Lindemann–Weierstrass theorem-The Daily Ledger

In transcendental number theory, the Lindemann–Weierstrass theorem is a result that is very useful in establishing the transcendence of numbers. It states the following: In other words, the extension field has transcendence degree over .

An equivalent formulation , is the following: This equivalence transforms a linear relation over the algebraic numbers into an algebraic relation over by using the fact that a symmetric polynomial whose arguments are all conjugates of one another gives a rational number.

Lindemann–Weierstrass theorem

The theorem is named for Ferdinand von Lindemann and Karl Weierstrass. Lindemann proved in 1882 that is transcendental for every non-zero algebraic number thereby establishing that is transcendental (see below). Weierstrass proved the above more general statement in 1885.

Lindemann–Weierstrass theorem

The theorem, along with the Gelfond–Schneider theorem, is extended by Baker's theorem, and all of these would be further generalized by Schanuel's conjecture.

Lindemann–Weierstrass theorem

Naming convention

The theorem is also known variously as the Hermite–Lindemann theorem and the Hermite–Lindemann–Weierstrass theorem. Charles Hermite first proved the simpler theorem where the exponents are required to be rational integers and linear independence is only assured over the rational integers, a result sometimes referred to as Hermite's theorem. Although that appears to be a special case of the above theorem, the general result can be reduced to this simpler case. Lindemann was the first to allow algebraic numbers into Hermite's work in 1882. Shortly afterwards Weierstrass obtained the full result, and further simplifications have been made by several mathematicians, most notably by David Hilbert and Paul Gordan.

Transcendence of and

The transcendence of and are direct corollaries of this theorem.

Suppose is a non-zero algebraic number; then is a linearly independent set over the rationals, and therefore by the first formulation of the theorem is an algebraically independent set; or in other words is transcendental. In particular, is transcendental. (A more elementary proof that is transcendental is outlined in the article on transcendental numbers.)

Alternatively, by the second formulation of the theorem, if is a non-zero algebraic number, then is a set of distinct algebraic numbers, and so the set is linearly independent over the algebraic numbers and in particular cannot be algebraic and so it is transcendental.

To prove that is transcendental, we prove that it is not algebraic. If were algebraic, i would be algebraic as well, and then by the Lindemann–Weierstrass theorem (see Euler's identity) would be transcendental, a contradiction. Therefore is not algebraic, which means that it is transcendental.

A slight variant on the same proof will show that if is a non-zero algebraic number then and their hyperbolic counterparts are also transcendental.

-adic conjecture

Modular conjecture

An analogue of the theorem involving the modular function was conjectured by Daniel Bertrand in 1997, and remains an open problem. Writing for the square of the nome and the conjecture is as follows.

Lindemann–Weierstrass theorem

Proof

The proof relies on two preliminary lemmas. Notice that Lemma B itself is already sufficient to deduce the original statement of Lindemann–Weierstrass theorem.

Preliminary lemmas

Proof of Lemma A. To simplify the notation set:

Then the statement becomes

Let be a prime number and define the following polynomials:

where is a non-zero integer such that are all algebraic integers. Define

Using integration by parts we arrive at

where is the degree of , and is the j-th derivative of . This also holds for s complex (in this case the integral has to be intended as a contour integral, for example along the straight segment from 0 to s) because

is a primitive of .

Consider the following sum:

In the last line we assumed that the conclusion of the Lemma is false. In order to complete the proof we need to reach a contradiction. We will do so by estimating in two different ways.

First is an algebraic integer which is divisible by p! for and vanishes for unless and , 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 the product of its conjugates (which is still non-zero), we would get that p divides , which is false.