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 ( e α 1 , … , e α n ) {\displaystyle \mathbb {Q} (e^{\alpha _{1}},\dots ,e^{\alpha _{n}})} has transcendence degree n over Q {\displaystyle \mathbb {Q} } .
An equivalent formulation from 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 {\displaystyle \mathbb {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). 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 αi 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 e and π
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 non-zero 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 non-zero algebraic number then sin(α), cos(α), tan(α) and their hyperbolic counterparts are also transcendental.
p-adic conjecture
Modular conjecture An analogue of the theorem involving the modular function j was conjectured by Daniel Bertrand in 1997, and remains an open problem. Writing q = e2πiτ for the square of the nome and j(τ) = J(q), 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:
n 0 = 0 , n i = ∑ k = 1 i m ( k ) , i = 1 , … , r n = n r , α n i − 1 + j = γ ( i ) j , 1 ≤ i ≤ r , 1 ≤ j ≤ m ( i ) β n i − 1 + j = c ( i ) . {\displaystyle {\begin{aligned}&n_{0}=0,&&\\&n_{i}=\sum \nolimits _{k=1}^{i}m(k),&&i=1,\ldots ,r\\&n=n_{r},&&\\&\alpha _{n_{i-1}+j}=\gamma (i)_{j},&&1\leq i\leq r,\ 1\leq j\leq m(i)\\&\beta _{n_{i-1}+j}=c(i).\end{aligned}}}
Then the statement becomes
∑ k = 1 n β k e α k ≠ 0. {\displaystyle \sum _{k=1}^{n}\beta _{k}e^{\alpha _{k}}\neq 0.}
Let p be a prime number and define the following polynomials:
f i ( x ) = ℓ n p ( x − α 1 ) p ⋯ ( x − α n ) p ( x − α i ) , {\displaystyle f_{i}(x)={\frac {\ell ^{np}(x-\alpha _{1})^{p}\cdots (x-\alpha _{n})^{p}}{(x-\alpha _{i})}},}
where ℓ is a non-zero integer such that ℓ α 1 , … , ℓ α n {\displaystyle \ell \alpha _{1},\ldots ,\ell \alpha _{n}} are all algebraic integers. Notice ℓ does not depend on p. Define
I i ( s ) = ∫ 0 s e s − x f i ( x ) d x . {\displaystyle I_{i}(s)=\int _{0}^{s}e^{s-x}f_{i}(x)\,dx.}
Using integration by parts we arrive at
I i ( s ) = e s ∑ j = 0 n p − 1 f i ( j ) ( 0 ) − ∑ j = 0 n p − 1 f i ( j ) ( s ) , {\displaystyle I_{i}(s)=e^{s}\sum _{j=0}^{np-1}f_{i}^{(j)}(0)-\sum _{j=0}^{np-1}f_{i}^{(j)}(s),}
where n p − 1 {\displaystyle np-1} is the degree of f i {\displaystyle f_{i}} , and f i ( j ) {\displaystyle f_{i}^{(j)}} is the j-th derivative of f i {\displaystyle f_{i}} . 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
− e s − x ∑ j = 0 n p − 1 f i ( j ) ( x ) {\displaystyle -e^{s-x}\sum _{j=0}^{np-1}f_{i}^{(j)}(x)}
is a primitive of e s − x f i ( x ) {\displaystyle e^{s-x}f_{i}(x)} . Consider the following sum:
