# About the $div(f)$ given a curve $C$

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 $$ord_{P_\infty}(x-e_i)=2?$$

• 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.
– Ahr
Aug 26, 2019 at 18:17
• Concerning the point at infinity, well you might want to change the affine chart and do the same thing.
– Ahr
Aug 26, 2019 at 18:19
• Isn't it apparent that $\text{ord}_{P_1}(x-e_1)=2\,\text{ord}_{P_1}(y)$? Aug 26, 2019 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, $$(x-e_1) \notin (y)^3$$ since no multiple of $$y^3$$ 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 $$\left.\frac{dz'}{dx'}\right|_{(0,0)} = \left.\frac{3x'^2 + a z'^2}{1-2ax'z'-3bz'^2}\right|_{(0,0)} = \frac{0}{1} = 0.$$ which shows that $$\frac{dz'}{dx'}$$ is finite at $$(0,0)$$. 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 $$x'^3 \cdot \frac{1}{1 - ax'z' - bz'^2} = z'$$, and no larger power of $$x'$$ divides $$z'$$ in the local ring at $$(0,0)$$ (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.
• Nice answer! May I ask you why is it that $x'$ is a uniformizer at $(0,0)$, and why we can ignore $-e_1$ (you may point me to some result as well)? Also, simplifying your last statement: the order of $z'$ is $3$ because the order of $x'^3$ is $3$ and the order of $1-ax'z'-bz'^2$ is $0$ since it doesn't vanish in $(0,0)$. Nov 6, 2021 at 22:22
• I explained in my answer the reasons why $x'$ is a uniformizer at $(0,0)$ and why we can ignore $-e_1$. I am not aware of any references for these results; they're rather simple consequences of the definitions. $x'$ is a uniformizer at $(0,0)$ because $x'$ is $0$ at this point and $dx'/dz'$ is non-zero. In general a non-zero derivative at a root guarantees order of vanishing equal to $1$ (just think about Taylor series). $-e_1$ can be ignored because in general if $f$ has a pole at a point then $f+c$ has the same order pole for any constant $c$ -- constants don't affect the value of infinity.