theorem on constructible numbersTheoremOnConstructibleNumbers2007-06-18 03:28:122007-06-23 20:18:26Theoremimmediately constructible from\usepackage{amssymb}
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\begin{thm*}
Let $\mathbb{F}$ be the field of constructible numbers and $\alpha\in\mathbb{F}$. Then there exists a nonnegative integer $k$ such that $[\mathbb{Q}(\alpha)\!:\!\mathbb{Q}]=2^k$.
\end{thm*}
Before proving this theorem, some preliminaries must be addressed.
First of all, within this entry, the following nonconventional definition will be used:
Let $S$ be a subset of $\mathbb{C}$ that contains a nonzero complex number and $\alpha\in\mathbb{C}$. Then $\alpha$ is \emph{immediately constructible from} $S$ if any of the following hold:
\begin{itemize}
\item $\alpha=a+b$ for some $a,b\in S$;
\item $\alpha=a-b$ for some $a,b\in S$;
\item $\alpha=ab$ for some $a,b\in S$;
\item $\alpha=a/b$ for some $a,b\in S$ with $b\neq 0$;
\item $\alpha=\sqrt{|z|}e^{\frac{i\theta}{2}}$ for some $z\in S$ with $z \neq 0$ and $\theta=\operatorname{arg}(z)$ with $0\le\theta <2\pi$.
\end{itemize}
The following lemmas are clear from this definition:
\begin{lemma}
Let $S$ be a subset of $\mathbb{C}$ that contains a nonzero complex number and $\alpha\in\mathbb{C}$. Then $\alpha$ is constructible from $S$ if and only if there exists a finite sequence $\alpha_1,\dots ,\alpha_n\in\mathbb{C}$ such that $\alpha_1$ is immediately constructible from $S$, $\alpha_2$ is immediately constructible from $S \cup \{\alpha_1\}$, $\dots$ , and $\alpha$ is immediately constructible from $S \cup \{\alpha_1,\dots ,\alpha_n\}$.
\end{lemma}
\begin{lemma}
Let $F$ be a subfield of $\mathbb{C}$ and $\alpha\in\mathbb{C}$. If $\alpha$ is immediately constructible from $F$, then either $[F(\alpha)\!:\!F]=1$ or $[F(\alpha)\!:\!F]=2$.
\end{lemma}
Now to prove the theorem.
\begin{proof}
By the first lemma, there exists a finite sequence $\alpha_1,\dots ,\alpha_n\in\mathbb{C}$ such that $\alpha_1$ is immediately constructible from $\mathbb{Q}$, $\alpha_2$ is immediately constructible from $\mathbb{Q} \cup \{\alpha_1\}$, $\dots$ , and $\alpha$ is immediately constructible from $\mathbb{Q} \cup \{\alpha_1,\dots ,\alpha_n\}$. Thus, $\alpha_2$ is immediately constructible from $\mathbb{Q}(\alpha_1)$, $\dots$ , and $\alpha$ is immediately constructible from $\mathbb{Q}(\alpha_1,\dots ,\alpha_n)$. By the second lemma, $[\mathbb{Q}(\alpha_1)\!:\!\mathbb{Q}]$ is equal to either $1$ or $2$, $[\mathbb{Q}(\alpha_1,\alpha_2)\!:\!\mathbb{Q}(\alpha_1)]$ is equal to either $1$ or $2$, $\dots$ , and $[\mathbb{Q}(\alpha_1, \dots ,\alpha_n,\alpha)\!:\!\mathbb{Q}(\alpha_1, \dots ,\alpha_n)]$ is equal to either $1$ or $2$. Therefore, there exists a nonnegative integer $m$ such that $[\mathbb{Q}(\alpha_1, \dots ,\alpha_n,\alpha)\!:\!\mathbb{Q}]=2^m$. Since $\mathbb{Q} \subseteq \mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\alpha_1, \dots ,\alpha_n,\alpha)$, it follows that there exists a nonnegative integer $k$ such that $[\mathbb{Q}(\alpha)\!:\!\mathbb{Q}]=2^k$.
\end{proof}