Divisors - Elliptic Curves: Number Theory and Cryptography - page 351

(2) Suppose T ∈ E [ n ] is such that e n ( S, T ) = 1 for all S ∈ E [ n ]. This means

that g ( P + S )= g ( P ) for all P and for all S ∈ E [ n ]. By Proposition 9.34,

there is a function h such that g = h ◦ n .Then

( h ◦ n ) n = g n = f ◦ n.

Since multiplication by n is surjective on E ( K ), we have h n = f . Therefore,

n div( h )=div( f )= n [ T ] − n [ ∞ ] ,

so div( h )=[ T ] .Th sproves

half of (2). The nondegeneracy in S follows immediately from (4) plus the
nondegeneracy in T .
(3) Let τ jT represent adding jT ,so f ◦ τ jT denotes the function P

− [ ∞ ]. By Theorem 11.2, we have T = ∞

→

f ( P + jT ). The divisor of f ◦ τ jT is n [ T − jT ] − n [ −jT ]. Therefore,

⎛ ⎞

n − 1 n − 1

⎝ ⎠ =

div f ◦ τ jT ( n [(1 − j ) T ] − n [ −jT ]) = 0 .

j =0 j =0

This means that n− 1

j =0 f

◦ τ jT is constant. The n th power of the function

n− 1

τ jT is the above product of f 's composed with multiplication by n ,
hence is constant. Since
⎛

j =0 g

◦

⎞

n

n − 1 n − 1

⎝ ⎠

g ◦ τ jT f ◦ n ◦ τ jT

=

j =0 j =0

n − 1

(since nT = T ) . = f ◦ τ jT ◦ n

j =0

Since we have proved that this last product is constant, it follows that n− 1 ◦

τ jT is constant (we are again using the connectedness of E in the Zariski
topology). Therefore, it has the same value at P and P + T ,so

j =0 g

n − 1 n − 1

g ( P + T + jT )= g ( P + jT ) .

j =0 j =0

Canceling the common terms (we assume P is chosen so that all terms are
finite and nonzero) yields

g ( P + nT )= g ( P ) .

Since nT = T , this means that

e n ( T,T )= g ( P + T )

=1 .

g ( P )