I have seen many textbooks state this result without proof.
$``$ If $E$ is the splitting field for the polynomial $f=x^p-1 \in \mathbb{Q}[X]$ where $p$ is prime, then the Galois group $Gal(E:Q)\simeq \mathbb{F}_{p}^{\times} \simeq\mathbb{Z}_{p-1} . "$
I would like to know how to prove this.
The things that I know from reading are that $E=Q(\zeta)$, where $\zeta$ is a root of the irreducible polynomial $x^{p-1} + x^{p-2} \dots + 1$. So $E$ a cyclotomic extension of Q of degree $p-1$ and that the roots of $x^p - 1$ are the $p^{th}$ roots of unity. Also that the Galois group is comprised of the automorphisms of $E$ that fix $\mathbb{Q}$ and permute the roots of $f$.
As a second question, if I am asked to prove that the Galois group $Gal(E/Q)$ is cyclic, when $E$ is the splitting field for some specific $f$ and $p$ is small, say $5$ or $7$, is there an easier way to prove this than first proving the more general statement above and then drawing the conclusion from that?
Any help will be much appreciated! Thanks.
$\endgroup$ 51 Answer
$\begingroup$If, as you say, you already know what you wrote there, then you've already proved what you need...almost.
Let $\;\zeta=e^{2\pi i/p}\;$ , then $\;\zeta\;$ is clearly a root of $\;f(x)=x^p-1\;$ . Observe (well, prove) now that
1) All the roots of $\;f(x)\;$ are different (hint: look at $\;f'(x)\;$ )
2) $\;\forall\,k\in\Bbb Z\;,\;\;\;\left(\zeta^k\right)^p=1\;$
Deduce then that $\;[\Bbb Q(\zeta):\Bbb Q]=p-1\;$ and that $\;\Bbb Q(\zeta)\;$ is the splitting field of $\;f(x)\;$ over the rationals. Also, from (2) it follows this extension is cyclic.
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