J i = ∑ k = 1 n β k I i ( α k ) = ∑ k = 1 n β k ( e α k ∑ j = 0 n p − 1 f i ( j ) ( 0 ) − ∑ j = 0 n p − 1 f i ( j ) ( α k ) ) = ( ∑ j = 0 n p − 1 f i ( j ) ( 0 ) ) ( ∑ k = 1 n β k e α k ) − ∑ k = 1 n ∑ j = 0 n p − 1 β k f i ( j ) ( α k ) = − ∑ k = 1 n ∑ j = 0 n p − 1 β k f i ( j ) ( α k ) {\displaystyle {\begin{aligned}J_{i}&=\sum _{k=1}^{n}\beta _{k}I_{i}(\alpha _{k})\\[5pt]&=\sum _{k=1}^{n}\beta _{k}\left(e^{\alpha _{k}}\sum _{j=0}^{np-1}f_{i}^{(j)}(0)-\sum _{j=0}^{np-1}f_{i}^{(j)}(\alpha _{k})\right)\\[5pt]&=\left(\sum _{j=0}^{np-1}f_{i}^{(j)}(0)\right)\left(\sum _{k=1}^{n}\beta _{k}e^{\alpha _{k}}\right)-\sum _{k=1}^{n}\sum _{j=0}^{np-1}\beta _{k}f_{i}^{(j)}(\alpha _{k})\\[5pt]&=-\sum _{k=1}^{n}\sum _{j=0}^{np-1}\beta _{k}f_{i}^{(j)}(\alpha _{k})\end{aligned}}}
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 | J 1 ⋯ J n | {\displaystyle |J_{1}\cdots J_{n}|} in two different ways. First f i ( j ) ( α k ) {\displaystyle f_{i}^{(j)}(\alpha _{k})} is an algebraic integer which is divisible by p! for j ≥ p {\displaystyle j\geq p} and vanishes for j < p {\displaystyle j<p} unless j = p − 1 {\displaystyle j=p-1} and k = i {\displaystyle k=i} , in which case it equals
ℓ n p ( p − 1 ) ! ∏ k ≠ i ( α i − α k ) p . {\displaystyle \ell ^{np}(p-1)!\prod _{k\neq i}(\alpha _{i}-\alpha _{k})^{p}.}
This is not divisible by p when p is large enough because otherwise, putting
δ i = ∏ k ≠ i ( ℓ α i − ℓ α k ) {\displaystyle \delta _{i}=\prod _{k\neq i}(\ell \alpha _{i}-\ell \alpha _{k})}
(which is a non-zero algebraic integer) and calling d i ∈ Z {\displaystyle d_{i}\in \mathbb {Z} } the product of its conjugates (which is still non-zero), we would get that p divides ℓ p ( p − 1 ) ! d i p {\displaystyle \ell ^{p}(p-1)!d_{i}^{p}} , which is false. So J i {\displaystyle J_{i}} is a non-zero algebraic integer divisible by (p − 1)!. Now
J i = − ∑ j = 0 n p − 1 ∑ t = 1 r c ( t ) ( f i ( j ) ( α n t − 1 + 1 ) + ⋯ + f i ( j ) ( α n t ) ) . {\displaystyle J_{i}=-\sum _{j=0}^{np-1}\sum _{t=1}^{r}c(t)\left(f_{i}^{(j)}(\alpha _{n_{t-1}+1})+\cdots +f_{i}^{(j)}(\alpha _{n_{t}})\right).}
Since each f i ( x ) {\displaystyle f_{i}(x)} is obtained by dividing a fixed polynomial with integer coefficients by ( x − α i ) {\displaystyle (x-\alpha _{i})} , it is of the form
f i ( x ) = ∑ m = 0 n p − 1 g m ( α i ) x m , {\displaystyle f_{i}(x)=\sum _{m=0}^{np-1}g_{m}(\alpha _{i})x^{m},}
where g m {\displaystyle g_{m}} is a polynomial (with integer coefficients) independent of i. The same holds for the derivatives f i ( j ) ( x ) {\displaystyle f_{i}^{(j)}(x)} . Hence, by the fundamental theorem of symmetric polynomials,
f i ( j ) ( α n t − 1 + 1 ) + ⋯ + f i ( j ) ( α n t ) {\displaystyle f_{i}^{(j)}(\alpha _{n_{t-1}+1})+\cdots +f_{i}^{(j)}(\alpha _{n_{t}})}
is a fixed polynomial with rational coefficients evaluated in α i {\displaystyle \alpha _{i}} (this is seen by grouping the same powers of α n t − 1 + 1 , … , α n t {\displaystyle \alpha _{n_{t-1}+1},\dots ,\alpha _{n_{t}}} appearing in the expansion and using the fact that these algebraic numbers are a complete set of conjugates). So the same is true of J i {\displaystyle J_{i}} , i.e. it equals G ( α i ) {\displaystyle G(\alpha _{i})} , where G is a polynomial with rational coefficients independent of i. Finally J 1 ⋯ J n = G ( α 1 ) ⋯ G ( α n ) {\displaystyle J_{1}\cdots J_{n}=G(\alpha _{1})\cdots G(\alpha _{n})} is rational (again by the fundamental theorem of symmetric polynomials) and is a non-zero algebraic integer divisible by ( p − 1 ) ! n {\displaystyle (p-1)!^{n}} (since the J i {\displaystyle J_{i}} 's are algebraic integers divisible by ( p − 1 ) ! {\displaystyle (p-1)!} ). Therefore
