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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.

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Isn't it just the complement of the zero locus of the $j^{\text{th}}$ coordinate function? –  Alex Youcis Sep 23 '13 at 11:02
2  
You mean principal open set, where "principal" is an adjective meaning "main". A principle is a noun describing a moral point of view. –  Georges Elencwajg Sep 23 '13 at 11:27
    
@GeorgesElencwajg, thank you very much. –  LJR Sep 23 '13 at 11:29
    
Dear LJR, congratulations on your positive attitude. English is a foreign language for me and since I am very grateful when someone corrects me, I thought I could help others too. And, just to show that I'm not only a grammatical pedant, I have answered your question :-) –  Georges Elencwajg Sep 23 '13 at 11:59
    
@GeorgesElencwajg: perhaps you know the topologist's mnemonic for remembering the distinction: "A principle bundle is one with moral fibre". A bit off-topic here, but maybe amusing enough to be forgiven. –  Asal Beag Dubh Sep 23 '13 at 12:34

1 Answer 1

up vote 2 down vote accepted

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 $1\leq j\leq i$.

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thank you very much. Let $n=3, i=2, j=0$. Then $U_i \cap U_j = \{[x_0, x_1, x_2]: x_0 \neq 0, x_2 \neq 0\}$ and $\phi_i(U_i \cap U_j) = \{[x_0/ x_2, x_1/ x_2] : x_0 \neq 0\}$. It seems that $\phi_i(U_i \cap U_j) = D(z_0) \neq D(z_1) = D(z_{j+1})$. I am confused. –  LJR Sep 23 '13 at 12:13
    
Of course, this is all just indexing. The point is that it's the principal open set associated to one of the coordinate functions. –  Alex Youcis Sep 23 '13 at 12:17
    
Dear LJR, for $n=3$, remember that $\mathbb P^3$ has coordinates $[x_0,x_1,x_2,x_3]$. You forgot $x_3$. Also remember that $x_0/x_2=z_1$, not $z_0$ which, once again, doesn't exist!! –  Georges Elencwajg Sep 23 '13 at 12:18
    
@GeorgesElencwajg, thank you very much. –  LJR Sep 23 '13 at 12:21
    
Dear LJR, you're welcome! –  Georges Elencwajg Sep 23 '13 at 12:29

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