Worked solution: Gelfond–Schneider transcendence proof via auxiliary functions (1934)

Frequently asked questions

Could it happen that

\Phi

vanishes to order exactly

L

everywhere and we can't find any nonzero derivative at all?

No — a nonzero entire function can only have isolated zeros of finite order; it cannot vanish to arbitrarily high (or infinite) order at a point without being identically zero, by the identity theorem for holomorphic functions. Since $\Phi\not\equiv0$ , some derivative at some chosen point must eventually be nonzero.

Source

Aleksandr Gelfond and Theodor Schneider (independently), 1934

Sur le septième problème de Hilbert

Step 4 of 7:

\Phi

is not identically zero: extracting a first nonzero derivative

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In plain words

Siegel's lemma guarantees the polynomial $G$ is not the zero polynomial, but it does not automatically guarantee that the resulting function $\Phi(z)=G(e^z,e^{\beta z})$ is not the zero function — that step needs the algebraic independence of $e^z$ and $e^{\beta z}$ established earlier. Because they are independent, a nonzero polynomial $G$ can never produce the identically zero function $\Phi$ .

So $\Phi$ genuinely has some smallest order $s\le L$ at which it fails to vanish at one of the chosen points — say at $z_0=0$ . This single nonzero value $\Phi^{(s)}(0)$ is the key number the rest of the proof will squeeze from both sides.

\Phi^{(s)}(z_0) \ne 0 \ \text{for some smallest } s \le L

Detailed analysis

Since $e^z$ and $e^{\beta z}$ are algebraically independent over $K$ (Step 2) and $G$ is a nonzero polynomial with coefficients in $K$ (Step 3, Siegel's lemma), the composite $\Phi(z)=G(e^z,e^{\beta z})$ cannot be the identically zero function — if it were, $G$ itself would witness a nontrivial algebraic relation between $e^z,e^{\beta z}$ , contradicting independence.

By construction $\Phi$ vanishes to order at least $L$ at each $z_0,\dots,z_m$ (Step 3), but it is a nonzero entire function, so it cannot vanish to infinite order everywhere. After relabeling the points $z_0,\dots,z_m$ if necessary, one may assume $\Phi$ 's vanishing order at $z_0$ is the smallest among these points, call it $s\le L$ ; then $\Phi^{(s)}(z_0)\ne0$ while $\Phi^{(t)}(z_0)=0$ for $t<s$ (Siu, proof of Main Theorem). Translating coordinates so $z_0=0$ for convenience, this single nonzero algebraic number $\Phi^{(s)}(0)\in K$ is the pivot of the whole argument: everything now comes down to bounding $|\Phi^{(s)}(0)|$ from both above and below.

The next two steps produce exactly such a pair of bounds: an upper bound from complex analysis (using that $\Phi$ is small near many points, via Jensen's formula), and a lower bound from algebraic number theory (using that nonzero algebraic integers cannot be too small).