# Let X be an integral scheme of finite type over an arbitrary field k . Let f : X &

Let X be an integral scheme of finite type over an arbitrary field $k$. Let $f:X\to {\mathbb{P}}^{1}$ be a dominant morphism. What is meant by the rational function of $f$? Fulton uses the notation $\text{div}\left(f\right)$ but it is not at all clear to me how to obtain a rational function from a dominant morphism to the projective line.
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soosenhc
I'm turning my comment into an answer. By definition of $f$, $f$ induces a morphism ${f}^{\mathrm{♯}}:K\left({\mathbb{P}}^{1}\right)\to K\left(X\right)$ (the function fields are the rings of stalks at the generic points). The rational function of $f$ is ${f}^{\mathrm{♯}}\left({t}_{0}/{t}_{1}\right)$, where ${t}_{0}/{t}_{1}\in K\left(X\right)$ generates that field. Depending on the conventions, it could be ${t}_{1}/{t}_{0}$ instead.
The basic idea is that ${\mathbb{A}}^{1}=\mathrm{Spec}k\left[t\right]$ is a dense open subset of ${\mathbb{P}}^{1}$ (in two different ways, hence the "convention" issue) and the map ${f}^{-1}\left({\mathbb{A}}^{1}\right)\to {\mathbb{A}}^{1}$ is given by a regular function on ${f}^{-1}\left({\mathbb{A}}^{1}\right)$ (the image of $t$), which is rational on $X$.