Part II, Chapter III — Analytical Summary

Gordan’s Elementary Umbral Proof of Hermite’s Theorem (§§ 1–3)

Blueprint of the Modular-Asymptotic Contradiction

Klein presents Paul Gordan's 1893 integral-free proof (1893) of Hermite's Theorem (1873): $e$ satisfies no polynomial $F(e) = \sum_{k=0}^n C_k e^k = 0$ with $C_k \in \mathbb{Z}$, $C_0 \neq 0$.

  1. Auxiliary Polynomial & Umbral Operator (§ 2): Define $h^r \mapsto r!$ after expanding in powers of $h$, and set

    \[\varphi(x) = \frac{x^{p-1}\bigl[(1 - x)(2 - x)\cdots(n - x)\bigr]^p}{(p - 1)!} \qquad \bigl(p \text{ prime},\; p > \max(n, |C_0|)\bigr).\]

  2. Simultaneous Arithmetic & Analytic Control (§ 2):

    • Modulo $p$ non-vanishing at $0$: Multiplicity $p - 1$ at $x = 0$ makes $\varphi(h) \equiv (n!)^p \not\equiv 0 \pmod p$.

    • Modulo $p$ vanishing at $1, \dots, n$: Multiplicity $p$ at $x = k$ shifts the lowest power of $h$ in $\varphi(h + k)$ to $h^p \mapsto p!$, forcing $\varphi(h + k) \equiv 0 \pmod p$.

    • Factorial damping as $p \to \infty$: Because $(p - 1)!$ outgrows $u^p$, $\sum_r |c_r|\,|x|^r \to 0$ as $p \to \infty$.

  3. Exponential Splitting (§ 3): Truncating $e^k \cdot r!$ at degree $r$ yields $e^k h^r = (h + k)^r + q_{r,k} k^r e^k$ ($0 < q_{r,k} < 1$), whence

    \[0 = \varphi(h) F(e) = \underbrace{C_0 \varphi(h) + \sum_{k=1}^n C_k \varphi(h + k)}_{\in\,\mathbb{Z},\;\equiv\, C_0(n!)^p \,\not\equiv\, 0 \pmod p} \;+\; \underbrace{\sum_{k=1}^n C_k \varepsilon_k}_{|\cdot|\,<\,1 \text{ for large } p},\]

    an impossible equality between a non-zero integer and a proper fraction.