Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The matter of explicitly finding the order of a rational function on an elliptic curve in the projective plane at infinity (i.e. at the point $(0, 1, 0)$) still seems unclear.

For example, Silverman (in The Arithmetic of Elliptic Curves) states that the order of the rational function $y$ on the elliptic curve \[ y^2 = (x - e_1)(x - e_2)(x - e_3), \] where $e_1$, $e_2$, and $e_3$ are distinct, is $-3$. That is, the function $y$ has a pole of order $3$ at $(0, 1, 0)$. I have no doubt that this is true; I'd like to know a simple way to see it, based on projective coordinates and independent of the fact that the sum of the orders of the zeros of $y$ is $3$ (which I understand).

share|improve this question
add comment

2 Answers

This fairly old-fashioned argument is the way I look at the situation. Homogenize $y^2=x^3+ax+b$ to $Y^2Z=X^3+aXZ^2+Z^3$, then dehomogenize by setting $Y=1$ to get $\zeta = \xi^3 +a\xi\zeta^2+\zeta^3$. The point you’re interested in is now $(0,0)$.

What does the curve look like there? Clearly $\xi$ is a local uniformizer, and $\zeta$ has a triple zero there. If you don’t see that right away, use the equation relating the two letters to see that $\zeta$ expands as a power series that starts $\zeta = \xi^3 + a\xi^7 +\cdots$. Now, what are the original $x$ and $y$ in terms of $\xi$ and $\zeta$? Yes: $x=\xi/\zeta$ and $y=1/\zeta$, and there you are.

share|improve this answer
add comment

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 $$ \operatorname{div}(y)=P_1+P_2+P_3-3\infty.$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.