Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 29 / 2

An integer version of Siegel lemma is as follows. For $M<N$ homogeneous linear equations $Az=0$, with integer matrix entries bounded in absolute value by $A_0\ge 1$, there is a nonzero integer solution satisfyingPut the displayed integer bound equal to $Q$ and evaluate the equations at every vector in $\{0,\dots ,Q{\}}^N$. There are $(Q+1{)}^N$ inputs and at most $(2N{A}_{0}Q+1{)}^M$ outputs. But $2N{A}_{0}Q+1\le 2N{A}_0(Q+1)$ and $(Q+1{)}^{N-M}>(2N{A}_0{)}^M$, so the number of inputs is larger. Two distinct inputs have identical outputs; their difference proves the claim. This is the pigeonhole proof of integer Siegel lemma. Rational coefficients are handled by clearing denominators. Over a number field of degree $d$, expanding in a rational basis gives at most $dM$ rational equations, so the same construction applies when $N>dM$, with the height of the resulting integer matrix controlling the bound.

Siegel's original arithmetic definition starts with an entire series $E(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$, with all $a_n$ in a fixed number field. For every $ϵ>0$ there must be positive integers $d_n$ such that $d_n{a}_0,\dots ,d_n{a}_n$ are algebraic integers, and both $d_n$ and the absolute values of every conjugate of $a_n$ are $O_{ϵ}(n^{ϵn})$. The growth condition ensures convergence everywhere. The stronger, now customary strict E-function definition uses bounds $C^{n+1}$ for a fixed $C$ and also includes the requirement of a nonzero linear differential equation over $\overline{\mathbb{Q}}(z)$. In the original formulation, that differential condition is explicitly part of the system hypothesis in the theorem below. These definitions should be distinguished; the exponentials used here satisfy even the stronger conditions.

The Siegel–Shidlovsky theorem states that if a vector of E-functions satisfies $f^{\prime }=A(z)f$, with $A(z)$ a matrix of rational functions over $\overline{\mathbb{Q}}$, then for every nonzero algebraic $\xi$ which is not a pole of $A$,In particular, algebraic independence of the functions implies algebraic independence of the values. The exclusion of zero is necessary: their Taylor values there are algebraic.

Take rationally independent algebraic ${\gamma }_1,\dots ,{\gamma }_s$ and $f_j(z)=e^{{\gamma }_{j}z}$. Their coefficients are ${\gamma }_j^n$, their conjugates grow exponentially, and powers of a fixed common denominator clear all coefficients through degree $n$. They satisfy the nonsingular diagonal system $f_j^{\prime }={\gamma }_j{f}_j$.

To check functional independence, a polynomial relation among these exponentials, after clearing rational-function denominators, would be ${\sum }_m{P}_m(z)e^{{\lambda }_{m}z}=0$ with distinct ${\lambda }_m={\sum }_j{m}_j{\gamma }_j$. Such an exponential polynomial cannot vanish identically. To isolate a nonzero term with exponent ${\lambda }_0$, apply ${\prod }_{\lambda \ne {\lambda }_0}(D-\lambda {)}^{\mathrm{deg}{P}_{\lambda }+1}$. All other terms vanish, while the remaining polynomial is acted on by operators $D+{\lambda }_0-\lambda$, each injective on polynomials because its nonzero constant preserves a nonzero leading coefficient. Thus the remaining term is nonzero, a contradiction.

The theorem at $\xi =1$ now makes $e^{{\gamma }_j}$ algebraically independent. For arbitrary distinct algebraic exponents ${\alpha }_i$, choose a rational basis of their span and divide its elements by a common denominator so that every ${\alpha }_i$ is an integer combination of the resulting ${\gamma }_j$. Then $e^{{\alpha }_i}$ are distinct Laurent monomials in $e^{{\gamma }_j}$. A linear relation among them becomes a nonzero polynomial relation after multiplying by a common monomial, contradicting independence. These particular E-functions therefore yield the full Lindemann–Weierstrass theorem.