Hyperelliptic Curves - Elliptic Curves: Number Theory and Cryptography - page 408

13.2 Divisors

We continue to assume that C is a hyperelliptic curve given by (13.2) over
an algebraically closed field K of characteristic not equal to 2.
In general, a line intersects C in 2 g + 1 points. Therefore, when g ≥ 2, we
cannot use the method from elliptic curves to make the points on C into a
group, since the line through two points intersects the curve in 2 g 1 additional

points, rather than in a unique third point. Instead, we form the group of
divisors of degree 0 modulo principal divisors (that is, modulo divisors of
functions on C ).
In order to discuss divisors of functions, we need to make precise the order
of vanishing of a function at a point. Let P =( a, b )beapointon C and let t
be a function that has a simple zero at P .If H ( x, y ) is a function on C ,write
H = t r G ,where G ( P ) =0 ,∞ .Then H has a zero of order r at P (if r< 0,
then H has a pole of order |r| ). If P =( a, b )with b =0,itcanbeshownthat
t = x − a has a simple zero at P .If b =0,then x − a has a double zero, but
t = y works since the function y has a simple zero. The intuition is that the
line x

−

a = 0 intersects the curve C nontangentially at ( a, b ) except when
b = 0, where it is a vertical tangent to the curve. Since tangency corresponds
to higher order vanishing (as in Section 2.4), we need to use y instead, since
the horizontal line y = 0 intersects C at ( a, 0) nontangentially.
The functions we will work with are polynomials in x and y .Since y 2 =
f ( x ), we can replace y 2 with f ( x ). By induction, any polynomial in x, y can
be reduced to a function of the form A ( x )+ B ( x ) y ,where A ( x )and B ( x )are
polynomials in x .
We need to consider two special forms of functions.

−

PROPOSITION 13.2

(a) L et A ( x )= j ( x − a j ) c j .Then
div( A ( x )) =
j

c j [ P j ]+[ w ( P j )] − 2[ ∞ ] ,

where P j = a j , f ( a j ) and w ( P j )= a j , − f ( a j ) .
(b) L et V ( x ) be a polynom ial.Let
f ( x ) − V ( x ) 2 =
j

( x − a j ) d j .

Then the function y

V ( x ) has divisor
div y

−

V ( x ) =
j

d j [( a j ,b j )] ] , − − [ ∞