I'm working through exercise 3.29 from Silverman's Arithmetic of Elliptic Curves:
Let $E$ be an elliptic curve over an algebraically closed field $K$ with Weierstrass coordinates $x$ and $y$, and base point $\mathcal{O}$. Fix a non-zero point $T \in E$, and write $$x(P + T) = f(x(P), y(P))$$ for some function $f \in K(E) = K(x, y)$ Prove that $f$ is a linear fractional transformation if and only if $T \in E[3]$, where a linear fractional transformation is a funcion of the form $$ \frac{\alpha x + \beta y + \gamma}{\alpha' x + \beta ' y + \gamma'} $$ then $T \in E[3]$.
I attempted the only if part of this question in the following way:
If $f$ has the above form then it has divisor $$ \text{div}(f) = [P_{1}] + [P_{2}] + [P_{3}] - [Q_{1}] - [Q_{2}] - [Q_{3}] $$ We can see that $[-T]$ must be a pole of $f$, so, say $Q_{1} = Q_{2} = -T$, and we also know that the poles must be colinear so $Q_{1} + Q_{2} + Q_{3} = \mathcal{O}$ i.e. $Q_{3} = 2T$. Thus $2T$ is a pole, and since $x$ has only one pole at $\mathcal{O}$ then we must have that $3T = 2T + T = \mathcal{O}$.
I am unsure of the validity of my method as I am unsure why $f$, which is the rational function $x$ composed with the translation map $\tau_{t}$ (which is an isomorphism of $E$), would have an order $3$ pole when $x$ has only an order $2$ pole. I am also unsure of the zeroes, since it makes sense to me that $$ \text{div}(f) = \text{div}(x \circ \tau_{T}) = [R - T] + [- R + T] - 2[-T] $$ where $R$ is such that $x(R) = 0$. But this would mean that the only way for the assumed form for $f$ to hold would be if the numerator and denominator share a (simple) zero, which would have to be at $2T$, and so $f$ wouldn't have a pole at $2T$ which ruins my argument.
Anyway I've confused myself enough here, can anyone help me with this muddle?
I think that this exercise is wrong.
The map $f : E \to \Bbb P^1$ has degree $2$ (just like $x$), and if $A$ and $-A$ are the two points whose $x$-coordinate is $0$ then its divisor is
$div(f) = [A-T] + [-A-T] - 2[-T] = ([A-T] + [-A-T] + [2T]) - (2[-T]+[2T])$
Let $L_1(x,y) = 0$ be an equation for the line joining $A-T$ and $-A-T$,
and $L_2(x,y)=0$ be an equation for the line joining $-T$ and $2T$.
They are linear, and $L_1/L_2$ has the same divisor as $f$, so they are the same function up to a multiplicative constant.
On the other hand, if you pick $g(x(P),y(P)) = y(P+T)$, then
$div(g) = [B_1-T] + [B_2-T] + [B_3-T] -3[-T]$, and this can only be written as a difference of divisors of two elements of $L(3\mathcal O)$ if $3T = \mathcal O$. Note that in this case, $L_2$ is the same line whether you look at $x$ or at $y$, and so what you can prove is that there is a matrix $A_T \in M_3(k)$ such that $[x(P+T) : y(P+T) : 1] = A_T [x(P) : y(P) : 1]$
Looking at the second part of the exercise (showing that translation by $T$ induces an isomorphism of $L(n\mathcal O)$ if and only if $T \in E[n]$), it seems that this is what the author intended