. Now form the splitting field $L_1$ of $f_1(x) = x^{n_1} - \alpha_1^{n_1}$ over $F$. Then $L_1$ is a normal extension of $F$, and since $\alpha_1^{n_1}\in F$, the preceding lemma implies that $G(L_1/F)$ is a solvable group. By assumption, $\alpha_2^{n_2}\in F(\alpha_1)\subseteq L_1$, and we form the polynomial
  $$
 f_2(x) = \prod_{\sigma\in G(L_1/F)} [x^{n_2} - \sigma(\alpha_2)^{n_2}].
$$

My question is, why is Fraleigh able to use the expression $\sigma(\alpha_2)^{n_2}$, when we don't necessarily know that $\alpha_2$ is in the domain of $\sigma$, $L_1$? Shouldn't we be using $\alpha_2^{n_2}$, which we know to be in the domain of $\sigma$, so then the definition of $f_2(x)$ would be
$$
 f_2(x) = \prod_{\sigma\in G(L_1/F)} [x^{n_2} - \sigma(\alpha_2^{n_2})]?
$$
I see that if $\sigma(\alpha_2)$ were defined, then by the properties of homomorphisms we would have $\sigma(\alpha_2^{n_2}) = \sigma(\alpha_2)^{n_2}$, but I don't think we know that there is a value for $\sigma(\alpha_2)$. Please advise.
 A: Abstract
The alternative definition you suggest for $f_2(x)$ is the correct one. I will support this claim by giving a complete proof of the theorem, following the same procedure as outlined in the book, but including all the major details omitted in the original proof.

Preliminaries
Let me first introduce some notation. For a subset $A\subseteq\bar{F}$, let $S(A)\subseteq\bar{F}$ be the set of elements generated by the elements of $A$, that is, all elements that can be written as a finite combination of sums, differences, products and quotients of the elements of $A$. It follows that if $A$ contains an element $a\neq 0$, then $S(A)$ is a field: $0 = a-a \in S(A)$, $1 = a/a \in S(A)$, all elements has additive and (except 0) multiplicative inverses, and $S(A)$ is closed under addition and multiplication. The remaining field properties are fulfilled by virtue of $S(A)$ being a subset of the field $\bar{F}$. It is clear that $S(A)$ must be the smallest subfield of $\bar{F}$ containing $A$, so we may for instance use this notation to express simple algebraic extentions $F(\alpha)$ as $S(F\cup\{\alpha\})$.
$S$ satisfies some simple equations; first, it is clear that if $A\subseteq B$, then $S(A)\subseteq S(B)$. It is also true that $S(S(A)) = S(A)$: On one hand, $A\subseteq S(A)$ implies $S(A)\subseteq S(S(A))$. Conversely, suppose $x\in S(S(A))$. Then $x$ can be expressed as a combination of a finite number of elements of $S(A)$ and the four aritmetic types. Each element of $S(A)$ appearing in the expression of $x$ may in turn be written as separate expressions using a finite number of elements of $A$. Then all these expressions can be substituted into the expression of $x$, and the resulting expression shows that $x\in S(A)$.
We now move on to the identity
$$S(S(A)\cup B) = S(A\cup B)$$
which we shall need in the proof. One inclusion follows from the inclusion $A\cup B\subseteq S(A)\cup B$. For the other, observe that $S(A)\subseteq S(A\cup B)$ and $B\subseteq S(B)\subseteq S(A\cup B)$ implies $S(A)\cup B\subseteq S(A\cup B)$, and therefore $S(S(A)\cup B)\subseteq S(S(A\cup B)) = S(A\cup B)$.
Finally, let $Z(f(x))\subseteq\bar{F}$ denote the set of zeroes of any given polynomial $f(x)\in\bar{F}[x]$. Observe that $Z$ satisfies the identity $Z(f(x)g(x)) = Z(f(x))\cup Z(g(x))$. Also note that the splitting field of a polynomial $f(x)\in F[x]$ over $F$ can now be expressed in a concise way as $S(F\cup Z(f(x)))$.

