Can't understand the multiplicity of poles in divisors

Question

I am studying the divisors in relation to elliptic curve cryptography.

Assume that elliptic curve is defined by $x^3 = y^2 + b$ and let's assume our rational function is $f(x,y) = x - P_x$. Let's fix some point $P = (P_x, P_y)$

Clearly, I understand that $f(x,y)$ is 0 at $P$ and $-P$, but I've read the following:

Poles is $2 * [O]$. The function goes to infinity as x goes to infinity, so we say the function is equal to infinity at O. There’s a technical reason why this infinity needs to be counted twice, so O gets added with a “multiplicity” of -2 (negative because it’s an infinity and not a zero, two because of this double counting). The technical reason is roughly this: because the equation of the curve is x³ = y² + b, y goes to infinity “1.5 times faster” than x does in order for y² to keep up with x³; hence, if a linear function includes only x then it is represented as an infinity of multiplicity 2, but if it includes y then it is represented as an infinity of multiplicity 3.

I still can't understand why multiplicity of pole is 2 whereas multiplicity of another rational function $ax + by + c = 0$ would be 3. I appreciate the explanations where not so much hard words to understand is mentioned. I am not so good at math, so is there an easy explanation ? I understand that in $x^3 = y^2 + b$, $y$ goes to infinity 1.5 times faster, but still.

Answer 1

The paragraph you've quoted doesn't provide an actual explanation for why $x$ has a pole of order $2$, so it's no wonder you still don't understand it. Here's what it does explain. Let $a$ and $b$ be the orders of pole of $x$ and $y$ at infinity, respectively. This means that, locally near the point at infinity, $x$ and $y$ have Laurent expansions that look like $$ x = \frac{c_{-a}}{t^a} + \frac{c_{-a+1}}{t^{a-1}} + \cdots \qquad \text{and} \qquad y = \frac{d_{-b}}{t^b} + \frac{d_{-b+1}}{t^{b-1}} + \cdots \, . $$

We have $x^3 - y^2 = b$, and since $b$ is a constant, it has no pole at infinity, i.e., a pole of order $0$. This means that all the terms in the Laurent expansions of $x^3$ and $y^2$ with negative exponents must cancel. In particular, their leading terms, which is $\frac{c_{-a}^3}{t^{3a}}$ for $x^3$ and $\frac{d_{-b}^2}{t^{2b}}$ for $y^2$ must cancel, so we have $c_{-a}^3 = -d_{-b}^2$ and $3a = 2b$.

The author then makes a leap, and apparently assumes that $a = 2$ (and thus $b = 3$). But that doesn't follow from what we've done so far---we could just as well have $a = 4$ and $b = 6$.

Here's a different way to see the order of pole of $x$ at infinity. Letting $E$ be the elliptic curve $y^2 = x^3 - b$, the map

\begin{align*} x: E &\to \mathbb{P}^1\\ (x,y) &\mapsto x \end{align*} is a degree $2$ map, i.e., it is generically $2$-to-$1$. This is because, for a given $x$-value $x_0$, there are $2$ solutions to the equation $y^2 = x_0^3 - b$, counted with multiplicity, and thus $2$ points in the fiber above $x_0$. This continues to hold true even for the point $\infty = [1:0] \in \mathbb{P}^1$. Since no point $(x,y)$ in the affine part of the elliptic curve maps to $\infty$ under $x$, then we must have that the point $\newcommand{\O}{\mathcal{O}} \O = [0:1:0]$ at infinity maps to $\infty$ with multiplicity $2$.

Thus $x$ has a pole of order $2$ at $\O$, and by the argument in the second paragraph, then $y$ must have a pole of order $3$ at $\O$.

Alternatively, one can compute in an affine patch near $\O$ at and explicitly see that $x$ has a pole of order $2$. See the second part of my answer here for this approach.