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(For context: the elements $x_1, ..., x_n$ are the roots of a generic $n$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 $σ$."


My question is rather simple,

Q1. What exactly "gives" a bijection?

Q2. Is the author in a sense defining an automorphism that has the property of mapping the set $x_1, ..., x_n$ into itself?


I would really appreciate any help/ideas.

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For simplicity I take $n=3$, and call the three roots $x,y,z$. An expression $e(x,y,z)$ in $E$ could be something like $ \sqrt{yz^3+5x^2}$ If the permutation of the roots is , say cyclic: $x\mapsto y\mapsto z\mapsto x$, then this leads to an expression for $e'$ viz. $\sqrt{zx^3+5y^2}$ This is also a radical extension: and the number of these radicals is finite as there are only a finite number of permutations from the finite number of radicals generating $E$.

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