Worked solution: Wantzel's algebraic impossibility proof for doubling the cube (1837)

Frequently asked questions

Does this argument use anything specific to cubes, or does it apply to any construction problem?

Nothing specific to cubes: this step only uses the geometry of lines and circles, so the same degree- $2^m$ criterion applies to any straightedge-and-compass problem, including angle trisection and regular polygons.

Source

Pierre Laurent Wantzel, 1837

Recherches sur les moyens de reconnaître si un problème de géométrie peut se résoudre avec la règle et le compas

Step 2 of 5: Wantzel's criterion: constructible numbers have degree

2^m

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In plain words

Every point a straightedge and compass can add to a picture comes from intersecting lines and circles, and in coordinates that always means solving an equation of degree at most $2$ . So building up a construction is like a game where each move either keeps the numbers you already have, or bolts on one new square root.

Because 'doubling' actions chain multiplicatively, however many moves are made, the total complexity of the numbers involved is always some power of $2$ — never a $3$ , a $5$ , or any other prime.

\mathbb{Q} = F_0 \subset F_1 \subset \cdots \subset F_k,\ [F_i{:}F_{i-1}]\in\{1,2\} \implies [\mathbb{Q}(\alpha){:}\mathbb{Q}] = 2^m

Detailed analysis

Fix coordinates with $F_0=\mathbb{Q}$ . As in the general theory of straightedge-and-compass constructions, each new constructible point is the intersection of two lines, a line and a circle, or two circles, all with coefficients in the current field $F_{i-1}$ ; solving the corresponding system produces coordinates in $F_{i-1}$ itself or in a quadratic extension $F_i=F_{i-1}(\sqrt{d_i})$ for some $d_i\in F_{i-1}$ . So a finite sequence of construction steps yields a tower $\mathbb{Q} = F_0 \subset F_1 \subset \cdots \subset F_k$ with $[F_i:F_{i-1}]\in\{1,2\}$ (Wantzel 1837, §I).

By the tower law, degrees multiply along the chain: $[F_k:\mathbb{Q}] = \prod_{i=1}^k [F_i:F_{i-1}] = 2^k$ . If a real number $\alpha$ is constructible, then $\alpha\in F_k$ for some such tower, so $\mathbb{Q}(\alpha)\subseteq F_k$ , and applying the tower law again shows $[\mathbb{Q}(\alpha):\mathbb{Q}]$ divides $2^k$ . A divisor of a power of $2$ is itself a power of $2$ , so $[\mathbb{Q}(\alpha):\mathbb{Q}] = 2^m$ for some integer $m\ge 0$ .

This necessary condition is the single tool the rest of the proof needs: to show $\sqrt[3]{2}$ is not constructible, it now suffices to compute $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]$ and check it is not a power of $2$ .

Terms in this step

Tower law (multiplicativity of degree): For a chain of fields $F_0\subset F_1\subset F_2$ , degrees multiply: $[F_2:F_0]=[F_2:F_1]\cdot[F_1:F_0]$ , so a long chain of small extensions can be measured all at once.

Knowledge used in this step

Galois theory

Common mistake. Degree

2^m

over

\mathbb{Q}

is necessary but not sufficient for constructibility; here it happens to be enough only because it will turn out that

3

fails to be a power of

2

at all.