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

Frequently asked questions

Where exactly does the assumed algebraicity of

\gamma=\alpha^\beta

come in, if this step is only about

\beta

It comes in through the field $K=\mathbb{Q}(\alpha,\beta,\gamma)$ : because $\gamma$ is assumed algebraic, $K$ is a genuine number field, which is essential for the later steps' size/height estimates. This step only establishes algebraic independence of the two functions, a prerequisite the later steps will exploit inside that field $K$ .

Source

Aleksandr Gelfond and Theodor Schneider (independently), 1934

Sur le septième problème de Hilbert

Step 2 of 7: Setting up: two exponentials that would collapse if

\beta

were rational

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In plain words

The proof begins with a clean observation about the two functions $e^z$ and $e^{\beta z}$ : they are 'algebraically independent' — meaning no nonzero polynomial relation $q(e^z,e^{\beta z})=0$ can hold identically — precisely because $\beta$ is irrational. If $\beta$ were rational, say $\beta=m/n$ , then $(e^{\beta z})^n=(e^z)^m$ would be exactly such a relation.

This independence is the seed of a contradiction: the entire proof will construct a polynomial-like combination of $e^z$ and $e^{\beta z}$ that is forced to vanish too often for an algebraically independent pair, unless it is the zero function outright — which will itself lead to a contradiction with how the coefficients were chosen.

e^z,\ e^{\beta z}\ \text{algebraically dependent} \iff \beta \in \mathbb{Q}

Detailed analysis

Following the argument as presented by Siu (Harvard Math 113 notes, 'Theorem of Gelfond–Schneider on Transcendental Numbers'), suppose for contradiction that $\alpha$ is algebraic with $\alpha\ne0,1$ , $\beta$ is algebraic irrational, and $\gamma=\alpha^\beta=e^{\beta\log\alpha}$ is algebraic. Let $K=\mathbb{Q}(\alpha,\beta,\gamma)$ , a number field (finite extension of $\mathbb{Q}$ ).

Consider the two entire functions $f_1(z)=e^z$ and $f_2(z)=e^{\beta z}$ . If they satisfied a nonzero polynomial identity $q(f_1(z),f_2(z))\equiv0$ over $K$ , writing $q(T_1,T_2)=\sum_{i,j} b_{ij}T_1^iT_2^j$ gives $\sum_{i,j}b_{ij}e^{(i+j\beta)z}\equiv0$ ; since the exponents $i+j\beta$ are pairwise distinct real numbers whenever $\beta$ is irrational (as $(i_1-i_2)=(j_2-j_1)\beta$ forces $i_1=i_2,j_1=j_2$ unless $\beta$ is rational), the only way for such a sum of distinct exponentials to vanish identically is for every $b_{ij}$ to be zero. So $f_1,f_2$ are algebraically independent over $K$ — this uses irrationality of $\beta$ essentially.

This sets up the machinery of the 'Main Theorem' used throughout: if two entire functions of controlled growth satisfy a common algebraic differential equation and are algebraically independent, then they cannot simultaneously take values in a fixed number field $K$ at too many points (Siu, Main Theorem, following Lang's Introduction to Transcendental Numbers, Theorem 1, p. 21). The next steps build exactly the auxiliary function needed to exploit this.

Terms in this step

Algebraically independent functions: Functions $f_1,\dots,f_n$ are algebraically independent over a field $K$ if no nonzero polynomial $q$ with coefficients in $K$ satisfies $q(f_1(z),\dots,f_n(z))\equiv0$ for all $z$ — i.e. there is no polynomial relation among their values that holds identically.