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I wonder why every birational transformation of $\mathbb{P}^n_k$ can be written as $[f_1:...:f_{n+1}]$, where the $f_i$ are homogeneous polynomials?

I wonder how to deduce this from the definition of rational map from On The Arithmetic of Elliptic Curves, Joseph H. Silverman:

Let $V_1$ and $V_2 \subseteq \mathbb{P}^n$ be projective varieties. A rational map from $V_1$ to $V_2$ is a map of the form $$\varphi : V_1 \rightarrow V_2, \qquad \varphi = [f_0,\ldots,f_n]$$

where the functions $f_0,...,f_n ∈ K(V_1)$ have the property that for every point $P ∈ V_1$ at which $f_0,...,f_n$ are all defined, $$\varphi (P) = [f_0(P),...,f_n(P)] \in V_2 \, .$$

6666
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1 Answers1

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Recall that $K(\Bbb P^n)\cong K(\frac{x_1}{x_0},\cdots,\frac{x_n}{x_0})$. Pick $f_0,\cdots,f_n\in K(\Bbb P^n)$ by the quoted definition. Clear denominators and homogenize with respect to $x_0$. This will give you that you may represent your map by homogeneous polynomials in $x_0,\cdots,x_n$.

KReiser
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  • I am worried about the pole of the $f_i$, for example $f=[x_1/x_0: x_0/x_2]$, then you get $f=[x_1x_2:x_0^2]$, then you can input $0$ for $x_0$, which can't happen for $f=[x_1/x_0: x_0/x_2]$. – 6666 Dec 31 '19 at 03:55
  • These maps are literally equal on the open set where the first was defined, so if you remember that rational maps are equivalence classes of open sets $U$ and morphisms $U\to Y$, per your previous question this is no bother at all. – KReiser Dec 31 '19 at 04:00
  • But $f_1/g_1\sim f_2/g_2$ if $f_1g2-f_2g_1\in I(P^n)=\emptyset$ so shouldn’t each rational function above has the unique formula? – 6666 Dec 31 '19 at 05:31
  • No. It's easy to check that both representatives from your first comment agree on closed points located in $D_+(x_0x_2)$: try it and see. I think you may be confusing a single rational function on $\Bbb P^n$, which can be described as a ratio of two polynomials of the same degree, with the components of a rational map, which have no such restriction. – KReiser Dec 31 '19 at 06:56