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§5. Torsion in Elliptic Curves over Q: Nagell–Lutz Theorem 117

over F 2 which is singular. Hence it has bad reduction at 2. Modulo 3 the situation is
better. From the derivative
2

(2y + 1 − x)y = 3x + y = y,
we see that (1, 0) is the only indeterminate value for y ,but (1, 0) is not on E(F 3 ).

We have the following “graph”
0
+1 ∗
0 ∗
−1 ∗ ∗ ∗
−1 0 +1

and from this we see that E (3) (F 3 ) is cyclic of order 6. Applying (5.1), we deduce that
E(Q) tors is cyclic of order 6 generated by (1, 1). The question of points of infinite
order is still pending.

(5.6) Example. In, respectively, 1(1.6), 1(2.3), and 1(2.4), we introduced three
curves

3
2
3
2
2
2
3
2

E : y + y = x − x, E : y + y = x − x , E : y + y = x + x .

On E and E the point (0, 0) has infinite order and on E it has order 5. Mod 2 all

three curves reduce to the same curve up to isomorphism, namely the curve called
E 3 in 3(6.4), and E 3 (F 2 ) is cyclic of order 5. Mod 3 we leave it to the reader to check
that all three curves are nonsingular and they have the following graphs:
0 0
+1 +1

E : 0 ∗ ∗ ∗ E : 0 ∗ ∗
−1 ∗ ∗ ∗ −1 ∗ ∗
−1 0 +1 −1 0 +1
0
+1 ∗
E : 0 ∗ ∗

−1 ∗ ∗
−1 0 +1
Hence E(F 3 ) = Z/7Z, E (F 3 ) = Z/5Z,and E (F 3 ) = Z/6Z.Any torsionin E(Q)

would have to inject into both the cyclic group of order 5, mod 2, and the cyclic group

of order 7, mod 3. Hence E(Q) tors = 0, and by the same principle, E (Q) tors = 0,

while E (Q) tors is cyclic of order 5.