MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know what is the $j$-invariant but I am asking about a general definition in sence of classical invariant theory. The following possible definition seems to be wrong: Consider an elliptic curve $C: y^2=x^3+ax+b.$ Acting by the transformation $$ \begin{array}{c} x=a_{1,1}x'+a_{1,2}y'+c_1,\\ y=a_{2,1}x'+a_{2,2}y'+c_2, \end{array} $$ we get new curve $C'.$ The invariant is a function of coefficients of the curves which is stable under above transformation. This definition is wrong becouse $C'$ now consists $y'^3$ and it isn't elliptic curve.

So, what is correct definition of an invariant of elliptic curve?

share|cite|improve this question
You are using a definition of "elliptic curve" which complots against you. – Mariano Suárez-Alvarez Apr 7 '11 at 18:08
The equation of $C'$ IS an elliptic curve, but it is not in the Weierstrass form. So the formula of $j$ is more complicated in this case. In Silvermann's book, transformation doesn't involve $y'$ in $x'$. – Auguste Hoang Duc Apr 7 '11 at 18:14
Oh, so the transformation must be of the restricted form x=a1,2x′+c1.? – Melania Apr 7 '11 at 18:33
@Mariano, Auguste. I understood, the definition of el.curve should be more general: $a_0y^2+a_1y=b_0 x^3+b_1 x^2 +b_2 x+b_3,$ and the transformation should be restricted $x=a_{1,1}x'+c_1,\\ y=a_{2,1}x'+a_{2,2}y'+c_2.$ Then we may define invariants as usual. Thanks – Melania Apr 7 '11 at 19:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.