40 2. Elliptic curves

Figure 5. Helmut Hasse (1898-1979).

E(Fp). If f(x0) is not a square modulo p, then there are no points in

E(Fp) with x-coordinate equal to x0. Hence:

Np ≈ p ·

1

2

· 2 +

1

2

· 0 + 1 = p + 1.

Notice that we have added 1 in order to account for the point at

infinity.

Remark 2.6.13. Suppose that E/Q is an elliptic curve that has bad

reduction at a prime p. How many points does the singular curve

E have over Fp? The reader can find the answer to this question in

Exercise 5.7.1.

Example 2.6.14. Let E/Q be the elliptic curve

y2

=

x3

+ 3. Its

minimal discriminant is ΔE = −3888 =

−24

·

35.

Thus, the only

primes of bad reduction are 2 and 3 and E/Fp is smooth for all p ≥ 5.

For p = 5, there are precisely 6 points on E(F5), namely

E(F5) = {O, (1,2),(1,3),(2,1),(2,4),(3,0)},