Let $\zeta\in \mathbb C$ be a primitive $7^{th}$ root of unity.

Show that there exists a $\sigma\in \operatorname{Gal}(\mathbb Q(\zeta)/\mathbb Q)$ such that $\sigma(\zeta)=\zeta^3$.

I already know that $\zeta$ is a root of $f(x)=x^6+x^5+x^4+x^3+x^2+x+1$ and that $f$ is irreducible (By applying Eisenstein's criterion on $f(x+1)$).

Also the powers of $\zeta$ are also roots of $f$. So $\mathbb Q(\zeta)$ is a splitting field of $f$. Now its clear that this extension is a Galois extension, since all the roots are different. But how to show the desired statement?


1 Answer 1


Note that $\langle \zeta\rangle$ is a finite, cyclic group of order $7$, so every non-trivial element has order $7$. In particular $\zeta^3$ is another primitive $7^{th}$ root of $1$, hence is another roots of the irreducible polynomial $\Phi_7(x)={x^7-1\over x-1}$. But then as this is irreducible by Eisenstein's criterion applied to $\Phi(x+1)$, we get that

$$\Bbb Q(\zeta)/\Bbb Q\cong \Bbb Q[x]/(\Phi_7(x)).$$

We now use the fact that the Galois group transitively permutes the roots of the irreducible polynomial in the quotient--if you haven't seen this before, see my addendum at the bottom--hence for any two roots, $r,s$, there is some $\sigma=\sigma_{r,s}$ such that $\sigma(r)=s$. Taking $r=\zeta$ and $s=\zeta^3$ we get the result.

The key observations for this are:

  • that both $\zeta$ and $\zeta^3$ are roots of the same, irreducible polynomial

  • that the Galois group permutes the roots of such polynomials transitively.

If you haven't seen the proof of the transitive action on roots, it's relatively straightforward: since $\Phi_7(x)$ is irreducible, we note that if $r,s$ are any two roots

$$\Bbb Q(r)/\Bbb Q\cong \Bbb Q[x]/(\Phi_7(x))\cong \Bbb Q(s)/\Bbb Q\qquad (*)$$

Then the automorphism of $\Bbb Q(r)/\Bbb Q$ is simply the composite of the isomorphisms. I.e. if the isomorphisms in $(*)$ are

$$\begin{cases}\varphi_r: \Bbb Q(r)\to \Bbb Q[x]/(\Phi_7(x)) \\ \varphi_s: \Bbb Q(s)\to \Bbb Q[x]/(\Phi_7(x)) \end{cases}$$

then we have that $\sigma_{r,s}=\varphi_s\circ \varphi_r^{-1}: \Bbb Q(s)\to \Bbb Q(r)$ is an isomorphism, but since $\Bbb Q(r)=\Bbb Q(s)$ is actually an equality for $r=\zeta, s=\zeta^3$ when we treat them as subfields of $\Bbb C$, we see that we may replace "iso" with "auto," and apply the definition of the Galois group as the group of all automorphisms of the field.

  • $\begingroup$ Thanks for the complete answer! That helped a lot. I just have forgotten this fact. (We proved this already, but in a different way). $\endgroup$ Jan 30, 2015 at 19:22
  • $\begingroup$ One more question: If I want to show that this $\sigma$ generates the Galois group it is sufficient to see that $\sigma^i(\zeta)=\sigma^j(\zeta)$ is different for $i\neq j$ with $i,j\in \{1,...,n\}$? Are there other possibilites? $\endgroup$ Jan 30, 2015 at 19:32
  • $\begingroup$ Really? I thought the Galois group has order 6? Since we are considering the splitting field of $x^6+x^5+x^4+x^3+x^2+x^1+1$ $\endgroup$ Jan 30, 2015 at 19:34
  • 1
    $\begingroup$ @Epsilondelta exponents of $\zeta$ are the same iff they are the same modulo $7$. Hence $$\sigma^i(\zeta)=\zeta^{3^ i}$$ In order for this to be the identity automorphism it is necessary and sufficient that $\sigma^i(\zeta)=\zeta$, since $\zeta$ generates the field. In order for this to be true, we need $3^i\equiv 1\mod 7$. So for $\sigma$ to have order $6$, you need to check that $6$ is the smallest, positive integer for which $3^i\equiv 1\mod 7$, but that's the definition of the order of an element of $\Bbb Z/7\Bbb Z^*$ as a group under multiplication. $\endgroup$ Jan 30, 2015 at 19:48
  • 1
    $\begingroup$ @epsilondelta you mean "abelian," its not obvious it's cyclic immediately, but the rest of your argument is the same as mine if you look back a few comments $\endgroup$ Jan 31, 2015 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.