I am currently reading Silverman's arithmetic of elliptic curves. In chapter II, reviewing divisor, there is an explicit calculation: Given $y^2 = (x-e_1)(x-e_2)(x-e_3)$

let $P_i = (e_i,0),$ and $ P_\infty$ be the point at infinity, with coordinate (0,1,0)

We have $div(y) = (P_1)+(P_2)+(P_3) - 3 (P_\infty)$

I am able to calculate div(y) at $P_i$ following his examples, but I don't know how to calculate the coefficient 3 at $P_\infty$, can anyone help me please?


2 Answers 2


Instead of concentrating on calculating the divisor of $y$, calculate first the divisor of $x-e_i$, for $i=1,2,3$.

First homogenize to $Y^2Z=(X-e_1Z)(X-e_2Z)(X-e_3Z)$. Let us calculate the divisor of $x-e_1$ (the divisors of $x-e_2$ and $x-e_3$ are calculated in the same manner). The original function $x-e_1$ is the function $(X-e_1Z)/Z$ in projective coordinates. But, from the equation of the curve we see that $$\frac{X-e_1Z}{Z} = \frac{Y^2}{(X-e_2Z)(X-e_3Z)}.$$ Thus, it is clear now that the function $(X-e_1Z)/Z$ has a double pole at $[X,Y,Z]=[0,1,0]$ and a double zero at $[e_1,0,1]$ because $e_1\neq e_2$, and $e_1\neq e_3$, as the curve is non-singular. Thus, $$\operatorname{div}((X-e_iZ)/Z) = \operatorname{div}(x-e_i) = 2P_i - 2\infty, $$ where $P_i = [e_i,0,1]$ and $\infty=[0,1,0]$. Hence, $$ \operatorname{div}(y^2)=\operatorname{div}(x-e_1) +\operatorname{div}(x-e_2) +\operatorname{div}(x-e_3)=2P_1+2P_2+2P_3-6\infty$$ and $$ \operatorname{div}(y)=P_1+P_2+P_3-3\infty.$$

Edit to add: Let me work this out in more detail. Let $K$ be a field, let $\overline{K}$ be a fixed algebraic closure, let $E/K$ be an elliptic curve given by $$ZY^2=X^3+AXZ^2+BZ^3=(X-e_1Z)(X-e_2Z)(X-e_3Z),$$ with $e_i\in K$ being distinct (notice the short Weierstrass equation implies $e_1+e_2+e_3=0$). Let $\overline{K}[E]$ be the function field of $E$. For $P\in E(\overline{K})$, we denote the ideal of functions in $\overline{K}[E]$ that vanish on $P$ by $M_P$ (strictly speaking we should define $M_P$ to be the maximal ideal in the localization $\overline{K}[E]_P$). For $f\in M_P$, we define $$\operatorname{ord}_P(f)=\sup\{d\geq 1: f\in M_P^d\}.$$ Some facts:

  • Let $P=P_1=[e_1,0,1]$. Then, $X-e_1Z$ and $Y$ belong to $M_{P}\subseteq \overline{K}[E]$, and moreover, they generate $M_{P}$ (because $\langle X-e_1Z,Y\rangle$ is maximal in $\overline{K}[E]$). But, in fact, $$(e_1-e_2)(e_1-e_3)(X-e_1Z)Z^2=ZY^2-(X-e_1Z)^3-(2e_1-e_2-e_3)(X-e_1Z)^2Z.$$ Since $Z$ does not vanish at $P=P_1=[e_1,0,1]$, and $(e_1-e_2)(e_1-e_3)$ is a non-zero constant, it follows that $X-e_1Z$ belongs to $M_P^2$. So $M_P/M_P^2$ is generated solely by $Y$ and, in particular, $\operatorname{ord}_P(Y)=1$ (note that if $Y$ was also in $M_P^2$, then $M_P/M_P^2$ would vanish, but this is impossible because $E$ is smooth and $\dim_{\overline{K}}(M_P/M_P^2)=\dim E=1$). The equation above also says that $$\operatorname{ord}_P(X-e_1Z)=\min\{2\operatorname{ord}_P(Y),2\operatorname{ord}_P(X-e_1Z),3\operatorname{ord}_P(X-e_1Z)\},$$ and since $\operatorname{ord}_P(X-e_1Z)>0$, the only non-contradictory statement is that $\operatorname{ord}_P(X-e_1Z)=2\operatorname{ord}_P(Y)=2$.

