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.

  • $\begingroup$ I don't get it. You begin your third parraph with "I know that... Also, that the Galois group...However, I am unsure how to prove this result". So you know it but you can't prove it? Or how can you know it but you can't prove it? So finally, what is the question here? $\endgroup$ – Timbuc Jun 14 '15 at 8:27
  • $\begingroup$ I stated the things that I think I know so that someone answering knows the extent or lack of what I am already familiar with. My question is how can I prove the statement in quotations? $\endgroup$ – aew1324 Jun 14 '15 at 9:19
  • 1
    $\begingroup$ @Timbuc: Nowadays, people "know" something when they have read it somewhere on the internet. $\endgroup$ – Martin Brandenburg Jun 14 '15 at 10:11
  • $\begingroup$ Okay. I don't know a single thing. Some help would still be appreciated though if anybody has any $\endgroup$ – aew1324 Jun 14 '15 at 10:22
  • $\begingroup$ @MartinBrandenburg Hehe...yes, I guess so. $\endgroup$ – Timbuc Jun 14 '15 at 10:22

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.

  • $\begingroup$ Shouldn't the degree of the splitting field for f(x) be p-1? $\endgroup$ – aew1324 Jun 14 '15 at 10:42
  • $\begingroup$ @aew1324 Of course, that was a typo. Thanks. $\endgroup$ – Timbuc Jun 14 '15 at 10:44
  • $\begingroup$ All good. Thanks for your help. I had been forgetting a few things that were preventing it all from clicking for me. $\endgroup$ – aew1324 Jun 14 '15 at 11:03
  • $\begingroup$ @aew1324 My pleasure. Observe that things get even easier if you show that all the roots of $\;f(x)\;$ are in fact a subgroup of the multiplicative group of the field $\;\Bbb C\;$ and thus a cyclic one. $\endgroup$ – Timbuc Jun 14 '15 at 11:05

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