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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 ?

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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.

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  • $\begingroup$ 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 $\endgroup$
    – Fotis
    Jun 5, 2023 at 20:00
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    $\begingroup$ @Fotis Yup, good catch. $\endgroup$
    – Arkady
    Jun 5, 2023 at 20:43
  • $\begingroup$ ah perfect, thank you, just making sure I'm not as lost as I think with divisors! $\endgroup$
    – Fotis
    Jun 5, 2023 at 21:02

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