= Solution
A finite <straightedge and compass construction> repeatedly draws the line through two available distinct points or the <circle> with an available center and an available distance as radius, and adjoins isolated intersections of two such lines or <circles>. A finite set is constructible when its points occur after finitely many such steps, starting with the specified points.
If the current coordinate <field> is $F$, intersections of two lines solve <linear equations> over $F$. A line and a <circle> lead to a <quadratic equation> over $F$. For two distinct <circles>, subtraction of their equations gives a line, reducing to the previous case. Thus each step adjoins coordinates in an extension of degree one or two. The <tower law> gives a finite tower of quadratic-or-trivial <field extensions> containing all output coordinates. If the final coordinate <field> contains $K_0$, then
$$
\boxed{[K:K_0]=2^r\quad\text{for some integer }r\ge0,}
$$
because its degree divides the power-of-two degree of that tower.
Two implicit qualifications are necessary for this degree statement. The output is obtained in finitely many steps; the aggregate of infinitely many individually constructible points need not generate a finite extension. Also, unless the output retains the original coordinates, $K$ need not contain $K_0$. In that case the degree statement applies to the compositum $K_0K$, rather than an undefined $[K:K_0]$. The heptagon application explicitly includes $S_0$, so neither issue obstructs it.
Here $K_0=\mathbb Q$, and a vertex adjacent to $(1,0)$ has first coordinate $c=\cos(2\pi/7)$, so $\mathbb Q(c)\subseteq K$. Let $\zeta=e^{2\pi i/7}$. The <cyclotomic polynomial> $1+X+\cdots+X^6$ is <irreducible>, giving $[\mathbb Q(\zeta):\mathbb Q]=6$. Since $\zeta+\zeta^{-1}=2c$, it satisfies $X^2-2cX+1$ over $\mathbb Q(c)$. The element $\zeta$ is nonreal while $\mathbb Q(c)\subset\mathbb R$, so this relative degree is exactly two. By the <tower law>,
$$
[\mathbb Q(c):\mathbb Q]=3.
$$
If the regular <regular heptagon> were constructible, this degree would divide $[K:\mathbb Q]=2^r$, impossible. Therefore \b[the <regular heptagon> is not constructible] from the given unit segment.
Back to article page
