Doubts in Silverman's AEC Chapter 3 Prop 1.5

Question

On Page $48$ of Silverman's Arithmetic of Elliptic Curves, he proves the theorem that the invariant differential associated to a Weierstrass equation for an elliptic curve is holomorphic and nonvanishing. I will write my understanding of the proof and point out a few doubts $\color{red}{\textrm{in red}}$ along the way. Please note that I am assuming results from Chapters $1$ and $2$.

Let $E : F(x,y) = y^2 + a_1xy + a_3y - x^3 - a_2x^2 - a_4 x -a_6 = 0$. Then the invariant differential is $\omega = \frac{dx}{F_y(x,y)} = - \frac{dy}{F_x(x,y)}$.

I need to show that $ord_P(\omega)$ = 0 when $P = (x_0,y_0) \in E$ or $P = O$, the point at infinity.

$\color{blue}{\textbf{Case 1 : } P = (x_0,y_0) \in E}$

We write $\omega = \frac{d(x-x_0)}{F_y(x,y)} = - \frac{d(y-y_0)}{F_x(x,y)}$. Let $t$ be a uniformizer at $P$. Then $ord_P(\omega) = ord_P(\frac{\omega}{dt}) = ord_P(\frac{d(x-x_0)}{dt})-ord_P(F_y(x,y))$. By Ch. $2$ Prop. $4.3 (b)$ , $ord_P(\frac{d(x-x_0)}{dt}) \geq 0$. So, if $ord_P(\omega) < 0$, then $ord_P(F_y(x,y))>0 \implies F_y(P)=0$. By similar argument, if $ord_P(\omega) < 0$, then $F_x(P)=0$. Thus, due to smoothness of $E$ at $P$, we must have $ord_P(\omega) \geq 0$.

Now consider the map $\phi : E \to \mathbb{P}^1$ given by $\phi([x,y,1]) = [x,1]$. Then, $\deg(\phi)=2$ because for a generic choice of $x$, there are two values of $y$ for which $F(x,y)=0$. $\color{red}{\textrm{How is this equivalent to the definition :} \deg(\phi)=[K(E) : \phi^*K(\mathbb{P}^1)]\textrm{ where } \phi^*(f) = f \circ \phi ?} $

Now, by Ch. $2$ Prop. $2.6$, for all $Q \in \mathbb{P}^1$, we have $\sum_{P\in \phi^{-1}(Q)}e_\phi(P) = \deg(\phi)$. Specifically for $Q = [x_0,1]$, say $P_1 = [x_0,y_1,1] \in E$ and $P_2 = [x_0,y_2,1] \in E$, not necessarily distinct. Since $e_\phi(P_i)= ord_{P_i}(\phi^*t_{\phi(P_i)}) = ord_{P_i}(x-x_0)$, we note that for any point $P \in E$, $ord_P(x-x_0)$ is either $1$ or $2$ and it is $2$ if and only if $F(x_0,y)$ has a double root i.e. $F_y(P) = 0$.

In any case, by Ch. $2$ Prop. $4.3(d)$, $ord_P(\omega) = ord_P(x-x_0)-ord_P(F_y)-1 = 0$.

$\color{blue}{\textbf{Case 2 : } P = O, \textbf{the point at infinity}}$

Silverman begins by stating that $ord_O(x)=-2$ and $ord_O(y)=-3$. $\color{red}{\textrm{How does this follow from the definition : }}$

$\color{red}{ord_P(f/g)=\sup\{d : f \in M_P^d\} - \sup\{d : g \in M_P^d\} \textrm{ where } M_P \textrm{ is the maximal ideal of } K[E]_P ?}$

Let $t$ be a uniformizer at $O$. Then there exist functions $f$ and $g$ not having a pole or a zero at $O$ such that $x=t^{-2}f, y = t^{-3}g$. Write $f'= \frac{df}{dt}$. This is regular at $P$ and hence so is $\omega$ by simple calculations. Thus $ord_O(\omega) = 0$.

Any help would be highly appreciated. Hints are also welcome.

