Algebraic Values of Transcendental Functions at Algebraic Points

In honor of Pi Day 2023, I’d like to discuss Hilbert’s 7th Problem, which in an oversimplified (and rather vague) form asks: under what circumstances can a transcendental function take algebraic values at algebraic points?

The connection with $\pi$ is that Lindemann proved in 1882 that the transcendental function $f(z) = e^z$ takes transcendental values at every nonzero algebraic number. Since $e^{\pi i} = -1$ by Euler’s formula, this proves that $\pi i$, and hence $\pi$ itself, is transcendental. In light of this theorem, it is natural to wonder what if anything is special here about the function $f(z) = e^z$ and the point $z=0$.

One thing that’s special about $z=0$ is that if $\alpha \neq 0$ is algebraic and $e^\alpha$ is also algebraic, then both $n\alpha$ and $e^{n \alpha}$ are algebraic for all $n \in {\mathbb Z}$, and these numbers are all distinct. So one might be led to speculate that if $f$ is a transcendental entire function then there are only finitely many algebraic numbers $\alpha$ for which $f(\alpha)$ is also algebraic.

Unfortunately, as Hilbert knew, this is completely false. For example, the function $f(z) = e^{2\pi iz}$ is transcendental but it takes the rational value 1 at every integer. In 1886, Weierstrass had given an example of a transcendental entire function that takes rational values at all rational numbers; later, in 1895, Stäckel showed that there is a transcendental entire function that takes rational values at all algebraic points. However, the functions of Weierstrass and Stäckel, are in some sense “pathological”; they have large growth rates and do not occur “in nature”. The challenge is to make this intuitive feeling more precise, and also to distinguish $e^z$ from $e^{2\pi iz}$.

One thing that is special about $e^z$, which is not shared by any of the other functions mentioned in the previous paragraph, is that it satisfies a linear differential equation with rational coefficients (namely $f'(z) = f(z)$). The existence of such a (not necessarily linear) differential equation turns out to be the key idea needed to generalize Lindemann’s theorem in a substantial way.

Another fruitful generalization is to rephrase our original question as an unlikely intersection problem: given two algebraically independent entire functions $f_1(z)$ and $f_2(z)$ satisfying suitable hypotheses, can we conclude that there are only finitely many complex numbers $\alpha$ such that $f_1(\alpha)$ and $f_2(\alpha)$ are simultaneously algebraic? This generalizes our original question by letting $f_1(z) = z$ and $f_2(z) = f(z)$.