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?

up vote 7 down vote accepted

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.

  • Thanks for the complete answer! That helped a lot. I just have forgotten this fact. (We proved this already, but in a different way). – Epsilondelta Jan 30 '15 at 19:22
  • 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? – Epsilondelta Jan 30 '15 at 19:32
  • Really? I thought the Galois group has order 6? Since we are considering the splitting field of $x^6+x^5+x^4+x^3+x^2+x^1+1$ – Epsilondelta Jan 30 '15 at 19:34
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    @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. – Adam Hughes Jan 30 '15 at 19:48
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    @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 – Adam Hughes Jan 31 '15 at 20:37

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