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Let $V/k$ be a variety. There seems to be a well-known correspondence between non constant elements of the function field $K(V)$ and dominant rational maps $V\dashrightarrow\mathbb{P}^1$. Furthermore, this correspondence seems to have good properties. However, I don't know exactly how to work with it.

I know there is a correspondence between dominant rational maps $f$ from $V$ to $\mathbb{P}^1$ and function field inclusions $K(\mathbb{P}^1)\to K(V)$. Since $K(\mathbb{P}^1)=K(T)$, I guess the correspondence should be given by $T\mapsto f$. However, I have some problems with this construction:

  1. I am not sure how it works explicitly. For example, $f^{-1}(0)$ and $f^{-1}(\infty)$ are supposed to correspond to the set of zeros and poles of the rational function $f\in K(X)$. I guess the correspondence I did above should be equivalent to pulling back $T\in K(\mathbb{P}^1)$ by the rational map $f$, in which case that would make sense. Am I right?

  2. If $k$ is not algebraically closed, not all elements of $K(X)\setminus k$ will be transcendental over $k$, so it seems like the correspondence above will not work. Is there a way to correct it? I have seen applications of the fact that don't seem to assume the base field is algebraically closed (for example, I'm reading a result in Fulton's Intersection Theory that uses this correspondence).

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  1. Yes, this is correct, you picked the right correspondence. Given a dominant rational map $f:X\to \Bbb P^1$, the map on function fields is $k(T)\to k(X)$ by $T\mapsto f^*(T)$, the pullback of $T$ along $f$; given a map of function fields $\varphi:k(T)\to k(X)$ the rational map is given by $[\varphi(T):1]$ (or maybe swapping the coordinates, depending on whether you like $\infty$ to be $[1:0]$ or $[0:1]$).
  2. If $k$ is not algebraically closed, the correspondence still works! In the one direction, a map of fields $\varphi:k(T)\to k(X)$ must sent $T$ to an element transcendental over $k$, since a field map is injective (if $m(Z)$ was the minimal polynomial of $\varphi(T)$ over $k$, then $\varphi(m(T))=0$, contradiction). In the other direction, if $f:X\to\Bbb P^1$ is dominant, then it cannot be the case that $f^*(T)$ is algebraic over $k$: the minimal polynomial of $f^*(T)$ would vanish on the image of $f$, contradicting the fact that $f$ is dominant.
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  • $\begingroup$ Thanks! In the last argument, you mean that being $p$ the minimal polynomial of $f^*(T)$, the image of $f$ will be contained in $V(p)$? How can you show that? $\endgroup$ Commented Oct 25, 2024 at 16:42
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    $\begingroup$ Exercise: if $f:X\to Y$ is a morphism of affine varieties, then $f^*(\alpha)=0$ for $\alpha\in k[Y]$ implies $f$ factors through $V(\alpha)\subset Y$. Now apply this to the case of a nonempty affine open where $f$ is defined and mapping in to the usual $\Bbb A^1\subset \Bbb P^1$; note that the inverse image of the complement of $V(p)$ is also an open in $X$, and as $X$ is irreducible, these two open sets must meet if they're both nonempty, but they don't by definition, so in fact $f(X)\subset V(p)$. $\endgroup$ Commented Oct 25, 2024 at 16:52

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