Pole Order of Weierstrass Coordinates

I'm trying to understand a proof in Silverman's The Arithmetic of Elliptic Curves.

Background: For an elliptic curve $E$, $x, y \in K(E)$ such that $\phi = [x, y, 1]: E \rightarrow C$ is a basepoint preserving isomorphism with a curve $C$ given by a Weierstrass equation are called Weierstrass coordinates. (By 'basepoint preserving' I mean that $\phi(O) = (0 : 1: 0)$, where $O$ is the basepoint of $E$. Put differently $\phi$ is an Isogeny if $C$ itself is viewed as an elliptic curve.)

In Silverman's proof about the relation between two pairs of Weierstrass coordinates (see The Arithmetic of Elliptic Curves, Ch. III Proposition 3.1) it is claimed that all Weierstraß coordinates $x, y$ have poles at the basepoint of orders 2 and 3, respectively.
I know that, since $\phi$ is an isomorphism, I can check this for the rational functions $X/Z$ and $Y/Z$ on $C$ instead. As $C$ is nonsingular, the line $Y = 0$ has 3 distinct intersections with C on the affine plane $Z \neq 0$, so $Y/Z$ has indeed order $-3$ at $(0 : 1 : 0)$.
But how can I proof that $X/Z$ has order $-2$ at the base point?

A similar question was asked here, but without a satisfactory answer (for me).

• Does this help? math.stackexchange.com/a/512825/14699 – Álvaro Lozano-Robledo May 25 '15 at 19:59
• This actually helped, thanks! – Martin Bidlingmaier May 26 '15 at 8:26
• OK, so I'm still struggling with $char K = 2$. In that case, the Weierstrass equation doesn't look as nice, in particular, I don't know whether $Y + a_1 X + a_3 Z$ vanishes at $(e_i : 0 : 1)$. (Is it even true for $char K = 2$ that the cubic on the right has non-zero discriminant?) – Martin Bidlingmaier May 26 '15 at 15:50
• Characteristic 2 is messy as always, and my link does not apply. One would have to deal with it using a model that is smooth. – Álvaro Lozano-Robledo May 28 '15 at 14:21

Let $E$ be an elliptic curve with basepoint $O$, defined over a field $k$. Let $x, y \in k(E)$ be Weierstrass coordinates, i.e. rational functions such that $$\phi = [x, y, 1] : E \rightarrow C$$ is an isomorphism with a curve $C$ given by a Weierstrass equation and $\phi(O) = (0 : 1 : 0)$.
Proof. Instead of calculating the poler orders of $x$ and $y$, we can check the pole orders of $X/Z = {\phi^*}^{-1}(x)$, $Y/Z = {\phi^*}^{-1}(y)$ in $(0 : 1 : 0)$ because $\phi$ is an isomorphism.
Notice that $(0 : 1 : 0)$ is in the affine open set $Y \neq 0$, so we dehomogenize the Weierstrass equation for $Y$ and get \begin{equation} \tag{1} Z + a_1 X Z + a_3 Z^2 = X^3 + a_2 X^2 Z + a_4 X Z^2 + a_6 Z^3. \end{equation} We're interested in the zero orders of $X$ and $Z$ in $(0, 0)$, which is the basepoint $(0 : 1 : 0)$ in affine coordinates. The ideal of functions $M_{(0, 0)} \subset K[X, Z]$ that vanish in $(0, 0)$ is generated by $X$ and $Z$.
Solving (1) for $Z$ yields \begin{equation*} \tag{2} Z = X^3 + a_2 X^2 Z + a_4 X Z^2 + a_6 Z^3 - a_1 X Z + a_3 Z^2, \end{equation*} so the zero order of $Z$ mod Weierstrass equation has to be at least $2$ because every summand on the right hand side is the product of at least 2 elements of $M_{(0, 0)}$. But then even $\operatorname{ord}_{(0, 0)} Z \geq 3$, since all summands have at least order 3.
As $C$ is non-singular, one of $X$ or $Z$ has to be a uniformizing parameter of order 1, so $\operatorname{ord}_{(0, 0)} X = 1$. Thus, the order of $Z$ is exactly 3 for otherwise, solving (2) for $X$ would imply $\operatorname{ord}_{(0, 0)} X^3 \gt 3$.
From the properties of valuations, it is immediate that \begin{equation*} \tag{3} \operatorname{ord}_{(0, 0)} X / Z = -2, \\ \operatorname{ord}_{(0, 0)} 1 / Z = -3, \end{equation*} but these are just the functions $X/Z$, $Y/Z \in K(C)$ in affine coordinates.