Elliptic curve over $K$ is defined as 'genus one smooth curve'. Using Rieman-Roch theorem, we can deduce Weierstrass equation form $Y^2Z+a_1XYZ+a_3YZ^2=X^3+a_2X^2Z+a_4XZ^2+a_6Z^3$ over $K$.

I wonder why coefficients's number are arranged $1,3,2,4,6$. Why not $1,2,3,4,5,6$?

Please tell me the background.

Thank you in advance.


1 Answer 1


The question is

I wonder why coefficients's number are arranged $1,3,2,4,6$. Why not $1,2,3,4,5,6$?

This is due to the homogeneous weights of the variables $\,X,Y,Z.\,$ Define

$$ X:=x\,t^2,\quad Y:=y\,t,\quad Z:=z\,t^4 \tag{1} $$

where $\,t\,$ is a formal variable counting the weight. The equation of the curve is

$$ Y^2Z+a_1XYZ+a_3YZ^2=X^3+a_2X^2Z+a_4XZ^2+a_6Z^3. \tag{2}$$

Divide both sides by $\,t^6\,$ and use equation $(1)$ to get

$$ y^2z +a_1xyz\,t +a_3yz^2t^3 = x^3 +a_2x^2z\,t^2 +a_4xz^2t^4 +a_6z^3t^6. \tag{3}$$

The $\,a_1,a_2,a_3,a_4,a_6\,$ subscripts come from the exponent of $\,t.\,$

An alternative, perhaps simpler approach, is to define (using Jacobian coordinates)

$$ X := x/u^2,\quad Y := y/u^3,\quad Z :=1 \tag {4} $$

where $\,u\,$ denotes weight with $\,x\,$ of weight $2$ and $\,y\,$ of weight $3$.

Multiply both sides of equation $(2)$ by $\,u^6\,$ and use equation $(4)$ to get

$$ y^2 +a_1xy\,u +a_3y\,u^3 = x^3 +a_2x^2\,u^2 +a_4x\,u^4 +a_6\,u^6. \tag{5}$$

This is close to equation $(3)$ with the same interpretation but now with exponents of $\,u\,$ instead.

  • $\begingroup$ Thank you so much! May I ask two questions? $1$. Why the weight of x,y,z is $2,1,4$? (How can I deduce it ?)$2$.Can't we deduce from non-homogeneous form $y^2+a1xy+a3y=x^3+a2x^2+a4x+a6$? In this case, y has weight(order at most)$3$, x has weight $2$, I believe. $\endgroup$
    – Pont
    Mar 6, 2021 at 3:08
  • $\begingroup$ @bellow Thanks for your helpful comment! I have added an alternative approach as you suggested. $\endgroup$
    – Somos
    Mar 6, 2021 at 3:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .