Lindemann–Weierstrass theorem

An equivalent formulation from Baker 1990, Chapter 1, Theorem 1.4, is the following: An equivalent formulation — If α1, ..., αn are distinct algebraic numbers, then the exponentials eα1, ..., eαn are linearly independent over the algebraic numbers.

by using the fact that a symmetric polynomial whose arguments are all conjugates of one another gives a rational number.

The theorem is named for Ferdinand von Lindemann and Karl Weierstrass.

Charles Hermite first proved the simpler theorem where the αi exponents are required to be rational integers and linear independence is only assured over the rational integers,[4][5] a result sometimes referred to as Hermite's theorem.

Lindemann was the first to allow algebraic numbers into Hermite's work in 1882.

[1] Shortly afterwards Weierstrass obtained the full result,[2] and further simplifications have been made by several mathematicians, most notably by David Hilbert[7] and Paul Gordan.

[8] The transcendence of e and π are direct corollaries of this theorem.

Suppose α is a non-zero algebraic number; then {α} is a linearly independent set over the rationals, and therefore by the first formulation of the theorem {eα} is an algebraically independent set; or in other words eα is transcendental.

, expp(αn) are p-adic numbers that are algebraically independent over

An analogue of the theorem involving the modular function j was conjectured by Daniel Bertrand in 1997, and remains an open problem.

[9] Writing q = e2πiτ for the square of the nome and j(τ) = J(q), the conjecture is as follows.

Modular conjecture — Let q1, ..., qn be non-zero algebraic numbers in the complex unit disc such that the 3n numbers are algebraically dependent over

Then there exist two indices 1 ≤ i < j ≤ n such that qi and qj are multiplicatively dependent.

Notice that Lemma B itself is already sufficient to deduce the original statement of Lindemann–Weierstrass theorem.

To simplify the notation set: Then the statement becomes Let p be a prime number and define the following polynomials: where ℓ is a non-zero integer such that

Consider the following sum: In the last line we assumed that the conclusion of the Lemma is false.

, in which case it equals This is not divisible by p when p is large enough because otherwise, putting (which is a non-zero algebraic integer) and calling

is obtained by dividing a fixed polynomial with integer coefficients by

appearing in the expansion and using the fact that these algebraic numbers are a complete set of conjugates).

is rational (again by the fundamental theorem of symmetric polynomials) and is a non-zero algebraic integer divisible by

for a sufficiently large C independent of p, which contradicts the previous inequality.

To reach a contradiction it suffices to see that at least one of the coefficients is non-zero.

This is seen by equipping C with the lexicographic order and by choosing for each factor in the product the term with non-zero coefficient which has maximum exponent according to this ordering: the product of these terms has non-zero coefficient in the expansion and does not get simplified by any other term.

Then let us assume that: We will show that this leads to contradiction and thus prove the theorem.

The proof is very similar to that of Lemma B, except that this time the choices are made over the a(i)'s: For every i ∈ {1, ..., n}, a(i) is algebraic, so it is a root of an irreducible polynomial with integer coefficients of degree d(i).

Since the product is over all the possible choice functions σ, Q is symmetric in

Thus, the evaluated polynomial is a sum of the form where we already grouped the terms with the same exponent.

Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof.

distinct algebraic numbers such that: As seen in the previous section, and with the same notation used there, the value of the polynomial at has an expression of the form where we have grouped the exponentials having the same exponent.

is a large enough positive integer, we get a non-trivial algebraic relation with rational coefficients connecting