---
title: Straightedge-and-Compass Constructions
module: Field Theory
moduleNumber: 10
lessonNumber: 2
order: 1002
summary: >
  The lengths a straightedge and compass can build from a unit form a field
  closed under square roots, and every constructible number lies in a tower of
  quadratic extensions. So its degree over the rationals is a power of two. That
  single obstruction settles three problems the Greeks left open: doubling the
  cube, trisecting a general angle, and squaring the circle are all impossible.
topics: [Field Theory]
sources:
  - book: Dummit & Foote
    ref: "Ch. 13 — Field Theory; §13.3 Classical Straightedge and Compass Constructions"
  - book: Judson
    ref: "Ch. 21 — Fields; §21.3 Geometric Constructions"
draft: false
---

Three geometry problems survived from antiquity without an answer for two
thousand years: build a cube of twice a given volume, cut an arbitrary angle in
three, and build a square equal in area to a given circle, each using only a
straightedge and a compass. Each was eventually shown impossible, and the proof
is not geometric. It translates ruler-and-compass steps into field operations,
notices that every step lands in an extension of degree at most $2$, and reads
the impossibility off the [tower law](/abstract-algebra/field-theory/field-extensions-and-algebraic-elements).

## Constructions as field operations

Fix a segment of length $1$. Using it to scale the axes, place every
construction in the Cartesian plane $\mathbb{R}^2$. A point is **constructible**
if it can be reached from the starting data by a finite sequence of the four
legal moves, and a real number is constructible if it is the coordinate of a
constructible point.

> **Definition (Constructible number).** A real number $\alpha$ is
> **constructible** if the point $(\alpha, 0)$ can be obtained from the points
> $(0,0)$ and $(1,0)$ by a finite sequence of the operations: draw the line
> through two constructed points, draw the circle centered at one constructed
> point through another, and mark intersections of these lines and circles.

Elementary geometry realizes each arithmetic operation as a construction. Given
lengths $a$ and $b$, laying off segments produces $a + b$ and $a - b$ directly;
similar triangles built from parallel lines produce the product $ab$ and the
quotient $a/b$; and a semicircle on a diameter of length $1 + a$ produces
$\sqrt{a}$ as the perpendicular half-chord.

$$
% caption: The intercept theorem: marking 1 and a on one ray, b on another, and
% drawing the parallel to the unit chord cuts off the product ab on the second ray.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\coordinate (O) at (0,0);
\coordinate (P1) at (1.4,0);
\coordinate (Pa) at (3.6,0);
\coordinate (Pb) at (0.99,0.99);
\coordinate (Pab) at (2.55,2.55);
\draw[black] (O) -- (4.2,0);
\draw[black] (O) -- (3.1,3.1);
\draw[acc, very thick] (P1) -- (Pb);
\draw[acc, very thick, dashed] (Pa) -- (Pab);
\fill[black!70] (P1) circle (1.4pt);
\fill[black!70] (Pa) circle (1.4pt);
\fill[black!70] (Pb) circle (1.4pt);
\fill[acc] (Pab) circle (1.8pt);
\node[font=\footnotesize, below] at (P1) {1};
\node[font=\footnotesize, below] at (Pa) {a};
\node[font=\footnotesize, left] at (Pb) {b};
\node[acc, font=\footnotesize, anchor=south east] at (Pab) {ab};
\node[font=\footnotesize, below left] at (O) {O};
\end{tikzpicture}
$$

