38 2. Elliptic curves (2) However E1 has bad reduction at p = 5, and the reduction is additive, since modulo 5 we can write the equation as ((y 0) 0 · (x 0))2 x3 and the unique slope is 0. (3) The elliptic curve E2 : y2 = x3 x2 +35 has bad multiplica- tive reduction at 5 and 7. The reduction at 5 is split, while the reduction at 7 is non-split. Indeed, modulo 5 we can write the equation as ((y 0) 2(x 0)) · ((y 0) + 2(x 0)) x3, the slopes being 2 and −2. However, for p = 7, the slopes are not in F7 (because −1 is not a quadratic residue in F7). Indeed, when we reduce the equation modulo 7, we obtain y2 + x2 x3 mod 7 and y2 + x2 can only be factored in F7[i] but not in F7. (4) Let E3 be an elliptic curve given by the model y2 + y = x3 x2 10x 20. This is a minimal model for E3 and its (minimal) discriminant is ΔE 3 = −115. The prime 11 is the unique prime of bad reduction and the reduction is split multiplicative. Indeed, the point (5,5) mod 11 is a singular point on E3(F11) and f(x, y) = y2 + y + x2 + 10x + 20 x3 = (y 5 5(x 5)) · (y 5 + 5(x 5)) (x 5)3. Hence, the slopes at (5,5) are 5 and −5, which are obviously in F11 and distinct. Proposition 2.6.8. Let K be a field and let E/K be a cubic curve given by y2 = f(x), where f(x) is a monic cubic polynomial in K[x]. Suppose that f(x) = (x α)(x β)(x γ) with α, β, γ K (an algebraic closure of K) and put D = β)2(α γ)2(β γ)2. Then E is non-singular if and only if D = 0. The proof of the proposition is left as an exercise (see Exercise 2.12.9). Notice that the quantity D that appears in the previous
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