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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).

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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.

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

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I expanded this answer here: – Álvaro Lozano-Robledo Nov 6 '15 at 21:18

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