Exercise 6.3.4 (Galois group third root of unity)

Question

Prove that $\mathrm{Gal}(\mathbb{Q}(e^{2\pi i /3}):\mathbb{Q}) = \{ \mathrm{id}, \kappa\}$, where $\kappa(z)=\bar z$.

Hassaan Naeem · Accepted Answer

extbf{Solution:}
We know that the identity is an automorphism of $\mathbb{Q}(e^{2\pi i /3})$ over $\mathbb{Q}$. 
By 	extbf{Lemma 1.1.2}, since $\mathbb{Q} \subset \mathbb{R}$ we have that $\kappa$ is also an automorphism of $\mathbb{Q}(e^{2\pi i /3})$ over $\mathbb{Q}$.
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Hence we have that $\{ id, \kappa\} \subseteq Gal(\mathbb{Q}(e^{2\pi i /3}):\mathbb{Q})$.
We also know that $\mathbb{Q}(e^{2\pi i /3}):\mathbb{Q} = \mathbb{Q}(\frac{-1+i\sqrt{3}}{2}):\mathbb{Q}=  \mathbb{Q}(i\sqrt{3}):\mathbb{Q}$
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We let $	heta \in Gal(\mathbb{Q}(e^{2\pi i /3}):\mathbb{Q})$. Since $	heta$ is a homomorphism we have that:
\begin{equation*}
    \begin{aligned}
        (	heta(i\sqrt{3}))^2 &= 	heta((i\sqrt{3})^2)\
        &= 	heta(-3)\
        &= -	heta(3)\
        &= -3
    \end{aligned}
\end{equation*}
Hence $	heta(i \sqrt{3}) = \pm i\sqrt{3}$. If $	heta(i\sqrt{3}) = i\sqrt{3}$ then $	heta = id$,
and if $	heta(i\sqrt{3}) = -i\sqrt{3}$ then $	heta=\kappa$. Hence $Gal(\mathbb{Q}(e^{2\pi i /3}):\mathbb{Q}) = \{id, \kappa\}$.

Richard Ganaye · Answer

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\begin{proof} Write $\omega = e^{2i\pi/3}$. Then $0 = \omega^3 -1 = (\omega - 1)(\omega^2 + \omega +1)$ and $\omega 
e 1$, thus $\omega^2 + \omega + 1 = 0$.

Here $\mathrm{id}$ and $\kappa$ are automorphisms of $\mathbb{Q}(e^{2\pi i /3})$ which fix the rationals. So $$\{ \mathrm{id}, \kappa\} \subset \Gal(\mathbb{Q}(\omega):\Q).$$

Now, let $\sigma$ be any element of  $\Gal(\Q(\omega) : \Q)$. From $\omega^2 + \omega + 1 = 0$, we deduce
$$\sigma(\omega)^2 + \sigma(\omega) + 1 = 0,$$
so $\sigma(\omega)$ is a root of $t^2 + t + 1 = (t-\omega)(t-\omega^2) = (t- \omega)(t- \overline{\omega})$. Therefore
$$\sigma(\omega) = \omega 	ext{ or } \sigma(\omega) = \overline{\omega}.$$
Moreover, $(1,\omega)$ is a basis of $\Q(\omega)$ over $\Q$, so every element of  $\Q(\omega)$ is of the form $\alpha = a + b \omega,\ a,b \in \Q$, and $\sigma(\alpha) = \sigma(a) + \sigma(b) \sigma(\omega) = a + b \sigma(\omega)$.
\begin{itemize}
\item If $\sigma(\omega) = \omega$, then $\sigma(\alpha) = a + b \omega = \alpha$, so $\sigma = \mathrm{id}$.

\item If $\sigma(\omega) = \overline{\omega}$, then $\sigma(\alpha) = a + b \overline{\omega} = \overline{\alpha}$, so $\sigma = \kappa$.
\end{itemize}
This shows that $\Gal(\mathbb{Q}(\omega):\Q) \subset  \{ \mathrm{id}, \kappa\}$, so
$$\Gal(\mathbb{Q}(\omega):\Q) =  \{ \mathrm{id}, \kappa\}.$$
\end{proof}

Exercise 6.3.4 (Galois group third root of unity)

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Exercise 6.3.4 (Galois group third root of unity)

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