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 =
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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