Transcendence theory

Transcendence theory proves that particular numbers are not algebraic numbers, or that several numbers satisfy no nonzero polynomial relation with algebraic coefficients. A recurring method constructs an auxiliary expression which is simultaneously too small analytically and too large arithmetically if the proposed values were algebraic.

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 $f_1,f_2$ are algebraically independent over $\mathbb{C}$ and have finite orders ${\rho }_1,{\rho }_2$. The set $S$ of points where every $f_j$ is finite and belongs to $K$ satisfies the displayed bound. The orders are growth orders for meromorphic functions. For $f_1=z$, $f_2=e^z$, the orders are zero and one and the functions are algebraically independent; an algebraic value of $e^{\alpha }$ at nonzero algebraic $\alpha$ would produce infinitely many such points $n\alpha$.

Suppose nonzero algebraic numbers ${\beta }_j$ form a Galois-stable set, with Galois-invariant integer weights $b_j$, and $k\ne 0$ is an integer. Choose $P\in \mathbb{Z}[X]$ vanishing at all ${\beta }_j$ with $P(0)\ne 0$, and a positive integer $c$ making all $c{\beta }_j$ algebraic integers. For a large prime $p$, put $D=(1+\mathrm{deg}P)p-1$ and $f_p(X)=c^D{X}^{p-1}P(X{)}^p/(p-1)!$, and let $F_p$ be the sum of its derivatives. The weighted sum $k{F}_p(0)+\sum b_j{F}_p({\beta }_j)$ is an integer, nonzero modulo $p$ when $p\nmid kcP(0)$. If the displayed exponential expression vanished, integrating $(e^{-z}{F}_p(z){)}^{\prime }=-e^{-z}{f}_p(z)$ along the finitely many segments to ${\beta }_j$ would bound that nonzero integer by $C^p/(p-1)!\to 0$. This contradiction proves the obstruction and supplies direct proofs for $e$ and $\pi$.

For algebraic $a\ne 0,1$ and algebraic irrational $b$, every value $\exp (b\lambda )$ with $e^{\lambda }=a$ is transcendental. Consequently any quotient of chosen nonzero logarithms of algebraic numbers is either rational or transcendental.

Assume $a$ and $a^b$ algebraic with algebraic irrational $b$. Apply Siegel lemma to construct a small-coefficient polynomial $P$ vanishing at the algebraic pairs $(s+tb,a^s(a^b{)}^t)$ on a large square grid. The function $P(z,e^{\lambda z})$ is a nonzero exponential polynomial. Its many initial zeros make later grid values exponentially tiny, whereas an algebraic norm estimate forbids such tiny nonzero values. Iterating extends the zero grid indefinitely. Its quadratic zero count in disks contradicts the linear zero-count growth of an exponential polynomial.

Mahler's method uses functional equations relating a function at $z$ and at $z^l$ to transport an auxiliary polynomial to points rapidly approaching zero. High vanishing order gives an upper bound which conflicts with an algebraic norm or height lower bound under assumed algebraicity of the values. Functional transcendence ensures that the auxiliary expression is not identically zero.

This lacunary series is analytic for $|z|<1$ and satisfies $f_l(z^l)=f_l(z)-z$. At roots of unity of $l$-power order, its radial tail diverges, giving a dense set of singularities on the unit circle and excluding algebraicity as a function. The Mahler method proves transcendence of its values at every algebraic $z$ with $0<|z|<1$. At $z=1/2$, reduced truncations have denominators $2^{l^m}$ and error less than twice their $-l$th power; for $l\ge 3$, the Roth theorem gives a shorter proof.

For $0<p<q$, truncating the Fredholm series at $n$ gives a rational with denominator at most $Q_n=q^{l^n}$ and positive error at most $2Q_n^{-l(1-\log p/\log q)}$ for large $n$. The displayed strict inequality first rules out a rational value by the rational separation bound, and then rules out an algebraic irrational value by Roth theorem. Reduced denominators are at most $Q_n$ and tend to infinity because the truncations are distinct. In particular $l\ge 3$ and $p<q^{\delta }$, $\delta <1/3$, imply transcendence.

