I would like to compute the Galois group of the Polynomial $f(x)=x^5-5x^4 +10 x^3 - 10 x^2 - 135 x + 131\in\mathbb{Q}[x] $

I already know that it is irreducible in $\mathbb{Q}[x]$ via Eisenstein's criterion, $ f(x-1)$ and $p=5$, but have no idea how to proceed.

Thank you very much!

  • $\begingroup$ $f(x+1)=x^5-140x-8$. Eisenstein does not apply. $\endgroup$ – Chris Eagle Jan 16 '13 at 10:47
  • $\begingroup$ Thank you, I meant f(x-1). I have corrected it in the post. $\endgroup$ – testrado Jan 16 '13 at 10:54
  • $\begingroup$ $f(x-1)=x^5-10x^4+40x^3-80x^2-60x+240$, so $p=2$ again fails. $p=5$ works, though. $\endgroup$ – Chris Eagle Jan 16 '13 at 10:56
  • $\begingroup$ Ok, sorry, I should really take more time to proofread in future. Sorry, that was only my second question I have asked until now. $\endgroup$ – testrado Jan 16 '13 at 10:58

As the polynomial is irreducible of degree $5$, the group is a subgroup of $S_5$ containing a $5$-cycle. If it has exactly $2$ non-real roots, then the group has a transposition coming from complex conjugation, and you should be able to take it from there. If it has $4$ non-real roots, it will take some more work.

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  • $\begingroup$ Note that I'm taking your word for it that the polynomial is irreducible. $\endgroup$ – Gerry Myerson Jan 16 '13 at 10:49
  • $\begingroup$ Yes, i corrected my statement in the question, it is irreducible. $\endgroup$ – testrado Jan 16 '13 at 10:56
  • $\begingroup$ Is there a way to find the group without kwnowing that it has exactly 2 non-real roots? (it has, but I don't know if we are really supposed to use a CAS) $\endgroup$ – testrado Jan 16 '13 at 11:03
  • $\begingroup$ Who said anything about a CAS? There is a simple theorem about a subgroup of $S_p$ containing a $p$-cycle and a transposition, when $p$ is prime. $\endgroup$ – Gerry Myerson Jan 16 '13 at 11:04
  • $\begingroup$ Oh ok, I see! Thanks for your help! $\endgroup$ – testrado Jan 16 '13 at 11:12

It has the same Galois group as does $f(x-1)=x^5−140x−8$.

The easiest way to show it is irreducible is to start with RRT, which rules out linear factors. Therefore, if it factors, then in every $\mathbb{Z}_p[x]$, it will always factor and there will never be an irreducible quartic term.

For $p=7$: $f(x-1)=x^5-1$, so that you initially get $(x-1)(x^4+x^3+x^2+x+1)$.

No further linear factor can be pulled since the quartic factor is never $0$ for $x\in\mathbb{Z}_7$.

Assuming $(x^2+ax+b)(x^2+cx+d)$, we have:

$a+c\equiv b+ac+d\equiv bc+ad\equiv bd\equiv 1$

We can eliminate $a$ and $d$.

$b+a(1-a)+b^{-1}\equiv b(1-a)+ab^{-1}\equiv 1$

Which after manipulation:

$ab-a^2b+1\equiv -ab^2+a\equiv b-b^2$

But you can check for each $b\in\mathbb{Z}_7$ that there is no answer for $a$.

So the quartic doesn't factor.

Therefore, $x^5-140x-8$ is irreducible in $\mathbb{Z}[x]$.

This means the Galois Group must contain $C_5$ as a subgroup. By observing through basic analysis that it has exactly three real roots, the Galois must also contain the complex conjugation automorphism, which will fix these roots and swap the two complex roots. This transposition and the guaranteed $5$-cycle generate all of $S_5$.

So the answer is $S_5$.

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  • $\begingroup$ To expand on the "basic analysis", the derivative, $5x^4-140$, has exactly two real roots, so by Rolle's Theorem, the function has at most three real roots. Also, $f(x-1)=x^5-140x-8$ is negative for large negative values of $x$, positive at $x=-1$, negative at $x=0$, and positive for large positive values of $x$, so, by the Intermediate Value Theorem, it has at least three real roots. Putting these together, the polynomial has exactly three real roots. $\endgroup$ – Gerry Myerson Aug 27 '17 at 3:17

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