Exercise 10.3.8

Proof.

We first prove that $\mathbb{Q}\subset O$ is a normal extension. Let $\alpha \in O$, and let $f(x)$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$. By Theorem 10.3.4, there are subfields

such that $\alpha \in F_n$ and $[F_i:F_{i-1}]=2$ or $3$ for $1\le i\le n$.

By Exercise 10.1.7, there exists a Galois closure $\mathbb{Q}\subset M$ of $\mathbb{Q}\subset F$ such that $M\subset ℂ$, so $\mathbb{Q}\subset F_n\subset M\subset ℂ$ and $\mathbb{Q}\subset M$ is a Galois extension. Note that $f$ splits completely in $M$, since $M$ is normal over $\mathbb{Q}$, $f$ is irreducible over $\mathbb{Q}$, and $\alpha \in F_n\subset M$ is a root of $f$.

Now let $\beta \in M$ be any root of $f$. By Proposition 5.1.8, there is $\sigma \in \mathrm{Gal}(M∕\mathbb{Q})$ such that $\sigma (\alpha )=\beta$. Applying $\sigma$ to the fields $\mathbb{Q}=F_0\subset \cdots \subset F_n\subset M$ gives

such that $[\sigma (F_i):\sigma (F_{i-1}]=[F_i:F_{i-1}]=2$ or $3$ for all $i$.

By Theorem 10.3.4, $\beta =\sigma (\alpha )\in \sigma (F_n)$ is an origami number, so $\beta \in O$, so we can conclude that $\mathbb{Q}\subset O$ is a normal extension.

$\text{∙}$

Suppose that $\alpha \in O$ and let $\mathbb{Q}\subset L$ be the splitting field of the minimal polynomial $f$ of $\alpha$ over $\mathbb{Q}$. Since $\mathbb{Q}\subset O$ is normal, $L\subset O$. By the theorem of the Primitive Element, we have $L=\mathbb{Q}(\gamma )$ for some $\gamma \in L$. Since $\gamma \in O$, there are subfields $$\mathbb{Q}=F_0^{\prime }\subset F_1^{\prime }\subset \cdots \subset F_{n-1}^{\prime }\subset F_m^{\prime }\subset ℂ$$

such that $\gamma \in F_m^{\prime }$ and $[F_i^{\prime }:F_{i-1}^{\prime }]=2$ or $3$ for $1\le i\le m$.

As $\mathbb{Q}\subset \mathbb{Q}(\gamma )\subset F_m^{\prime }$, by the Tower Theorem,

$[L:\mathbb{Q}]=[\mathbb{Q}(\gamma ):\mathbb{Q}]$ divides $[F_m^{\prime }:\mathbb{Q}]=2^u{3}^v,u,v\in \mathbb{N}$, so

$[L:\mathbb{Q}]=2^a{3}^b$ for some integer $a,b\ge 0$.

$\text{∙}$

Conversely, suppose that $[L:\mathbb{Q}]=2^a{3}^b$ for some integer $a,b\ge 0$.

Since $\mathbb{Q}\subset L$ is Galois, then $G=\mathrm{Gal}(L∕\mathbb{Q})$ satisfies $|G|=[L:\mathbb{Q}]$ is of the form

$$|G|=2^a{3}^b.$$

By Burnside’s $p^n{q}^m$ Theorem (Theorem 8.1.8), $G$ is solvable, so we have subgroups

$$\{e\}=G_m\subset G_{m-1}\subset \cdots \subset G_1\subset G_0=G=\mathrm{Gal}(L∕\mathbb{Q})$$

such that $G_i$ is normal in $G_{i-1}$ of index 2 or 3, since $|G|=2^a{3}^b$. The Galois Correspondence Theorem gives

$$\mathbb{Q}=L_{G_0}\subset L_{G_1}\subset \cdots \subset L_{G_m}=L,$$

where $[L_{G_i}:L_{G_{i-1}}]=2$ or 3 for all $i$.

By Theorem 10.3.4, $\alpha \in L$ is an origami number.