$$
% caption: A right triangle inscribed in the semicircle on diameter 1+a has
% altitude the square root of a, so a compass and straightedge extract square roots.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% semicircle on diameter from (0,0) to (5,0), split 1 and a at x=1.4
\draw[black] (0,0) -- (5,0);
\draw[black] (2.5,0) ++(180:2.5) arc (180:0:2.5);
\coordinate (O) at (0,0);
\coordinate (P) at (1.4,0);
\coordinate (E) at (5,0);
% altitude at P up to circle: h = sqrt(1.4 * 3.6) = sqrt(5.04) approx 2.245
\coordinate (T) at (1.4,2.245);
\draw[acc, very thick] (P) -- (T);
\draw[black] (O) -- (T) -- (E);
\fill[acc] (T) circle (1.8pt);
\node[font=\footnotesize] at (0.7,-0.32) {1};
\node[font=\footnotesize] at (3.2,-0.32) {a};
\node[acc, font=\footnotesize, anchor=west] at (1.55,1.1) {square root of a};
\draw (1.4,0.28) -- (1.68,0.28) -- (1.68,0);
\end{tikzpicture}
$$

So constructible numbers are closed under the four field operations and under
square roots. Two consequences fix the algebraic setting.

- **The constructible numbers form a subfield of $\mathbb{R}$** containing
  $\mathbb{Q}$, since starting from $1$ the field operations reach every
  rational.
- **A construction cannot escape a quadratic step.** Lines through constructed
  points have equations with coefficients in the field $F$ generated so far, and
  intersecting two lines stays in $F$. A circle contributes an equation
  $(x - h)^2 + (y - k)^2 = r^2$ with $h, k, r \in F$; intersecting it with a line
  gives a quadratic for one coordinate, and intersecting two circles reduces, by
  subtracting the equations, to a line meeting a circle. Every new coordinate
  therefore lies in an extension of $F$ of degree $1$ or $2$.

## The degree obstruction

Iterating the second point along a construction produces a tower in which each
step is quadratic.

> **Proposition (Constructible degree).** If $\alpha \in \mathbb{R}$ is
> constructible from a subfield $F \subseteq \mathbb{R}$, then $\alpha$ lies in a
> field $K \supseteq F$ with $[K : F] = 2^m$ for some $m \ge 0$. In particular
> $[F(\alpha) : F]$ is a power of $2$.

> **Proof.** A finite construction visits fields
> $$
> F = F_0 \subseteq F_1 \subseteq \cdots \subseteq F_m = K,
> \qquad [F_{i+1} : F_i] \le 2,
> $$
> so by the tower law $[K : F] = \prod_i [F_{i+1} : F_i]$ is a power of $2$.
> Since $F(\alpha)$ sits between $F$ and $K$, its degree $[F(\alpha) : F]$
> divides $[K : F]$ and is itself a power of $2$.[^df-prop23]

$$
% caption: Each straightedge-and-compass move adjoins at most a square root, so a
% constructible number sits atop a tower of degree-2 steps and has 2-power degree.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node (F0) at (0,0) {Q};
\node (F1) at (0,1.4) {F1};
\node (F2) at (0,2.8) {F2};
\node (F3) at (0,4.2) {K};
\draw[thick] (F0) -- (F1) node[midway, right=2pt, font=\footnotesize] {2};
\draw[thick] (F1) -- (F2) node[midway, right=2pt, font=\footnotesize] {2};
\draw[thick] (F2) -- (F3) node[midway, right=2pt, font=\footnotesize] {2};
\node[acc, font=\footnotesize, align=left] at (3.4,2.1)
  {each step adjoins one\\square root, so the\\degree of K over Q is\\2 times 2 times 2 = 8};
\end{tikzpicture}
$$

In particular every constructible number is algebraic over $\mathbb{Q}$, since it
lies in a finite extension. The obstruction is one-directional: a $2$-power degree
is necessary for constructibility, not sufficient. Some numbers of degree
$4$ over $\mathbb{Q}$ are not constructible, because the obstruction lies in the
splitting field rather than in the degree of the number itself; the sharp
criterion is that $\alpha$ is constructible exactly when the splitting field of
its minimal polynomial has $2$-power degree, which the Galois theory of later
lessons makes precise. What matters for the three classical problems is the crude
contrapositive: if $[\mathbb{Q}(\alpha) : \mathbb{Q}]$ is not a power of $2$, then
$\alpha$ is not constructible.

