# 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 that

Lindemann–Weierstrass theorem — if α1, ..., αn are algebraic numbers which are linearly independent over the rational numbers ℚ, then eα1, ..., eαn are algebraically independent over ℚ;

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:

An equivalent formulation — If α1, ..., αn are distinct algebraic numbers, then the exponentials eα1, ..., eαn are linearly independent over the algebraic numbers.

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.

## 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,[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].

## Transcendence of e and π

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

Suppose α is a nonzero 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 {e0eα} = {1, eα} is linearly independent over the algebraic numbers and in particular eα cannot be algebraic and so it is transcendental.

The proof that π is transcendental is by contradiction. 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.

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.

p-adic Lindemann–Weierstrass Conjecture. — Suppose p is some prime number and α1, ..., αn are p-adic numbers which are algebraic and linearly independent over ℚ, such that | αi |p < 1/p for all i; then the p-adic exponentials expp1), . . . , exppn) are p-adic numbers that are algebraically independent over ℚ.

## Modular conjecture

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.

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

${\displaystyle \left\{J(q_{1}),J'(q_{1}),J''(q_{1}),\ldots ,J(q_{n}),J'(q_{n}),J''(q_{n})\right\}}$

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

## Lindemann–Weierstrass theorem

Lindemann–Weierstrass Theorem (Baker's reformulation). — If a1, ..., an are non-zero algebraic numbers, and α1, ..., αn are distinct algebraic numbers, then[9]

${\displaystyle a_{1}e^{\alpha _{1}}+\cdots +a_{n}e^{\alpha _{n}}\neq 0.}$

### Proof

#### Preliminary lemmas

Lemma A. — Let c(1), ..., c(r) be non-zero integers and, for every k between 1 and r, let {γ(k)1, ..., γ(k)m(k)} be the roots of a polynomial with integer coefficients Tk (x) = v(k) xm(k) + ... + u(k) (with v(k), u(k) ≠ 0). If γ(k)i ≠ γ(u)v whenever (ki) ≠ (uv), then

${\displaystyle c(1)\left(e^{\gamma (1)_{1}}+\cdots +e^{\gamma (1)_{m(1)}}\right)+\cdots +c(r)\left(e^{\gamma (r)_{1}}+\cdots +e^{\gamma (r)_{m(r)}}\right)\neq 0.}$

Proof of Lemma A. To simplify the notation set:

{\displaystyle {\begin{aligned}&n_{0}=0,&&n=n_{r}\\&n_{i}=\sum \nolimits _{k=1}^{i}m(k),&&i=1,\ldots ,r\\&\alpha _{n_{i}+j}=\gamma (i+1)_{j},&&0\leqslant i

Then the statement becomes:

${\displaystyle \sum _{k=1}^{n}\beta _{k}e^{\alpha _{k}}\neq 0.}$

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

${\displaystyle f_{i}(x)={\frac {\ell ^{np}(x-\alpha _{1})^{p}\cdots (x-\alpha _{n})^{p}}{(x-\alpha _{i})}},}$

where is a non-zero integer such that ${\displaystyle \ell \alpha _{1},\ldots ,\ell \alpha _{n}}$ are all algebraic integers. Define:[10]

${\displaystyle I_{i}(s)=\int _{0}^{s}e^{s-x}f_{i}(x)\,dx.}$

Using integration by parts we arrive at:

${\displaystyle I_{i}(s)=e^{s}\sum _{j=0}^{np-1}f_{i}^{(j)}(0)-\sum _{j=0}^{np-1}f_{i}^{(j)}(s),}$

where ${\displaystyle np-1}$ is the degree of ${\displaystyle f_{i}}$, and ${\displaystyle f_{i}^{(j)}}$ is the j-th derivative of ${\displaystyle f_{i}}$. 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

${\displaystyle -e^{s-x}\sum _{j=0}^{np-1}f_{i}^{(j)}(x)}$

is a primitive of ${\displaystyle e^{s-x}f_{i}(x)}$.

