9.2. TRANSCENDENTAL NUMBERS 201

You see that this is unchanged from switching two variables. For example, switch x1 andx2. By assumption, nothing changes in the terms after the first two. The first term thenbecomes

h(x2,x2,x1...,xn)xr2 = h(x2,x1,x2, ...,xn)xr

2

and the second term becomes

h(x1,x2,x1, ...,xn)xr1 = h(x1,x1,x2, ...,xn)xr

1

which are the same two terms, just added in a different order. The situation works the sameway with any other pair of variables. ■

Recall that every algebraic number is a root of a polynomial having integer coefficients.

Lemma 9.2.3 Let Q(x) = vxm+ · · ·+u have roots β 1, ...,β m listed according to multiplic-ity and let the coefficients be integers. Let

f (x)≡ v(m+1)pQp (x)xp−1

(p−1)!(9.6)

a polynomial of degree n = pm+ p−1. Then

n

∑j=0

f ( j) (0) = vp(m+1)up +m1 (p) p (9.7)

m

∑i=1

n

∑j=0

f ( j) (β i) = m2 (p) p (9.8)

where m1 (p) ,m2 (p) are integers and p will be a large prime.

Proof: First consider 9.7. f (x) = v(m+1)p(vxm+···+u)pxp−1

(p−1)! . Then f j (0) = 0 unless j ≥p−1 because otherwise, that xp−1 term will result in some xr,r > 0 and everything is zerowhen you plug in x = 0. Now say j = p−1. Then it is clear that you get a (p−1)! whichcancels the denominator and letting x = 0, you get the integer f (p−1) (0) = upv(m+1)p. Sowhat if j > p−1?

d j

dx j

((vxm + · · ·+u)p xp−1)

=j

∑r=0

(ji

)di

dxi ((vxm + · · ·+u)p)d j−i

dx j−i xp−1

and, since eventually x = 0, only j− i = p−1 is of interest, so i = j− p+1 where j ≥ pas just mentioned. Since i ≥ 1, there will be a factor of p and a factor of (p−1)! fromd j−i

dx j−i xp−1. Thus when x = 0, this reduces to m1 (p) p(p−1)! and so this yields 9.7.Next consider 9.8 which says that ∑

mi=1 ∑

nj=0 f ( j) (β i) = m2 (p) p. The factorization of

Q(x) is v(x−β 1) · · ·(x−β m) . Replace Q(x) with its factorization in 9.6 to get

f (x)(p−1)! = vpv(m+1)p ((x−β 1)(x−β 2) · · ·(x−β m))p xp−1 (9.9)

First notice that (p−1)! f ( j) (β i) = 0 unless j ≥ p. Thus all terms in computing

f ( j) (β i)(p−1)!