I think this is going to be a silly question. I'm happy with the following fact:
If $\alpha : X \to Y$ is a non-constant morphism of irreducible curves, then $\alpha$ induces an embedding of field $k(Y) \hookrightarrow k(X)$ such that $[ k(X) : k(Y) ] = \mathrm{deg}(\alpha)$ is finite.
I have the following question:
Let $\phi = (1:f) : \mathbb P^1 \to \mathbb P^1 $ be a morphism given by a non-constant polynomial $f \in k[t] \subset k(\mathbb P^1)$. Prove that $\mathrm{deg}(\phi) = \mathrm{deg}\ f$.
I can't see why this is true. $\phi$ is a non-constant morphism of smooth irreducible curves, right? Why isn't $\mathrm{deg}(\phi) = 1$? (both domain and codomain have the same function field...)
Thanks