Let us consider the following sum:

{\displaystyle {\begin{aligned}J_{i}&=\sum _{k=1}^{n}\beta _{k}I_{i}(\alpha _{k})\\&=\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)\\&=\sum _{k=1}^{n}\left(\beta _{k}e^{\alpha _{k}}\sum _{j=0}^{np-1}f_{i}^{(j)}(0)\right)-\sum _{k=1}^{n}\left(\beta _{k}\sum _{j=0}^{np-1}f_{i}^{(j)}(\alpha _{k})\right)\\&=\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})\\&=-\sum _{k=1}^{n}\sum _{j=0}^{np-1}\beta _{k}f_{i}^{(j)}(\alpha _{k})&&\sum _{k=1}^{n}\beta _{k}e^{\alpha _{k}}=0\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 ${\displaystyle |J_{1}\cdots J_{n}|}$ in two different ways.

First ${\displaystyle f_{i}^{(j)}(\alpha _{k})}$ is an algebraic integer which is divisible by p! for ${\displaystyle j\geqslant p}$ and vanishes for j < p unless j = p − 1 and k = i, in which case it equals

${\displaystyle \ell ^{np}(p-1)!\prod _{k\neq i}(\alpha _{i}-\alpha _{k})^{p}.}$

This is not divisible by p (if p is large enough) because otherwise, putting

${\displaystyle \delta _{i}=\prod _{k\neq i}(\ell \alpha _{i}-\ell \alpha _{k})}$

(which is an algebraic integer) and calling ${\displaystyle d_{i}}$ the product of its conjugates, we would get that p divides ${\displaystyle \ell ^{p}(p-1)!d_{i}^{p}}$ (and ${\displaystyle d_{i}}$ is a non-zero integer), so by Fermat's little theorem p would divide ${\displaystyle \ell (p-1)!d_{i}}$, which is false.

So ${\displaystyle J_{i}}$ is a non-zero algebraic integer divisible by (p&nbdp;− 1)!. Now

${\displaystyle J_{i}=-\sum _{j=0}^{np-1}\sum _{t=0}^{r-1}c(t+1)\left(f_{i}^{(j)}(\alpha _{n_{t}+1})+\cdots +f_{i}^{(j)}(\alpha _{n_{t+1}})\right).}$

Since each ${\displaystyle f_{i}(x)}$ is obtained by dividing a fixed polynomial with integer coefficients by ${\displaystyle (x-\alpha _{i})}$, it is of the form

${\displaystyle f_{i}(x)=\sum _{m=0}^{np-1}g_{m}(\alpha _{i})x^{m},}$

where ${\displaystyle g_{m}}$ is a polynomial (with integer coefficients) independent of i. The same holds for the derivatives ${\displaystyle f_{i}^{(j)}(x)}$.

Hence, by the fundamental theorem of symmetric polynomials,

${\displaystyle f_{i}^{(j)}(\alpha _{n_{t}+1})+\cdots +f_{i}^{(j)}(\alpha _{n_{t+1}})}$

is a fixed polynomial with integer coefficients evaluated in ${\displaystyle \alpha _{i}}$ (this is seen by grouping the same powers of ${\displaystyle \alpha _{n_{t}+1},\dots ,\alpha _{n_{t+1}}}$ appearing in the expansion and using the fact that ${\displaystyle \alpha _{n_{t}+1},\ldots ,\alpha _{n_{t+1}}}$ are a complete set of conjugates). So the same is true of ${\displaystyle J_{i}}$, i.e. it equals ${\displaystyle G(\alpha _{i})}$, where G is a polynomial with integer coefficients which is independent of i.

Finally ${\displaystyle J_{1}\dots J_{n}=G(\alpha _{1})\dots G(\alpha _{n})}$ is an integer number (again by the fundamental theorem of symmetric polynomials), it is non-zero (since the ${\displaystyle J_{i}}$'s are) and it is divisible by ${\displaystyle (p-1)!^{n}}$ and therefore:

${\displaystyle |J_{1}\cdots J_{n}|\geqslant (p-1)!^{n}.}$

However one clearly has:

${\displaystyle |I_{i}(\alpha _{k})|\leqslant |\alpha _{k}|e^{|\alpha _{k}|}F_{i}(|\alpha _{k}|),}$

where Fi is the polynomial whose coefficients are the absolute values of those of fi (this follows directly from the definition of ${\displaystyle I_{i}(s)}$). Thus

${\displaystyle |J_{i}|\leqslant \sum _{k=1}^{n}\left|\beta _{k}\alpha _{k}\right|e^{|\alpha _{k}|}F_{i}\left(\left|\alpha _{k}\right|\right)}$

and so by the construction of the ${\displaystyle f_{i}}$'s we have ${\displaystyle |J_{1}\cdots J_{n}|\leq C^{p}}$ for a sufficiently large C independent of p, which contradicts the previous inequality. This proves Lemma A.∎

Lemma B. — If b(1), ..., b(n) are non-zero integers and γ(1), ..., γ(n), are distinct algebraic numbers, then

${\displaystyle b(1)e^{\gamma (1)}+\cdots +b(n)e^{\gamma (n)}\neq 0.}$

Proof of Lemma B: Assuming

${\displaystyle b(1)e^{\gamma (1)}+\cdots +b(n)e^{\gamma (n)}=0,}$

we will derive a contradiction, thus proving Lemma B.

