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

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.

• 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, 2020 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, 2020 at 7:54

This file here contains sage solutions for this question:

https://github.com/narodnik/elliptic-curves-washington-solutions/blob/master/11/exercises.md

# 11.1

## a

sage: R.<x, y> = QQ[]
sage: I = ideal(y^2 - x^3 - x, y^4 + 1)
sage: I.variety()
[]
sage: J = ideal(y^2 - x^3 - x, x^2 + 1)
sage: J.variety()
[]


We can also just note that $$(x^3 - x)^4 + 1 = 0$$ is impossible in $$\mathbb{Q}$$.

## b

sage: I = ideal(y^2 - x^3 - x, y^4)
sage: I.variety()
[{y: 0, x: 0}]


## c

$$x^3 - x = a$$ for some $$a$$ has 3 solutions, but there are 4 possible values for $$y$$, so total solutions are 12.

sage: R.<x, y> = QQbar[]
sage: I = ideal(y^2 - x^3 + x, y^4 + 1)
sage: I.variety()
[{y: -0.7071067811865475? - 0.7071067811865475?*I, x: -1.161541399997252? + 0.3411639019140097?*I},
{y: -0.7071067811865475? - 0.7071067811865475?*I, x: 0.?e-19 - 0.6823278038280193?*I},
{y: -0.7071067811865475? - 0.7071067811865475?*I, x: 1.161541399997252? + 0.3411639019140097?*I},
{y: -0.7071067811865475? + 0.7071067811865475?*I, x: -1.161541399997252? - 0.3411639019140097?*I},
{y: -0.7071067811865475? + 0.7071067811865475?*I, x: 0.?e-19 + 0.6823278038280193?*I},
{y: -0.7071067811865475? + 0.7071067811865475?*I, x: 1.161541399997252? - 0.3411639019140097?*I},
{y: 0.7071067811865475? - 0.7071067811865475?*I, x: -1.161541399997252? - 0.3411639019140097?*I},
{y: 0.7071067811865475? - 0.7071067811865475?*I, x: 0.?e-19 + 0.6823278038280193?*I},
{y: 0.7071067811865475? - 0.7071067811865475?*I, x: 1.161541399997252? - 0.3411639019140097?*I},
{y: 0.7071067811865475? + 0.7071067811865475?*I, x: -1.161541399997252? + 0.3411639019140097?*I},
{y: 0.7071067811865475? + 0.7071067811865475?*I, x: 0.?e-19 - 0.6823278038280193?*I},
{y: 0.7071067811865475? + 0.7071067811865475?*I, x: 1.161541399997252? + 0.3411639019140097?*I}]


$$y^2 = x^3 - x = x(x^2 - 1)$$. But $$(x^2 + 1)^3 \implies x^2 = -1 \implies y^2 = 0 \implies y = 0$$. So $$y$$ is fixed, and there are 2 values for $$x = i, -i$$.

sage: I = ideal(y^2 - x^3 + x, (x^2 + 1)^3)
sage: I.variety()
[{y: 1 - 1*I, x: 1*I},
{y: 1 + 1*I, x: -1*I},
{y: -1 - 1*I, x: -1*I},
{y: -1 + 1*I, x: 1*I}]

# Try to find gradient at P = (i, 1 - i) for C and f
sage: x = I
sage: y = 1 - I
sage: y^2 == x^3 - x
True
sage: (3*x^2 - 1)/(2*y)
-I - 1
sage: (x^2 + 1)^3
0
sage: var("X Y")
(X, Y)
sage: diff(((X^2 + 1)^3).expand())(X=x)
0


Lets look at $$(y^4 + 1)$$. There are 4 possible values for $$y$$, and for each $$y$$, there are 3 possible values for $$x$$. Essentially we can look at $$\div{(y^4 + 1)} = \div{(\prod (y - y_i))}$$.

Let $$y = \pm \frac{\sqrt{2}}{2} \pm \frac{\sqrt{2}}{2}$$ and $$x$$ by the solutions for $$y^2 = x^3 - x$$ giving us 12 points. Then $$\div{(y - y_i)} = [P_1] + [P_2] + [P_3] - 3[\infty]$$. Adding them up should give us a divisor with 12 unique points total. By Bezout's these all have multiplicity of 1.

Px Py
-0.00000 + -0.68233i -0.70711 + -0.70711i
-1.16154 + 0.34116i -0.70711 + -0.70711i
1.16154 + 0.34116i -0.70711 + -0.70711i
-0.00000 + 0.68233i -0.70711 + 0.70711i
-1.16154 + -0.34116i -0.70711 + 0.70711i
1.16154 + -0.34116i -0.70711 + 0.70711i
-0.00000 + 0.68233i 0.70711 + -0.70711i
-1.16154 + -0.34116i 0.70711 + -0.70711i
1.16154 + -0.34116i 0.70711 + -0.70711i
-0.00000 + -0.68233i 0.70711 + 0.70711i
-1.16154 + 0.34116i 0.70711 + 0.70711i
1.16154 + 0.34116i 0.70711 + 0.70711i

For the denominator, note that $$(x^2 + 1) = (x - i)(x + i)$$ Giving us 4 values total (since $$y^2 = x^3 - x$$). Each one of these factors is a vertical line that cuts in $$+y, -y$$, giving us the divisor $$\textrm{div}(x^2 + 1) = [(i, 1 - i)] + [(i, -1 + i)] + [(-i, 1 + i)] + [(-i, -1 - i)] - 4[\infty]$$ $$\textrm{div}((x^2 + 1)^3) = 3[(i, 1 - i)] + 3[(i, -1 + i)] + 3[(-i, 1 + i)] + 3[(-i, -1 - i)] - 12[\infty]$$

Lastly for $$g$$, we look at $$y^4 = 0 \implies y = 0$$ and get 2 points $$x = \pm 1, y = 0$$.

$$\div(y^4) = 4 \div(y) = 4[(1, 0)] + 4[(-1, 0)] - 8[\infty]$$