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74 3. Elliptic Curves and Their Isomorphisms

3
Case 1. j = 0or12 ,orequivalently a 4 a 6 = 0. Then we see that E and E ¯
¯
are isomorphic only if the quotient a 4 ¯a 6 /¯a 4 a 6 is a square u −2 . Hence E and E are
isomorphic over any field extension of k containing the square root of the quotient
¯
a 4 ¯a 6 /¯a 4 a 6 . Further, specializing to E = E, we have that the automorphism group
Aut(E) ={+1, −1}, the group of square roots of 1.
3 2
Case 2. j = 12 , or equivalently a 6 = 0. The basic example is the curve y =
3
x − x. Then E and E are isomorphic if and only if the quotient a 4 /¯a 4 is a fourth
¯
4
¯
power u . Hence E and E are isomorphic over any field extension of k containing a
¯
fourth root of the quotient a 4 /¯a 4 . Further, specializing to E = E,wehavethatthe
automorphism group Aut(E) ={+1, −1, +i, −i}, the group of fourth roots of unity.
2
Case 3. j = 0, or equivalently a 4 = 0. The basic example is the curve y =
3
x − 1. Then E and E are isomorphic if and only if the quotient a 6 /¯a 6 is a sixth
¯
6
power u . Hence E and E are isomorphic over any field extension of k containing
¯
a sixth root of the quotient a 6 /¯a 6 . Further, specializing to E = E,wehavethatthe
¯
2
2
automorphism group Aut(E) ={+1, −1, +ρ, −ρ, +ρ , −ρ }, the group of sixth
2
roots of unity where ρ + ρ + 1 = 0.
¯
¯
It is natural to ask (1) j(E) = j(E) implies E and E are isomorphic for k
3
algebraically closed, and (2) whether all values in k, besides 0 and 12 ,are j values
of some elliptic curve. When a 4 a 6 = 0and k is algebraically closed, we can rescale
3
2
x and y so that the Weierstrass equation has the form y = 4x − cx − c. In terms
of c we calculate with (3.4)
c 3 3 c 3 3
3
j = 12 = 12 = 12 J (12 = 1728).
3
c − 27c 2 c − 27
From the relation J = c/(c − 27) we can solve for c in terms of j as
J j
c = 27 = 27 .
J − 1 j − 1728
Thus we have the proposition.
¯
¯
(4.3) Proposition. Two elliptic curves E and E over k are isomorphic over k if and
¯
only if j(E) = j(E). The curve with classical Weierstrass equation
j j
2 3
y = 4x − 27 x − 27
j − 1728 j − 1728
has j-invariant equal to the parameter j in the formula for the coefficients.
This is another version of the result (3.7) where for all j values unequal to 0 and
3
12 an elliptic curve E with given j = j(E). These curves are in a family of elliptic
curves over the twice punctured plane with fibre over j equal to an elliptic curve with
j value equal to the given j.