# Computing order of point at infinity when computing principal divisor

Let the elliptic curve $y^2=x^3+1 \pmod{13}$ and the rational function $g = \frac{x^2}{y}$. Compute the principal divisor of $g$ on the above curve.

What I've done. First homogenize both (the curve and the function) and obtain $\frac{x^2}{yz} = 0$ and $y^2z-x^3-z^3=0$.

Zeros are obtain solving $x^2 = 0$ and $y^2z-x^3-z^3=0$. This leads to the points $\mathcal{O}(0:1:0)$ with order $n_{\mathcal{O}}$ and points $P(0:1:1)$ and $Q(0:-1:1)$ both of order $2$ (i.e note that $x$ is a uniformizer and $g = x^2 \cdot \frac{1}{yz}$ and $\frac{1}{yz}$ neither has a pole nor a zero in points $P$ and $Q$.)

Poles are obtain solving $yz = 0$ and $y^2z-x^3-z^3=0$. This leads to the points points $\mathcal{O}(0:1:0)$ with order $m_{\mathcal{O}}$ and points $R(-1:0:1)$, $S(4:0:1)$ and $T(-3:0:1)$ all three of them with order $-1$ (i.e. note that $y$ is a uniformizer and $g=y^{-1} \cdot \frac{x^2}{z}$ and $\frac{x^2}{z}$ neither has a pole nor a zero in $R,S$ and $T$).

Hence by definition $$div(g) = n_{\mathcal{O}}(\mathcal{O}) + 2(P) + 2(Q) - m_{\mathcal{O}}(\mathcal{O}) - (R) - (S) - (T)$$

It is well know that $deg(div(g)) = 0$ in this case, so I expect that $n_{\mathcal{O}} = m_{\mathcal{O}} - 1$.

Question: How to compute $n_{\mathcal{O}}$ and $m_{\mathcal{O}}$ using uniformizers ?

Let $$k=\mathbb F_{13}$$. The trick with these kinds of problems is always to work in affine charts. Since we already have the equation in the $$xy$$-chart, notice that the only way the function $$\frac{x^2}{y}$$ (or really it's image in the coordinate ring $$A=k[x,y]/\langle y^2-x^3-1 \rangle$$) vanishes on the $$xy$$ chart on your curve is if $$x=0$$ so that $$y=\pm 1$$ (which are distinct mod $$13$$.)

Let's start and zoom in at the point $$p_1=(0,1)$$, i.e. localize the coordinate ring $$A$$ at the ideal $$m_1=(x-0,y-1)=(x,y-1)$$. The maximal ideal in the local ring $$A_{m_1}$$ is $$(x,y-1)$$ but we twist $$y^2-1-x^3=0$$ into $$y-1=\frac{x^3}{y+1}$$ and since $$y+1$$ does not vanish at our point, we are allowed to invert it in $$A_{m_1}$$. So the ideal $$(x,y-1)$$ in $$A_{m_1}$$ is actually secretly just $$(x)$$ so $$x$$ is a uniformizer which is just fancy speak for $$\text{ord}_{p_1}(x)=1$$.

So $$\text{ord}_{p_1}(\frac{x^2}{y})=\text{ord}_{p_1}(x^2)-\text{ord}_{p_1}(y)=2\text{ord}_{p_1}(x)-0=2$$

Next up, the same song and dance with $$p_2=(0,-1)$$ with the main point being $$y+1=\frac{x^3}{y-1}$$ shows that $$\text{ord}_{p_2}(\frac{x^2}{y})=2$$

Similarly, for poles we focus on when $$y=0$$ so $$x=-1,-3,4$$ so play the same uniformizer game we did above.

Now the only point on the curve missing from our chart must happen when $$z=0$$ so plug in $$z=0$$ into your homogeneous equation to get $$x^3=0$$ so that $$x=0,y=1,z=0$$. So lets work in the $$x,z$$ chart to get $$z-x^3-z^3$$ and our function $$\frac{x^2}{y}$$ becomes $$\frac{x^2}{z}$$ and the only point we haven't dealt with is $$(0,0)$$ (This is important to keep track of, otherwise you will end up double counting points, which is bad). So localize $$k[x,z]/(z-x^3-z^3)$$ at $$(x,z)$$. But since the equation here can be twisted into $$z=\frac{x^3}{(1-z^2)} ,$$ we have $$x$$ is the king, I mean uniformizer. So $$\text{ord}_{(0,0)}(\frac{x^2}{z})=2-3=-1$$ so pole of order $$1$$.

This is really wordy but the point is once you learn this, it's pretty routine.

Now I should stop procrastinating and do my own work.

• This might be a little late to ask this question, but shouldn't we have $\operatorname{ord}_{(0,0)}(\frac{x^2}{z})=2-3=-1$ since $x$ is a uniformizer and $z=ux^3$ for $u=\frac{1}{1-z^2}$ a unit. Or is there something I'm missing Jun 5, 2023 at 20:00
• @Fotis Yup, good catch. Jun 5, 2023 at 20:43
• ah perfect, thank you, just making sure I'm not as lost as I think with divisors! Jun 5, 2023 at 21:02