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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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    $\begingroup$ Do you mean $Y_i = Y\cap U_i$? $\endgroup$ – bradhd Oct 9 '12 at 13:39
  • $\begingroup$ Yes, thanks, i correct! $\endgroup$ – Manos Oct 9 '12 at 13:39
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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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  • $\begingroup$ $Y$ is the vanishing set of the homogeneous elements of $I$. $\endgroup$ – Manos Oct 9 '12 at 14:07
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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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  • $\begingroup$ So $Y_i$ is a projective variety itself, equal to its intersection with $U_i$, thus an affine variety. $\endgroup$ – Manos Oct 9 '12 at 13:52
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    $\begingroup$ $Y_i$ 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. $\endgroup$ – bradhd Oct 9 '12 at 13:54
  • $\begingroup$ Right, cause $Y_i$ is open in $Y$. To be a projective variety we need it to be closed. $\endgroup$ – Manos Oct 9 '12 at 13:56
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Not sure if I am misunderstanding your question, but to show that $Y \cap U_i$ is homeomorphic to an affine variety, as you noted, $Y \cap U_i$ is open in the irreducible set $Y$, so $Y \cap U_i$ is itself an irreducible topological space. Also, $Y \cap U_i$ is closed in $U_i$, because $Y$ is closed in $\mathbb{P}^n$.

Now we have established $Y \cap U_i$ is an irreducible closed subset of $U_i$, so the homeomorphism $\phi: U_i \rightarrow \mathbb{A}^n$ sends $Y \cap U_i$ to an irreducible closed subset of $\mathbb{A}^n$, i.e. an affine variety.

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