366 APPENDIX A. CLASSIFICATION OF REAL NUMBERS
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 A.3.3 Let Q(x) = vxm + · · ·+ u have integer coefficients with roots β 1, ...,β mlisted according to multiplicity. Let
f (x)≡ v(m+1)pQp (x)xp−1
(p−1)!(1.6)
a polynomial of degree n = pm+ p−1. Then
n
∑j=0
f ( j) (0) = vp(m+1)up +m1 (p) p (1.7)
m
∑i=1
n
∑j=0
f ( j) (β i) = m2 (p) p (1.8)
where m1 (p) ,m2 (p) are integers and p will be a large prime.
Proof: First consider 1.7. 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 1.7.Next consider 1.8 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 1.6 to get
f (x)(p−1)! = vpv(m+1)p ((x−β 1)(x−β 2) · · ·(x−β m))p xp−1 (1.9)
First notice that (p−1)! f ( j) (β i) = 0 unless j ≥ p. Thus all terms in computing
f ( j) (β i)(p−1)!
for j ≥ p have a factor of p!. If you have
g(x,β 1, · · · ,β m)≡ vpv(m+1)p ((x−β 1)(x−β 2) · · ·(x−β m))p xp−1,
it is symmetric in the β i so all derivatives with respect to x are also symmetric in these β iby Lemma A.3.2. By the same lemma, for j ≥ p
m
∑i=1
d j
dx j
(g(·,β 1, · · · ,β m)(β i)
1(p−1)!
)=
m
∑i=1
f ( j) (β i)