I'm currently studying Washington's book about elliptic curves and stumbled upon this exercise:
Let $E: y^2 = x^3-x$ over $ \mathbb{Q}$ elliptic curve.
Let $ f(x,y) = (y^4+1)/(x^2+1)^3$ and find $div(f)$ over the algebraic closure of $\mathbb{Q}$.
Now, $f$ does not have zeros or poles in $E(\mathbb{Q})$, so these coordinates must lie in the algebraic closure.
After setting $y^4+1 = 0$ resp. $(x^2+1)^3 = 0$, I get points with complex coordinates.
What I struggle with is finding the order of these points, so I can compute $div(f) = \sum{n_p[P]}$.
Is this just tedious computation, where I have to find a uniformizer at each point $P$, or this there some trick I'm missing?
Best Answer
We can rewrite $f$ in terms of just $x$ by using the equation of our curve: up to using the equality $y^2=x^3-x$, we get that $f=\frac{(x^3-x)^2+1}{(x^2+1)^3}$. Now things should be pretty straightforwards, since it's easy to tell when this either vanishes or has a pole. There's a full solution under the following spoiler text so you can give it a go yourself using the hint and then check your work.
In general, it is often the case that local rings of points on an elliptic curve have a nice choice of uniformizer. See for instance here.