A nonzero integer polynomial of bidegree at most $N$ can cancel the first $(N+1{)}^2-1$ Taylor coefficients of the Fredholm series. Functional transcendence makes the resulting analytic function nonzero. If $\alpha$ and $f_l(\alpha )$ were both algebraic, the relation $f_l({\alpha }^{l^r})=f_l(\alpha )-{\sum }_{j<r}{\alpha }^{l^j}$ would keep the auxiliary values in one fixed number field. Their Absolute logarithmic Weil height is at most $N{l}^{r}h(\alpha )(1+1/(l-1))+O_N(r+1)$, whereas their chosen absolute value is $O_N(|\alpha {|}^{T_N{l}^r})$. The Liouville height inequality forces $T_N=O(N)$, contradicting its quadratic lower bound for sufficiently large $N$.

In the modern strict sense an E-function satisfies a nonzero linear differential equation over $\overline{\mathbb{Q}}(z)$, has algebraic coefficients $a_n$ with all conjugates bounded by $C^{n+1}$, and has common positive denominators $d_n\le C^{n+1}$ making $d_n{a}_0,\dots ,d_n{a}_n$ algebraic integers. Siegel's original arithmetic growth condition permits, for each $ϵ>0$, bounds of size $O_{ϵ}(n^{ϵn})$ instead. The strict and original growth formulations should be distinguished rather than asserted equivalent. Exponentials with algebraic exponent satisfy the stronger conditions and a first-order differential equation.

Work in the compositum of the two coefficient fields. Conjugate coefficient bounds add, while multiplying the common denominators clears both coefficient lists. For Siegel's original bounds $O_{ϵ}(n^{ϵn})$, use $ϵ/2$ in each denominator estimate. A nonzero linear differential equation for the sum follows because its derivatives lie in the sum of the two finite-dimensional derivative spaces over the rational function field. The stricter exponential arithmetic bounds are preserved as well.

Let a vector of E-functions satisfy $f^{\prime }=A(z)f$ with $A\in {\mathrm{Mat}}_s(\overline{\mathbb{Q}}(z))$. At a nonzero algebraic point $\xi$ which is not a pole of $A$, the transcendence degree of the values over $\overline{\mathbb{Q}}$ equals the transcendence degree of the functions over $\overline{\mathbb{Q}}(z)$. In particular algebraically independent functions give algebraically independent values. The exponential system $f_j=e^{{\gamma }_{j}z}$, for rationally independent algebraic ${\gamma }_j$, recovers the Lindemann–Weierstrass theorem by evaluating at one.

For distinct algebraic numbers ${\alpha }_1,\dots ,{\alpha }_s$, the numbers $e^{{\alpha }_1},\dots ,e^{{\alpha }_s}$ are linearly independent over $\overline{\mathbb{Q}}$. Equivalently, exponentials of numbers linearly independent over $\mathbb{Q}$ are algebraically independent over $\overline{\mathbb{Q}}$. The special case of a nonzero algebraic exponent gives a transcendental exponential. In particular $e^{i\pi }=-1$ forces $\pi$ to be transcendental.

Under an assumed algebraicity of $i\pi$, form the product ${\prod }_j(1+e^{{\alpha }_j})$ over its conjugates; its one known vanishing factor makes the whole product zero and regroup the subset sums as $k+\sum b_j{e}^{{\beta }_j}=0$, with nonzero exponents and Galois-invariant integral weights. After clearing their denominators, use $f_p(z)=(cz{)}^{p-1}{\prod }_j(cz-c{\beta }_j{)}^p/(p-1)!$. The sum of its derivatives at each ${\beta }_j$ is divisible by $p$ in the algebraic integers, whereas at zero its first nonzero derivative is a fixed nonzero residue for all but finitely many primes. Summing with the weights gives a nonzero rational integer. Integrating the derivative of $e^{-z}\sum f_p^{(j)}(z)$ expresses that integer as a quantity bounded by $C^p/(p-1)!$, which tends to zero. This is the contradiction.