Divisors: function composition with multiplication by n map on elliptic curve Washington's Weil pairing proof gives a divisor in section 11.2
$$\textrm{div}(f) = n[T] - n[\infty]$$
But then he says let $f \circ n$ denote applying the multiplication by n map, then applying f. Obviously the zeroes of this function will be all $T' : nT' = T$, like this:
$$\textrm{div}(f \circ n) = n \left( \sum_{T' : nT' = T} [T'] \right) - n \left( \sum_{R \in E[n]} [R] \right)$$
I've searched everywhere but cannot find an answer why the order of these points are n.
The book simply says that it follows from the previous equation, and other proofs of the weil pairing are similarly vague.
Why is the order of the points equal to n? How do they get from $\textrm{div}(f)$ to $\textrm{div}(f \circ n)$?
For example why is $\textrm{ord}_{T'}(f \circ n) = n$? How do think about the order of multiplication by n, and how that affects f?
 A: The fact that the zeroes and poles of $f \circ [n]$ have the same orders as those of $f$ follows from the facts that: (i) ramification index is multiplicative with respect to composition; and (ii) $[n]$ has ramification index $1$ at each of these points $T'$.
Recall that, given a nonconstant morphism $\varphi: C_1 \to C_2$ of smooth curves and a point $P \in C_1$, the ramification index $e_\varphi(P)$ of $\varphi$ at $P$ is
$$
\DeclareMathOperator{\ord}{ord}
e_\varphi(P) := \ord_P(\varphi^* t)
$$
where $t$ is a local uniformizer at $\varphi(P)$. In the case where $f: C \to \mathbb{P}^1$, $f(P) = 0$, and $t$ is the usual affine coordinate on $\mathbb{P}^1$ at $0$, then $f^* t$ is just $f$ itself, so
$$
e_f(P) = \ord_P(f^* t) = \ord_P(f)
$$
is the order of vanishing of $f$ at $P$.
The two facts above are given in Silverman's The Arithmetic of Elliptic Curves Proposition II.2.6, parts (a) and (c):
Proposition. Let $\varphi: C_1 \to C_2$ be a nonconstant morphism of smooth curves.
(a) For every $Q \in C_2$,
$$
\sum_{P \in \varphi^{-1}(Q)} e_\varphi(P) = \deg(\varphi).
$$
(c) Given another nonconstant map $\psi: C_2 \to C_3$ of smooth curves, then for all $P \in C_1$, we have
$$
e_{\psi \circ \varphi}(P) = e_\varphi(P) e_\psi(\varphi(P)) \, .
$$
Assuming $n$ is relatively prime to the characteristic of the base field $K$, there are $n^2$ distinct points $T'$ such that $[n]T' = T$ over $\overline{K}$. Since $[n]$ has degree $n^2$, then $e_{[n]}(T') = 1$ for every such $T'$ by part (a). Now applying part (c), we have
\begin{align*}
e_{f \circ [n]}(T') = e_{[n]}(T') e_{f}(T) = 1 \cdot n = n \, .
\end{align*}