Proof of Theorem 56.4
First we use that $K$ is an extension by radicals to establish the existence of a subset $\{\alpha_1,...,\alpha_r\}\subseteq\bar{F}$ such that $K = F(\alpha_1,...,\alpha_r)$ where $\alpha_1^{n_1}\in F$ and $\alpha_i^{n_i}\in F(\alpha_1,...,\alpha_{i-1})$ for $1<i\leq r$. Let $K_i$ denote the extension field $F(\alpha_1,...,\alpha_i)$, or equivalently, $S(F\cup\{\alpha_1,\dots,\alpha_i\})$.
We will prove the theorem by constructing a sequence of finite normal extensions $L_i$ of $F$ such that $K_i\leq L_i$ and $G(L_i/F)$ is solvable. In particular, the last field in the sequence $L = L_r$ will have a solvable Galois group $G(L/F)$. The subgroup $G(L/E)$ is normal since $L$ is a normal extension of $E$ ($L$ is separable over $E$ by 51.9, and a splitting field over $E$ since it is a splitting field over $F\leq E$). Hence by the Galois theorem 56.3 there is an isomorphism $G(E/F)\simeq G(L/F)/G(L/E)$. But $G(L/F)/G(L/E)$ is solvable by exercise 29 of section 35, and therefore $G(E/F)$ is solvable, which is what we were to prove.
Now to the definition of $L_i$: Let $L_0 = F$, and define the rest recursively by the formula
\begin{equation}
L_i = S(L_{i-1}\cup Z(f_i(x)))
   \quad\text{where}\quad 
f_i(x) = \prod_{\sigma\in G(L_{i-1}/F)}[x^{n_i}-\sigma(\alpha_i^{n_i})]
\end{equation}
We will show that $L_i$ satisfies the following three properties:

*

*$K_i\leq L_i$

*$L_i$ is a finite normal extension of $F$

*$G(L_i/F)$ is solvable

The first propery implies that $\alpha_i^{n_i}\in L_{i-1}$, which means the definition of $f_i(x)$ makes sense. The other two properties are precisely the ones we needed to deduce that $G(E/F)$ is solvable. Hence we are done after showing these three properties.

*

*$L_0 = K_0 = F$ by definition. Assume by induction that $K_{i-1}\leq L_{i-1}$. Then
$$\begin{eqnarray*}
K_i &=& S(F\cup\{\alpha_1,\dots,\alpha_i\}) = S(S(F\cup\{\alpha_1,\dots,\alpha_{i-1}\})\cup\{\alpha_i\}) \\\\
&=& S(K_{i-1}\cup\{\alpha_i\})\leq S(L_{i-1}\cup\{\alpha_i\})\leq S(L_{i-1}\cup Z(x^{n_i}-\alpha_i^{n_i}))\\
&\leq& S(L_{i-1}\cup\left(\bigcup_\sigma Z(x^{n_i}-\sigma(\alpha_i^{n_i}))\right)) = S(L_{i-1}\cup Z(\prod_\sigma [x^{n_i}-\sigma(\alpha_i^{n_i})]))\\
&=& S(L_{i-1}\cup Z(f_i(x))) = L_i
\end{eqnarray*}$$


*$L_i$ is a finite extension of $L_{i-1}$, which we as induction hypothesis may assume is a finite extension of $F$. But then $[L_i:F] = [L_i:L_{i-1}][L_{i-1}:F]$ is also finite, so $L_i$ is a finite extension of $F$. $L_i$ is a separable extension of $F$ since $F$ is perfect (Theorem 51.13). Thus it remains only to prove that $L_i$ is a splitting field over $F$. Assume by induction that $L_{i-1}$ is a splitting field over $F$. This means there exists a set of polynomials $g_j(x)\in F[x]$ for $j\in J$ such that
$$L_{i-1} = S(F\cup\left(\bigcup_{j\in J}Z(g_j(x))\right))$$
Since
$$ L_i = S(L_{i-1}\cup Z(f_i(x))) = S(F\cup\left(\bigcup_{j\in J}Z(g_j(x))\right)\cup Z(f_i(x))) $$
we see that $L_i$ is a splitting field over $F$ if $f_i(x)$ happens to be an element of $F[x]$. We will show that this indeed is the case. Observe first that any ring homomorphism $\phi:R\rightarrow R'$ induces a homomorphism $\bar{\phi}:R[x]\rightarrow R'[x]$ such that $\bar{\phi}(c_0+c_1x+\cdots+c_nx^n) = \phi(c_0)+\phi(c_1)x+\cdots+\phi(c_n)x^n$. Now, since each $\tau\in G(L_{i-1}/F)$ is a ring homomorphism (automorphism) $\tau:L_{i-1}\rightarrow L_{i-1}$, it has an induced homomorphism $\bar{\tau}:L_{i-1}[x]\rightarrow L_{i-1}[x]$. Applied to $f_i(x)$ this yields
$$ \bar{\tau}(f_i(x)) = \bar{\tau}\left(\prod_\sigma[x^{n_i}-\sigma(\alpha_i^{n_i})]\right) = \prod_\sigma\bar{\tau}(x^{n_i}-\sigma(\alpha_i^{n_i})) = \prod_\sigma(x^{n_i}-\tau\sigma(\alpha_i^{n_i})) $$
where the last expression is just a rearrangment of the factors in $f_i(x)$, so $\bar{\tau}(f_i(x)) = f_i(x)$. On the other hand, if $c_0+c_1x+\cdots+c_mx^m$ is the expression of $f_i(x)$ after multiplying out all factors, we get
$$ f_i(x) = \bar{\tau}(f_i(x)) = \bar{\tau}(c_0+c_1x+\cdots+c_mx^m) = \tau(c_0)+\tau(c_1)x+\cdots+\tau(c_m)x^m $$
which means $\tau(c_k)=c_k$ for all $\tau\in G(L_{i-1}/F)$. The elements of $L_i$ with this property are precisely those contained in $F$: by assumption $L_{i-1}$ is a normal extension of $F$, so $(L_{i-1})_{G(L_{i-1}/F)} = F$ by the Galois theorem. Hence $c_k\in F$ for all $k$ and consequently $f_i(x)\in F[x]$.


