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78 3. Elliptic Curves and Their Isomorphisms

and for the coefficients

3
4
4
u ¯a 3 = a 3 , u ¯a 4 = a 4 + sa 3 + s ,
6
6
2
2
u ¯a 6 = a 6 + s a 4 + ta 3 + s + t .
Then E and E are isomorphic if and only if:
¯
3
(1) the quotient a 3 /¯a 3 is a cube u ,
4
4
(2) the separable quartic equation s + a 3 t + a 4 + u ¯a 4 = 0in s has a solution, and
2
6
2
6
(3) the quadratic equation t +a 3 t +(s +s a 4 +a 6 +u ¯a 6 ) = 0in t has a solution.
Hence E and E are isomorphic over any field extension of k containing the roots
¯
to the equations (1), (2), and (3). Further, specializing to E = E,wehavethatthe
¯
automorphism group Aut(E) is a certain group of order 24 provided that k contains
the roots of the above equations.
(6.3) Remark. The group Aut(E) can be described two ways either as SL 2 (F 3 )or
as the units in the integral quaternions ±1, ±i, ± j, ±k,(±1 ± i ± j ± k)/2.
Now we consider the special case of k = F 2 , the field of two elements and write
down all the elliptic curves. There are five up to isomorphism (over F 2 ) two with
j = 1 and three with j = 0.
3
3
2
Case 1. j = 1. Since y + xy = f (x) with f (x) = x and f (x) = x + x 2
have a singularity at (0, 0), we have four equations related by replacing y by y + 1
pairwise:
3
2
2
3
2
2

∼
E 1 : y + xy = x + x + 1 = E : y + xy = x + x + x,
1
2 3 2 3
E 2 : y + xy = x + 1 ∼ 2
= E : y + xy = x + x.
We can “graph” the elliptic curves E 1 and E 2 over F 2 .

The groups E 1 (F 2 ) = Z/2Z ={0,(0, 1)} and E 2 (F 2 ) = Z/4Z ={0,(0, 1), (1, 0),
(1, 1)}.
2
3
Case 2. j = 0. Since the general form of such a curve is y + y = x +a 4 x +a 6 ,
there are four possibilities of which two are isomorphic. Also we can replace y by
y + x or x by x + 1 to obtain forms up to isomorphism of the same elliptic curve.
2
3
2
2
3
∼
E 3 : y + y = x + x = E : y + y = x + x ,

3