What is the rational function associated to a dominant morphism to $\mathbb{P}^1$?

Question

Let X be an integral scheme of finite type over an arbitrary field $k$. Let $f:X \rightarrow \mathbb{P}^1$ be a dominant morphism. What is meant by the rational function of $f$? Fulton uses the notation $\text{div}(f)$ but it is not at all clear to me how to obtain a rational function from a dominant morphism to the projective line.

Answer 1

I'm turning my comment into an answer. By definition of $f$, $f$ induces a morphism $f^\sharp: K(\mathbb{P}^1) \rightarrow K(X)$ (the function fields are the rings of stalks at the generic points). The rational function of $f$ is $f^\sharp(t_0/t_1)$, where $t_0/t_1 \in K(X)$ generates that field. Depending on the conventions, it could be $t_1/t_0$ instead.

The basic idea is that $\mathbb{A}^1=\operatorname{Spec} k[t]$ is a dense open subset of $\mathbb{P}^1$ (in two different ways, hence the "convention" issue) and the map $f^{-1}(\mathbb{A}^1)\rightarrow \mathbb{A}^1$ is given by a regular function on $f^{-1}(\mathbb{A}^1)$ (the image of $t$), which is rational on $X$.