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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.

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  • $\begingroup$ The case you are considering is galois. This only permute roots of irreducibles transitively.(i.e. $\sqrt{3},\sqrt{2}$ will have a disjoint orbit from $\pm i$ under permutation. Similarly for $\sqrt{3}$. $\endgroup$ – user45765 Sep 22 '17 at 14:22
  • $\begingroup$ An algebraic extension $K$ of $\mathbb{Q}(x_1,\ldots,x_n)$ is a field of functions in $n$ variables. With a permutation $\sigma \in S_n$ you can permute the $x_i$ to obtain a (possibly different) function field $K_\sigma$. Let $L = \prod_{\sigma \in S_n} K_\sigma$ (the field generated by each $K_\sigma$). It is an algebraic extension of $\mathbb{Q}(x_1,\ldots,x_n)$ and now $S_n$ embeds in $\text{Gal}(L/\mathbb{Q})$. Example : $K = \mathbb{Q}(x_1,x_2,\sqrt{x_1^2+1}), L = \mathbb{Q}(x_1,x_2,\sqrt{x_1^2+1},\sqrt{x_2^2+1})$. $\endgroup$ – reuns Sep 22 '17 at 14:28
  • $\begingroup$ @user45765. I apologize for taking so long. By saying that they will have a disjoint orbit from $i$, you mean that it doesn't matter how many permutations we apply to $2^{1/2}$ , it will never be equal to $i$, right? $\endgroup$ – Leo Sep 22 '17 at 15:04
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    $\begingroup$ @Leo That is the point. Since the ground field $Q$ is fixed by this galois action, the polynomial of $x^2-2$ remains and it will never be $x^2+1$ under any condition. $\endgroup$ – user45765 Sep 22 '17 at 18:29
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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.

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  • $\begingroup$ I understand that 3 days have passed, I alologize for that. I understand everything except the conclusion (that there cannot be any polynomial relation between them of the sort [I'm] concerned about). $\endgroup$ – Leo Sep 25 '17 at 15:40
  • $\begingroup$ I understand that the theorem is talking about general undefined variables, while in my example I'm giving them values. But how is it that one can prove that there won't exists such relations in general? $\endgroup$ – Leo Sep 25 '17 at 15:41
  • $\begingroup$ Well, that's pretty much just the definition of "field of rational functions". If $\mathbb{Q}(x_1,\dots,x_n)$ is a field of rational functions in the variables $x_1,\dots,x_n$, that means it is just the field of fractions of the ring $\mathbb{Q}[x_1,\dots,x_n]$ of formal polynomial expressions in the variables with coefficients in $\mathbb{Q}$. So if two polynomials in them are equal, they must be the same polynomial. $\endgroup$ – Eric Wofsey Sep 25 '17 at 16:00

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