| J 1 ⋯ J n | ≥ ( p − 1 ) ! n . {\displaystyle |J_{1}\cdots J_{n}|\geq (p-1)!^{n}.}
However one clearly has:
| I i ( α k ) | ≤ | α k | e | α k | F i ( | α k | ) , {\displaystyle |I_{i}(\alpha _{k})|\leq {|\alpha _{k}|}e^{|\alpha _{k}|}F_{i}({|\alpha _{k}|}),}
where Fi is the polynomial whose coefficients are the absolute values of those of fi (this follows directly from the definition of I i ( s ) {\displaystyle I_{i}(s)} ). Thus
| J i | ≤ ∑ k = 1 n | β k α k | e | α k | F i ( | α k | ) {\displaystyle |J_{i}|\leq \sum _{k=1}^{n}\left|\beta _{k}\alpha _{k}\right|e^{|\alpha _{k}|}F_{i}\left(\left|\alpha _{k}\right|\right)}
and so by the construction of the f i {\displaystyle f_{i}} 's we have | J 1 ⋯ J n | ≤ C p {\displaystyle |J_{1}\cdots J_{n}|\leq C^{p}} for a sufficiently large C independent of p, which contradicts the previous inequality. This proves Lemma A. ∎
Proof of Lemma B: Assuming
b ( 1 ) e γ ( 1 ) + ⋯ + b ( n ) e γ ( n ) = 0 , {\displaystyle b(1)e^{\gamma (1)}+\cdots +b(n)e^{\gamma (n)}=0,}
we will derive a contradiction, thus proving Lemma B. Let us choose a polynomial with integer coefficients which vanishes on all the γ ( k ) {\displaystyle \gamma (k)} 's and let γ ( 1 ) , … , γ ( n ) , γ ( n + 1 ) , … , γ ( N ) {\displaystyle \gamma (1),\ldots ,\gamma (n),\gamma (n+1),\ldots ,\gamma (N)} be all its distinct roots. Let b(n + 1) = ... = b(N) = 0. The polynomial
P ( x 1 , … , x N ) = ∏ σ ∈ S N ( b ( 1 ) x σ ( 1 ) + ⋯ + b ( N ) x σ ( N ) ) {\displaystyle P(x_{1},\dots ,x_{N})=\prod _{\sigma \in S_{N}}(b(1)x_{\sigma (1)}+\cdots +b(N)x_{\sigma (N)})}
vanishes at ( e γ ( 1 ) , … , e γ ( N ) ) {\displaystyle (e^{\gamma (1)},\dots ,e^{\gamma (N)})} by assumption. Since the product is symmetric, for any τ ∈ S N {\displaystyle \tau \in S_{N}} the monomials x τ ( 1 ) h 1 ⋯ x τ ( N ) h N {\displaystyle x_{\tau (1)}^{h_{1}}\cdots x_{\tau (N)}^{h_{N}}} and x 1 h 1 ⋯ x N h N {\displaystyle x_{1}^{h_{1}}\cdots x_{N}^{h_{N}}} have the same coefficient in the expansion of P. Thus, expanding P ( e γ ( 1 ) , … , e γ ( N ) ) {\displaystyle P(e^{\gamma (1)},\dots ,e^{\gamma (N)})} accordingly and grouping the terms with the same exponent, we see that the resulting exponents h 1 γ ( 1 ) + ⋯ + h N γ ( N ) {\displaystyle h_{1}\gamma (1)+\dots +h_{N}\gamma (N)} 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 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. ∎
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:
a ( 1 ) e α ( 1 ) + ⋯ + a ( n ) e α ( n ) = 0. {\displaystyle a(1)e^{\alpha (1)}+\cdots +a(n)e^{\alpha (n)}=0.}
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 x 11 , … , x 1 d ( 1 ) , … , x n 1 , … , x n d ( n ) , y 1 , … , y n {\displaystyle x_{11},\dots ,x_{1d(1)},\dots ,x_{n1},\dots ,x_{nd(n)},y_{1},\dots ,y_{n}}
Q