According to 1 and 2 (quadratic example) the only permutation of a symmetric polynomial in terms of the distinct rational roots of a monomial is trivial.

For example, let $$p(t) = (t-1)(t-2) = t^2-3t+2$$ be the monomial in question, with roots $x_1 = 1$ and $x_2 = 2$.

The elementary symmetric polynomials in terms of these roots are: $$x_1+x_2 = 3$$ and $$x_1 x_2 = 2$$

These polynomials are equal under two permutations: identity and transposition, just like the case for distinct irrational roots: $$x_2+x_1 = 3$$ and $$x_2 x_1 = 2$$

What does the approach in these sources use to justify, then, that the permutation group of the roots of $p$ is trivial?

I understand why this is the case when the problem is generalized to automorphisms on field extensions, where the splitting field over the rationals of a polynomial with rational roots is still the rationals, and hence only admits the trivial field automorphism. However, how does this translate to the classical case where permutations of symmetric polynomials in terms of the roots are considered instead?

Edit: Clarification:

Starting from a formulation using field automorphisms, I understand why the Galois group is trivial.

As far as I'm aware, however, the formulation in terms of symmetric polynomials of the roots came much earlier; so the question is not why is the Galois group trivial for this case, to which the answer using a modern approach is clear to me, but rather what assumptions are being made in the sources, that I am missing, that justify the only allowed permutations being trivial in the older symmetric polynomial approach?

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    $\begingroup$ Not every permutation of the roots comes from a field automorphism. If your field extension is trivial, there are no nontrivial automorphisms, so there is only the identity permutation. In particular, there is no field automorphism of $\Bbb Q$ that permutes $1$ and $2$, like there is a field automorphism of the transcendental extension $\Bbb Q(a,b)$ that permutes $a$ and $b$. $\endgroup$ – arctic tern Jun 28 '17 at 4:45

... because that means that there is no field extension after all, i.e., the roots are already in the ground field.

  • $\begingroup$ Right, but how does this tie in to the consideration of just permutations of the variables in the symmetric polynomials, where field extensions aren't used? Why does transposing the variables count when they are irrational, but not when they are rational? $\endgroup$ – hussain_kj Jun 28 '17 at 4:44
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    $\begingroup$ @hussain_kj The permutations must come from field automorphisms, and a field automorphism cannot permute distinct rational numbers like it can permute irrational or transcendental ones. $\endgroup$ – arctic tern Jun 28 '17 at 4:47
  • $\begingroup$ Do you mean that we only count permutations that come from field automorphisms as members of the Galois group? I don't see why the permutations of symmetric polynomials in general would need to. $\endgroup$ – hussain_kj Jun 28 '17 at 4:54
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    $\begingroup$ A galois group is, essentially by definition, a (sub)group of automorphisms of a field. $\endgroup$ – Ravi Jun 28 '17 at 5:34

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