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I'm studying the proof of Proposition 1.5 on silverman "The Arithmetic of Elliptic Curves" :

Proposition: Let $E$ be an elliptic curve. Then the invariant differential $\omega$ associated to a Weierstrass equation for $E$ is holomorphic and nonvanishing, i.e., $\mathrm{div}(\omega)=0.$

Let $P=(x_0,y_0)\in E$ and $$E:F(x, y)=y^2+a_1xy+a_3y−x^3−a_2x^2−a_4x−a_6=0,$$ so $$\omega=\frac{d(x-x_0)}{F_y(x,y)}=-\frac{d(y-y_0)}{F_x(x,y)}$$ Thus $P$ cannot be a pole of $ω,$ since otherwise $F_y(P)=F_x(P)=0,$ which would say that $P$ is a singular point of $E.$

I do not understand why the map $$E\longrightarrow \mathbb{P}^1\;\;\;[x,y,1]\longrightarrow [x,1]$$ is of degree $2$, and why this implies that $\mathrm{ord}_P(x−x_0)\leq2,$ and we have equality $\mathrm{ord}_P(x−x_0)=2$ if and only if the quadratic polynomial $F(x_0,y)$ has a double root ?

Thanks.

Med
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1 Answers1

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The map $x: E \to \mathbb P^1$ is of degree $2$ because for a generic choice of $x$, there are two values of $y$ for which $[x,y,1] \in E$, i.e. there are two values of $y$ with $F(x,y)=0$. Indeed, $F$ is a quadratic polynomial in $y$.

The map will be ramified above $x_0 \in \mathbb A^1$ if there is only one point in the fibre above $x_0$, which is to say only one value of $y_0$ with $F(x_0, y_0)=0$. This means precisely that the polynomial $F(x_0, y)$ has a double root $y_0$, since it's a quadratic polynomial with a single root.

As for the claim about $\text{ord}_P(x-x_0)$, it follows from the fact that, for a smooth curve and a rational function $f$, and a point $P$ which is not a pole of $f$, $\text{ord}_P(f-f(P))$ equals the ramification degree of $f$ at $P$, viewed as a rational map to $\mathbb P^1$.

Bruno Joyal
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