# For every $n$, there exists a polynomial of degree $n$ with Galois Group $S_n$ [duplicate]

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.

## marked as duplicate by Dietrich Burde abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 21 '17 at 18:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

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.

• 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 – Daniele1234 Aug 21 '17 at 18:17
• Could you please explain how on the wikipedia page they are able to claim that Gal(f(x, s)/Q(s)) is doubly transitive? – Daniele1234 Aug 23 '17 at 9:10