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

Cooper Doyle 2022-07-04 Answered
Let X be an integral scheme of finite type over an arbitrary field k. Let f : X P 1 be a dominant morphism. What is meant by the rational function of f? Fulton uses the notation 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.
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Answers (1)

soosenhc
Answered 2022-07-05 Author has 16 answers
I'm turning my comment into an answer. By definition of f, f induces a morphism f : K ( P 1 ) K ( X ) (the function fields are the rings of stalks at the generic points). The rational function of f is f ( t 0 / t 1 ), where t 0 / t 1 K ( X ) generates that field. Depending on the conventions, it could be t 1 / t 0 instead.
The basic idea is that A 1 = Spec k [ t ] is a dense open subset of P 1 (in two different ways, hence the "convention" issue) and the map f 1 ( A 1 ) A 1 is given by a regular function on f 1 ( A 1 ) (the image of t), which is rational on X.
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