Equal divisors of Weil-pairing I try to understand the approach of Silverman to the Weil pairing.
In The Arithmetic Of Elliptic Curves he defines the Weil pairing in section III.8 with the help of two divisors:
\begin{equation*}
div(f)= m(T)-m(O) \hspace{0.5cm}\text{and} \hspace{0.5cm}div(g)= [m]^*(T)-[m]^*(O)= \sum_{R\in E[m]}(T'+R)-(R),
\end{equation*}
where $[m]T'=T$ and $E[m]$ being the m-torsion points.
Later on he uses the fact that $f \circ [m]$ and $g^m$ have the same divisors to define the Weil pairing. I can't relate to this fact because I'm not understanding the difference between $[m]^*(T)-[m]^*(O)$ and $m(T)-m(O)$. Can anybody tell me how $[m]^*(T)-[m]^*(O)$ would be defined?
Thank you really much!
 A: I'll assume the ground field is algebraically closed of characteristic zero.
$[m]^*(T)$ is the pullback of the divisor $(T)$ under the operation $[m]$ which is multiplication by $m$ on the elliptic curve $E$, This means that
$[m]^*(T)=(P_1)+(P_2)+\cdots+(P_{m^2})$ where the $P_i$ are the $m^2$ points
solving $[m]P=T$, that is the $m$-fold addition $P+\cdots+P$ on the elliptic curve
equals $T$. The case $T=O$ gives $[m]^*(O)$. So $[m]^*(T)-[m]^*(O)$
is a rather different beast from $m(T)-m(O)$, which is a sum of $m$ copies of $(T)$
minus a sum of $m$ copies of $(O)$.
A: To answer Anuj's question in the comment, this is quite a general construction. If $f:X\to Y$ is any morphism of smooth projective varieties, it induces a pull-back map on divisors,
$$ f^*:\operatorname{Div}(Y)\longrightarrow \operatorname{Div}(X). $$
This map respects linear equivalence, so it induces a map
$$ f^*:\operatorname{Pic}(Y)\longrightarrow \operatorname{Pic}(X). $$
If $X=Y$, and if you manage to identify the linear equivalence class of $f^*X$ as some other divisor that you know in $\operatorname{Pic}(X)$, then you've acquired interesting geometric information relating the map $f$ and the geometry of the variety $X$.
For example, on an abelian variety (a higher dimensional elliptic curve), there is a multiplication-by-$m$ map $[m]:A\to A$, and an enormously useful formula says that
$$ [m]^*D \sim \frac{m^2+1}{2} D + \frac{m^2-1}[-1]^*D. $$
