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§7. The Canonical Height and Norm on an Elliptic Curve 139

Proof. Since we must have h E (P) = h E (−P), we consider the proper map

(1/2)hq = h : E(k) → R. Then h (P) = (1/2)h(q(P)) satisfies h (−P) = h (P).

h (2 P) which exists by (5.8) and (7.1).
Now form the limit h E (P) = lim n→∞ 2 −2n n
Then by (7.1), it follows that h E satisfies (1) and (2). Moreover, h E is proper ince h
is proper and it is positive.
To prove that h E is quadratic on E(k), we have only to check the weak parallelo-
gram law by (5.5) and (5.6) in view of (2). For this we use the following commutative
diagram:
u ✲
E × E E × E
q × q ❅ q × q
❘

✠ ❅
P 1 × P 1 θ θ P 1 × P 1
❅ s ❄ ❄ s
❅
❘ f ✲ ✠

P 2 P 2

where s((w, x), (w , x )) = (ww ,wx + w x, xx ), u(P, Q) = (P + Q, P − Q),
and f (α, β, γ ) = ( f 0 (α, β, γ ), f 1 (α, β, γ ), f 2 (α, β, γ )) such that
2 2
f 0 (α, β, γ ) = β − 4αγ, f 1 (α, β, γ ) = 2β(bα + γ) + 4cα ,
2
f 2 (α, β, γ ) = (γ − bα) − 4cαβ.
To check the commutativity of the above diagram we consider P = 1: x : y and
Q = 1: x : y with x and x unequal. Then we have

θ(P, Q) = s(q × q)(P, Q) = s(1: x, 1: x ) = 1: (x + x ) : xx .
From 1(1.4) we calculate
2 2 2

P + Q = (x − x ) : (y − y ) − (x + x )(x − x ) : ∗,
2 2 2

P − Q = (x − x ) : (y + y ) − (x + x )(x − x ) : ∗,
and hence the composite with the upper arrow in the diagram takes the following
form:
2
4
2
2
2

θu(P, Q) = (x − x ) :2(x − x ) (y + y − (x + x )(x − x ) ) :
2
4
2
2
2
2
2 2

(y − y ) + (x − x ) (x + x ) − 2(x − x ) (x + x )(y + y )
2
2
2

= (x − x ) :2(xx + x x + b(x + x ) + 2c) :
2

((xx − b) − 4c(x + x ))

= f 0 (1, x + x , xx ) : f 1 (1, x + x , xx ) : f 2 (1, x + x , xx ),
2
where as homogeneous forms of degree 2 in three variables f 0 (α, β, γ ) = β −
2
2
4αγ, f 1 (α, β, γ ) = 2βγ + 2bαβ + 4cα ,and f 2 (α, β, γ ) = (γ − bα) − 4cαβ.
From this commutative diagram we calculate