Finding the poles (and their order) of a function on an elliptic curve

Question

Let $f(x, y) = y-\lambda x - \nu$ and $E$ be the curve $y^2=x^3 + Ax + B$. I am trying to understand how to show directly that $f$ has a pole of order 3 at $\mathcal{O}$ (i.e., without relying on the fact that the degree of the divisor of $f$ is 0). The "proof" from Example 3.0.3 of "Pairings for Beginners" says to homogenize the equations, and then use the facts that $$\left( \frac{\lambda X+\nu Z}{Z}\right)^2 = (X/Z)^3 + A (X/Z) + B$$ and that the right-hand side has a pole of order 3 when $Z=0$.

I fail to see where the displayed equation above comes from. We want to show that the (homogenized) function $F(X,Y,Z) = Y/Z - \lambda (X/Z) - \nu$ evaluates to $\infty$ at the point $[0:1:0]$, yet it seems that the above equation is instead derived by taking $F(X,Y,Z)=0$ and then plugging into the equation for $E$.

Note: I have read a few other answers on stackexchange (as well as textbook examples) showing how to compute the order of the pole at $\mathcal{O}$ for the special cases $f(x,y)=x$ and $f(x,y)=y$; I have not found a source that describes (algorithmically) how to do it in the general case.

Answer 1

Here's the quick and dirty way to see that $f = y-\lambda x - \nu$ has a pole of order $3$ at infinity. Let $v = v_\infty$ be the order of vanishing at $\infty$. Then $v(1) = 0$, and by the special cases you mentioned, $$ v(x) = -2 \, , \qquad v(y) = -3 \, . $$ By properties of discrete valuations (see here), then \begin{align*} v(f) &= v(y-\lambda x - \nu) = \min\{v(y), v(-\lambda x), v(\nu)\} = \min\{-3,-2,0\} = -3 \end{align*} for $\lambda$ and $\nu$ nonzero. Thus $f$ has a pole of order $3$ at $\infty$.

All right, and now the longer method with the change of coordinates. Let $X,Y,Z$ be the homogeneous coordinates of $\mathbb{P}^2$, so $x = X/Z$ and $y = Y/Z$. In terms of these homogeneous coordinates, the equation for $E$ is $Y^2 Z = X^3 + A X Z^2 + B Z^3$. Note that $Y \neq 0$ near the point $\infty = [0:1:0]$, so we can take $w = X/Y$ and $z = Z/Y$ as affine coordinates in a neighborhood of $\infty$. Dividing both sides of the homogeneous equation by $Y^3$, we find that in this neighborhood $E$ has the affine equation \begin{align*} z = \frac{Z}{Y} = \frac{Y^2 Z}{Y^3} = \frac{X^3 + A X Z^2 + B Z^3}{Y^3} = w^3 + A w z^2 + B z^3 \, . \end{align*} Observe that in these new local coordinates, $\infty$ is the point $(w,z) = (0,0)$. Writing $f$ in terms of these coordinates, we have \begin{align*} f &= y - \lambda x - \nu = \frac{Y}{Z} - \lambda \frac{X}{Z} - \nu = \frac{Y}{Z} - \lambda \frac{X/Y}{Z/Y} - \nu = \frac{1}{z} - \lambda \frac{w}{z} - \nu \, . \end{align*}

Either by the calculations in this post, or by noting that \begin{align*} z = Z/Y = 1/y \qquad w = X/Y = \frac{X/Z}{Y/Z} = x/y \, , \end{align*} then $v(z) = 3$ and $v(w) = -2 - (-3) = 1$. Writing \begin{align*} f &= \frac{1}{z} - \lambda \frac{w}{z} - \nu = \frac{1}{z}\left(1 - \lambda w - \nu z \right) \, , \end{align*} we see that \begin{align*} v(f) &= v\left(\frac{1}{z}\left(1 - \lambda w - \nu z \right)\right) = v(1/z) + v\left(1 - \lambda w - \nu z \right) = -3 + 0 = -3 \, , \end{align*} since the other factor $1 - \lambda w - \nu z$ doesn't vanish at $(w,z) = (0,0)$.