Prop III.1.5: the invariant differential w associated with WF is s.t. div(w) = 0.

Question

In pp.48, Proposition III.1.5 of Silverman "The Arithmetic of Elliptic Curves," we consider a curve $E$ in Weierstrass form and the differential $$\omega = \frac{dx}{F_y} = -\frac{dy}{F_x}$$ For a point $P = (x_0, y_0)$, we have that the map $\phi : E \to \mathbb{P}^1 : [x:y:1] \mapsto [x:1]$ has degree 2. I understand this vaguely ($\#\phi^{-1}([x:1]) = 2$). But he then concludes from this that $\text{ord}_P(x-x_0) \leq 2$ w.r.t. $E$. Why is this true? I don't understand the connection between $\phi$ and $x-x_0$.

Answer 1

Recall from Example II.2.2 that there is a bijection between $K(E) \cup \{\infty\}$, where $K(E)$ is the function field of $E$, and morphisms $E \to \mathbb{P}^1$. Identifying morphisms $E \to \mathbb{P}^1$ with elements of $K(E) \cup \{\infty\}$ via this bijection, the map $\phi$ you've given is simply $x \in K(E)$.

Let $P = (x_0, y_0) \in E$ and $Q = x_0 \in \mathbb{P}^1$, so $x(P) = Q$. (We're really working in affine patches of both $E$ and $\mathbb{P}^1$.) Letting $t$ be the coordinate on this affine open of $\DeclareMathOperator{\ord}{ord} \mathbb{P}^1$, then $t - x_0$ is a local uniformizer at $Q=x_0$.

By the definition of ramification index, we have $$ e_x(P) = \ord_P(x^*(t - t_0)) = \ord_P(x - x_0) \, . $$ Proposition II.2.6 then implies that $$ e_x(P) \leq \sum_{R \in x^{-1}(Q)} e_{x}(R) = \deg(x) = 2 \, , $$ so $\ord_P(x - x_0) \leq 2$.

You wrote that the fact that $x$ (or $\phi$) has degree $2$ means that $\#\phi^{-1}([x_0:1]) = 2$. But this isn't quite true. It's true generically, either in the precise sense using generic points, or the colloquial sense of "for almost all $x_0$". However, writing $E: y^2 = f(x)$, if we choose $x_0$ such that $f(x_0)=0$, then $\phi^{-1}([x_0 : 1])$ consists only of the point $(x_0, 0)$. Proposition II.2.6 says that if we weight the points of the fiber $\phi^{-1}(x_0)$ by their ramification indices, then this weighted counted does equal the degree of the map. And for $x_0$ with $f(x_0) = 0$, the point $(x_0, 0)$ is a ramification point of $x$ with index $2$.

(I wrote a related answer about this same proposition here; you might also find it helpful.)