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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