Let us choose a polynomial with integer coefficients which vanishes on all the ${\displaystyle \gamma (k)}$'s and let ${\displaystyle \gamma (1),\ldots ,\gamma (n),\gamma (n+1),\ldots ,\gamma (N)}$ be all its distinct roots. Let b(n + 1) = ... = b(N) = 0.

The product

${\displaystyle \prod _{\sigma \in S_{N}}(b(1)e^{\gamma (\sigma (1))}+\cdots +b(N)e^{\gamma (\sigma (N))}),}$

vanishes by assumption, but by expanding it we obtain a sum of terms of the form ${\displaystyle e^{h_{1}\gamma (1)+\cdots +h_{N}\gamma (N)}}$ multiplied by integer coefficients. Since the product is symmetric, we have that, for any ${\displaystyle \tau \in S_{n}}$, ${\displaystyle e^{h_{1}\gamma (\tau (1))+\cdots +h_{N}\gamma (\tau (N))}}$ has the same coefficient as ${\displaystyle e^{h_{1}\gamma (1)+\dots +h_{N}\gamma (N)}}$.

Thus (after having grouped the terms with the same exponent) we see that the set of the 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 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:

${\displaystyle a(1)e^{\alpha (1)}+\cdots +a(n)e^{\alpha (n)}=0.}$

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 a polynomial with integer coefficients, we denote its degree by d(i). Let us denote the roots of this polynomial a(i)1, ..., a(i)d(i), with a(i)1 = a(i).

Let σ be a function which chooses one element from each of the sequences (1, ..., d(1)), (1, ..., d(2)), ..., (1, ..., d(n)), such that for every 1 ≤ i ≤ n, σ(i) is an integer between 1 and d(i). Then according to our assumption:

${\displaystyle \prod \nolimits _{\{\sigma \}}\left(a(1)_{\sigma (1)}e^{\alpha (1)}+\cdots +a(n)_{\sigma (n)}e^{\alpha (n)}\right)=0}$

where the product is over all possible choices. The product vanishes because one of the choices is just σ(i) = 1 for all i, for which the term vanishes according to our assumption above.

By expanding this product we get a sum of the form:

${\displaystyle b(1)e^{\beta (1)}+b(2)e^{\beta (2)}+\cdots +b(N)e^{\beta (N)}=0.}$

for some non-zero integer N, some distinct algebraic β(1), ..., β(N) (these are indeed algebraic because each is a sum of α's which are algebraic themselves), and b(1), ..., b(N) are polynomial in a(i)j (i in 1, ..., n and j in 1, ..., d(i)) with integer coefficients.

Since the product is over all possible choices, each of b(1), ..., b(N) is symmetric in a(i)1, ..., a(i)d(i) for every i; therefore each of b(1), ..., b(N) is a polynomial with integer coefficients in elementary symmetric polynomials of the sets {a(i)1, ..., a(i)d(i)} for every i. Each of the latter is a rational number (as in the proof of Lemma B).

Thus b(1), ..., b(N) ∈ Q, and 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 + eiπ + eiπ ≠ 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

${\displaystyle a_{n}e^{n}+\cdots +a_{0}e^{0}\neq 0.}$

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

## Notes

1. ^ a b
2. ^ a b Weierstrass 1885, pp. 1067–1086,
3. ^ Hermite 1873, pp. 18–24.
4. ^ Hermite 1874
5. ^
6. ^ Hilbert 1893, pp. 216–219.
7. ^ Gordan 1893, pp. 222–224.
8. ^ Bertrand 1997, pp. 339–350.
9. ^ (in French) french Proof's Lindemann-Weierstrass (pdf)
10. ^ Up to a factor, this is the same integral appearing in the proof that e is a transcendental number, where β1 = 1, ..., βm = m. The rest of the proof of the Lemma is analog to that proof.