equivalent statements of Lindemann-Weierstrass theorem

Proposition 1.

The following versions of the Lindemann-Weierstrass TheoremMathworldPlanetmath are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath:

  1. 1.

    If α1,,αn are linearly independentMathworldPlanetmath algebraic numbersMathworldPlanetmath over , then eα1,,eαn are algebraically independentMathworldPlanetmath over .

  2. 2.

    If α1,,αn are distinct algebraic numbers over , then eα1,,eαn are linearly independent over .


(12). Write Ai=eαi for each i=1,,n. Suppose 0=r1A1++rnAn, where ri. Moving r1A to the LHS and by multiplying a common denominator we can assume that r1A1=r2A2++rnAn where ri. We want to show that r1==rn=0. We induct on n. The case when n=1 is trivial because A1 is never 0 and therefore 0=r1A1 forces r1=0.

By induction hypothesis, suppose the statement is true when n<k. Now suppose n=k. If α1,,αk are linearly independent over , A1,,Ak are algebraically independent and certainly linearly independent over . So suppose α1,,αk are not linearly independent over . Without loss of generality, we can assume s1α1=s2α2++skαk, where si and s10. By multiplying a common denominator we can further assume that si and s1>0. Then


Since r1A1=r2A2++rnAn, we get

(r1s1)(A2s2Aksk) = (r1s1)A1s1
= (r1A1)s1
= (r2A2++rkAk)s1
= g(A2,,Ak),

where g(x2,,xk)=(r2x2++rkxk)s1[x2,,xk]. PartitionPlanetmathPlanetmath the numbers sis into non-negative and negative ones, so that, say, sc(1),,sc(l) are non-negative and sd(1),,sd(m) are negative, then


If we define


then f(x2,,xk)[x2,,xk] and f(A2,,Ak)=0. By the induction hypothesis, f=0. It is not hard to see that r1==rk=0 and therefore A1,,Ak are linearly independent.

(12). We first need two lemmas:

Lemma 1. Given 2., if α0 is algebraic over , then eα is transcendental over .


Suppose f(eα)=0 where f(x)=r0+r1x++rnxn[x]. Then we have

0 = r0+r1eα++rn(eα)n
= r0e0+r1eα++rnenα.

Since α0, 0,α,,nα are all distinct, 1,eα,,enα are linearly independent by the hypothesisMathworldPlanetmathPlanetmath. Thus, r0=r1==rn=0 and we have f(x)=0, which means that eα is transcendental over . ∎

Lemma 2. Given 2., if α and β are linearly independent and algebraic over , then eα is transcendental over (eβ).


Let A=eα and B=eβ. Suppose f(A)=0 where f(x)(B)[x]. We want to show that f(x)=0. Write


where each ri(x)=pi(x)/qi(x) with pi(x), qi(x)0[x]. Let Q(x)=q1(x)qn(x). So Q(B), being the productPlanetmathPlanetmathPlanetmath of the denominators qi(B)0, is non-zero. Multiply f(x) by Q(B) we get a new polynomialPlanetmathPlanetmath g(x) such that


where each Ri(x)=ri(x)Q(x)=pi(x)Q(x)/qi(x)[x]. Now, g(A)=f(A)Q(B)=0. So

0 = R0(B)+R1(B)A++rn(B)An
= j=0m0a0jBj+j=0m1a1jBjA++j=0mnanjBjAn
= j=0m0a0jejβ+j=0m1a1jejβ+α++j=0mnanjejβ+nα,

where each aij. Now, the exponents in the above equation are all distinct, or else we would end up with α and β being linearly dependent, contrary to the assumptionPlanetmathPlanetmath. Therefore, by 2 (Lindemann-Weierstrass Version 2), all eiβ+jα are linearly independent, which means each aij=0. This implies that g(x)=0. But g(x)=f(x)Q(B) and Q(B)0, we must have f(x)=0. ∎

Now onto the main problem. We proceed by inductionMathworldPlanetmath on the number of linearly independent algebraic elements over . The case when n=1 is covered in Lemma 1, since a linearly independent singleton is necessarily non-zero. So suppose α1,,αk are linearly independent and algebraic over . Then each pair αk,αi are independent and algebraic over , ik. Thus eαk is transcendental over (eαi) for all ik. This means that eαk is transcendental over (eα1,,eαk-1).

Now let Ai=eαi for all i=1,,k. Suppose f(A1,,Ak)=0 where f[x1,,xk]. To show the algebraic independence of the Ais, we need to show that f=0. Rearranging terms of f and we have


If we let g(x)=f(A1,,Ak-1,x), we see that g(x)(A1,,Ak-1)[x] and g(Ak)=f(A1,,Ak-1,Ak)=0. Since eαk is transcendental over (A1,,Ak-1), we must have g(x)=0. This implies that each gj(A1,,Ak-1)=0. But then A1,,Ak-1 are algebraically independent by the induction hypothesis, we must have each gj=0. This means that f=0. ∎

Title equivalent statements of Lindemann-Weierstrass theorem
Canonical name EquivalentStatementsOfLindemannWeierstrassTheorem
Date of creation 2013-03-22 18:05:18
Last modified on 2013-03-22 18:05:18
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 4
Author CWoo (3771)
Entry type Result
Classification msc 12D99
Classification msc 11J85