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As part of a homework problem, I need to show that the value of $e_m(P,Q)$ is independent of the choice of a point $S \in E[m] \setminus \{\mathcal{O},P,-Q,P-Q\}$, where $E[m]$ is the collection of all points on an elliptic curve $E : Y^2 = X^3 + AX + B$ of order $m$.

The problem provides a hint to fix the points $P$ and $Q$ and to consider a function of $S$ $$F(S) = \left.{\frac{f_P(Q+S)}{f_P(S)}}\middle/{\frac{f_Q(P-S)}{f_Q(S)}}\right.$$ and to compute the divisor of $F$ and to use the fact that every nonconstant function on $E$ has at least one zero.

What I know is that $f_P$ is defined to be a function that has divisor (formal sum) $m[P] - m[\mathcal{O}]$, and $f_Q$ is defined to be a function that has divisor $m[Q] - m[\mathcal{O}]$. The gap that I do not know how to fill is the evaluation of $f_P$ (or $f_Q$) at a point other than $P$ (or $Q$).

I believe what I am most confused about is how to represent the rational function $f_P$ in a way other than what its formal sum is. Could someone explain this?

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If $f_P(S)$ is a function with divisor $m[P]-m[\mathcal{O}]$, then $f_P$ has a zero at $P$ and a pole at $\mathcal{O}$. Then, for a fixed constant $T$, the function $h_{P,T}(S)=f_P(T+S)$ is simply a translation of $f_P$ by $T$. Thus, $h_{P,T}$ has a zero when $T+S=P$, i.e., at $S=P-T$ and a pole at $T+S=\mathcal{O}$, i.e., at $S=-T$. Moreover, the multiplicities of these zeros and poles coincide for $h$ and $f$. Therefore the divisor of $h_{P,T}$ must be $m[P-T]-m[-T]$.

Does this help?

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  • $\begingroup$ This does help me understand this concept more. Then if $f$ and $g$ are two rational functions and I take $h = f/g$, is it reasonable to believe that the zeros of $h$ consists of the zeroes of $f$ and the poles of $g$ and the poles of $h$ consist of the poles of $f$ and the zeroes of $g$? $\endgroup$
    – Alex
    Mar 5, 2014 at 5:02
  • $\begingroup$ Yes the divisor of a quotient is the difference of divisors. $\endgroup$ Mar 5, 2014 at 11:52

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