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The motivation behind this question is to find an explicit example of a family of polynomials of degree $n$ whose Galois group over $\mathbb{Q}$ is the symmetric group $S_n$.

I thought that $f_n=x^n-x+1$ would fit the bill. Although the answer is no in general, it is not so bad; here is the guess (conjecture seems a little too much):

A. For $n>2$, $f_n$ is irreducible over $\mathbb{Q}[x]$ iff $n\not\equiv 2 \pmod{6}$. In that case, $\mathrm{Gal}(f_n,\mathbb{Q})=S_n$.

The case $n=6r+2$ leads us to another family with similar properties. Set $h=x^6+x^5-x^3-x^2$ and let $g_{r}=h\cdot(x^{6(r-1)}+x^{6(r-2)}+\cdots+x^6+1)+1$ for $r\geq 1$. Notice that $g_r$ has degree $6r$. Here are the first two: $$ g_1=x^6+x^5-x^3-x^2+1, \qquad g_2 = x^{12}+x^{11}-x^9-x^8+x^6+x^5-x^3-x^2+1. $$ These polynomials come from the factorization $f_{6r+2}=(x^2-x+1)\cdot g_r$. Then we have another guess:

B. For $r>1$, $g_r$ is irreducible over $\mathbb{Q}[x]$ and $\mathrm{Gal}(g_r,\mathbb{Q})=S_{6r}$.

The main interest is A. Anyhow: Is A or B known? Any reason why this should be true?

Using SAGE, I was able to verify that A holds for $n\leq 11$ and B holds for $r=1$.

Thanks in advance.

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    $\begingroup$ It is known that $r_n(x)=x^n-x-1$ is irreducible. If $n$ is odd, then $r_n(-x)=-f_n(x)$, so the irreducibility of $f_n$ follows for odd $n$. $\endgroup$ – Jyrki Lahtonen Jun 17 '19 at 13:58
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    $\begingroup$ See here for more about the irreducibilify of $g_r(x)$. $\endgroup$ – Jyrki Lahtonen Jun 17 '19 at 14:01
  • $\begingroup$ Actually Keith Conrad's argument also shows $f_n$ to be irreducible unless $n\equiv2\pmod 6$. He left it as an exercise! $\endgroup$ – Jyrki Lahtonen Jun 17 '19 at 14:06
  • $\begingroup$ @JyrkiLahtonen nice catch with $r_n(x)$. The paper by Ljunggren pointed out by Michael Stoll settles the irreducibility question for $g_r$ (it uses reciprocal polynomials as well). Thank you! $\endgroup$ – rts Jun 18 '19 at 11:15
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For (A) you are asking if $x^n-x+1$ is irreducible over $\mathbf Q$ with Galois group $S_n$ over $\mathbf Q$ when $n \not\equiv 2 \bmod 6$ (or for $n=2$, a trivial case). The irreducibility for $n \not\equiv 2 \bmod 6$ follows by the same arguments used in my answer at Irreducibility of $x^n-x-1$ over $\mathbb Q$ to show $x^n-x-1$ is irreducible over $\mathbf Q$ for all $n \geq 2$; see the exercise at the end of that answer.

Once we know $x^n-x+1$ is irreducible over $\mathbf Q$, its Galois group over $\mathbf Q$ is $S_n$ as a special case of Theorem 1 of Osada's paper The Galois groups of the polynomials $X^n+aX^l+b$, J. Number Theory 25, pp. 230-238, 1987 (in that theorem use $l = 1$, $a_0 = 1$, $b_0 = -1$, $c = 1$). The proof uses inertia groups and other ideas from algebraic number theory.

Another reference for the same method of calculating the Galois group is Serre's book "Topics in Galois Theory" (section 4.4), which you can find at a link in Vesselin Dimitrov's answer at http://mathoverflow.net/questions/177789/is-xn-x-1-irreducible. He also mentions Osada's paper. There is a follow-up paper by Osada, which you can find at https://projecteuclid.org/download/pdf_1/euclid.tmj/1178228289.

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  • $\begingroup$ Got it. Also I took a look at your notes on $x^n-x-1$. Thank you for the references and for your detailed answer! $\endgroup$ – rts Jun 18 '19 at 10:37
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In the special case where we work in the ring $\mathbb{Q}[x]$, we have that an irreducible polynomial $f\in\mathbb{Q}[x]$ of degree $n$ has a Galois group isomorphic to a subgroup of $A_n$ if and only if the discriminant of $f$, $\Delta(f)$, is a square; I must quickly note that this also holds when we replace $\mathbb{Q}$ for some other field not of characteristic $2$ and when also $f$ is separable (this all holds in $\mathbb{Q}$). Now let the discriminant of a polynomial of the form $f:=x^n+px+q\in\mathbb{Q}(p,q)[x]$ with $n>1$ be (fairly) easy to compute, namely, $$(-1)^{n(n-1)/2}(n^nq^{n-1}+(1-n)^{n-1}p^{n}).$$ Maybe this helps a little bit.

EDIT: so in other words, you can tell when the Galois group is not a subgroup of $A_n$, namely, when the discriminant of the irreducible polynomial $f$ is not a square.

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    $\begingroup$ The Galois group is a subgroup of $A_n$ if and only if the discriminant is a square. It does not have to be all of $A_n$. For example, the minimal polynomial of $2\cos(2\pi/11)=\zeta_{11}+\zeta_{11}^{-1}$ is $f(x)=x^5+x^4-4x^3-3x^2+3x+1$. We have $\Delta(f)=11^4$ which is a square. The Galois group is cyclic of order five. $\endgroup$ – Jyrki Lahtonen Jun 17 '19 at 18:23
  • $\begingroup$ @JyrkiLahtone Ah, yes, that's true. You can only tell it's not $S_n$ when $\Delta(f)=\square$. I changed it now. $\endgroup$ – Algebear Jun 17 '19 at 21:19
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    $\begingroup$ Your edit right now is wrong: if the discriminant is not a square the Galois group might be $S_n$. Consider an irreducible cubic over $\mathbf Q$ with negative discriminant. $\endgroup$ – KCd Jun 17 '19 at 21:56
  • $\begingroup$ @KCd okay, at least it is not a subgroup of $A_n$ then. $\endgroup$ – Algebear Jun 18 '19 at 19:26

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