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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?

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2 Answers 2

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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.

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  • $\begingroup$ 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, 2020 at 7:23
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    $\begingroup$ 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, 2020 at 7:54
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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 $x = \pm 1, y = 0$.

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

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