*We shall construct a tower of fields $L_{i-1} = M_0\leq M_1\leq\cdots\leq M_m = L_i$ such that all $M_j$ are normal extensions of $M_0$, and such that the Galois groups $G(M_{j+1}/M_j)$ are solvable. The series
$$ \{e\} = G(M_m/M_m)\leq G(M_m/M_{m-1})\leq\cdots\leq G(M_m/M_0) = G(L_i/L_{i-1})\leq G(L_i/F) $$
is then subnormal (in the non-strict inequality sense): Note first that since all
fields $M_j$ are normal over a common field $M_0$, each $M_j$ must be normal over all subfields $M_k$ in the tower. In particular we have that both $M_m$ and $M_{j+1}$ are normal extensions of $M_j$. But then it follows from the Galois theorem that $G(M_m/M_{j+1})$ is a normal subgroup of $G(M_m/M_j)$. Also, since both $L_i$ and $L_{i-1}$ are normal extensions of $F$ (property 2), we can use the Galois theorem to conclude that $G(L_i/L_{i-1})$ is a normal subgroup of $G(L_i/F)$. Next, by construction, each quotient group
$$G(M_m/M_j)/G(M_m/M_{j+1})\simeq G(M_{j+1}/M_j)$$
is solvable. The last quotient group $G(L_i/F)/G(L_i/L_{i-1})\simeq G(L_{i-1}/F)$ is also solvable, if we assume by induction that $G(L_{i-1}/F)$ is solvable. Thus, since each consequtive pair of groups has a solvable quotient group, we may by excercise 7 insert simple abelian refinements between each pair. The resulting series (after deleting any repeated entries) is the desired solution of $G(L_i/F)$.
We now turn to the definition of $M_j$: Let
$$ M_j = S(M_{j-1}\cup Z(x^{n_i}-\sigma_j(\alpha_i^{n_i}))) $$
where $\{\sigma_1,\dots,\sigma_m\} = G(L_{i-1}/F)$ is an enumeration of the elements of $G(L_{i-1}/F)$. We immediately see that $M_j$ is the splitting field of $x^{n_i}-\sigma_j(\alpha_i^{n_i})$ over $M_{j-1}$, and since this is exactly the kind of extension discussed in the preceding lemma 56.3, it follows that $G(M_j/M_{j-1})$ is solvable. Thus what remains to be shown is that each $M_j$ is a normal extension of $M_0$ and that $M_m = L_i$. Showing that $M_j$ is a normal extension of $M_0$ amounts to showing that $M_j$ is a splitting field over $M_0$. By expanding the recursive definition of $M_j$, and using the identity $S(S(A)\cup B) = S(A\cup B)$ repeatedly, we find that
$$ M_j = S(M_0\cup\bigcup_{k=1}^j Z(x^{n_i}-\sigma_k(\alpha_i^{n_i}))) $$
This shows that $M_j$ is the splitting field of $\{x^{n_i}-\sigma_1(\alpha_i^{n_i}),\dots,x^{n_i}-\sigma_j(\alpha_i^{n_i})\}$ over $M_0$. For $j=m$ we have that
$$ M_m = S(M_0\cup\bigcup_{k=1}^m Z(x^{n_i}-\sigma_k(\alpha_i^{n_i}))) = S(M_0\cup Z(\prod_{k=1}^m[x^{n_i}-\sigma_k(\alpha_i^{n_i})])) = S(L_{i-1}\cup Z(f_i(x))) $$
so $M_m = L_i$, completing the proof.

