# The invariant differential associated to a Weierstrass form for an elliptic curve is holomorphic and nonvanishing

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?

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 $$\bar{K} \setminus \{ \text{x-coordinates of 2-torsion points} \}$$)