Cryptography Reference
In-Depth Information
Using the fact that
b k +1 = ( p
2
1) / 2
k +1
=
2
(( p
1) / 2)!
( k + 1)!((( p
1) / 2)
k
1)!
= (( p 1) / 2) k
k +1
2
b k ,
we find that the coecient of T k
is
4((( p − 1) / 2) − k ) 2
(2 k +1) 2 b k = p ( p − 2 4 k ) b k 0(mod p ) .
This proves the claim.
Suppose now that H p ( λ )=0with λ
=0,
we have λ =0 , 1. Write H p ( T )=( T−λ ) r G ( T ) for some polynomial G ( T )with
G ( λ ) = 0. Suppose r ≥ 2. In (4.5), we have ( T − λ ) r− 1 dividing the last term
and the middle term, but only ( T − λ ) r− 2 divides the term 4 T (1 − T ) H p ( T ).
Since the sum of the three terms is 0, this is impossible, so we must have
r = 1. Therefore, λ is a simple root. ( Technical point: Since the degree of
H p ( T ) is less than p ,wehave r<p , so the first term of the derivative
F p .Since H p (0)
=0and H p (1)
H p ( T )= r ( r
λ ) r− 2 G ( T )+2 r ( T
λ ) r− 1 G ( T )+( T
λ ) r G ( T )
1)( T
does not disappear in characteristic p . Hence ( T − λ ) r− 1 does not divide the
first term of (4.5).)
REMARK 4.39 The differential equation 4.5 is called a Picard-Fuchs
differential equation . For a discussion of this equation in the study of
families of elliptic curves in characteristic 0, see [24]. Once we know that
H p ( T ) satisfies this differential equation, the simplicity of the roots follows
from a characteristic p version of the uniqueness theorem for second order
differential equations. If λ is a multiple root of H p ( T ), then H p ( λ )= H p ( λ )=
0. Such a uniqueness theorem would say that H p ( T ) must be identically 0,
which is a contradiction. Note that we must avoid λ =0 , 1 because of the
coe cient T (1 − T )for H p ( T ).
COROLLARY 4.40
Let p ≥ 5 be prime.Thenumberof j ∈ F p that occur as j -invariantsof
su persingular elliptic curves is
p
12 + p ,
where p =0 , 1 , 1 , 2 if p ≡ 1 , 5 , 7 , 11 (mod 12) , respectively.
 
Search WWH ::




Custom Search