In the Arithmetic of Elliptic Curves we define the divisor associated to a function $f\in K(C)^*$ for a given a curve $C$ as follows: $$div(f)=\sum_{P\in C}ord_P(f)(P)$$

where $ord_P(f)$ is the max $d$ for which $f\in M_p^d$.

I don't get to understand the following example. Given the curve $$C:y^2=(x-e_1)(x-e_2)(x-e_3)$$

in some $K$ with $char(K)\neq 2$, we want to find $div(x-e_i)$ for each $i$. The book states that if we denote $P_i=(e_i,0)\in C$, then $$div(x-e_i)=2(P_i)-2(P_{\infty})$$

but I don't really see why. Taking the definition we have $$div(x-e_i)=\sum_{P\in C}ord_{P}(x-e_i)(P)$$

I understand that for each $P_i$ we have $ord_{P_i}(x-e_i)=1$ (and not $2$), because if it were $2$ then $(x-e_i)\in M_{P_i}^2$ and that would mean we could write $(x-e_i)=f_1f_2$ with $f_i\in M_{P_i}$, if I'm not mistaken, but then that is not possible because $(x-e_i)$ is already degree 1. So I don't see how it could be 2, and that's what I'm trying to understand, why do we have $ord_{P_i}(x-e_i)=2?$ And finally, why is it that $ord_{P_\infty}(x-e_i)=2?$

  • $\begingroup$ This is a local question, so you wanna look at what happens when you localise at $P_i$, say $i=1$ for simplicity, now you know by general theorem that the local ring at this point should be a DVR (because your variety is normal of dimension 1), you have two candidates to be the uniformizer at this point, $y$ and $(x-e_1)$, and one of them has to be the uniformizer. But as $(x-e_0)(x-e_2)$ is invertible, it is clear that $y$ has to be the uniformizer and that the order of $(x-e_1)$ is thus 2. $\endgroup$ – Ahr Aug 26 '19 at 18:17
  • $\begingroup$ Concerning the point at infinity, well you might want to change the affine chart and do the same thing. $\endgroup$ – Ahr Aug 26 '19 at 18:19
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    $\begingroup$ Isn't it apparent that $\text{ord}_{P_1}(x-e_1)=2\,\text{ord}_{P_1}(y)$? $\endgroup$ – Lord Shark the Unknown Aug 26 '19 at 18:25

At $P_1$, the function $y$ is a uniformizer, since $\frac{dx}{dy}$ is finite at $P_1$. Note that $(x-e_1) \in (y)^2$ since $(x-e_1) = y^2 \cdot \frac{1}{(x-e_2)(x-e_3)}$ (here $(x-e_2)(x-e_3)$ is invertible in the local ring at $P_1$); however, $(x-e_1) \notin (y)$ since no multiple of $y$ equals $x-e_1$ (proof left as an exercise for you). Hence $\operatorname{ord}_{P_1} (x-e_1) = 2.$

To get the order at $\infty$, you need to change (projective) coordinates. Assume for now the elliptic curve has short Weierstrass form $y^2 = x^3 + ax + b$ (the more general case is left as an exercise for you). Set as usual $y = Y/Z$ and $x = X/Z$. Then the point at infinity has coordinates $(X:Y:Z) = (0:1:0)$, so we need to normalize at $Y$. Set $x' = X/Y$ and $z' = Z/Y$. Then the new curve equation under this change of coordinates is $z' = x'^3 + a x' z'^2 + b z'^3$. The point at $\infty$ has coordinates $(x',z') = (0,0)$. Note that $x'$ is a uniformizer at $(0,0)$ since $$ \left.\frac{dz'}{dx'}\right|_{(0,0)} = \left.\frac{3x'^2 + a z'^2}{1-2ax'z'-3bz'^2}\right|_{(0,0)} = \frac{0}{1} = 0. $$ We also have $$ \operatorname{ord}_{\infty} (x-e_1) = \operatorname{ord}_{(0,0)} \left(\frac{x'}{z'} - e_1\right). $$ We can ignore the $-e_1$ term, since $x'/z' \to \infty$ as $(x',z') \to (0,0)$. So we need to calculate $$ \operatorname{ord}_{(0,0)} x' - \operatorname{ord}_{(0,0)} z'. $$ Obviously $\operatorname{ord}_{(0,0)} x' = 1$ since $x'$ is a uniformizer. As for $\operatorname{ord}_{(0,0)} z'$, observe that $x'^3 = z' (1 - ax'z' - bz'^2)$, and no smaller power of $x'$ is a multiple of $z'$ (proof left as an exercise for you), so that $\operatorname{ord}_{(0,0)} z' = 3$. The difference of these two terms is $-2$, as desired.


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