Q:

How to show that $\phi_i(U_i \cap U_j)$ is a principal open set $D(f)$ in $A^n$ for some funtion $f$?

Let $A^n$ be the affine variety defined by $k^n$, $k$ is a field, and $U_i=\{(x_0, \ldots, x_n) \in \mathbb{P}^n \mid x_i \neq 0\}$. Define $\phi: U_i \to A^n$ by
$$
\phi_i(x_0, \ldots, x_n) = (\frac{x_0}{x_i}, \ldots, \frac{x_{i-1}}{x_{i}}, \frac{x_{i+1}}{x_{i}}, \ldots, \frac{x_n}{x_i}).
$$
How to show that $\phi_i(U_i \cap U_j)$ is a principal open set $D(f)$ in $A^n$ for some function $f$? Here $D(f)=\{x \in A^n \mid f(x) \neq 0 \}$ Thank you very much. I think that 
$$
U_i \cap U_j = \{(x_0, \ldots, x_n) \in \mathbb{P}^n \mid x_i \neq 0, x_j \neq 0 \}.
$$
If $i=j$, then
$$
\phi_i(U_i \cap U_j) = \phi_i(U_i) = \{(\frac{x_0}{x_i}, \ldots, \frac{x_{i-1}}{x_{i}}, \frac{x_{i+1}}{x_{i}}, \ldots, \frac{x_n}{x_i}) \mid x_0, \ldots, x_{i-1}, x_{i+1}, \ldots, x_n \in A \}.
$$
How to show that $\phi_i(U_i)$ is some $D(f)$? Thank you very much.

A:

I am answering this question because it hides a slight combinatorial problem.    
a) The image $\phi_i(U_i)$ is all of $\mathbb A^n$, so can be written $D(1)$, which answers your last question.
b) The description of $\phi_i(U_i\cap U_j)$ for $i\neq j$ however must be split into two cases.
Call $(z_1,\cdots,z_n)$ the coordinates in $\mathbb A^n$. Then we have:
$\bullet$ $\phi_i(U_i\cap U_j)=D(z_{j+1})$ if $j\lt i$
$\bullet \bullet$ $\phi_i(U_i\cap U_j)=D(z_j)$ if $j\gt i$  
Try it with $n=3, i=2, j=0$ and $j=3$.
In particular you will see that you can't have  $\phi_i(U_i\cap U_j)=D(z_j)$ for $j=0$ since there is no $z_0$ !  
Edit
Maybe I should have emphasized that $z_j=x_j/x_i$ for $j\gt i\geq 0$ and $z_j=x_{j-1}/x_i$ for $1\leq j\leq i$.

