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 ΔE3 =

−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 = (α −

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