Appendix
Here I will solve the two exercises that were referenced in the proof. To simplify the proofs that the exercises ask for, let me start by introducing a correspondence theorem which does not appear in the book:
Theorem:  Let $N$ be a normal subgroup of a group $G$, and let $\gamma: G\rightarrow G/N$ be the canonical homomorphism. Then the map $H\mapsto\gamma[H]$ is a bijective correspondence between the subgroups of $G$ containing $N$ and the subgroups of $G/H$.
Proof: Recall that $G/N$ is constructed by creating a partition of the elements of $G$, letting the elements of $G/N$ be the various bins in this partition. The bins in $G/N$ are the subsets of the form $xN = \{xn\,|\,n\in N\}$ for $x\in G$. The group operation on these bins is defined thus: $(xN)(yN) = (xy)N$. A subgroup of $G/N$ is a subcollection of these bins having the property of being closed under group operation and element inversion. Let $H'$ be any subgroup of $G/N$, and let $H\subseteq G$ be the union of all the elements of each bin in $H'$. If $x,y\in H$ then $xN,yN\in H'$, so $(xy)N\in H'$, but then $xy\in H$. Similarly $x\in H\Rightarrow xN\in H'$ $\Rightarrow x^{-1}N\in H'\Rightarrow x^{-1}\in H$. This shows that $H$ is a subgroup of $G$. Note also that $H$ contains $eN = N$ and satisfies $\gamma[H] = H'$, hence the correspondence is surjective. To see injectivity we will begin in the opposite end and start with an arbitrary subgroup $H\leq G$ containing $N$. Since $N$ is necessarily a normal subgroup of $H$, we can partition $H$ into a group of bins $H/N$. The bins in this group are of the form $xN$ for $x\in H$, which evidently form a subset of the bins in $G/N$, since these are all of the form $xN$ for $x\in G$. Hence $H/N$ is a subgroup of $G/N$. Note now that we can recover the mapped subgroup $H$ from the image $\gamma[H] = H/N$ by taking the union of all elements of each bin in $H/N$ (just observe that $xN\subseteq H$ for each bin $xN\in H/N$). This implies injectivity, and the correspondence is proven.
$$\tag*{$\Box$}$$
If $N$ is a normal subgroup of $G$ and $H$ is a subgroup containing $N$, we saw in the proof of the above correspondence theorem that $H/N$ is a subgroup of $G/N$, and that its corresponding subgroup in $G$ is $H$. We will therefore in what follows write the corresponding subgroup of $H$ as $H/N$.
The first exercise I'm going to solve is not litteraly the one given in the book, but rather the natural lemma from which the result in the book will follow:
$$ $$
Exercise 56.7    Let $N$ be a normal subgroup of $G$. If $G/N$ is solvable, then there exists a refinement $N=H_0<H_1<\cdots<H_n=G$ such that $H_{i+1}/H_i$ are simple abelian for $0\leq i<n$.
Proof: All subgroups of $G/N$ are of the form $H/N$ where $H$ is a subgroup of $G$ containing $N$. A solution of $G/N$ may thus be written
$$ N/N=H_0/N<H_1/N<\cdots <H_n/N=G/N $$
We shall show that the series
$$ N=H_0<H_1<\cdots <H_n=G $$
is a simple abelian refinement of $N<G$. Observe first that $H_i$ is a normal subgroup of $H_{i+1}$ since $H_i/N$ is a normal subgroup of $H_{i+1}/N$; this follows by lemma 34.3, which in plain text states that the subgroup correspondence theorem we proved above maps normal subgroups to normal subgroups. But then $H_{i+1}/H_i$ is simple abelian by the third isomorphism theorem 34.7:
$$ H_{i+1}/H_i\simeq (H_{i+1}/N)/(H_i/N) $$
hence $N=H_0<H_1<\cdots <H_n=G$ is a refinement with the desired propery.
$$\tag*{$\Box$}$$
$$ $$
Exercise 35.29    Prove that a homomorphic image of a solvable group is solvable.
Proof: Let $G$ be a solvable group. The homomorphic image of $G$ is isomorphic to $G/N$ where $N$ is the kernel of the homomorphism (1st isomorphism theorem), so we shall prove that $G/N$ is solvable for normal subgroups $N\leq G$. If $N=\{e\}$ or $N=G$ the group $G/N$ is obviously solvable, so suppose $\{e\}<N<G$. By theorem 35.16, this series can be refined to a composition series $\{e\}=H_0<\cdots <H_n=G$, and by the comment following definition 35.18 the factor groups $H_{i+1}/H_i$ must be abelian. Let $H_k = N$. Then $H_i/N$ are subgroups of $G/N$ for $k\leq i\leq n$, which means we have a series
$$ N/N=H_k/N<H_{k+1}/N<\cdots <H_n/N=G/N $$
This is subnormal by the fact that $H_i$ is a normal subgroup of $H_{i+1}$, since this implies $H_i/N$ is a normal subgroup of $H_{i+1}/N$, by lemma 34.3. But then it follows from the third isomorphism theorem that $(H_{i+1}/N)/(H_i/N)\simeq H_{i+1}/H_i$, which means $(H_{i+1}/N)/(H_i/N)$ is simple abelian, hence the above series is a solution of $G/N$.
$$\tag*{$\Box$}$$
