Hyperelliptic Curves - Elliptic Curves: Number Theory and Cryptography - page 413

Replace D with D + g [ ∞ ]. Then

( D + g [ ∞ ]) = ( K− D − g [ ∞ ]) + 1 ≥ 1 ,

since ( K−D − g [ ∞ ]) ≥ 0. This means that there is a function F

=0such

that

div( F )+ D + g [ ∞ ] ≥ 0 .

Let D 1 = div( F )+ D , which is in the same divisor class as D .Then D 1 +
g [ ]to D 1 makes

all coe cients nonnegative, and since deg( D 1 ) = 0, it follows easily that the
only point in D 1 with negative coe cients is [

∞ ] ≥ 0 and deg( D 1 ) = 0. Since adding a multiple of [ ∞

] and that there are at most g
other points in the sum. If D 1 contains both [ P ]and[ w ( P )] for some P ,then
subtracting an appropriate multiple of the principal divisor [ P ]+[ w ( P )]

∞

]
removes either [ P ]or[ w ( P )] from D 1 and leaves the other with a nonnegative
coecient. Therefore, we may assume that D 1 is reduced, and hence D 1 is
the required divisor.
We now show that D 1 is unique. Suppose D

− 2[ ∞

− D 1 =div( F )and D − D 2 =

div( G )withboth D 1 and D 2 reduced. Then

D 1 + w ( D 2 )= D + w ( D ) − div( F ) − w (div( G )) ,

which is principal, since D + w ( D ) is principal (Proposition 13.3) and w
applied to a principal divisor yields a principal divisor (Exercise 13.4). Write
D 1 + w ( D 2 )=div( H ). Then
div( H )+2 g [ ∞ ]= D 1 + g [ ∞ ] + w D 2 + g [ ∞ ] ≥ 0 ,

so H ]) (see Section 11.5).

The Riemann-Roch theorem says that

∈L (2 g [ ∞

(2 g [ ∞ ]) − ( K− 2 g [ ∞ ]) = 2 g − g +1= g +1 .

Since deg( K− 2 g [ ∞ ]) = − 2 < 0 by Corollary 11.16, we have ( K− 2 g [ ∞ ]) = 0

]) = g +1. Since x j

by Proposition 11.14. Therefore, (2 g [ ∞ ∈L (2 j [ ∞ ]), the

set

1 ,x,x 2 ,...,x g

{ }

gives g + 1 functions in L (2 g [ ∞ ]). They are linearly independent since they
have poles of distinct orders. Therefore, they form a basis of L (2 g [ ∞ ]). This
means that every element can be written as a polynomial in x of degree at
most g .
We conclude that D 1 + w ( D 2 )=div( H ), where H is a polynomial in x .As
we showed earlier, this means that
D 1 + w ( D 2 )=
j

c j ([ P j ]+[ w ( P j )] − 2[ ∞ ])