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I need help to find the Galois Group of $x^5 +1$. I know that it has a 5-cycle and a 4 cycle and is not a subgroup of $A_5$. Thanks!

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    $\begingroup$ but that has x+1 as a factor $\endgroup$ – user58512 Mar 3 '13 at 23:19
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    $\begingroup$ It might be helpful to note that $x^5+1$ divides $x^{10}-1$. So there's no chance the Galois group could be non-abelian. $\endgroup$ – JSchlather Mar 3 '13 at 23:22
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the polynomial factors as $$(x+1)(x^4-x^3+x^2-x+1)$$ and that quartic has discriminant $5^3$ and Galois group $C_4$.

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    $\begingroup$ Which formula did you use to compute the discrimant of the quartic? $\endgroup$ – User123456 Mar 3 '13 at 23:37
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    $\begingroup$ And how did you figure out the Galois group? $\endgroup$ – Jyrki Lahtonen Mar 4 '13 at 6:52
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Notice that $x^5+1=(x-1)(x^4-x^3+x^2-x+1)$ so we're really concerned with the fourth degree term. This in fact equivalent to the 5th cyclotomic polynomial. To see this note that the automorphism of $\mathbb Q[x]$ sending $x$ to $-x$ maps $(x^4-x^3+x^2-x+1)$ to $x^4+x^3+x^2+x+1$. So the splitting field of $x^5+1$ is in fact the 5th cyclotomic field.

The analysis of the Galois group of such fields is easy. Let $\zeta$ be a $p$-th primitive root of unity then the automorphisms of $\mathbb Q(\zeta)$ are generated by $\zeta \mapsto \zeta^i$ for $0<i<p$. We see in particular that $\mathrm{Gal}(\mathbb Q(\zeta)/\mathbb Q) \cong (\mathbb Z_p)^\times$.

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    $\begingroup$ @JyrkiLahtonen Yeah, I think you're right. I guess the roots of $x^4-x^3+x^2-x+1$ are $\zeta,\zeta^3,\zeta^7$ and $\zeta^9$. I'll try to fix this in the morning, I'm too tired right now. $\endgroup$ – JSchlather Mar 4 '13 at 7:11
  • $\begingroup$ It is the $10^{\text{th}}$ cyclotomic polynomial, not the $5^{\text{th}}$. $\endgroup$ – Kenny Lau Jan 2 at 3:13

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