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 ℚ(eα1, ..., eαn) has transcendence degree n over ℚ.

An equivalent formulation (Baker 1990, Chapter 1, Theorem 1.4), is the following.

This equivalence transforms a linear relation over the algebraic numbers into an algebraic relation over ℚ: 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. Lindemann proved in 1882 that eα is transcendental for every non-zero algebraic number α, thereby establishing that π is transcendental (see below).^[1] Weierstrass proved the above more general statement in 1885.^[2]

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

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 αi exponents are required to be rational integers and linear independence is only assured over the rational integers,^[3][4] a result sometimes referred to as Hermite's theorem.^[5] Although apparently a rather 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.^[1] Shortly afterwards Weierstrass obtained the full result,^[2] and further simplifications have been made by several mathematicians, most notably by David Hilbert^[6] and Paul Gordan.^[7]

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. In particular, e1 = e is transcendental. (A more elementary proof that e is transcendental is outlined in the article on transcendental numbers.)

Alternatively, by the second formulation of the theorem, if α is a nonzero algebraic number, then {0, α} is a set of distinct algebraic numbers, and so the set {e0, eα} = {1, eα} is linearly independent over the algebraic numbers and in particular eα cannot be algebraic and so it is transcendental.

To prove that π is transcendental, we prove that it is not algebraic. If π were algebraic, πi would be algebraic as well, and then by the Lindemann–Weierstrass theorem eπi = −1 (see Euler's identity) would be transcendental, a contradiction. Therefore π is not algebraic, which means that it is transcendental.

A slight variant on the same proof will show that if α is a nonzero algebraic number then sin(α), cos(α), tan(α) and their hyperbolic counterparts are also transcendental.

An analogue of the theorem involving the modular function j was conjectured by Daniel Bertrand in 1997, and remains an open problem.^[8] Writing q = e2πiτ for the nome and j(τ) = J(q), the conjecture is as follows.

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

Proof of Lemma A. To simplify the notation set:

Then the statement becomes

Let p be a prime number and define the following polynomials:

Using integration by parts we arrive at

Consider the following sum:

This is not divisible by p when p is large enough because otherwise, putting

Hence, by the fundamental theorem of symmetric polynomials,

However one clearly has:

Proof of Lemma B: Assuming

we will derive a contradiction, thus proving Lemma B.

The polynomial

So we are in the situation of Lemma A. 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. This proves Lemma B. ∎

We turn now to prove the theorem: Let a(1), ..., a(n) be non-zero algebraic numbers, and α(1), ..., α(n) distinct algebraic numbers. 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). Let us denote the distinct roots of this polynomial a(i)1, ..., a(i)d(i), with a(i)1 = a(i).

By multiplying the equation with an appropriate integer factor, we get an identical equation except that now b(1), ..., b(N) are all integers. Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof. ∎

Note that Lemma A is sufficient to prove that e is irrational, since otherwise we may write e = p / q, where both p and q are nonzero integers, but by Lemma A we would have qe − p ≠ 0, which is a contradiction. Lemma A also suffices to prove that π is irrational, since otherwise we may write π = k / n, where both k and n are integers) and then ±iπ are the roots of n2x2 + k2 = 0; thus 2 − 1 − 1 = 2e0 + e**iπ + e−iπ ≠ 0; but this is false.

Similarly, Lemma B is sufficient to prove that e is transcendental, since Lemma B says that if a0, ..., a**n are integers not all of which are zero, then

Lemma B also suffices to prove that π is transcendental, since otherwise we would have 1 + e**iπ ≠ 0.