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It has been a while now that I've been trying to understand a proof about Galois theory. I have tried asking about it on this website, although it doesn't seem that I explained my self very well. I'm going to attempt to be far more specific on what I'm confused about, and give an example of a case that the proof doesn't seem to rule out.

(For context, $x_1,...,x_n$ refer to the roots of a $n$th degree polynomial)

Theorem. "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 $σ$."


It might be because I don't have much experience with Galois theory, but the fact that "this gives a bijection" is definitely unclear to me. I do understand that $σ$ will map each element in $ℚ(x_1,...,x_n)$ to another element of the same field, but how could prove that this mapping is bijective? For instance, how could one prove that it is impossible for the following to happen?

Let's say that $(x_1)^3+7x_2=(x_3)^2$, yet $(σx_1)^3+7(σx_2)≠(σx_3)^2$, meaning that $σ$ has mapped the same element into two different places.


This has been troubling me for a while now, and I can't find a way of proving it nor any 'complete' proof in the internet. I would incredibly appreciate it if someone could help me understand.

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    $\begingroup$ In case someone wants to read the text from where the theorem and proof came from: math.jhu.edu/~smahanta/Teaching/Spring10/Stillwell.pdf $\endgroup$ – Leo Sep 20 '17 at 9:26
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    $\begingroup$ The claim "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 $x_1,...,x_n$." is false in general. As you pointed out yourself, if the roots $x_i$ have some hidden relations, all automorphisms of all extension fields must respect those relations. Need to know more about the context of that claim to make sense out of it $\endgroup$ – Jyrki Lahtonen Sep 20 '17 at 9:29
  • $\begingroup$ @JyrkiLahtonen. Do you mean that there could be cases where there won't exist automorphisms for each permutation $σ$ of $x_1, ... , x_n$? $\endgroup$ – Leo Sep 20 '17 at 9:34
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    $\begingroup$ I took a quick look. For this to make sense it must be the case that the author assumes $x_1,x_2,\ldots,x_n$ to be algebraically independent variables. Given that, all permutations of $x_i$s are automorphisms of $\Bbb{Q}(x_1,x_2,\ldots,x_n)$. A clue pointing in this direction is the section title "generic equation of degree $n$" (or some such). $\endgroup$ – Jyrki Lahtonen Sep 20 '17 at 9:35
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    $\begingroup$ Leo, I would guess that it still does prove the Abel-Ruffini theorem. After all, Abel-Ruffini is about the generic quintic (or higher). There are special quintics solvable in radicals. For example $x^5-2$. $\endgroup$ – Jyrki Lahtonen Sep 22 '17 at 16:21

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