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Then
div( y + V )=
j
d j ([( a j , −b j )] [ ]) =
j
d j ([ w ( a j ,b j )] [ ]) .
Since
div( f −V 2 )=div( y + V )+div( y−V )=
j
d j ([( a j ,b j )]+[( a j , −b j )] 2[ ]) ,
V 2 = j ( x
a j ) d j , by part (a). Therefore, every root of
we must have f
V 2 yields a term in div( y
f
V ). This completes the proof.
Part (a) of the proposition has a converse. If D = c j [ P j ] is a divisor, let
w ( D )= c j [ w ( P j )].
PROPOSITION 13.3
Let D be a divisor of degree 0. T hen D + w ( D ) isaprincipaldivisor; infact,
itisthe divisor of a rationalfunction in x .
Write D = j c j [ P j ], where possibly some P j is .Sincedeg D =
PROOF
0, we have j c j =0,so D = j c j ([ P j ] [ ]). If some P j = ,that
term can now be omitted, so we may assume P j
=
for all j . Therefore,
D + w ( D )= j c j ([ P j ]+[ w ( P j )]
2[
]), which is the divisor of a polynomial
in x , by Proposition 13.2.
A divisor of the form D = j c j ([ P j ] [ ]), with P j =( a j ,b j ), is called
semi-reduced if the following hold:
1. c j
0 for all j
2. if b j =0,then c j =0or1
3. if [ P j ]with b j
= 0 occurs in the sum (that is, c j > 0), then [ w ( P j )] does
not occur.
If, in addition, j c j ≤ g ,then D is called reduced .
Proposition 13.2 implies that div ( y − V ( x )) is semi-reduced for every poly-
nomial V ( x ).
Let D 1 = j c j ([ P j ] [ ]) and D 2 = j d j ([ P j ] [ ]) be two divisors
with c j 0and d j 0. Define
gcd( D 1 ,D 2 )=
j
min
{
c j ,d j }
([ P j ]
[
]) .
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