morphisms between algebraic curves

Question

I have a cubic $(E)$: $y^2=ax^3+bx^2+cx+d$ over rational field.

Setting: $d=0, \: \: a'=-2a,\:\: b'=a^2-4b$, we find that $(E)$ is isomorphic to her "little sister" $(E'): \: y^2=x(x^2+a'x+b')$.

For example, with $a=8, \: \: b=1$: \begin{equation} \displaystyle (E): y^2=x^3+8x+1\:\: \text{and}\:\: (E'): y^2=x(x^2-16x+60)=x(x-6)(x-10)\: \: \text{are isomorphic}. \end{equation}

Let's introduce now a quartic curve $H_{(1,a,b)}:\: v^2=u^4+au^2+b$ in $(u,v)$ rational coefficients.

$(E')$ and $H_{(1,a,b)}$ are also isomorphic: set $u=y/(2x)$, $v=(x^2-b')/(4x)$...

We can establish the homogeneous version $E'_p:\:y^2z=x^3+a'x^2z+b'xz^2 \subset \mathbb{P}^2(1,1,1)_{(x:y:z)}$ and $H_p: \:v^2=u^4+au^2w^2+bw^4 \subset \mathbb{P}^2(1:2:1)_{(u:v:w)}$ of $E'$ and $H_{(1,a,b)}$ respectively.

Note: weight $2$ is allocated to the $v$ coefficient of $H_p$ and weight $1$ to the $u,w$ coefficients. With some little help, I have established a "solidified" isomorphism between the two projective versions $E'_p$ and $H'_p$:

\begin{equation} \displaystyle \varphi:\:\:E'_p \longrightarrow H_p\\ (x:y:z)_{E'_p} \longmapsto (y:y^2-a'x^2-2b'xz:2x)_{H_p} \end{equation}

Let us now consider $p_{\infty}=(0:1:0)_{E'_p}$, the neutral of $(E'_p)$ and $p_0=(0:0:1)_{E'_p}$ a $2$-torsion point of $(E'_p)$.

I have been told that there are particular problems with $\varphi\big(p_{\infty}\big)=\varphi \big((0:1:0)_{E'_p}\big)=(1:1:0)_{H_p}$ and $\varphi(p_0)=(0:0:0)_{H_p}$. I've also been told that those problems are classical when we study morphisms between algebraic curves but I don't see what does it consist of except for the point $(0:0:0)_{H_p}$ wich is FORBIDDEN in the projective plane... But what about $\varphi(p_{\infty})$ ?

A little help would be greatly appreciated. I thank you in advance for any suggestions.

Answer 1

$$U=\{ (x:y:1), y^2=x^3+a'x^2+b'x\}$$ is an affine subset of $E_p'$ which gives the rational map $\varphi:U \to H$ $$\varphi(x:y:1) = (\frac{y}{2x}:\frac{x^2-b'}{4x}:1)=(y:x^3-b'x:2x)=(1:\frac{x^3-b'x}{y^2}:\frac{2x}{y})$$

$x,y$ are well-defined polynomial functions $U\to \overline{\Bbb{Q}}$ (in contrary to the homogeneous coordinates).

• At $(0:0:1)$ they both have a zero, for $b' \ne 0$ then $y^2=(x^2+a'x+b')x$ means $y^2/x = x^2+a'x+b'$ is regular and non-zero at $(0:0:1)$ so $y$ has a simple zero and $x$ has a double zero.

  Thus $$\frac{x^3-b'x}{y^2}=\frac{x^2-b'}{x^2+a'x+b'},\qquad\frac{2x}{y}=\frac{2y}{x^2+a'x+b'}$$ both are regular rational functions on some open containing $(0:0:1)$.

Regular means you can plug directly the value for $x,y$ in the quotient of polynomials, the denominator doesn't vanish so you get a well-defined value. Equivalently it means the rational function is in some localization of the coordinate ring, localization at a few points away from those of interest.

Whence around $(0:0:1)$ the rational map is $$\varphi(x:y:1)=(1:\frac{x^2-b'}{x^2+a'x+b'}:\frac{2y}{x^2+a'x+b'}),\qquad \varphi(0:0:1)=(1:-1:0)$$

It is given by some regular functions on $V=U-\{(e_1:0:1),(e_2:0:1)\}$ so it is a morphism $V \to \varphi(V)\subset H$, then the question is what is its image, if it is injective, what is its inverse.