Lindemann–Weierstrass theorem

Lindemann–Weierstrass theorem
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.
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.
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.
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.
So is a non-zero algebraic integer divisible by (p − 1)!. Now
Since each is obtained by dividing a fixed polynomial with integer coefficients by , it is of the form
where is a polynomial (with integer coefficients) independent of i. The same holds for the derivatives .
Hence, by the fundamental theorem of symmetric polynomials,
is a fixed polynomial with rational coefficients evaluated in (this is seen by grouping the same powers of appearing in the expansion and using the fact that these algebraic numbers are a complete set of conjugates). So the same is true of , i.e. it equals , where G is a polynomial with rational coefficients independent of i.
Finally is rational (again by the fundamental theorem of symmetric polynomials) and is a non-zero algebraic integer divisible by (since the 's are algebraic integers divisible by ). Therefore
However one clearly has:
where is the polynomial whose coefficients are the absolute values of those of fi (this follows directly from the definition of ). Thus
and so by the construction of the 's we have for a sufficiently large C independent of p, which contradicts the previous inequality. This proves Lemma A. ∎
Proof of Lemma B: Assuming
we will derive a contradiction, thus proving Lemma B.
Let us choose a polynomial with integer coefficients which vanishes on all the 's and let be all its distinct roots. Let b(n + 1) = ... = b(N) = 0.
The polynomial
vanishes at by assumption. Since the product is symmetric, for any the monomials and have the same coefficient in the expansion of P.
Thus, expanding accordingly and grouping the terms with the same exponent, we see that the resulting exponents form a complete set of conjugates and, if two terms have conjugate exponents, they are multiplied by the same coefficient.
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 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. ∎
Final step
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).
Let S be the functions σ which choose one element from each of the sequences (1, ..., d(1)), (1, ..., d(2)), ..., (1, ..., d(n)), so that for every 1 ≤ i ≤ n, σ(i) is an integer between 1 and d(i). We form the polynomial in the variables
Since the product is over all the possible choice functions σ, Q is symmetric in for every i. Therefore Q is a polynomial with integer coefficients in elementary symmetric polynomials of the above variables, for every i, and in the variables yi. Each of the latter symmetric polynomials is a rational number when evaluated in .
The evaluated polynomial vanishes because one of the choices is just σ(i) = 1 for all i, for which the corresponding factor vanishes according to our assumption above. Thus, the evaluated polynomial is a sum of the form
where we already grouped the terms with the same exponent. So in the left-hand side we have distinct values β(1), ..., β(N), each of which is still algebraic (being a sum of algebraic numbers) and coefficients .
The sum is nontrivial: if is maximal in the lexicographic order, the coefficient of is just a product of a(i)j's (with possible repetitions), which is non-zero.
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 non-zero integers, but by Lemma A we would have qe − p ≠ 0, which is a contradiction. Lemma A also suffices to prove that is irrational, since otherwise we may write = k / n, where both k and n are integers) and then ±i are the roots of n2x2 + k2 = 0; thus 2 − 1 − 1 = 2e0 + ei + e−i ≠ 0; but this is false.
Similarly, Lemma B is sufficient to prove that e is transcendental, since Lemma B says that if a0, ..., an are integers not all of which are zero, then
Lemma B also suffices to prove that is transcendental, since otherwise we would have 1 + ei ≠ 0.
Equivalence of the two statements
Baker's formulation of the theorem clearly implies the first formulation. Indeed, if are algebraic numbers that are linearly independent over , and
is a polynomial with rational coefficients, then we have
and since are algebraic numbers which are linearly independent over the rationals, the numbers are algebraic and they are distinct for distinct n-tuples . So from Baker's formulation of the theorem we get for all n-tuples .
Now assume that the first formulation of the theorem holds. For Baker's formulation is trivial, so let us assume that , and let be non-zero algebraic numbers, and 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. Here, as proved above, are rational numbers, not all equal to zero, and each exponent is a linear combination of with integer coefficients. Then, since and are pairwise distinct, the -vector subspace of generated by is not trivial and we can pick to form a basis for For each , we have
For each let be the least common multiple of all the for , and put . Then are algebraic numbers, they form a basis of , and each is a linear combination of the with integer coefficients. By multiplying the relation
by , where is a large enough positive integer, we get a non-trivial algebraic relation with rational coefficients connecting , against the first formulation of the theorem.
Related result
A variant of Lindemann–Weierstrass theorem in which the algebraic numbers are replaced by the transcendental Liouville numbers (or in general, the U numbers) is also known.
See also
Gelfond–Schneider theorem
Baker's theorem; an extension of Gelfond–Schneider theorem
Schanuel's conjecture; if proven, it would imply both the Gelfond–Schneider theorem and the Lindemann–Weierstrass theorem
Notes
References
Further reading
External links
Articles containing proofs
E (mathematical constant)
Exponentials
Pi
Theorems in number theory
Transcendental numbers