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In order to find a short Weierstrass model for the curve over $\Bbb Q$:

$$C:y^2 = x^4+4x^3-4 \tag 1$$

I used this answer to "Birational Equivalence of Diophantine Equations and Elliptic Curves". It yields the Elliptic curve

$$E:y^2=x^3 + 1296x -46656\tag 2$$

over $\Bbb Q$ and two birational transformations $f:C\to E$ and $g:E\to C$:

$$f: \binom xy\mapsto\binom {18(x^2 + 2x - y)}{108(x^3 + 3x^2 - xy - y)}\tag 3$$

$$g:\binom xy\mapsto\begin{pmatrix} \dfrac{-6x + y}{6(x - 36)} \\ \dfrac{-2x^3 + 108x^2 + y^2 - 432y}{36(x - 36)^2} \end{pmatrix} \tag 4$$

I verified that $g\circ f= f\circ g = id$.

The trouble I am having is this: $f$ maps the rational point $(1,1)_C$ to $$P=(36,216)_E$$ but $g$ fails to map $P$ back to $C$ because the denominators of (4) are zero for $x=36$.

[The answer allowed me to handle the case $(36,216)_E$, but the case $(36,-216)_E$ is still not solved because denominators in (4) are zero but numerators are non-zero.]

How can this be? $P$ is a regular point of $E$, so $g(P)$ should be the regular point $(1,1)_C$. Multiplying through by the denominators of (4) and trying to go projective didn't work either, because the result was $(0:0:0)$ as $g_y(P)=0/0$.

Addendum 1: I got the mapping $g:E\to C$ from this answer as linked above. The $x$-component of a point in $C$ is

$$x=\frac{2G-3bH+9(bc-6d)}{12H-9(3b^2-8c)} \tag 5$$

where the quartic is $C: y^2=x^4+bx^3+cx^2+dx+e$ and $(H,G)\in E$. There's a unique $H'$ that renders the denominator of (5) to zero: $H'=9(3b^2-8c)/12$, and there's a unique $G'$ that renders the numerator to zero: $G'=(3bH'-9(bc-6d))/2$.

Now if that specific $(H',G')\in E$ (which is the case for the curves $E$ and $C$ in question), then also $(H',-G')\in E$. But $(H',-G')$ will give a denominator =0 and a numerator ≠0 in (5), so how to treat that case?

emacs drives me nuts
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1 Answers1

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Let's scale $E$ to get smaller numbers in the maps: $$f(x,y)=\left(\frac{x^2+2x-y}2,\frac{x^3+3x^2-xy-y}2\right)$$ $$g(x,y)=\left(\frac{y-x}{x-1},\frac{-2x^3+3x^2+y^2-2y}{(x-1)^2}\right)$$ $C$ remains as before and the new $E$ is $y^2=x^3+x-1$; $(1,1)_C\mapsto(1,1)_E$.

Notice that when $(1,1)$ is input to $g$ the numerator is also zero. This means taking the limit along the tangent to $E$ at $(1,1)$ will produce the correct image in $C$; in this case the tangent is $y=2x-1$ and $$\lim_{y\to2x-1,x\to1}\frac{y-x}{x-1}=\lim_{y\to2x-1,x\to1}\frac{-2x^3+3x^2+y^2-2y}{(x-1)^2}=1$$

Parcly Taxel
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  • Thank you! That is annoying because special casing makes symbolic computation messy. What I did is computing the tangent at E and use that equation to eliminate $x$ (or $y$) in the birational. This is a function in only 1 (or none) variable $y$ (or $x$) and Sage does the computation so that there's no need to take explicit limits. For example substituting $y=2x-1$, your 1st limit becomes $\frac{x-1}{x-1}=1$ where Sage performs the cancellation for me. – emacs drives me nuts Aug 07 '22 at 09:07
  • ...what I am wondering is whether there is a representation of $g$ such that the zero can be shortened out? The representation of $g$ is only determined modulo $y^2-x^3-ax-b$, so maybe there's a representation with a common factor? The current $g$ has no common factor and still may yield $0/0$, which is annoying. – emacs drives me nuts Aug 07 '22 at 09:11
  • I am still having trouble... It works for $(36,216)_E$, but when I am using the same method on $(36,-216)_E$ then I am getting $$g_x = \frac{-18x + 216}{6x - 216}$$ so the denominator is 0 for $x=36$ but the numerator is not. How to handle that case? – emacs drives me nuts Aug 07 '22 at 09:43
  • This relates to $(1,-1)_E$ on your curve, where $g_x$ becomes $g_x = -2/0$ so it cannot work for your curve, either? – emacs drives me nuts Aug 07 '22 at 09:56