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§5. Isomorphism Classification in Characteristic 3 75

§5. Isomorphism Classification in Characteristic 3

Let k be a field of characteristic 3 in this section. For an elliptic curve E over k we
can choose coordinates x and y giving the Weierstrass model in the form
dx
2 3 2
y = x + a 2 x + a 4 x + a 6 and ω =− .
y
By (3.1) and (3.3) and using3=0,wesee that
2
b 2 = a 2 , b 4 =−a 4 , b 6 = a 6 , b 8 =−a + a 2 a 6
4
and

2
3
3
3
2 2
c 4 = a , c 6 =−a , = a a − a a 6 − a .
2 2 2 4 2 4
(5.1) Conditions for Smoothness. As in (4.1) the curve E is smooth if and only if
2
3
the cubic polynomial g(x) = x + a 2 x + a 4 x + a 6 and its derivative g (x) have

no common zero. Since and the discriminant of g(x) are equal up to a nonzero
constant, we deduce again that E is smooth if and only if = 0.
Further, j = j(E) is given by
c 3 4 a 6 2
j = = 2 2 3 3 .
a a − a a 6 − a
2 4 2 4
(5.2) Isomorphisms Between Two Curves with the Same j-Invariant. Suppose E
¯
and E are two elliptic curves defined over k with Weierstrass equations y 2 =
2
3
2
3
2
x + a 2 x + a 4 x + a 6 and y = x +¯a 2 x +¯a 4 x +¯a 6 such that j = j(E) = j(E).
¯
If f : E → E is an isomorphism, then its form is determined by whether j = 0or
¯
3
j = 0 = 12 .
Case 1. j = 0 or equivalently a 2 = 0. By completing the square in both Weier-
3
strass equations, we can assume that a 4 =¯a 4 = 0. Then j(E) =−a /a 6 =
2
3
−¯a /¯a 6 = j(E), and the following hold:
¯
2
2
2
3
xf = u ¯x, yf = u ¯y, and a 2 = u ¯a 2 .
2
Thus E and E are isomorphic if and only if the quotient a 2 /¯a 2 is a square u .Hence
¯
E and E are isomorphic over any field extension of k containing the square root of
¯
the quotient a 2 /¯a 2 . Further, specializing to E = E, we have that the automorphism
¯
group Aut(E) ={+1, −1}, the group of square roots of 1.
Case 2. j = 0, or equivalently a 2 = 0. Then = a 4 and ω = dy/a 4 , and the
following hold for the isomorphism f :
2
4
3
xf = u ¯x + r, yf = u ¯y, a 4 = u ¯a 4 ,
and