A.3. TRANSCENDENTAL NUMBERS 371

where A(k) signifies the roots of Qk (x) ,{

α (k)1 , · · · ,α (k)mk

}. Thus, by the symmetric

polynomial theorem applied to the commutative ring Q [A(1) , · · · ,A(n−1)], the abovepolynomial is of the form

µ(r)

∑l=0

xµ(r)−l∑kl

Bkl (A(1) , · · · ,A(n−1))skl

11 · · ·s

kln

µ(r)

where the sk is one of the elementary symmetric polynomials in{α (n)1 , · · · ,α (n)mn

}and Bkl is symmetric in α (k)1 ,α (k)2 , · · · ,α (k)mk

for each k ≤ n−1 and

Bkl ∈Q [A(1) , · · · ,A(n−1)] .

Now do to Bkl what was just done to Bl featuring A(n−1) this time, and continue tilleventually you obtain for the coefficient of xµ(r)−l a large sum of rational numbers times aproduct of symmetric polynomials in A(1) ,A(2) , etc. By Theorem A.2.7 applied repeat-edly, beginning with A(1) and then to A(2) and so forth, one finds that the coefficient ofxµ(r)−l is a rational number and so the β (r) j for j ≤ µ (r) are algebraic numbers and rootsof a polynomial which has rational coefficients, namely the one in 1.17, hence roots of apolynomial with integer coefficients. Now 1.16 contradicts Corollary A.3.6.

Note this lemma is sufficient to prove Lindermann’s theorem that π is transcendental.Here is why. If π is algebraic, then so is iπ and so from this lemma, e0 +eiπ ̸= 0 but this isnot the case because eiπ =−1.

The next theorem is the main result, the Lindermann Weierstrass theorem. It replacesthe integers bi in the above lemma with algebraic numbers.

Theorem A.3.8 Suppose a(1) , · · · ,a(n) are nonzero algebraic numbers and sup-pose

α (1) , · · · ,α (n)

are distinct algebraic numbers. Then

a(1)eα(1)+a(2)eα(2)+ · · ·+a(n)eα(n) ̸= 0

Proof: Suppose a( j)≡ a( j)1 is a root of the polynomial

v jxm j + · · ·+u j

where v j,u j ̸= 0. Let the roots of this polynomial be a( j)1 , · · · ,a( j)m j. Suppose to the

contrary thata(1)1 eα(1)+a(2)1 eα(2)+ · · ·+a(n)1 eα(n) = 0

Then consider the big product

∏(i1,··· ,in)

ik∈{1,··· ,mk}

(a(1)i1 eα(1)+a(2)i2 eα(2)+ · · ·+a(n)in eα(n)

)(1.18)