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§4. Finiteness of the Index (E(k) :2E(k)) 129

Proof. We factor multiplication by 2 as

ϕ 2 ϕ
E = E[a, b] → E = E[−2a, a − 4b] → E.

By 4(5.7) two different maps α induce isomorphisms E(Q)/ϕ E (Q) → im(α)

and E (Q)/ϕE(Q) → im(α).Also ϕ induces an isomorphism E (Q)/ϕE(Q) →

ϕE (Q)/ϕ ϕE(Q). Since we have two-stage filtration ϕ ϕE(Q) = 2E(Q) ⊂

ϕ E (Q) ⊂ E(Q), it follows that the index (E(Q) :2E(Q)) is the product of the

orders of the two im(α) where in one case α is defined on E(Q) and on E (Q) in the

other case. Now we apply the previous proposition, (3.1), to derive the theorem.
(3.3) Remark. With the notations of the previous theorem we see that r + s + 2is
an upper bound on the rank of E(Q). Since “most” curves over Q have rank 0 or 1,
this is not a very good upper bound. Using the methods of the next chapter we will
derive improved bounds.

§4. Finiteness of the Index (E E E(k k k):2E E E(k k k))

Throughout this section, we use the notations of 5(1.1) where R is a factorial ring
with field of fractions k. Observe that if c is a square in k, then ord p (c) is an even
number for all irreducibles p, and, conversely, if ord p (c) is an even number for all
irreducibles p, then c is a square times a unit in R.
Let (x, y) be a point on the elliptic curve E(k) defined by a factored Weierstrass
2
equation y = (x − r 1 )(x − r 2 )(x − r 3 ). By Theorem 1(4.1) we know that (x, y) ∈
2E(k) if and only if all x − r i are squares for i = 1, 2, 3. In particular, ord p (x − r i )
is even for such points. For an arbitrary point on E(k) we have the following 2-
divisibility property.

(4.1) Proposition. Let E be an elliptic curve defined by
2
y = (x − r 1 )(x − r 2 )(x − r 3 ),
where r 1 ,r 2 , and r 3 are in R. If (x, y) is a point on E(k), then ord p (x − r i ) is even
for all irreducibles pnot dividing any elements r i − r j for i = j.

Proof. Let p be an irreducible not dividing r i − r j , or, equivalently p such that
ord p (r i − r j ) = 0 for i = j.Iford p (x − r i )< 0 for one i, then for all j = 1, 2, 3,
we have ord p (x) = ord p (x − r i ) = ord p (x − r j ) since each ord p (r j ) ≥ 0. Thus it
follows that

2
2 ord p (y) = ord p (y ) = ord p ((x − r 1 )(x − r 2 )(x − r 3 )) = 3 ord p (x),
and hence ord p (x) = ord p (x − r j ) is even for each j.
We cannot have both ord p (x − r i )> 0 and ord p (x − r j )> 0 for i = j since
r j − r i = (x − r i ) − (x − r j ) and ord p (r i − r j ) = 0. Hence, if ord p (x − r i )> 0, for
one root r i , then we have the relation