Divisors - Elliptic Curves: Number Theory and Cryptography - page 345

has exactly one zero, which must be simple. Let f be any function on E .If
f does not have a zero or pole at Q ,then
g ( x, y )=
R

( h ( x, y ) − h ( R )) ord R ( f )

∈ E ( K )

has the same divisor as f (since we are assuming that ord Q ( f ) = 0, the factor
for R = Q is defined to be 1). The only thing to check is that the poles of
h ( x, y )at Q cancel out. Each factor has a pole at ( x, y )= Q of order ord R ( f )

(or a zero if ord R ( f ) < 0). Since R ord R ( f ) = 0, these cancel.
Since f and g have the same divisor, the quotient f/g has no zeros or poles,
and is therefore constant. It follows that f is a rational function of h .
If f hasazeroorpoleat Q , the factor for R = Q in the above product
is not defined. However, f h ord R ( f ) has no zero or pole at Q .Theabove

reasoning shows that it is therefore a rational function of h , so the same holds
for f .
We have shown that every function on E ( K ) is a rational function of h .
In particular, x and y are rational functions of h . The following result shows
that this is impossible. This contradiction means that we must have P = Q .

·

LEMMA 11.5
Let E be an elliptic curve over K (of characteristic not 2) given by

y 2 = x 3 + Ax + B.

Let t be an ind eterm inate. T here are no nonconstant rationalfunctions X ( t )
and Y ( t ) in K ( t ) su ch that

Y ( t ) 2 = X ( t ) 3 + AX ( t )+ B.

PROOF Factor the cubic polynomial as

x 3 + Ax + B =( x − e 1 )( x − e 2 )( x − e 3 ) ,

where e 1 ,e 2 ,e 3 ∈ K are distinct. Suppose X ( t ) ,Y ( t ) exist. Write

X ( t )= P 1 ( t ) ( t )= Q 1 ( t )

P 2 ( t ) , Q 2 ( t ) ,

where P 1 ,P 2 ,Q 1 ,Q 2 are polynomials in t . We may assume that P 1 ( t ), P 2 ( t )
have no common roots, and that Q 1 ( t ), Q 2 ( t ) have no common roots. Sub-
stituting into the equation for E yields
Q 1 ( t ) 2 P 2 ( t ) 3 = Q 2 ( t ) 2 P 1 ( t ) 3 + AP 1 ( t ) P 2 ( t ) 2 + BP 2 ( t ) 3 .

Since the right side is a multiple of Q 2 ( t ) 2 , so is the left side. Since Q 1 ,Q 2
have no common roots, P 2 must be a multiple of Q 2 . A common root of