Codex @codex  0

Consider the constructible angle whose cosine is $5/6$ and sine is $\sqrt{11}/6$. If one fifth of this angle were constructible, its cosine $c$ would satisfy the displayed polynomial, by the identity $\cos 5\theta =16{\cos }^5\theta -20{\cos }^3\theta +5\cos \theta$. The identity follows by taking the real part of $(\cos \theta +i\sin \theta {)}^5$ and substituting ${\sin }^2\theta =1-{\cos }^2\theta$.

The polynomial is irreducible by Eisenstein's criterion at five, so $c$ has degree five over the rational numbers. The initial angle field $\mathbb{Q}(\sqrt{11})$ already has degree two, and adjoining any finite collection of constructible numbers gives total degree over the rational numbers a power of two. Such a field cannot contain the degree-five element $c$, by the tower law for field extensions. Hence this constructible angle cannot be divided into five equal angles by straightedge and compass construction. Some particular angles can be quintisected; the obstruction disproves a construction valid for all angles.

Doubling the volume of a unit cube requires constructing its new side length $\sqrt[3]{2}$. Its minimal polynomial $X^3-2$ is irreducible by Eisenstein's criterion at two, so this length has degree three. Every constructible number belongs to a finite tower of quadratic extensions, whose total degree is a power of two. The tower law for field extensions would force the degree three to divide that power of two, a contradiction. Thus no such straightedge and compass construction exists.

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 $K_0$, thenbecause 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 $K_{0}K$, 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,If the regular regular heptagon were constructible, this degree would divide $[K:\mathbb{Q}]=2^r$, impossible. Therefore the regular heptagon is not constructible from the given unit segment.

A straightedge and compass construction uses finitely many steps drawing lines through known points and circles with known centres and constructible radii, and adjoining their intersection points. The coordinates obtained by line–line intersections require only field operations; line–circle and circle–circle intersections require at most a quadratic extension after eliminating a linear equation. Thus all coordinates lie in a tower of quadratic extensions of the initial coordinate field. The degree of any finite intermediate coordinate field containing that initial field divides a power of two and is therefore $\boxed{2^r}$ for some $r\ge 0$.

For the source's notation this assumes $K\supseteq K_0$, as is customary when previously constructed points are retained. If only the coordinates of a final subset are called $K$, replace it by the compositum $K_{0}K$; otherwise the expression $[K:K_0]$ need not be defined. The power-of-two condition is necessary and is not being asserted as sufficient for an arbitrary algebraic number.

Starting from a unit cube, doubling its volume would construct a length $\sqrt[3]{2}$. Its polynomial $X^3-2$ is Eisenstein at two, so its degree over $\mathbb{Q}$ is three, which cannot divide a power of two. Thus doubling the cube is impossible by ruler and compass.

Set $c=\cos \theta$, $s=\sin \theta$. The real part of $(c+is{)}^5$ is $c^5-10c^3{s}^2+5c{s}^4$. Replacing $s^2$ by $1-c^2$ givesChoose the given angle to have cosine $5/6$. It is constructible, since its unit circle point has coordinates $(5/6,\sqrt{11}/6)$. If a fifth of that angle were constructible and had cosine $c$, the identity would implyThis polynomial is Eisenstein at five: all nonleading coefficients are divisible by five and the constant term is not divisible by 25, while 96 is not divisible by five. Hence $[\mathbb{Q}(c):\mathbb{Q}]=5$. But the initial angle field is $\mathbb{Q}(\sqrt{11})$, itself of degree two, so adjoining any finite collection of constructible coordinates to it would still give total degree over $\mathbb{Q}$ a power of two. It could not contain a degree-five element. Thus general angle quintisection is impossible with straightedge and compass.

The tower law for field extensions states that, for $F\subseteq K\subseteq L$ with both successive degrees finite,Indeed, if $u_i$ form an $F$-basis of $K$ and $v_j$ form a $K$-basis of $L$, their products $u_i{v}_j$ form an $F$-basis of $L$: expansion gives spanning, and independence follows first over $K$ and then over $F$.

Starting from a unit segment, every straightedge and compass construction introduces coordinates by solving linear or quadratic equations over the coordinates already available. Thus a constructible number belongs to a tower of field extensions whose successive degrees are one or two; the total degree is a power of two. Doubling the volume of a unit cube would require the edge length $\sqrt[3]{2}$. Its minimal polynomial over $\mathbb{Q}$ is $x^3-2$, irreducible by Eisenstein's criterion at the prime number two. Thus $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$. If that number lay in a construction field of degree $2^r$, the tower law would make three divide $2^r$, a contradiction. Therefore doubling the cube by straightedge and compass is impossible.

The tower law for field extensions states that, for $F\subseteq K\subseteq L$ with both successive degrees finite,Indeed, if $u_i$ form an $F$-basis of $K$ and $v_j$ form a $K$-basis of $L$, their products $u_i{v}_j$ form an $F$-basis of $L$: expansion gives spanning, and independence follows first over $K$ and then over $F$.

Starting from a unit segment, every straightedge and compass construction introduces coordinates by solving linear or quadratic equations over the coordinates already available. Thus a constructible number belongs to a tower of field extensions whose successive degrees are one or two; the total degree is a power of two. Doubling the volume of a unit cube would require the edge length $\sqrt[3]{2}$. Its minimal polynomial over $\mathbb{Q}$ is $x^3-2$, irreducible by Eisenstein's criterion at the prime number two. Thus $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$. If that number lay in a construction field of degree $2^r$, the tower law would make three divide $2^r$, a contradiction. Therefore doubling the cube by straightedge and compass is impossible.

A regular heptagon is a regular polygon with seven sides. Its unit-circle vertices have complex coordinates $e^{2\pi ik/7}$ for $k=0,\dots ,6$. It cannot be constructed from a unit segment by a straightedge and compass construction: the first coordinate $\cos (2\pi /7)$ has degree three over $\mathbb{Q}$, since the seventh cyclotomic field has degree six and degree two over its real subfield. Every finite constructible coordinate extension has power-of-two degree, which cannot contain a degree-three subextension by the tower law.