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here is my attempt at a solution.
First of all, I had to show $f(x)$ was irreducible on $\mathbb{Q}$. It suffices to show it is irreducible $\mod 3$. But $f(x)=x^6+1= x^6+7 \mod 3$. We can apply eisenstein on $\mathbb{Z}/3\mathbb{Z}$ with $p=7$. So $f(x)$ is irreducible on $\mathbb{Q}$.

We calculate the discriminant and get $D(f(x))=5^6 2^{16}3^6 $ so it is a square. Thus it now suffices to show we have a 3-cycle.

$f(x)=x^6+24x-20 \equiv x^6+3x+1 \mod 7 \equiv (x+1)(x^5+4x^4+x^2+4x+5) \mod 7 \equiv (x+1)(x^5+4x^4-6x^2+4x-2) \mod 7$ But the last polynomial of degree 5 is irreducible in $\mathbb{Z}/7\mathbb{Z}$ by eisenstein with $p=2$. So we have (1,5) i.e. a 5-cycle.

$f(x)= x^6+24x-20 \equiv x^6+2x+2 \mod 11.$ Irreducible by eisenstein with $p=2$.

$f(x)= x^6+24x-20 \equiv x^6-2x+6 \mod 13.$ Irreducible by eisenstein with $p=2$.

$f(x)= x^6+24x-20 \equiv x^6+7x+14 \mod 17.$ Irreducible by eisenstein with $p=7$.

Finally,

$f(x)=x^6+24x-20 \equiv x^6+x+3 \mod 23 \equiv (x+7)(x+12)(x+21)(x^3+6x^2+13x+16) \mod 23$

So we have (1,1,1,3), i.e. we have a 3-cycle.

Thus, Gal$(f(x))= \mathbb{A_6}$

Am I correct? If not, how can I solve this?

usere5225321
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    The "Eisenstein on $\mathbb{Z}/3\mathbb{Z}$" step would seem to be nonsense. The same reasoning would show $x^6 - 1 = x^6 + 5 \pmod{3}$ is irreducible, which it clearly isn't. – Daniel Schepler Mar 30 '17 at 19:44
  • Oh, and also, even the original $x^6 + 1 = (x^2 + 1) (x^4 - x^2 + 1)$ is reducible in any ring, so $f(x)$ is reducible over $\mathbb{Z}/3\mathbb{Z}$. – Daniel Schepler Mar 30 '17 at 19:46
  • so how do I show irreducibility mod 3, 7,11,13,17? – usere5225321 Mar 30 '17 at 19:48
  • What is correct is, that $f(x)$ is irreducible over $\mathbb{Q}$, and has Galois group $A_6$. The same is true for $g(x)=x^6-24x-20$, or $x^6-30x+25$, or $x^6+30x+25$, etc. – Dietrich Burde Mar 30 '17 at 19:48
  • Okay, so how do I solve this problem ? @DietrichBurde – usere5225321 Mar 30 '17 at 19:50
  • That would be a new question, right? You could use the solution given here. – Dietrich Burde Mar 30 '17 at 19:50
  • Edited my question! – usere5225321 Mar 30 '17 at 19:52
  • If you have Mathematica or some such that will do modular factorization, then you see that you get a $1+5$ factorization modulo $7$ and $2+4$ modulo $29$. This forces your polynomial to be irreducible. Then the modulo $23$ stuff shows the presence of a 3-cycle. Therefore the groupis at least triply transitive. IIRC that forces it to be either $A_6$ or $S_6$. Then the discriminant settles that it is $A_6$. – Jyrki Lahtonen Mar 30 '17 at 20:21
  • It seems difficult to prove the irreducibility without a calculator. And to determine the galois group will probably be even more difficult. – Peter Mar 30 '17 at 20:23
  • @JyrkiLahtonen I will not be able to use Mathematica on my exam, so I have to learn to do these kinds of problems without it – usere5225321 Mar 30 '17 at 20:23
  • @Peter I am only allowed to use a basic calculator. – usere5225321 Mar 30 '17 at 20:24
  • @usere5225321 To be honest, I would have no idea. – Peter Mar 30 '17 at 20:26
  • Rational root test tells that the polynomial has no linear factors, so if you are given that modulo 7 that quintic is irreducible, then it must be irreducible. Mod 23 you can do by hand, but coming up with the idea to user mod 23 particularly is asking a bit more :) – Jyrki Lahtonen Mar 30 '17 at 20:40
  • @JyrkiLahtonen I do not see why the irreducibility of the quintic mod 7 will imply that $f(x)$ is irreducible. Also, how do I show this quintic is irreducible mod 7 without a computer – usere5225321 Mar 30 '17 at 20:47
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    If mod 7 factorization is 5+1, then that implies that the factorization over $\Bbb{Z}$ has to be 5+1 or 6+0 (i.e. irreducible). 5+1 over $\Bbb{Z}$ implies a rational root, so that possibility can be excluded. I don't know how to prove that any of the modular quintic factors are irreducible. By hand, and in reasonable time. – Jyrki Lahtonen Mar 30 '17 at 21:00
  • It is easy to show that the sextic has two real zeros, so complex conjugation gives a permutation of cycle type $(2,2)$. I don't see how that would help either. – Jyrki Lahtonen Mar 30 '17 at 21:03

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