204 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA
Corollary 9.2.6 Let K and ci for i = 1, · · · ,n be nonzero integers. For each k between 1and n let {β (k)i}
mki=1 be the roots of a polynomial with integer coefficients,
Qk (x)≡ vkxmk + · · ·+uk
where vk,uk ̸= 0. Then
K + c1
(m1
∑j=1
eβ (1) j
)+ c2
(m2
∑j=1
eβ (2) j
)+ · · ·+ cn
(mn
∑j=1
eβ (n) j
)̸= 0. (∗)
Proof: Let Kk be nonzero integers which add to K. It is certainly possible to obtain thissince the Kk are allowed to change sign. They only need to be nonzero. Also let αk be as inthe above lemma such that α
mk+2k vmk+1
k uk = U some integer. Thus, replacing each Qk (x)with αkvkxmk + · · ·+αkuk, it follows that for each large prime p,(αkv)p(mk+1) (αku)p =(
αmk+2k vmk+1
)p=U p. From now on, use the new Qk (x).
Defining fk (x) and Ik (s) as in Lemma 9.2.4,
fk (x)≡v(m+1)pQp
k (x)xp−1
(p−1)!
and as before, let p be a very large prime number. It follows from Lemma 9.2.4 that foreach k = 1, · · · ,n,
ck
mk
∑i=1
Ik (β (k)i) =
(Kk + ck
mk
∑i=1
eβ (k)i
)deg( fk)
∑j=0
f ( j)k (0)
−
(Kk
deg( fk)
∑j=0
f ( j)k (0)+ ck
mk
∑i=1
deg( fk)
∑j=0
f ( j)k (β (k)i)
)This is exactly the same computation as in the beginning of that lemma except one addsand subtracts Kk ∑
deg( fk)j=0 f ( j)
k (0) rather than K ∑deg( fk)j=0 f ( j)
k (0) where the Kk are chosen suchthat their sum equals K and the term on the left converges to 0 as p→∞. By Lemma 9.2.4,
ck
mk
∑i=1
Ik (β (k)i) =
(Kk + ck
mk
∑i=1
eβ (k)i
)(U p +Nk p)
−Kk (U p +Nk p)− ckN′k p
=
(Kk + ck
mk
∑i=1
eβ (k)i
)U p−KU p +Mk p
where Mk is some integer. Now add.
m
∑k=1
ck
mk
∑i=1
Ik (β (k)i) =U p
(K +
m
∑k=1
ck
mk
∑i=1
eβ (k)i
)−KmU p +Mp
If K +∑mk=1 ck ∑
mki=1 eβ (k)i = 0, then if p > max(K,m,U) you would have −KmU p +Mp
an integer so it cannot equal the left side which will be small if p is large. Therefore, ∗follows. ■
Next is an even more interesting Lemma which follows from the above corollary.