Proof involving rational functions on an elliptic curve I am working through understanding the proof I posted below, and I have questions about 3 parts. (Note, the proof assumes results developed before, but I don't have questions on those. My questions are on things in the proof that are "common knowledge" mathematics but beyond my reach.)
Q1) In the proof they claim that if $\frac{\alpha(x)}{\beta(x)} \in k(x,y)$ is a rational function on $E$ with $\gcd (\alpha(x), \beta(x)) = 1$.
Then if the degree of $\alpha(x)$ > degree of $\beta(x)$, somehow this implies that $\frac{\alpha(x)}{\beta(x)} \notin \mathscr{O_\infty}$. Where $\mathscr{O_\infty}$ is the local ring of rational functions defined at $\infty$. Why is this the case?
Q2) Why in the DVR $\mathscr{O_\infty}$ , $x$ has order $-2$: It is expressed as $x = u\pi^{-2}$ for a unit $u$ and uniformizer $\pi$. 
Q3) Why simply because $\gamma(\infty) = \infty$ this implies that $v_{\infty}(\frac{\alpha(x)}{\beta(x)}) < 0$ where I think $v_{\infty}$ is the valuation or in other words the exponent of $\pi$ in the representation of $\frac{\alpha(x)}{\beta(x)}$.

 A: Q1: The behaviour of $f(x)=\frac{\alpha(x)}{\beta(x)}$ at $\infty$ can be studied as behaviour of $g(x):=f(1/x)$ at $x=0$. If $\deg\alpha>\deg\beta$, then $\alpha(1/x)$ has a higher order pole than $\beta(1/x)$, meaning that $g(x)$ has a pole at $x=0$ and $f(x)$ is note defined at $x=\infty$.
Q2: I'll give some illustrative heuristics: At $\infty$, the lower degree parts in the Weierstrass form $y^2=x^3+ax+b$ become irrelevant, or argue more precisely by taking reciprocals as above and considering the origin. There $y^2=x^3$ has a cusp and $E$ is described by $(x,y)=(t^2,t^3)$, which makes $x$ a square of a function vanishing at the origin. Going back to $\infty$, it is of course $\frac1x$ that is the square of something vanishing at $\infty$
Q3: I suppose you are confused only because the valuation is taken at $\infty$ instead of at a finte place. In the end, positive valuation means a zero at that place, negative means a pole at that place, and zero valuation means a finite value at that place. It is the same at $\infty$: If  $\gamma(\infty)=0$, then $\gamma$ has a zero at $\infty$ and is in $\mathfrak m_\infty$, i.e,, $v_\infty(\gamma)>0$. Vice versa, if $\gamma(\infty)=\infty$, it has a pole at $\infty$, i.e., is the reciprocal of something having a zero there, hence $v_\infty(\gamma)<0$. (And only if $\gamma(\infty)$ were a non-zero finite value, we'd have $v_\infty(\gamma)=0$).
