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§3. Basic Descent Formalism 163

has no solution mod p in this case, and, hence, no solution over the integers. This
means that for p ≡±3(mod8)

∗ 2

im(α on E ) ={1, 4p} mod Q .
The above discussion together with the exact sequence in 4(5.7) shows that cer-
tain curves have only finitely many rational points, and with the theory of 5 we are
able to determine completely the Mordell group of rational points.
(2.5) Theorem. For a prime number p ≡ 3(mod8) the groups of rational points
on the elliptic curves with equations

2 3 2 3
y = x − px and y = x + 4px
are all equal to the groupwith two elements {0,(0, 0)}. In particular, the rank of
these groups is zero.
Proof. The fact that the groups are finite follows from the above analysis of the quar-
tic curves and their rational points together with 4(5.7) showing that E(Q)/2E(Q)
is a group of two elements with nonzero element of the class of (0, 0).
It remains to show that there is no odd torsion in E(Q). For this we use the
2
6 3
3
Nagell–Lutz theorem, 5(5.1). Since the discriminant of y = x + bx is −2 b ,
the above curves have bad reduction only at the primes p and 2. Mod 3 the curves
3
2
3
2
become either the curve given by y = x − x or by y = x + x. In both cases
there are exactly four elements on these curves over F 3 . Thus by 5(5.1) there is no
odd torsion, for it would have to map injectively into a group with four elements.
2
3
We are left only with the case p = 3. The curves with equations y = x − 3x
2
3
3
2
and y = x + 12x both reduce to y = x + 2x mod 5, and this curve has only
two points over F 5 . Again there is no odd torsion. This completes the proof of the
theorem.
§3. Basic Descent Formalism
In this section we return to the Mordell–Weil theorem asserting that E(k) is finitely
generated for an elliptic curve E over a number field k. We use the notations of
6(8.2) for the family V (k) of absolute values v of k, and we also use the fact that
multiplication by any m is surjective E(k) → E((k), see 12(3.6). By (2.1) and (2.2)
the short exact sequence of Gal(k/k)-modules
¯
¯
¯
0 → m E(k) → E(k) → E(k) → 0
leads to the long exact sequence for G = Gal((k/k)
m 1 1 m 1
E(k) → E(k) → H k, m E(k) → H k, E(k) → H k, E(k)
¯
¯
¯
which reduces to the short exact sequence
1
1

¯
¯
0 → E(k)/mE(k) → H k, m E(k) → m H k, E(k) → 0.