# coefficients of $Y^2Z+a_1XYZ+a_3YZ^2=X^3+a_2X^2Z+a_4XZ²+a_6Z³$ over $K$

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.

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.

• 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.
– Pont
Mar 6, 2021 at 3:08
• @bellow Thanks for your helpful comment! I have added an alternative approach as you suggested. Mar 6, 2021 at 3:46