### A constructive tower

The positive direction is equally concrete: a number built from rationals by a
finite chain of square roots is constructible, and that chain is the
tower itself. The regular pentagon is the clean case. Its central quantity is
$\alpha = 2\cos\tfrac{2\pi}{5}$, a root of $x^2 + x - 1 = 0$, so
$$
\alpha = \frac{-1 + \sqrt 5}{2},
\qquad [\mathbb{Q}(\alpha) : \mathbb{Q}] = 2.
$$
The single quadratic step $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt 5)$ carries
$\alpha$, and the construction mirrors it: build $\sqrt 5$ as a semicircle
altitude, do one rational division and subtraction, and the pentagon's vertex
coordinate is in hand. A number needing two nested roots, such as
$\sqrt{2 + \sqrt 2}$, sits atop a two-step tower
$\mathbb{Q} \subseteq \mathbb{Q}(\sqrt 2) \subseteq \mathbb{Q}(\sqrt{2 + \sqrt 2})$
of degrees $2$ and $2$, so it is constructible with degree $4$ over $\mathbb{Q}$.
Each nested radical is one more quadratic floor.

## The three impossibilities

> **Theorem.** None of doubling the cube, trisecting a general angle, or squaring
> the circle is possible with straightedge and compass.

Each collapses to a degree that is not a power of $2$.

> **Worked example (doubling the cube).** A cube of twice the unit volume has
> edge $\sqrt[3]{2}$, so the task is to construct that length. Over $\mathbb{Q}$
> the polynomial $x^3 - 2$ is irreducible by Eisenstein at $2$, so it is the
> minimal polynomial of $\sqrt[3]{2}$ and
> $$
> [\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 3.
> $$
> Since $3$ is not a power of $2$, $\sqrt[3]{2}$ is not constructible.

> **Worked example (trisecting $60^\circ$).** Constructing an angle $\theta$ is
> equivalent to constructing $\cos\theta$. The triple-angle identity
> $\cos\theta = 4\cos^3\tfrac{\theta}{3} - 3\cos\tfrac{\theta}{3}$ turns
> trisection into a cubic for $\cos\tfrac{\theta}{3}$. Take $\theta = 60^\circ$,
> where $\cos\theta = \tfrac12$. Writing $\beta = \cos 20^\circ$,
> $$
> 4\beta^3 - 3\beta = \tfrac12
> \quad\Longrightarrow\quad
> 8\beta^3 - 6\beta - 1 = 0.
> $$
> Substituting $\alpha = 2\beta$ clears denominators to $\alpha^3 - 3\alpha - 1 = 0$.
> By the rational root test the only candidates for a root of $x^3 - 3x - 1$ are
> $\pm 1$, and $(\pm 1)^3 - 3(\pm 1) - 1 = \mp 2 - 1 \ne 0$, so the cubic has no
> rational root; a cubic without one is irreducible, giving
> $[\mathbb{Q}(\alpha) : \mathbb{Q}] = 3$. The angle $60^\circ$ is constructible
> (an equilateral triangle produces it), but $20^\circ$ is not, so no procedure
> trisects every angle. No finite sequence of quadratic steps reaches degree $3$.

**Squaring the circle.** A square equal in area to the unit circle has side
$\sqrt{\pi}$, whose constructibility would make $\pi$ constructible and in
particular algebraic. But $\pi$ is transcendental, so
$[\mathbb{Q}(\pi) : \mathbb{Q}]$ is infinite, let alone a power of $2$.[^df-thm24]

