# Viewing a quasi-projective variety as an affine variety

Let $Y$ be a projective variety of $\mathbb{P}^n$. Let $U_i$ be the open set of $\mathbb{P}^n$ corresponding to $x_i \neq 0$. Define $Y_i = Y \cap U_i$. Why can we view $Y_i$ as an affine variety? We know that $U_i$ is isomorphic to $\mathbb{A}^n$, so we can view $Y_i$ as a subset of an affine variety. (i assume that variety implies irreducible, following Hartshorne). Also, $Y_i$ is open in $Y$, so it is irreducible and dense in $Y$. That makes it a quasi-projective variety. But why is it an algebraic set of $\mathbb{A}^n$?

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Do you mean $Y_i = Y\cap U_i$? –  Brad Oct 9 '12 at 13:39
Yes, thanks, i correct! –  Manos Oct 9 '12 at 13:39

For simplicity of notation, let's assume $i=0$. Write $Y$ as the homogeneous vanishing set of a homogeneous ideal $I\subset k[x_0,\ldots,x_n]$, so $Y = \{[a_0:\cdots:a_n]: f(a_0,\ldots,a_n)=0 \forall f\in I\}$. Then

$Y_0 = \{[1:a_1:\cdots:a_n]:f(1,a_1,\ldots,a_n)=0 \forall f\in I\}$ is the vanishing set in $\mathbb{A}^n$ of $I_0 := \{f(1,x_1,\ldots,x_n): f\in I\}$. Similarly for $i\neq 0$.

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$Y$ is the vanishing set of the homogeneous elements of $I$. –  Manos Oct 9 '12 at 14:07

Well, if I understand well, $Y=\{{\bf x} \in k^{n+1} \mid f({\bf x})=0\}$ for a (family of) homogeneous polynomial(s) $f\in k[x_0,..,x_n]$.

Then $Y_i$ will correspond to the $f(x_0,..,\overbrace{1}^{i\text{-th}},..x_n)$ polynomial(s).

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So $Y_i$ is a projective variety itself, equal to its intersection with $U_i$, thus an affine variety. –  Manos Oct 9 '12 at 13:52
$Y_i$ is not necessarily projective. For example, you could take $Y = \mathbb{P}^n$ in which case $Y_i=U_i\cong\mathbb{A}^n$, which is not projective. –  Brad Oct 9 '12 at 13:54
Right, cause $Y_i$ is open in $Y$. To be a projective variety we need it to be closed. –  Manos Oct 9 '12 at 13:56