Answer 1

1. I think you've more or less figured out the answer to this question: the key is the formula \begin{align} \label{2.6} \deg(\phi) = \sum_{P \in \phi^{-1}(Q)} e_{\phi}(P) \tag{$\ast$} \end{align} from Silverman Proposition II.2.6 that you mentioned. For a generic choice of $\newcommand{\A}{\mathbb{A}} x_0 \in \A^1$, there are two distinct $y$-values $y_1, y_2$ such that $F(x_0, y_1) = 0$ and $F(x_0, y_2) = 0$. Let $P_1 = (x_0, y_1)$ and $P_2 = (x_0, y_2)$. Since $F(x_0,y)$ has two distinct roots, then $F_y(x_0,y) \neq 0$, so $e_{\phi}(P_i) = 1$ for $i = 1,2$. Taking $Q = x_0$, then (\ref{2.6}) gives $$ \deg(x) = e_{x}(P_1) + e_{x}(P_2) = 1 + 1 = 2 \, . $$

2. Here are two ways to see that $\newcommand{\ord}{\operatorname{ord}} \ord_O(x) = -2$ and $\ord_O(y) = -3$, one using the proposition above again, and the other working explicitly with affine charts.

(i) Let $\newcommand{\P}{\mathbb{P}} Q = \infty \in \P^1$. Since $x$ and $y$ have no poles on the affine part of $E$, then the only place they could have poles is at $O$. Thus $x^{-1}(\infty) = \{O\}$ and $y^{-1}(\infty) = \{O\}$ (or they could be empty). Then (\ref{2.6}) gives \begin{align*} 2 = \deg(x) = \sum_{P \in x^{-1}(Q)} e_{x}(P) = e_x(O) \, . \end{align*} Recall that $e_{\phi}(P) = \ord_P(\phi^*u)$ where $u$ is a local uniformizer at $\phi(P)$. Letting $t$ be the usual affine coordinate on $\P^1$ (at $0 = [0:1]$), then $1/t$ is a uniformizer at $\infty$. Then $$ e_x(O) = \ord_O(x^*(1/t)) = \ord_O(1/x) = -\ord_O(x) $$ so $\ord_O(x) = -2$, as desired. A similar computation shows $\ord_O(y) = -3$.

(ii) The equation \begin{align} y^2 + a_1 x y + a_3 y - x^3 - a_2 x^2 - a_4 x - a_6 = 0 \end{align} for $E$ is really just an equation on the affine open of $\P^2$, with projective coordinates $X,Y,Z$, where $Z \neq 0$. The homogenized equation for $E$ in $\P^2$ is $$ Y^2 Z + a_1 X Y Z + a_3 Y Z^2 - X^3 - a_2 X^2 Z - a_4 X Z^2 - a_6 Z^3 = 0 $$

In terms of these projective coordinates, we have $x = X/Z$ and $y = Y/Z$. The point $O = [0:1:0]$ doesn't lie in this affine open, but does lie in the patch where $Y \neq 0$. Letting $w=X/Y$ and $z=Z/Y$ be affine coordinates on this affine open, then $E$ is given by the local equation \begin{align} z + a_1 w z + a_3 z^2 - w^3 - a_2 w^2 z - a_4 w z^2 - a_6 z^3 = 0 \, . \end{align} or solving for $z$, \begin{align} \label{wz} z = -a_1 w z - a_3 z^2 + w^3 + a_2 w^2 z + a_4 w z^2 + a_6 z^3 \tag{$\dagger$} \, . \end{align} Note that $O = [0:1:0]$ corresponds to the point $(w,z) = (0,0)$ on this affine open. The local ring at $O$ has maximal ideal $M = (w,z)$. From (\ref{wz}), we see that $z \in M^2$. Thus $w$ must be a local uniformizer at $O$, so $\ord_O(w) = 1$. Collecting the terms of (\ref{wz}) with a factor of $z$, we have \begin{align*} w^3 = z + a_1 w z + a_3 z^2 - a_2 w^2 z - a_4 w z^2 - a_6 z^3 = z(1 + a_1 w + a_3 z - a_2 w^2 - a_4 w z - a_6 z^2) \, . \end{align*} This second factor doesn't vanish at $(0,0)$, hence $$ \ord_O(z) = \ord_O(w^3) = 3 \, . $$

Note that on this affine open we can write $x$ and $y$ as \begin{align*} x = \frac{X}{Z} = \frac{X/Y}{Z/Y} = \frac{w}{z} \qquad \text{and} \qquad y = Y/Z = \frac{1}{Z/Y} = \frac{1}{z} \, . \end{align*} Then \begin{align*} \ord_O(x) &= \ord_O(w/z) = \ord_O(w) - \ord_O(z) = 1 - 3 = -2\\ \ord_O(y) &= \ord_O(1/z) = -\ord_O(z) = -3 \, , \end{align*} as desired.