Part I, Chapter I — Analytical Summary

Step-by-Step Logical Architecture of Klein’s Degree-$2^h$ Theorem (§§ 1–15)

Logical Chain of Proof

Without invoking abstract Galois group machinery prematurely, Klein proves Wantzel's (1837) theorem entirely by elementary polynomial algebra over a domain of rationality $K_0$:

  1. Radical Order & Algebraic Independence (§§ 1–4): Every constructible quantity $x$ has a finite radical order $\mu$ and depends on $m$ square roots $Q_1, \dots, Q_m$, pruned from the bottom up so that no $Q_k$ is a rational function of the remaining $m - 1$ radicals over $K_0$.

  2. Rationalized Normal Form (§§ 5–6): Successive conjugation $(c - d Q_k)$ eliminates every $Q_k$ from denominators ($c + d Q_k \neq 0$ by independence), expressing $x = \alpha + \beta Q_\mu$ and ultimately expanding $x$ as a $K_0$-linear combination of multilinear products $Q_{i_1}\cdots Q_{i_r}$.

  3. The $2^m$ Conjugate Branches & $F(x) = 0$ (§§ 7–9): Flipping $\pm Q_k$ across all $m$ independent square roots yields $2^m$ formal conjugate values $x_1, \dots, x_{2^m}$. Because flipping each highest-order radical permutes the linear factors $(x - x_i)$, the symmetric product $F(x) = \prod_{i=1}^{2^m} (x - x_i)$ has coefficients in $K_0$ and degree $2^m$.

  4. Conjugate Propagation Under Any Rational Polynomial (§ 10): If $f(x) \in K_0[x]$ satisfies $f(x_1) = 0$, writing $f(\alpha + \beta Q_\mu) = A + B Q_\mu = 0$ forces $B = 0$ (otherwise $Q_\mu = -A/B$ violates § 4) and hence $A = 0$, so $f(\alpha - \beta Q_\mu) = A - B Q_\mu = 0$. Descending through all $m$ radicals proves $f(x_i) = 0$ for all $i = 1, \dots, 2^m$.

  5. Irreducible Minimal Polynomial & Power Factorization (§§ 11–14): The monic minimal polynomial $\varphi(x) \in K_0[x]$ of $x_1$ divides every polynomial vanishing at $x_1$. Repeatedly dividing $F(x)$ by $\varphi(x)$ yields $F(x) = [\varphi(x)]^\nu$, whence $2^m = \deg(\varphi)\cdot\nu$ and $\deg(\varphi) = 2^h$.

  6. Contrapositive Obstruction & Galois Sufficiency Caveat (§ 15): Any problem whose irreducible polynomial has degree $M \neq 2^h$ is impossible with straightedge and compasses. Conversely, $M = 2^h$ is necessary but not sufficient in general (e.g., an $S_4$ or $A_4$ quartic of degree $4 = 2^2$ fails because its cubic resolvent is irreducible).