Any smooth projective curve $X$ of genus one admits a morphism $f:X\to\mathbb{P}^1$ of degree 2, which by the Riemann-Hurwitz formula is ramified at exactly four points. Given a $4$-tuple $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}\subset\mathbb{P}^1$, I would like to give an explicit such morphism ramified exactly at $\alpha_1,...,\alpha_4$. How can I do this?

For instance, for $\{\alpha_1,\alpha_2,\alpha_3,\infty=[1:0]\}\subset\mathbb{P}^1$ we can consider \begin{equation} X=\{[x:y:z]\in\mathbb{P}^2\mid zy^2=(x-\alpha_1z)(x-\alpha_2z)(x-\alpha_3z)\}, \end{equation} which is the closure in $\mathbb{P}^2$ of the affine curve $\{(x,y)\in\mathbb{C}^2\mid y^2=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)\}$, and the morphism $F:X\to\mathbb{P}^1$ given by \begin{equation} F([x:y:z]) = \left\{\begin{array}{lr} \left[1:0\right] & \text{if } [x:y:z]=[0:1:0]\\ \left[x:z\right] & \text{otherwise} \end{array}\right\}, \end{equation} which looks like $(x,y)\mapsto x$ on the open chart $\mathbb{C}^2$. This construction satisfies all my requirements (in particular, $X$ is smooth).

However, if I take $\{\alpha_1,..,\alpha_4\}\subset\mathbb{P}^1\setminus\{\infty\}$ and I consider the curve \begin{equation} Y=\{[x:y:z]\in\mathbb{P}^2\mid z^2y^2=(x-\alpha_1z)(x-\alpha_2z)(x-\alpha_3z)(x-\alpha_4z)\},\end{equation} then $Y$ has a singular point at $[0:1:0]$, and the corresponding morphism $Y\to\mathbb{P}^1$ is ramified at $\alpha_1,...,\alpha_4$ $\textbf{and}$ $\infty=[1:0]$, so this construction doesn't satisfy my requirements. So what is the correct construction in this case?

  • 2
    $\begingroup$ Move one of the four points on $\mathbb{P}^1$ to $\infty$ using an automorphism of $\mathbb{P}^1$. Reduce to previous case. (In fact you can even set three points to $0,1,\infty$). $\endgroup$ – Jake Levinson Dec 2 '16 at 6:04
  • $\begingroup$ But then what is the branched covering associated to the arbitrary original 4-tuple? How does an automorphism of $\mathbb{P} ^1$ induce an isomorphism of the correspoding elliptic curves? $\endgroup$ – Marco Flores Dec 2 '16 at 6:39
  • $\begingroup$ Mostly, an automorphism of $\Bbb P^1$ will not be liftable to an elliptic curve. After all, the projective line has a $3$-dimensional automorphism group, while a curve of genus $1$ has only a $1$-dimensional automorphism group. And if you think of your genus-one curve as an Abelian variety, it has only finitely many automorphisms. $\endgroup$ – Lubin Dec 2 '16 at 13:48
  • $\begingroup$ It doesn't need to lift to an automorphism of the elliptic curve. The point is that once you've constructed the map $E \to \mathbb{P}^1$, you can compose with an automorphism $\mathbb{P}^1 \to \mathbb{P}^1$. That will move the branch points. $\endgroup$ – Jake Levinson Dec 2 '16 at 18:42
  • $\begingroup$ Actually, after following through the details, it looks like it boils down to an equation like this: $y^2(a_4 z-x) = a_4(x-a_1 z)(x-a_2z)(x-a_3z)$. Note that as $a_4 \to \infty$, the equation specializes to the first one you stated (after dividing out a factor of $a_4$). $\endgroup$ – Jake Levinson Dec 5 '16 at 13:48

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