206 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

the coefficients of which are elementary symmetric polynomials in the

β (r)i , i≤ µ (r)

Thus the coefficients are symmetric polynomials in the α (k)1 ,α (k)2 , · · · ,α (k)mkfor each

k. Say the polynomial is of the form

µ(r)

∑l=0

xn−lBl (A(1) , · · · ,A(n))

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)mkfor 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 9.1.6 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 9.17, hence roots of apolynomial with integer coefficients. Now 9.16 contradicts Corollary 9.2.6. ■

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 9.2.8 Suppose a(1) , · · · ,a(n) are nonzero algebraic numbers and suppose

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