Linear Algebra, Theory And Applications, 2012a

452 FIELDS AND FIELD EXTENSIONS
Letting
f (x) = v(m−1)p Q p (x) x p−1
(p − 1)!
and I (s) be defined in terms of f (x) as above, it follows,
m∑
lim I (β i )=0
and
p→∞
i=1
n∑
f (j) (0) = v p(m−1) u p + m 1 (p) p
j=0
m∑
i=1 j=0
n∑
f (j) (β i )=m 2 (p) p
where m i (p) is some integer.
Proof: Let p be a prime number. Then consider the polynomial f (x) ofdegreen ≡
pm + p − 1,
f (x) = v(m−1)p Q p (x) x p−1
(p − 1)!
From (6.3)
⎛
⎞
m∑
m∑ n∑
n∑
c I (β i )=c ⎝e β i
f (j) (0) − f (j) (β i ) ⎠
i=1
(
m∑
= K + c
i=1
e β i
i=1
j=0
)
∑ n
f (j) (0) − K
j=0
j=0
j=0
n∑
m∑ n∑
f (j) (0) − c f (j) (β i ) (6.5)
i=1 j=0
Claim 1: lim p→∞ c ∑ m
i=1 I (β i)=0.
Proof: This follows right away from the definition of I ( )
β j and the definition of f (x) .
∣
∣I ( ∫
)∣ 1 ∣
β j ∣ ≤ ∣β j f ( ) ∣
tβ j e
β j −tβ j dt
0
∫ 1
|v| (m−1)p ∣ ∣Q ( )∣
tβ j ∣
p ∣ ∣ t
p−1 ∣β j p−1
≤
dt
0 ∣
(p − 1)!
∣
which clearly converges to 0. This proves the claim.
The next thing to consider is the term on the end in (6.5),
n∑
m∑ n∑
K f (j) (0) + c f (j) (β i ) (6.6)
j=0
i=1 j=0
The idea is to show that for large enough p it is always an integer. When this is done, it
can’t happen that K + c ∑ m
i=1 eβ i = 0 because if this were so, you would have a very small
number equal to an integer. Now
⎛
⎞p
Q(x)
v (m−1)p ⎜
{ }} {
⎟
⎝v (x − β 1 )(x − β 2 ) ···(x − β m ) ⎠ x p−1
f (x) =
(p − 1)!
= vmp ((x − β 1 )(x − β 2 ) ···(x − β m )) p x p−1
(p − 1)!
(6.7)