Exercise 1.1.10 (Conjugacy of Complex K-Tuples)

Question

Suppose that $(z_1, ..., z_k)$ and $(z_1', ..., z_k')$ are conjugate. Show that $z_i$ and $z_i'$ are conjugate, for each $i \in \{1, ..., k\}$

Hassaan Naeem · Accepted Answer

\paragraph{Solution:} By 	extbf{Definition 1.1.9} have that:
    \begin{equation*}
        \begin{aligned}
            p(z_1, ..., z_k) & = 0 \Longleftrightarrow p(z_1', ..., z_k')=0
        \end{aligned}
    \end{equation*}
When $k=1$ we have that:\
\begin{center}
$p(z_1)=0 \Longleftrightarrow p(z_1') = 0 \implies$  $z_1$ and $z_1'$ are conjugate
\end{center}
\
Similarly, for any $k:$
\centerline{$p(z_i)=0 \Longleftrightarrow p(z_i') = 0 \implies$  $z_i$ and $z_i'$ are conjugate}

Richard Ganaye · Answer

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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} Suppose that $(z_1, ..., z_k)$ and $(z_1', ..., z'_k)$ are conjugate over $\Q$. Let $p \in \Q[t]$ be any polynomial such that $p(z_i) = 0$, where $1\leq i \leq k$. We will prove that $p(z'_i) = 0$. Consider le polynomial $P(t_1,\ldots,t_k) = p(t_i) \in \Q[t_1,\ldots,t_k]$. Then $P(z_1,\ldots,z_k) = 0$, thus $P(z'_1,\ldots,z'_k) = 0$. But $p(z'_i) = P(z'_1,\ldots,z'_k)$, and so $p(z'_i) = 0$. This proves that $z_i$ and $z_i'$ are conjugate.
\end{proof}

Exercise 1.1.10 (Conjugacy of Complex K-Tuples)

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Exercise 1.1.10 (Conjugacy of Complex K-Tuples)

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