I have read the book the Arithmetic of Elliptic curves of J.H Silverman (we can read in http://www.pdmi.ras.ru/~lowdimma/BSD/Silverman-Arithmetic_of_EC.pdf)
In chapter 3, the proof of proposition 1.5
I don't understand why the map they consider $\phi:E\rightarrow \mathbb{P}^1$ like above has degree one. By the difinition of $\deg$ of mmorphism, we must find the degree of filed extension $[K(E):\phi^*\left(K(\mathbb{P}^1\right)]$, whereas $K(C)$ is the function field of $C$ over an
algebraic closure field and $\phi^*:K(\mathbb{P}^1)\rightarrow K(E), f\mapsto f\circ \phi$. I can't imaginary what $\phi^*(K(\mathbb{P}^1))$ is. Somebody can help me?
Best Answer
The more intuitive way of viewing the degree of a morphism is thinking about its fibres. Let's just assume $\phi : C_1 \to C_2$ is a separable morphism between smooth projective curves - then it is a theorem that for all but finitely many $P \in C_2$ that the size of the fibre $\phi^{-1}(P)$ is equal to the degree of $\phi$ (in Silverman, this is Prop II.2.6).
The map you have $E \to \mathbb{P}^1$ which records the $x$-coordinate of a point has precisely $2$ points in the fibre of $[x_0,1]$, except when $x_0$ is the $x$-coordinate of a $2$-torsion point.
In particular $\phi$ must have degree $2$ (there are infinitely many elements of $\bar{K} \setminus \{ \text{$x$-coordinates of $2$-torsion points} \}$)