Exercise 6.2.9 (Splitting field example)

Question

In Example 6.2.8(iii), I said that $\mathbb{Q}(\xi, \omega\xi, \omega^2\xi) = \mathbb{Q}(\xi,\omega)$.
Why is that true?

Hassaan Naeem · Accepted Answer

extbf{Solution:}
$\omega=e^{2\pi i /3} = \frac{-1+i\sqrt{3}}{2}$, and hence $\omega^2 = \frac{-1-i\sqrt{3}}{2}$. and hence we see that $\omega^2$ is a rational multiple
of $\omega$, hence $\mathbb{Q}(\xi, \omega\xi, \omega^2\xi) = \mathbb{Q}(\xi,\omega)$.
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As an aside, 
we know that $\mathbb{Q}(\xi)= \left\{a + b\xi + c\xi^2: a,b,c \in \mathbb{Q} ight\}$, and hence $\{1, \xi, \xi^2\}$
forms a basis for $\mathbb{Q}(\xi):\mathbb{Q}$. Similarly, $\{1, \omega\}$ forms a basis for $\mathbb{Q}(\xi, \omega): \mathbb{Q}(\xi)$.
By 	extbf{Theorem 5.1.17 (Tower Law)(i)} we then have that $\{1, \xi, \xi^2, \omega, \xi\omega, \xi^2\omega\}$ forms a basis for $\mathbb{Q}(\xi, \omega):\mathbb{Q}$.
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Richard Ganaye · Answer

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof} 
The field $\mathbb{Q}(\xi,\omega)$  contains $\xi$ and $\omega$, thus it contains $\xi, \omega\xi, \omega^2\xi$.

So  $\Q\subset  \Q(\xi,\omega)$, and $\{\xi, \omega\xi, \omega^2\xi\} \subset \Q(\xi,\omega)$. Since $\Q(\xi, \omega\xi, \omega^2\xi) $ is the smallest subfield of $\C$ containing $\Q$ and $\{\xi, \omega\xi, \omega^2\xi\}$, we obtain
 $$\Q(\xi, \omega\xi, \omega^2\xi) \subset \Q(\xi,\omega).$$
 
 Conversely, $\xi \in \mathbb{Q}(\xi, \omega\xi, \omega^2\xi)$, and $\omega = (\omega \xi) / \xi \in \mathbb{Q}(\xi, \omega\xi, \omega^2\xi)$. Thus $\Q \subset \mathbb{Q}(\xi, \omega\xi, \omega^2\xi)$, and $\{\omega, \xi\} \subset \mathbb{Q}(\xi, \omega\xi, \omega^2\xi)$. Since $\mathbb{Q}(\xi,\omega)$ is the smallest field satisfying these two properties,
 $$\mathbb{Q}(\xi,\omega) \subset \mathbb{Q}(\xi, \omega\xi, \omega^2\xi).$$
 We have proved
 $$\mathbb{Q}(\xi, \omega\xi, \omega^2\xi) = \mathbb{Q}(\xi,\omega).$$
\end{proof}

Exercise 6.2.9 (Splitting field example)

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Exercise 6.2.9 (Splitting field example)

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