# divisor of function on elliptic curve over algebraic closure of Q

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?

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.
The numerator vanishes exactly when $$x^3-x=\pm i$$, and one can check that this gives six distinct possible values for $$x$$. Since none of these values satisfy $$x^3-x$$, this means that each possible value gives two distinct $$y$$-coordinates where it vanishes, so the divisor of zeros is a sum of twelve points (which I hope you'll excuse me for not writing out explicitly). The denominator can be written as $$(x+i)^3(x-i)^3$$, which vanishes to order 3 at both $$x=i$$ and $$x=-i$$, so the divisor of poles is $$3(i,\sqrt{-2i})+3(i,-\sqrt{-2i})+3(-i,\sqrt{2i})+3(-i,-\sqrt{2i})$$.
• Thanks, this makes sense. A question regarding the order of the zeros of $f$: Since we get 12 distinct points, and $deg(y^4+1)*deg(y^2-(x^3-x)) = 12$, can I argue with Bézout (there should be 12 intersection points with multiplicities) that each point has order 1? Or do I have to argue with the uniformizer for each point? May 14 '20 at 7:23
• I think that should be fine. You may also argue based on our rewriting of $f$: the numerator splits as a product of distinct linear factors $x-\alpha_i$, and no $\alpha_i$ is zero. So for any $y$ so that $(\alpha_i,y)$ is on the curve, $f$ vanishes to order 1 at that point by the link provided. May 14 '20 at 7:54