Number Theory: An Introduction to Mathematics

4 Mordell’s Theorem 561

Proof Supposehˆhas the properties (i),(ii). Then, by (ii), 4nhˆ(P) =hˆ( 2 nP)and
hence, by (i), 4nhˆ(P)−h( 2 nP)is bounded. Dividing by 4n, we see thath( 2 nP)/ 4 n→
hˆ(P)asn→∞. This proves uniqueness.
To prove existence, chooseCas in the statement of Proposition 8. Then, for any
integersm,nwithn>m>0,

| 4 −nh( 2 nP)− 4 −mh( 2 mP)|=

∣

∣

∣

∣

∑n−^1

j=m

{ 4 −j−^1 h( 2 j+^1 P)− 4 −jh( 2 jP)}

∣

∣

∣

∣

≤

n∑− 1

j=m

4 −j−^1 |h( 2 j+^1 P)− 4 h( 2 jP)|

≤

n∑− 1

j=m

4 −j−^1 C< 4 −mC/ 3.

Thus the sequence{ 4 −nh( 2 nP)}is a fundamental sequence and consequently conver-
gent. If we denote its limit byhˆ(P), then clearlyhˆ( 2 P)= 4 hˆ(P). On the other hand,
by takingm=0 and lettingn→∞in the preceding inequality we obtain

|hˆ(P)−h(P)|≤C/ 3.

Thushˆhas both the required properties.

The valuehˆ(P)is called thecanonical heightof the rational pointP. The formula
forhˆ(P)shows that, for allP∈E,

hˆ(−P)=hˆ(P)≥ 0.

Moreover, by Proposition 9(i), for anyr>0 there exist only finitely many elements
PofEwithhˆ(P)≤r.
It will now be shown that the canonical height satisfies theparallelogram law:

Proposition 10Fo r a l l P 1 ,P 2 ∈E,

hˆ(P 1 +P 2 )+hˆ(P 1 −P 2 )= 2 hˆ(P 1 )+ 2 hˆ(P 2 ).

Proof It is sufficient to show that there exists a constantC′ >0suchthat,forall
P 1 ,P 2 ∈E,

h(P 1 +P 2 )+h(P 1 −P 2 )≤ 2 h(P 1 )+ 2 h(P 2 )+C′. (∗)

For it then follows from the formula in Proposition 9 that, for allP 1 ,P 2 ∈E,

hˆ(P 1 +P 2 )+hˆ(P 1 −P 2 )≤ 2 hˆ(P 1 )+ 2 hˆ(P 2 ).

But, replacingP 1 byP 1 +P 2 andP 2 byP 1 −P 2 ,wealsohave

hˆ( 2 P 1 )+hˆ( 2 P 2 )≤ 2 hˆ(P 1 +P 2 )+ 2 hˆ(P 1 −P 2 )

and hence, by Proposition 9(ii),

2 hˆ(P 1 )+ 2 hˆ(P 2 )≤hˆ(P 1 +P 2 )+hˆ(P 1 −P 2 ).