  • Similarly, one deduces that at $P_i$, for any $i=1,2$, or $3$, the function $X-e_iZ$ vanishes to order $2$, and $Y$ vanishes to order $1$.

  • Now let $P=\infty=[0,1,0]$. The ideal $M_\infty$ is generated by $X$ and $Z$. It follows that $X-e_iZ\in M_\infty$ for any $i=1,2,3$. But $$ZY^2 = (X-e_1Z)(X-e_2Z)(X-e_3Z)$$ and the fact that $Y\not\in M_\infty$, implies that $Z$ itself belongs to $M_\infty^3$. We conclude that $M_\infty/M_\infty^2$ is generated by $X$, and so $\operatorname{ord}_\infty(X)=1$, and this implies that $\operatorname{ord}_\infty(Z)=3$, and $$\operatorname{ord}_\infty(X-e_iZ)=\min\{\operatorname{ord}_\infty(X),\operatorname{ord}_\infty(Z)\}=\min\{1,3\}=1,$$ for $i=1,2,3$.

  • Hence, $\operatorname{ord}_\infty((X-e_iZ)/Z)=1-3=-2$, so $(X-e_iZ)/Z$ has a pole of order $2$ at $\infty$. Also, $\operatorname{ord}_\infty(Y/Z)=0-3=-3$, i.e., $Y/Z$ has a pole of order $3$ at $\infty$.

  • Putting it all together in terms of divisors we find: $$\operatorname{div}(Y/Z)=(P_1)+(P_2)+(P_3)-3(\infty), \text{ and } \operatorname{div}((X-e_iZ)/Z) = 2(P_i)-2(\infty).$$

  • $\begingroup$ Dear Alvaro, can you please explain why the function $\dfrac{X-e_1Z}{Z} = \dfrac{Y^2}{(X-e_2Z)(X-e_3Z)}$ has a double pole at $[0,1,0]$? I'm studying from Silverman's book and according to his definitions in page 18, I found a uniformizer at $[0,1,0]$ to be $x= \dfrac{X}{Y}$, but then it is not clear to me how to prove that the above rational function is in the ideal $M_{[0,1,0]}^{-2} = \langle x^{-2} \rangle_{K[E]_{[0,1,0]}}$ $\endgroup$ Mar 17, 2014 at 21:19
  • 1
    $\begingroup$ @AdriánBarquero, I added a much more detailed explanation. I hope this helps. $\endgroup$ Mar 18, 2014 at 3:34
  • $\begingroup$ Dear Alvaro, thank you so much for this very detailed and clearly explained answer. I really appreciate your help. $\endgroup$ Mar 18, 2014 at 20:40
  • $\begingroup$ Can I ask where the equation $(e_1−e_2)(e_1−e_3)(X−e_1Z)Z^2=ZY^2−(X−e_1Z)^3−3(e_2+e_3)(X−e_1Z)^2Z$ comes from? I've tried calculating several times, but fail to get this equation to hold. $\endgroup$
    – Mmhmm
    Mar 3, 2015 at 5:44
  • $\begingroup$ Dear @Mmhmm, thank you for pointing out my mistake. I am not sure how I got that equation wrong... the point is to write the Taylor expansion of $(x-e_1)(x-e_2)(x-e_3)$ around the point $x=e_1$. I believe the correct expression should be $(e_1-e_2)(e_1-e_3)(X-e_1Z)Z^2=ZY^2-(X-e_1Z)^3-(2e_1-e_2-e_3)(X-e_1Z)^2Z.$ I've fixed this in the answer above. $\endgroup$ Mar 3, 2015 at 21:22

The divisor of a rational function on an elliptic curve always has degree 0.

Alternatively, you could change coordinates so that the point at infinity is an affine point. Recall that in projective coordinates, $y = Y/Z$, and your elliptic curve has equation

$$ Y^2 Z = (X - e_1Z)(X - e_2 Z)(X-e_3 Z)$$

$\infty$ has coordinates $(X:Y:Z) = (0:1:0)$. If you dehomogenize by setting $Y=1$, then $\infty$ is simply the origin in the affine $(X/Y,Z/Y)$-plane.


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