Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 20 / 3

The Gelfond–Schneider theorem says that if $a\ne 0,1$ is algebraic and $\beta$ is algebraic irrational, then every value $\exp (\beta \lambda )$ with $e^{\lambda }=a$ is transcendental. Fix such a nonzero logarithm $\lambda$, and suppose $\gamma =e^{\beta \lambda }$ is algebraic. The field $K=\mathbb{Q}(a,\beta ,\gamma )$ then contains all values needed in Schneider lattice proof of the Gelfond–Schneider theorem.

Here is an outline including the parameter balance. For a large $L$, choose polynomial degrees $M\asymp L\log L$, $N\asymp L/\log L$, with constants large enough that $(M+1)(N+1)>[K:\mathbb{Q}]L^2$. ImposeExpand these $K$-linear equations in a rational basis and clear denominators. Siegel lemma gives a nonzero integer polynomial $P$ with $\log H(P)=O(L^2/\log L)$. This follows because each coefficient in the grid equations has arithmetic height $O(M\log L+NL)$ and the ratio of unknowns to constraints stays bounded away from one. The exponential polynomial $F(z)=P(z,e^{\lambda z})$ is not identically zero: distinct exponentials are independent over the polynomial ring, as one sees by successively applying powers of $D-j\lambda$ to kill all but one term.

The irrationality of $\beta$ makes the $L^2$ grid points distinct. Their zeros make subsequent grid values extremely small. More precisely, if a whole $S\times S$ grid has been obtained, take an outer radius $R\asymp S\log S$. Growth gives $\log {\max }_{|z|=R}|F(z)|\le \log H(P)+O(M\log R+NR)$. The product form of Schwarz's lemma contributes $(C/\log S{)}^{S^2}$ at the next grid points. Thereforefor large initial $L$, unless the value is zero. But every such value is algebraic in the same $K$, since $e^{\lambda (s+t\beta )}=a^s{\gamma }^t$. Its arithmetic height is $O(\log H(P)+M\log S+NS)$, so the Liouville lower bound is incompatible with the displayed upper bound for a nonzero value. Induction extends the zero grid indefinitely.

A fixed nonzero exponential polynomial has only $O(R)$ zeros in a disc of radius $R$, by its exponential-type growth and Jensen's formula centered at a nonzero value. The grids give $S^2$ distinct zeros in a disc of radius $O(S)$, a contradiction. Thus $a^{\beta }$ is transcendental for every choice of logarithm, completing the Schneider lattice proof of the Gelfond–Schneider theorem.

A standard form of the Schneider-Lang theorem is this. Let $f_1,\dots ,f_s$ be meromorphic functions with $f_j^{\prime }\in K[f_1,\dots ,f_s]$ for a fixed number field $K$. Suppose $f_1,f_2$ have algebraic independence over $\mathbb{C}$ and have finite growth orders ${\rho }_1,{\rho }_2$. There are at most $[K:\mathbb{Q}]({\rho }_1+{\rho }_2)$ points at which all the functions are finite and all their values lie in $K$.

Take $f_1(z)=z$ and $f_2(z)=e^z$, of orders zero and one. Their derivatives belong to $K[z,e^z]$. They have algebraic independence: a polynomial relation ${\sum }_j{p}_j(z)e^{jz}=0$ is impossible by exponential-polynomial independence, or by letting real $z$ tend to infinity and considering the largest $j$.

If nonzero algebraic $\alpha$ had algebraic $e^{\alpha }$, put $K=\mathbb{Q}(\alpha ,e^{\alpha })$. At each of the distinct points $z=n\alpha$, both $z$ and $e^z=(e^{\alpha }{)}^n$ belong to $K$. Infinitely many points contradict the theorem's finite bound. Therefore