2

I have a quartic in $(u,v): v^2=u^4+au^2+b$ and I want to transform it in a cubic form in $(X,Y)$ and then in a cubic Weierstrass form in $(x,y)$.

I know by the Nagell algorithm, that I have to introduce $G(u)=u^2+a/2$, so that I can write: \begin{equation} \displaystyle v^2=u^4+au^2+b=G(u)^2+c \end{equation}

By Nagell, I have to consider $v^2-G(u)^2=c$ and set $X=v+G(u)$.

$$ \big(v-G(u)\big) \overbrace{\big(v+G(u)\big)}^{X}=c \qquad\qquad \text{so that} \qquad\qquad \left\{\begin{array}{cclc} v+G(u) &=& X&(1) \\ v-G(u)&=& c/X & (2) \\ \end{array} \right.$$

Then $(1)-(2)$ gives \begin{equation} \displaystyle 2G(u)=X-c/X \Longleftrightarrow 2u^2=X-c/X-a \end{equation}

Multiplying this equation by $X^2$ gives:

\begin{equation} \displaystyle 2(uX)^2=X^3-aX^2-cX \end{equation}

Set $Y=uX$, we then have: \begin{equation} \displaystyle 2Y^2=X^3-aX^2-cX \end{equation}

Do you know which rational substitution I can use to transform the above cubic in $(X,Y)$ in a cubic Weierstrass form: $y^2=x^3+a'x^2+b'x \quad \text{with} \quad a'=-2a \quad \text{and} \quad b'=a^2-4b$ ?

I thank you in advance for any suggestions.

beginarray
  • 475

1 Answers1

2

Hint Substituting $$X = \lambda X' , \qquad Y = \mu Y' ,$$ gives $$2 \lambda^2 (Y')^2 = \mu^3 (X')^3 + \cdots.$$ To be able to cancel the coefficients of $(Y')^2, (X')^3$ and produce an equation of the form $$(Y')^2 = (X')^3 + \cdots ,$$ we need $$2 \lambda^2 = \mu^3 ,$$ and we can parameterize the solutions of this equation by $$(\lambda, \mu) = (2 \tau^3, 2 \tau^2) .$$

Travis Willse
  • 99,363