theorem on constructible numbers

Theorem 1.

Let $\mathrm{F}$ be the field of constructible numbers and $\alpha \mathrm{\in }\mathrm{F}$. Then there exists a nonnegative integer $k$ such that $\mathrm{[}\mathrm{Q}\mathrm{(}\alpha \mathrm{)}\mathrm{:}\mathrm{Q}\mathrm{]}\mathrm{=}{\mathrm{2}}^k$.

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 $ℂ$ that contains a nonzero complex number and $\alpha \in ℂ$. Then $\alpha$ is immediately constructible from $S$ if any of the following hold:

• $\alpha =a+b$ for some $a,b\in S$;

• $\alpha =a-b$ for some $a,b\in S$;

• $\alpha =ab$ for some $a,b\in S$;

• $\alpha =a/b$ for some $a,b\in S$ with $b\ne 0$;

• $\alpha =\sqrt{|z|}{e}^{\frac{i\theta }{2}}$ for some $z\in S$ with $z\ne 0$ and $\theta =\arg (z)$ with .

The following lemmas are clear from this definition:

Lemma 1.

Let $S$ be a subset of $\mathrm{C}$ that contains a nonzero complex number and $\alpha \mathrm{\in }\mathrm{C}$. Then $\alpha$ is constructible from $S$ if and only if there exists a finite sequence ${\alpha }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\alpha }_n\mathrm{\in }\mathrm{C}$ such that ${\alpha }_{\mathrm{1}}$ is immediately constructible from $S$, ${\alpha }_{\mathrm{2}}$ is immediately constructible from $S\mathrm{\cup }\mathrm{\{}{\alpha }_{\mathrm{1}}\mathrm{\}}$, $\mathrm{\dots }$ , and $\alpha$ is immediately constructible from $S\mathrm{\cup }\mathrm{\{}{\alpha }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\alpha }_n\mathrm{\}}$.

Lemma 2.

Let $F$ be a subfield of $\mathrm{C}$ and $\alpha \mathrm{\in }\mathrm{C}$. If $\alpha$ is immediately constructible from $F$, then either $\mathrm{[}F\mathrm{(}\alpha \mathrm{)}\mathrm{:}F\mathrm{]}\mathrm{=}\mathrm{1}$ or $\mathrm{[}F\mathrm{(}\alpha \mathrm{)}\mathrm{:}F\mathrm{]}\mathrm{=}\mathrm{2}$.

Now to prove the theorem.

Proof.

By the first lemma, there exists a finite sequence ${\alpha }_1,\mathrm{\dots },{\alpha }_n\in ℂ$ such that ${\alpha }_1$ is immediately constructible from $\mathbb{Q}$, ${\alpha }_2$ is immediately constructible from $\mathbb{Q}\cup \{{\alpha }_1\}$, $\mathrm{\dots }$ , and $\alpha$ is immediately constructible from $\mathbb{Q}\cup \{{\alpha }_1,\mathrm{\dots },{\alpha }_n\}$. Thus, ${\alpha }_2$ is immediately constructible from $\mathbb{Q}({\alpha }_1)$, $\mathrm{\dots }$ , and $\alpha$ is immediately constructible from $\mathbb{Q}({\alpha }_1,\mathrm{\dots },{\alpha }_n)$. 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 $2$, $\mathrm{\dots }$ , and $[\mathbb{Q}({\alpha }_1,\mathrm{\dots },{\alpha }_n,\alpha ):\mathbb{Q}({\alpha }_1,\mathrm{\dots },{\alpha }_n)]$ is equal to either $1$ or $2$. Therefore, there exists a nonnegative integer $m$ such that $[\mathbb{Q}({\alpha }_1,\mathrm{\dots },{\alpha }_n,\alpha ):\mathbb{Q}]=2^m$. Since $\mathbb{Q}\subseteq \mathbb{Q}(\alpha )\subseteq \mathbb{Q}({\alpha }_1,\mathrm{\dots },{\alpha }_n,\alpha )$, it follows that there exists a nonnegative integer $k$ such that $[\mathbb{Q}(\alpha ):\mathbb{Q}]=2^k$. ∎

Title	theorem on constructible numbers
Canonical name	TheoremOnConstructibleNumbers
Date of creation	2013-03-22 17:16:28
Last modified on	2013-03-22 17:16:28
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	10
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 12D15
Related topic	ConstructibleNumbers
Related topic	ClassicalProblemsOfConstructibility
Defines	immediately constructible from