pole at infinity of differential on elliptic curve

Question

Let $K$ be a field and let $C$ be the elliptic curve given by the equation $y^2=P(x)$ where $P(x)=x^3+ax+b$ (plus the point at infinity)...and $P(x)$ has no repeated roots. We have $2y dy = P'(x) dx$. Many times I have read that the form $$\omega = \frac{dx}{y}=\frac{2 dy}{P'(x)} $$ extends to a regular form at the point at infinity, and that the form $x\omega$ has a (double?) pole at infinity. How do I verify this?

Answer 1

No matter what form my answer may take, it will be in irremediably old-fashioned language. I’ll leave it to you to translate my words into the language that to you is most comprehensible.

First notice that if your curve is given by $F(x,y)=0$, then any polynomial expression in $x$ and $y$ will have no poles in the finite plane: any poles are on the line at infinity.

In particular, $x$ and $y$ have all their poles on the line at infinity; the only point of our curve that’s on this line is $\Bbb O$, namely $(0,1,0)$ in projective coordinates.

Now $x$ and $y$ clearly have, respectively, two and three zeros in the finite plane, so that $x$ must have a double pole at $\Bbb O$, and $y$ a triple.

Choose now any local uniformizing parameter $t$ at $\Bbb O$; from what I’ve said before, $t=x/y$ will do. The important thing is that when expanded locally in terms of $t$, you get $x=at^{-2}+\text{(higher)}$, $y=bt^{-3}+\text{(higher)}$, with $a$ and $b$ both nonzero elements of $K$. Since $dx=\bigl(-2at^{-3}+\text{(higher)}\bigr)dt$, you see that at $\Bbb O$, $dx/y$ has the form $\bigl(-2a/b+\text{(higher)}\bigr)dt$, in other words regular, without zero.

Notice that the argument doesn’t work in characteristic $2$; but an elliptic curve can’t be written in form $y^2=P(x)$ there anyway.