I am interested in describing the Galois group for the polynomial $f(x) = x^5 - 1$ over $\mathbb{Q}$. The roots of this polynomial are the fifth roots of unity $\{1, \omega, \omega^2, \omega^3, \omega^4 \}$ and the field extension that will serve as a splitting field for this polynomial is $K = \mathbb{Q}(\omega)$. Its basis will consist of the five fifth roots of unity.

If my understanding is correct, the Galois group $Gal(K/\mathbb{Q})$ will consist of automorphisms $\pi$ of $K$ that fix $\mathbb{Q}$ and it will be isomorphic to some subgroup of $S_5$ (since we have five basis elements for $K$). I think one possible automorphism is the following:

$\pi_1(1) = 1$ since the ground field is fixed.

$\pi_1(\omega) = \omega^2$

$\pi_1(\omega^2) = \omega$

$\pi_1(\omega^3) = \omega^4$

$\pi_1(\omega^4) = \omega^3$

Which would correspond to a transposition $(23)(45)$ in $S_5$

Does that seem legit? To construct my other automorphisms, can I send $\omega$ to any power of $\omega$ I wish? Or are there some restrictions?

I'm sorry if this is a dumb question, but Galois theory is new for me and I am studying on my own.

  • 1
    $\begingroup$ You really want to look at $x^4+x^3+x^2+x+1$ which is irreducible over $\Bbb{Q}$, $x^5-1$ is not. The Galois Group is a subgroup of $S_4$ And yes the Galois Group will send $\omega$ to any other primitive root. $\endgroup$
    – sharding4
    Jul 1, 2017 at 3:28
  • 2
    $\begingroup$ If $\pi_1(\omega)=\omega^2$ then $\pi_1(\omega^2)=\pi(\omega)^2=\omega^4$. $\endgroup$ Jul 1, 2017 at 3:31

1 Answer 1


The Galois extensions defined by $K_n=\Bbb Q(\zeta_n)$ where $\zeta_n= \exp(2\pi i/n)$ is a root of $x^n-1=0$ is called the $n$-th cyclotomic field. Its Galois group is isomorphic to $(\Bbb Z/n\Bbb Z)^\times$ and each automorphism has the form $\sigma_a:\zeta_n\mapsto\zeta_n^a$ where $a$ is coprime to $n$. There is a huge literature on these fields, see for instance Washington's Introduction to Cyclotomic Fields (GTM Springer).


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