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This yields a bijection between triples in Z n 1 n 2 and pairs of triples, one in
Z n 1 and one in Z n 2 . It is not hard to see that primitive triples for Z n 1 n 2
correspond to pairs of primitive triples in Z n 1 and Z n 2 .Moreover,
y 2 z ≡ x 3 + Axz 2 + Bz 3
(mod n 1 n 2 )
y 2 z
x 3 + Axz 2 + Bz 3
(mod n 1 )
⇐⇒
y 2 z
x 3 + Axz 2 + Bz 3
(mod n 2 )
Therefore, there is a bijection
ψ : E ( Z n 1 n 2 ) −→ E ( Z n 1 ) ⊕ E ( Z n 2 ) .
It remains to show that ψ is a homomorphism. Let P 1 ,P 2
E ( Z n 1 n 2 )andlet
P 3 = P 1 + P 2 . This means that there is a linear combination of the outputs
of formulas I, II, III that is primitive and yields P 3 . Reducing all of these
calculations mod n i (for i =1 , 2) yields exactly the same result, namely the
primitive point P 3 (mod n i )isthesumof P 1 (mod n i )and P 2 (mod n i ).
This means that ψ ( P 3 )= ψ ( P 1 )+ ψ ( P 2 ), so ψ is a homomorphism.
COROLLARY 2.33
Let E be an elliptic curve over Q given by
y 2 = x 3 + Ax + B
with A, B ∈ Z .Let n be a positive odd integer such that gcd( n, 4 A 3 +27 B 2 )=
1 .Representthe elem entsof E ( Q ) as primitive triples ( x : y : z ) P 2 ( Z ) .
Themap
red n : E ( Q )
−→
E ( Z n )
( x : y : z )
( x : y : z )(mod n )
is a group hom om orphism .
PROOF
E ( Q )and P 1 + P 2 = P 3 ,then P 3 is a primitive point
that can be expressed as a linear combination of the outputs of formulas I, II,
III. Reducing all of the calculations mod n yields the result.
If P 1 ,P 2
Corollary 2.33 can be generalized as follows.
COROLLARY 2.34
Let R be a ring and let I be an ideal of R . A ssu m e that both R and R/I
satisfy conditions (1) and (2) on page 66. Let E be given by
y 2 z = x 3 + Axz 2 + Bz 3
 
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