$E$ be an elliptic curve and from Silverman's AEC book(Chapter 3, Section3) where we show using Riemann-Roch theorem that E is isomorphic to a projective curve(smooth) given by a Weierstrass equation.
Recall map $\phi: E \longrightarrow \mathbb{P}^2$ given as $ P \mapsto [x(P):y(P):1]$.
We want to show that it's degree is $1$ by first showing that degree of the map $h: E \longrightarrow \mathbb{P}^1$ is $2$. And I'm stuck in this last part.
For $\infty \in \mathbb{P}^1$, since the only points mapping to it are the poles of $x$ (why?) we have
$$ [K(E):K(x)] = e_{\phi}(O) = ord_O(h^*t_{\infty}) = ord_O(\frac{1}{x}) = 2$$
Can some one also explain the second last equality above?
Thank you.
Remember that $x$ was chosen so that $\{1,x\}$ would be a basis for $\mathcal L( 2(O))$. We don't really need to do work to show that $x$ has a pole of order $2$ at $O$. It was chosen so that $\{1,x\}$ is a basis for $\mathcal L(2(O))$.
This tells us that $div(x) \geq -2(O)$ and in fact we must have $div(x) = -2(O) + D$ with D positive and not supported at $(O)$ or else $\{1,x\}$ couldn't span $\mathcal L (2(O))$. From that, we can see that $x$ has no other poles, and that the pole it has is one of exactly order 2.
For your last equality: give $\mathbb P^1$ coordinates $[s:t]$. To find a uniformizer at infinity, dehomogenize to $[1:t/s]$ and so the uniformizer at $\infty$ in these coordinates is just $t/s$. Then: $$\phi^* (t/s) = (t/s)\circ \phi = (t/s) \circ [x:1] = 1/x$$