3.5 Constructible numbers revisited

Earlier (1.36) we showed that a real number $\alpha$ is constructible if and only if it is contained in a subfield of $\mathbb{R}{}$ of the form $\mathbb{Q}[\sqrt{a_{1}},\ldots,\sqrt{a_{r}}]$ with each $a_{i}$ a positive element of $\mathbb{Q}{}[\sqrt{a_{1}},\ldots,\sqrt{a_{i-1}}]$ . In particular

equation (7) (7)

\alpha\text{\ constructible }\implies[\mathbb{Q}[\alpha]\colon\mathbb{Q}]=2^{s% }\text{\ some }s.

Now we can prove a partial converse to this last statement.

Theorem 3.23

If $\alpha$ is contained in a subfield of $\mathbb{R}{}$ that is Galois of degree $2^{r}$ over $\mathbb{Q}$ , then it is constructible.

Proof.

Suppose $\alpha\in E\subset\mathbb{R}{}$ where $E$ is Galois of degree $2^{r}$ over $\mathbb{Q}$ , and let $G=\operatorname{Gal}(E/\mathbb{Q})$ . Because finite $p$ -groups are solvable (GT, 6.7), there exists a sequence of groups

\{1\}=G_{0}\subset G_{1}\subset G_{2}\subset\cdots\subset G_{r}=G

with $G_{i}/G_{i-1}$ of order $2$ . Correspondingly, there will be a sequence of fields,

E=E_{0}\supset E_{1}\supset E_{2}\supset\cdots\supset E_{r}=\mathbb{Q}{}

with $E_{i-1}$ of degree $2$ over $E_{i}$ . The next lemma shows that $E_{i}=E_{i-1}[\sqrt{a_{i}}]$ for some $a_{i}\in E_{i-1}$ , and $a_{i}>0$ because otherwise $E_{i}$ would not be real. This proves the theorem. _□

Lemma 3.24

Let $E/F$ be a quadratic extension of fields of characteristic $\neq 2$ . Then $E=F[\sqrt{d}]$ for some $d\in F$ .

Proof.

Let $\alpha\in E$ , $\alpha\notin F$ , and let $X^{2}+bX+c$ be the minimal polynomial of $\alpha$ . Then $\alpha=\frac{-b\pm\sqrt{b^{2}-4c}}{2}$ , and so $E=F[\sqrt{b^{2}-4c}]$ . _□

Corollary 3.25

If $p$ is a prime of the form $2^{k}+1$ , then $\cos\frac{2\pi}{p}$ is constructible.

Proof.

The field $\mathbb{Q}[e^{2\pi i/p}]$ is Galois over $\mathbb{Q}$ with Galois group $G\simeq(\mathbb{Z}/p\mathbb{Z})^{\times}$ , which has order $p-1=2^{k}$ . The field $\mathbb{Q}{}[\cos\frac{2\pi}{p}]$ is contained in $\mathbb{Q}[e^{2\pi i/p}]$ , and therefore is Galois of degree dividing $2^{k}$ (fundamental theorem 3.16 and 1.20). As $\mathbb{Q}{}[\cos\frac{2\pi}{p}]$ is a subfield of $\mathbb{R}{}$ , we can apply the theorem. _□

Thus a regular $p$ -gon, $p$ prime, is constructible if and only if $p$ is a Fermat prime, i.e., of the form $2^{2^{r}}+1$ . For example, we have proved that the regular $65537$ -polygon is constructible, without (happily) having to exhibit an explicit formula for $\cos\frac{2\pi}{65537}$ .

Remark 3.26

The converse to (7) is false; in particular, there are nonconstructible algebraic numbers of degree $4$ over $\mathbb{Q}{}$ . The polynomial $f(X)=$ $X^{4}-4X+2\in\mathbb{Q}{}[X]$ is irreducible, and we’ll show below (LABEL:cg8a) that the Galois group of a splitting field $E{}$ for $f$ is $S_{4}$ . Each root of $f(X)$ lies in an extension of degree $2^{2}$ of $\mathbb{Q}{}$ . If the four roots of $f(X)$ were constructible, then all the elements of $E$ would be constructible (1.36a), but if $H$ denotes a Sylow $2$ -subgroup of $S_{4}$ , then $E^{H}$ has odd degree over $\mathbb{Q}{}$ , and so no element of $E^{H}\smallsetminus\mathbb{Q}{}$ is constructible.³³ 3 It is possible to prove this without appealing to the Sylow theorems. If a root $\alpha$ of $f(X)$ were constructible, then there would exist a tower of quadratic extensions $\mathbb{Q}[\alpha]\supset M\supset\mathbb{Q}$ . By Galois theory, the groups $\operatorname{Gal}(E/M)\supset\operatorname{Gal}(E/\mathbb{Q}[\alpha])$ have orders $12$ and $6$ respectively. As $\operatorname{Gal}(E/\mathbb{Q})=S_{4}$ , $\operatorname{Gal}(E/M)$ would be $A_{4}$ . But $A_{4}$ has no subgroup of order $6$ , a contradiction. Thus no root of $f(X)$ is constructible. (Actually $\operatorname{Gal}(E/\mathbb{Q}[\alpha])=S_{3}$ , but that does not matter here.)