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The problem:

Let p be a prime number. Let$\;\omega$ be a pth root of unity.

Prove$\;Gal(\Bbb Q(\omega)/\Bbb Q)$ is cyclic.

The same basic question is here: Galois Group of $x^p-1$ cyclic

but the answer just points out that$\;Gal(\Bbb Q(\omega)/\Bbb Q)$ is isomorphic to the multiplicative group of integers modulo p.

In another post, the issue of the proof that the multiplicative group of integers modulo p is cyclic is brought up: Proof that the following multiplicative groups modulo m are cyclic

The poster said they'd already proven it for p being prime. Another said the proof is quite long.

Is there another approach to the initial question of proving the Galois group is cyclic? The book I'm working on (Pinter's Abstract Algebra) didn't have anything that complicated wrt this subject.

Shaun
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Alan Ox
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1 Answers1

5

  • $\Bbb{F}_p$ is the field with $p$-elements, $\Bbb{F}_p^\times$ is a group with $p-1$ elements,

  • All its elements satisfy $a^m=1$ for some $m$ dividing $p-1$.

  • Let $k$ be the smallest integers such that all the elements satisfy $a^k=1$. We have that $k \le p-1$.

  • Assume that $k<p-1$. All the elements are roots of $x^k-1\in \Bbb{F}_p[x]$, but this polynomial has at most $k$ distinct roots, contradicting that $\Bbb{F}_p^\times$ has $p-1$ elements.

  • Whence $k=p-1$ and we get some element of order $p-1$ ie. $\Bbb{F}_p^\times$ is cyclic.

  • $\omega$ is a root of $f=\frac{x^p-1}{x-1}$, thus any automorphism of $\Bbb{Q}(\omega)$ must send $\omega$ to another root of $f$ ie. to $\omega^a$ for some $a\in 1\ldots p-1$. As an automorphism is determined by where it sends $\omega$ this makes $Gal(\Bbb{Q}(\omega)/\Bbb{Q})$ a subgroup of the cyclic group $\Bbb{F}_p^\times$.

  • It remains to prove that $f$ is irreducible ($f(x+1)$ is Eisenstein at $p$) so that every $a$ gives an automorphism and $Gal(\Bbb{Q}(\omega)/\Bbb{Q})\cong \Bbb{F}_p^\times$.

reuns
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