$$
% caption: Each classical problem reduces to a target number whose degree over Q
% is 3 or infinite, never a power of 2, which is the exact obstruction.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node[draw, minimum width=32mm, minimum height=10mm, align=center] (a) at (0,2.2) {double the cube};
\node[draw, minimum width=32mm, minimum height=10mm, align=center] (b) at (0,0.9) {trisect 60 degrees};
\node[draw, minimum width=32mm, minimum height=10mm, align=center] (c) at (0,-0.4) {square the circle};
\node[minimum width=26mm, align=center] (a2) at (4.3,2.2) {cube root of 2\\degree 3};
\node[minimum width=26mm, align=center] (b2) at (4.3,0.9) {cos 20 degrees\\degree 3};
\node[minimum width=26mm, align=center] (c2) at (4.3,-0.4) {pi\\degree unbounded};
\draw[->, acc, thick] (a) -- (a2);
\draw[->, acc, thick] (b) -- (b2);
\draw[->, acc, thick] (c) -- (c2);
\node[draw=acc, thick, fill=acc!8, minimum width=26mm, minimum height=13mm, align=center] (out) at (8.6,0.9) {not a power of 2\\so impossible};
\draw[->, black, thick] (a2) -- (out);
\draw[->, black, thick] (b2) -- (out);
\draw[->, black, thick] (c2) -- (out);
\end{tikzpicture}
$$

The proof of one impossibility also produces a positive fact worth recording:
the trigonometric functions of an integer-degree angle are constructible exactly
when the angle is a multiple of $3^\circ$. The equilateral triangle gives
$60^\circ$, the regular pentagon gives $72^\circ$, and the difference and
half-angle formulas assemble $3^\circ$; but $1^\circ$ and $2^\circ$ are out of
reach, since either would let the addition formulas produce the impossible
$20^\circ$.

## The marked ruler

The negative results are specific to the straightedge, an unmarked edge. A
**ruler** carries marks, and the extra move of sliding a marked ruler until a
fixed distance falls between two curves reaches beyond quadratic steps. An
Archimedean construction with a marked ruler trisects any angle, and a related
one doubles the cube. Both classical impossibilities disappear once the tool is
allowed to solve certain cubics, which shows the theorem is a statement about
the algebra the two idealized instruments generate, degree-$2$ steps only,
rather than about geometry in general.

## Constructible regular polygons

The regular $n$-gon is constructible exactly when $\cos\tfrac{2\pi}{n}$ is a
constructible number, so the question is again one of degree, now for a root of
unity. The pentagon works because $\alpha = 2\cos\tfrac{2\pi}{5}$ satisfies
$x^2 + x - 1 = 0$, degree $2$. The regular $7$-gon and $9$-gon fail:
$2\cos\tfrac{2\pi}{7}$ satisfies an irreducible cubic, and constructing the
$9$-gon would trisect $120^\circ$.

| Regular $n$-gon | governing degree | constructible? |
| --- | --- | --- |
| $3$ (triangle) | $1$ | yes |
| $4$ (square) | $1$ | yes |
| $5$ (pentagon) | $2$ | yes |
| $6$ (hexagon) | $1$ | yes |
| $7$ (heptagon) | $3$ | no |
| $9$ (enneagon) | $3$ | no |
| $17$ | $2^3 = 8$ | yes |

The full criterion belongs to the theory of cyclotomic fields: the degree of
$\mathbb{Q}(\cos\tfrac{2\pi}{n})$ over $\mathbb{Q}$ is $\tfrac12\varphi(n)$, and
constructibility of the $n$-gon amounts to that degree being a power of $2$,
which happens precisely when $n$ is a power of $2$ times distinct Fermat primes.
The degree $\tfrac12\varphi(n)$ comes from the
[cyclotomic polynomials](/abstract-algebra/field-theory/separable-and-cyclotomic-extensions);
the $17$-gon entry, Gauss's discovery, follows from $\varphi(17) = 16 = 2^4$.

[^df-prop23]: **Dummit & Foote**, §13.3, Proposition 23 — a number obtained by compass and straightedge from $F$ generates an extension of $2$-power degree, via the tower law applied to the quadratic steps.
[^df-thm24]: **Dummit & Foote**, §13.3, Theorem 24 — impossibility of the three classical constructions; the transcendence of $\pi$ (Lindemann) is quoted, not proved there. See also Judson §21.3 for the same three arguments.