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§6. Isomorphism Classification in Characteristic 2 77

(6.1) Conditions for Smoothness. Unlike characteristics different from 2 we have
to treat the two cases j = 0and j = 0 separately.
Case 1. j = 0 or equivalently a 1 = 0. Under a change x to x + c, the term
2
2
y +a 1 xy +a 3 becomes y +a 1 xy +(a 1 c+a 3 )y, and for a 1 nonzero we can choose
3
3
a 3 = 0. Changing x to a x and y to a y allows us to normalize a 1 = 1, and a linear
1 1
change of the variable allows us to choose a 4 = 0. The normal form becomes
dx
2 3 2
y + xy = x + a 2 x + a 6 and ω = .
x
Then b 2 = 1, b 4 = b 6 = 0, and b 8 = a 6 , and, moreover, c 4 = 1and = a 6 = 1/j.
2
The partial derivatives f x = y + x and f y = x have only x = y = 0 as a common
zero, and this lies on the curve if and only if a 6 = = 0. Hence E is smooth if and
only if = 0.
Case 2. j = 0 or equivalently a 1 = 0. By completing the cube, we can choose
the normal form of the cubic to be
dx
2 3
y + a 3 y = x + a 4 x + a 6 and ω = .
a 3
4
2
2
Then b 2 = b 4 = 0, b 6 = a ,and b 8 = a , and, moreover, = a and j = 0. Since
3 4 3
2
the partial derivative f x = x + a 4 and f y = a 3 , it follows that the curve is smooth
if and only if a 3 = 0 or equivalently = 0.
(6.2) Isomorphisms Between Two Curves with the Same j-Invariant. Suppose E
¯
¯
and E are two elliptic curves defined over k such that j = j(E) = j(E).If
¯
f : E → E is an isomorphism, then its form is determined for j = 0or j = 0 =
3
12 .
Case 1. j = 0 or equivalently a 1 = 0. Using the form of the Weierstrass equa-
¯
tions in (6.1), Case 1 for E and E,wehave
xf =¯x, yf =¯y + s ¯x,
2
and for the coefficients ¯a 2 = a 2 +s +s and ¯a 6 = a 6 . Then E and E are isomorphic if
¯
and only if the difference ¯a 2 −a 2 is of the form s 2 +s. Hence E and E are isomorphic
¯
over any field extension of k containing a solution to the quadratic equation
2
s + s =¯a 2 − a 2 .
¯
Further, specializing to E = E, we have that the automorphism group Aut(E) =
{0, 1} under addition.
Case 2. j = 0 or equivalently a 1 = 0. Using the form of the Weierstrass equa-
tions in (6.1), Case 2 for E and E,wehave
¯
3
2
2
xf = u ¯x, yf = u y + su ¯x + t,