Gelfond–Schneider theorem

Remarks

The proof builds an auxiliary function $F(z) = \sum_{i,j} c_{ij} e^{(i + j b) z \log a}$ with integer coefficients $c_{ij}$, not all zero, chosen so that $F$ vanishes at many integer points. If $a^b$ were algebraic, the values of $F$ and its derivatives at integers would be algebraic numbers of controlled size. An analytic estimate and a lower bound for nonzero algebraic numbers then force too many zeros, which is a contradiction. Examples: $2^{\sqrt{2}}$ is transcendental, and so is $e^{\pi}$, since $e^{\pi} = (-1)^{-i}$ with $\log(-1) = i\pi$. Another consequence: $\log 3 / \log 2$ is transcendental, since it is irrational and $2^{\log 3 / \log 2} = 3$. The hypothesis $b \notin \Q$ is necessary: $2^{1/2} = \sqrt{2}$ is algebraic. The theorem answers Hilbert's seventh problem. Baker's theorem generalizes it to linear forms in several logarithms.

Going deeper

For a student

An equivalent form: if $\alpha$ and $\beta$ are nonzero algebraic numbers and $\log \alpha$ and $\log \beta$ are linearly independent over $\Q$, then they are linearly independent over the algebraic numbers. This is the case of two logarithms in Baker's theorem. Together with the Lindemann–Weierstrass theorem, it is one of the classical results of transcendence theory. The question whether $e + \pi$ or $e \pi$ is transcendental is still open.