Part I, Chapter I — Critical & Symbolic Notes

Galois-Theoretic Refinement of § 15 and Exact Quartic Resolvent Tests in geometor.klein

Why Klein’s Independence Hypothesis (§ 4) Is Essential

Klein’s proof of § 10 ($A + B Q_\mu = 0 \implies A = B = 0$) hinges directly on § 4: no $Q_k$ may lie in the field generated by the remaining radicals. Consider, for instance, a hidden dependency:

\[Q_1 = \sqrt{2}, \quad Q_2 = \sqrt{3}, \quad Q_3 = \sqrt{6} = Q_1 Q_2.\]

If one failed to eliminate $Q_3 = Q_1 Q_2$, toggling the sign of $Q_3$ independently of $Q_1, Q_2$ would turn $x = Q_1 Q_2 - Q_3 = 0$ into $2\sqrt{6} \neq 0$, violating conjugate invariance under $f(x) = x = 0$! In Chapter IV, § 6, Klein returns explicitly to § 4 when simplifying Gauss's $17$-gon tower: among the four radicals $\sqrt{17}$, $\sqrt{x_1^2 + 4}$, $\sqrt{x_2^2 + 4}$, and $\sqrt{y_1^2 - 4y_4}$, the identity

\[\sqrt{x_1^2 + 4}\,\sqrt{x_2^2 + 4} = 2\sqrt{17}\]

shows that $\sqrt{x_2^2 + 4}$ is rationally dependent upon $\sqrt{17}$ and $\sqrt{x_1^2 + 4}$, reducing $m$ from $4$ to $4$ independent radicals only after accounting for $\sqrt{y_1^2 - 4y_4}$ (so $2^4 = 16 = \deg \Phi_{17}$).

Modern Galois Criterion vs. Degree $2^h$ (§ 15)

In modern field theory (Évariste Galois), if $\varphi(x) \in K_0[x]$ is irreducible with splitting field $L$ over $K_0$:

• Necessity of $\deg \varphi = 2^h$: Tower multiplicativity $[K_m : K_0] = 2^m = [K_m : K_0(x_1)]\,[K_0(x_1) : K_0]$ immediately gives $\deg \varphi = [K_0(x_1) : K_0] = 2^h$.

• Exact Necessary & Sufficient Condition: $x_1$ is constructible by square roots over $K_0$ if and only if the degree $[L : K_0] = |\mathrm{Gal}(L / K_0)|$ of the normal closure $L$ is a power of $2$ (a finite $2$-group).

For degree $M = 4 = 2^2$, let $\varphi(x) = x^4 + p x^2 + q x + r \in \mathbb{Q}[x]$ be a depressed irreducible quartic. In geometor.klein.constructibility.cubic_resolvent() and analyze_constructibility(), we form the exact cubic resolvent

\[R(t) = t^3 - 2p\,t^2 + (p^2 - 4r)\,t + q^2 \in \mathbb{Q}[t].\]

• Constructible $D_4 / V_4 / C_4$ Quartic: For $\varphi(x) = x^4 - 2$ ($p = q = 0, r = -2$), $R(t) = t^3 + 8t = t(t^2 + 8)$ has the rational root $t = 0$, so $|\mathrm{Gal}| = 8 = 2^3$ and $x = \sqrt{\sqrt{2}}$ is constructible (tower depth $2$).

• Non-Constructible $S_4$ Quartic: For $\varphi(x) = x^4 - x - 1$ ($p = 0, q = -1, r = -1$), $\deg \varphi = 4 = 2^2$, yet $R(t) = t^3 + 4t + 1$ is irreducible over $\mathbb{Q}$ ($R(\pm 1) \neq 0$), so $\mathrm{Gal}(\varphi) \cong S_4$ of order $24 = 2^3 \cdot 3$—verified symbolically as non-constructible by analyze_constructibility(x**4 - x - 1)!