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This question already has an answer here:

I'm trying to prove that one can always construct an nth degree polynomial with Galois Group $S_n.$ I've proven that one can always construct an nth degree polynomial with Galois Group Sn over the field $F_0(s_1,...,s_n)$ where $s_1,...,s_n$ are the elementary symmetric polynomials. I've also seen conditions which a polynomial must have to have Galois Group $S_n,$ but this doesn't prove existence. Does anyone have a proof that there is at least one $n^{th}$ degree polynomial with Galois Group $S_n?$

Thanks

EDIT: I am considering only polynomials over Q (rationals), not other fields in general.

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marked as duplicate by Dietrich Burde abstract-algebra Aug 21 '17 at 18:57

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Hilbert showed that all symmetric and alternating groups can be realized as Galois groups of polynomials with coefficients in $\mathbb{Q}$.

A modern exposition of his argument is given in the Wikipedia page on the Inverse Galois problem, see the section Symmetric and Alternating Groups.

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  • $\begingroup$ This is interesting, but I would like to prove it without using already known theorems and lemnas which I can't prove myself. I would rather have a proof 'from scratch' if you see what I am getting at $\endgroup$ – Daniele1234 Aug 21 '17 at 18:17
  • $\begingroup$ Could you please explain how on the wikipedia page they are able to claim that Gal(f(x, s)/Q(s)) is doubly transitive? $\endgroup$ – Daniele1234 Aug 23 '17 at 9:10

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