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Let $f$ be the rational function on $\mathbb{P}^2$ given by $f = x_1/x_0$. Find the set of points where $f$ is defiend and describe the corresponding regular function.

My question: I know that the set of points where $f$ is defined is $\{[1: x_1: x_2] \in \mathbb{P}^2\}$. Isn't the corresponding regular function also just $f$ itself, or the regular function $g = x_2/x_0$ (both works and both maps $\mathbb{P}^2$ to $\mathbb{A}^1$)? So what is this problem trying to get at?

Now think of this function as a rational map from $\mathbb{P}^2$ to $\mathbb{A}^1$. Embed $\mathbb{A}^1$ in $\mathbb{P}^1$, and let $\phi: \mathbb{P}^2 \to \mathbb{P}^1$ be the resultiong rational map. Find the set of points where $\phi$ is defined, and describe the corresponding morphism.

My question: Since $\phi$ is $f \circ \epsilon$ where $\epsilon$ is the embedding, why isn't $\phi$ defined on the same set as $f$ is defined? I'm also not sure what I'm trying to solve for here.

nekodesu
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    " So what is this problem trying to get at? " The problem is to find at which points a rational map is regular. This means being very precise about domains and targets (=codomains) of actual functions deduced from "rational maps" which are not maps at all ! The terminology is alas very confusing, but we are stuck with it for historic reasons. – Georges Elencwajg May 24 '18 at 22:39

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(a) The rational function $f=\frac{x_1}{x_0}\in k(x_0,x_1,x_2)$ is regular on the open subset $U_0=\{[x_0:x_1:x_2]\in \mathbb P^2\vert x_0\neq 0]\}\subset \mathbb P^2$ and defines on that open subset the regular function $$f_U:U\to \mathbb A^1:[x_0:x_1:x_2]\mapsto\frac{x_1}{x_0}$$ (b) The rational map $\phi:\mathbb P^2--\to \mathbb P^1$ induced by $f$ is regular on the open subset $V=\mathbb P^2\setminus \{[0:0:1]\} \subset \mathbb P^2$ and defines on that open subset the regular function $$f_V:V\to \mathbb P^1:[x_0:x_1:x_2]\mapsto [x_0:x_1]$$ (c) Of course we have $U\subset V\subset \mathbb P^2$ and thus letting $$\epsilon:\mathbb A^1\hookrightarrow \mathbb P^1:x_1\mapsto [1:x_1]$$ be the canonical embedding we have the formula $$\epsilon \circ f_U=f_V\vert U: U\to \mathbb P^1$$ This is not very surprising: increasing the target of $f_U$ from $\mathbb A^1$ to $\mathbb P^1$ allows $f_U$, which is regular on $U$, to be extended to the function $f_V$, regular on the larger set $V$.

(d) Notice carefully that there is no way to extend the regular map $$f_V:V=\mathbb P^2\setminus \{[0:0:1]\}\to \mathbb P^1$$ to a regular map $\mathbb P^2 \to \mathbb P^1$: the point $[0:0:1]$ is said to be a point of indeterminacy for the rational map $\phi:\mathbb P^2--\to \mathbb P^1$.

Georges Elencwajg
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