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76 3. Elliptic Curves and Their Isomorphisms

6
3
u ¯a 6 = a 6 + ra 4 + r .
Then E and E are isomorphic if and only if the quotient a 4 /¯a 4 is a fourth power u 4
¯
6
3
and u ¯a 6 − a 6 is of the form r + ra 4 . Hence E and E are isomorphic over any field
¯
where a 4 /¯a 4 is a fourth power and there is a solution for the cubic equation for r.
This can always be realized by going to a separable extension of degree dividing 12.
Further, specializing to E = E, we have that the automorphism group is a semi-
¯
direct product Aut(E) = (Z/4) × (Z/3) where the cyclic group of order 4 acts
nontrivially on the normal subgroup of order 3. The only nontrivial action is for a
generator of Z/4 to carry every element to its inverse. The automorphisms of E are
parametrized by pairs (u,r), where u is in +1, −1, +i, −i and r satisfies
3
(i) r + a 4 r = 0 if u =+1, −1,
3
(ii) r + a 4 r + 2a 6 = 0 if u =+i, −i.
2
3
4
For the curve y = x −x the automorphisms are given by (u,r) with u = 1and
3
r −r = 0, i.e., r in F 3 . The action of (u,r) on the curve E is given by (u,r)(x, y) =
(x + r, uy) in terms of points on the curve.
3
The problem of realizing all j values, different from 0 = 12 , has been solved in
(3.7). This is the case a 2 = 0and when k is algebraically closed, we can rescale x
2
2
3
and y so that the cubic equation has the form y = x + x + a 6 or in characteristic
3
2
3 the form y + xy = x + a 6 .Inbothcases j(E) =−1/a 6 .
(5.3) Proposition. The curves with normal cubic equations
1 1
2 3 2 2 3
y = x + x − or y + xy = x −
j j
have j-invariant equal to the parameter j appearing in the formula for the coefficient
for all nonzero j.
Exercises
2
1. Find the relation between for the curve defined by the cubic equation y = g(x) =
3
2
x + a 2 x + a 4 x + a 6 , and the discriminant of the cubic polynomial g(x).
2. Find an elliptic curve in characteristic 3 with j value 0.

§6. Isomorphism Classification in Characteristic 2

Let k be a field of characteristic 2 in this section. For an elliptic curve E over k the
invariant differential has the form ω = dx/(a 1 x + a 3 ) so that either a 1 or a 3 = 0.
2
2
Also by (3.4) using 2 = 0, we see that b 2 = a and c 4 = b , and therefore
1 2
12
c 4 = a 4 and j = a / .
1 1
3
In particular, a 1 = 0ifand onlyif j = 0 = 12 .