TITLE: Is there a flaw on "Galois for beginners"?
QUESTION [9 upvotes]: I guess that the following question is only for those who by any chance have read the text "Galois for beginners" by John Stillwell.  Although the text is short and there might be some remarkable people that might be able to read it and understand it fast enough. In any case, I'm thankful to anyone who answers.
The aim of the text is to prove that there is no solution by radicals for polynomials of degree 5 or higher, and the issue I'm having is that the author seems to have made a mistake on a proof along the way. Reading this text -as brief as it is- is proving to be really hard for me, since I know little about abstract algebra, but it is the shortest 'proof' I've been able to find. 
What I wish to know, is if this mistake the author made will at the end ruin the whole proof (meaning that I would have lost my time) or if even with this flaw, the author is still able to prove Abel-Ruffini Theorem.

The flaw:
(For context, $x_1,...,x_n$ refer to the roots of a generic $n$th degree polynomial)
Theorem (page $4$ of the pdf.). "for each radical extension $E$ of $ℚ(x_1,...,x_n)$ there is a radical extension $E′/E$ with automorphisms extending all permutations $σ$ of $x_1,...,x_n$."
Proof.
 "for each adjoined element, represented by the expression $e(x_1,...,x_n)$, and each permutation $σ$ of $x_1,...,x_n$ adjoin the element $e(σx_1,...,σx_n)$. Since there are only finitely many permutations of $σ$ then, the resulting field $E'⊇E$ is also a radical extension of $ℚ(x_1,...,x_n)$.
This gives a bijection (also called $σ$ ) of $E'$ sending each $f(x_1,...,x_n)∈E'$ (a rational function of $x_1,...x_n$ and the adjoined elements) to $f(σx_1,...,σx_n)$, and this bijection is clearly an automorphism of $E′$, extending the permutation $σ$."
This theorem should hold for a generic polynomial $P$, but it doesn't, since something like this might happen:
Let's say that $(x_1)^2+x_2^2=(x_3)^2$, yet $(σx_1)^3+(σx_2)^2≠(σx_3)^2$, meaning that $σ$ has mapped the same element into two different places, meaning that $σ$ can not be an automorphism. For a specific example let $x_1=i$, $x_2=3^{1/2}$ and $x_3=2^{1/2}$. 
So the theorem doesn't hold for all polynomials.

Will this at the end ruin the proof? Or is just insignificant to consider the polynomials in which the flaw occurs?
I will really appreciate any help/thoughts.

REPLY [4 votes]: The term "general" has a technical meaning here.  Namely, we are considering a polynomial whose coefficients are purely formal variables, not specific numbers.  More precisely, we are starting with the field $\mathbb{Q}(a_{n-1},\dots,a_0)$ of rational functions in $n$ variables $a_{n-1},\dots,a_0$, and we are considering the polynomial $$x^n+a_{n-1}x^{n-1}+\dots+a_0$$ whose coefficients are these variables.  We then formally extend this field to a field $\mathbb{Q}(x_1,\dots,x_n)$ in which this polynomial has $n$ roots $x_1,\dots,x_n$.  It turns out that this extension field is also a field of rational functions, this time with $x_1,\dots,x_n$ as the variables.  So since $x_1,\dots,x_n$ are just formal variables, there cannot be any polynomial relation between them of the sort you are concerned about.