Divisors - Elliptic Curves: Number Theory and Cryptography - page 346

P 2 and P 1 + AP 1 P 2 + BP 2 would be a root of P 1 .Since P 1 and P 2 have
no roots in common, this is impossible. Therefore, Q 2 must be a multiple of
P 2 . Therefore, P 2 and Q 2 are multiples of each other, hence are constant
multiples of each other. By adjusting P 1 and Q 1 if necessary, we may assume
that

P 2 = Q 2 .
Canceling these from the equation yields

Q 1 = P 1 + AP 1 P 2 + BP 2 =( P 1 − e 1 P 2 )( P 1 − e 2 P 2 )( P 3 − e 3 P 2 ) .

Suppose i = j and P 1 − e i P 2 and P 1 − e j P 2 have a common root r .Then r is
arootof

e j ( P 1 − e i P 2 ) − e i ( P 1 − e j P 2 )=( e j − e i ) P 1 (11.2)

and of

( P 1 − e i P 2 ) − ( P 1 − e j P 2 )=( e j − e i ) P 2 . (11.3)

Since e j − e i = 0, this means that r is a common root of P 1 and P 2 ,which
is a contradiction. Therefore P 1 − e i P 2 and P 1 − e j P 2 have no common roots
when i = j . Since the product

( P 1 − e 1 P 2 )( P 1 − e 2 P 2 )( P 1 − e 3 P 2 )

is t he square of a polynomial, each factor must be a square of a polynomial
in K [ t ] (it might seem that each factor is a constant tim es a square, but
all constants are squares in the algebraically closed field K , hence can be
absorbed into the squares of polynomials).
Since P 2 = Q 2 , we find that P 2 must also be a square of a polynomial.

LEMMA 11.6
Let P 1 and P 2 be polynom ialsin K [ t ] withn o commonroots. Suppose there
are four pairs ( a i ,b i ) , 1

≤ i ≤ 4 ,with a i ,b i ∈ K satisfying

1. for each i ,atleast on e of a i ,b i isnonzero
2. if i = j ,then there does not exist c ∈ K × with ( a i ,b i )=( ca j ,cb j )

3. a i P 1 + b i P 2 is a square of a polynom ialfor 1 ≤ i ≤ 4 .

Then P 1 ,P 2 are constant polynom ials.

PROOF The assumpti on s imply that any tw o o f the vectors ( a i ,b i )are
linearly independent over K and therefore span K 2 . Suppose that at least
one of P 1 ,P 2 is nonconstant. We may assume that P 1 ,P 2 are chosen so that

Max(deg( P 1 ) , deg( P 2 )) > 0