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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!

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

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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Note that I'm taking your word for it that the polynomial is irreducible. – Gerry Myerson Jan 16 '13 at 10:49
Yes, i corrected my statement in the question, it is irreducible. – testrado Jan 16 '13 at 10:56
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) – testrado Jan 16 '13 at 11:03
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. – Gerry Myerson Jan 16 '13 at 11:04
Oh ok, I see! Thanks for your help! – testrado Jan 16 '13 at 11:12

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