Divisors - Elliptic Curves: Number Theory and Cryptography - page 375

1andlet A and B be two finite abelian groups (written additively)
such that na =0forall a

Let n ≥

∈ A and nb =0forall b ∈ B .Let , : B × A → μ n

be a bilinear pairing. If we fix a ∈ A ,then

ψ a : b −→ b, a

gives a homomorphism from B to μ n .LetHom B, μ n ) denote the set of
all group homomorphisms from B to μ n . We can make Hom( B, μ n )intoan
abelian group by defining the product of α, β ∈ Hom( B, μ n )by( α · β )( b )=

α ( b ) · β ( b ) for all b ∈ B .

LEMMA 11.26
If B is a finitegroup(written additively) such that nb =0 for all b ∈ B ,then
#Hom( B, μ n )=# B .

PROOF First, suppose B = Z m ,with m | n .If α ∈ Hom( B, μ n ), then

α (1) m = α (1 + ··· +1)= α (0) = 1 .

So α (1) is one of the m elements in μ m ⊆ μ n . Since 1 generates Z m ,the

value of α (1) determines α ( b ) for all b . Moreover, any choice of α (1) ∈ μ m

α (1) b . Therefore, there is a
bijection between Hom( Z m ,μ n )and μ m , so #Hom( Z m ,μ n )= m =# B .
Now consider an arbitrary finite abelian group B . By Theorem B.3 (Ap-
pendix B), B

determines a well-defined homomorphism by b →

Z m 1 ⊕···⊕ Z m s .Since nb =0forall b ∈ B ,wemusthave

m i | n for all i .Thereisamap

Φ : Hom( Z m 1 ,μ n ) ⊕···⊕ Hom( Z m 1 ,μ n ) −→ Hom ( Z m 1 ⊕···⊕ Z m s ,μ n ) ,

(11.14)

where the isomorphism maps the s -tuple ( α 1 ,α 2 ,...,α s ) to the homomor-
phism given by

( b 1 ,b 2 ,...,b s ) −→ α 1 ( b 1 ) α 2 ( b 2 ) ··· α s ( b s ) .

The map

α −→ ( α 1 ,α 2 ,...,α 2 ) ,

where α i ( b i )= α (0 , 0 ,...,b i ,..., 0), is the inverse of Φ, so Φ is a bijection.
Since the group on the left side of (11.14) has order m 1 m 2 ···m s =# B ,we
obtain the lemma.

Part (b) of the next lemma makes our task easier since it allows us to deduce
nondegeneracy in one argument from nondegeneracy in the other.