Worked solution: Wantzel's algebraic impossibility proof via field extensions (1837)

Step 3 of 7: Multiplying degrees along the tower: Wantzel's criterion

In plain words

Degrees of field extensions multiply along a tower, much like exchange rates multiply along a chain of currency conversions: two steps that each double something combine to quadruple it. Since every single step in a construction tower is a doubling (or does nothing), the whole tower's total size is a power of two.

Any constructible number $\alpha$ sits somewhere inside such a tower, so the smallest field containing it, $\mathbb{Q}(\alpha)$ , must have a size that divides a power of two — and a divisor of a power of two is itself a power of two.

[F_k:\mathbb{Q}] = \prod_{i=1}^k [F_i:F_{i-1}] = 2^k \implies [\mathbb{Q}(\alpha):\mathbb{Q}] = 2^m

Detailed analysis

By the tower law for field extensions, degrees multiply: $[F_k:\mathbb{Q}] = \prod_{i=1}^{k} [F_i:F_{i-1}]$ . Since Step 2 showed every factor is $1$ or $2$ , this product equals $2^k$ for some integer $k\ge 0$ — a power of $2$ regardless of how many construction steps were used.

Now suppose a real number $\alpha$ is constructible, so $\alpha \in F_k$ for some such tower, hence $\mathbb{Q}(\alpha)\subseteq F_k$ . Applying the tower law again to $\mathbb{Q}\subset\mathbb{Q}(\alpha)\subset F_k$ gives $[F_k:\mathbb{Q}] = [F_k:\mathbb{Q}(\alpha)]\cdot[\mathbb{Q}(\alpha):\mathbb{Q}]$ , so $[\mathbb{Q}(\alpha):\mathbb{Q}]$ divides $2^k$ . The only divisors of a power of $2$ are powers of $2$ , so $[\mathbb{Q}(\alpha):\mathbb{Q}] = 2^m$ for some integer $m\ge 0$ (Wantzel 1837, §I).

This is Wantzel's necessary condition for constructibility, and it says nothing about sufficiency: an irreducible degree- $4$ polynomial with Galois group $S_4$ , for instance, has roots of degree $4=2^2$ that are still not constructible. What it does give is a sharp obstruction — if an equation's degree over $\mathbb{Q}$ is not a power of $2$ , its roots are certainly not constructible, and that is exactly the tool the next steps apply to $60^\circ$ .

Terms in this step

Tower law (multiplicativity of degree): For a chain of fields $F_0\subset F_1\subset F_2$ , the degrees multiply: $[F_2:F_0]=[F_2:F_1]\cdot[F_1:F_0]$ . It lets a long chain of small extensions 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: an irreducible degree-

4

polynomial can still have non-constructible roots if its Galois group is not built from repeated square roots.