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 $3(i,\sqrt{-2i})+3(i,-\sqrt{-2i})+3(-i,\sqrt{2i})+3(-i,-\sqrt{2i})$.

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.

  • $\begingroup$ 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? $\endgroup$
    – Tylwyth
    May 14 '20 at 7:23
  • 1
    $\begingroup$ 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. $\endgroup$
    – KReiser
    May 14 '20 at 7:54

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