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)},
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