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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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.

soosenhc · Accepted Answer

I&#039;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 &quot;convention&quot; 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.

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

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