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I have a very simple question about quasiprojective varieties. I'm reading "Basic Algebraic Geometry" of Shafarevich.

Definition: A quasiprojective variety is an open subset of a closed projective set. A quasiprojective sub-variety $ Y \subset X$ is a subset $ Y \subset X$ such that $Y$ is itself a quasiprojective variety.

I want to prove that $ Y \subset X $ is a questiprojective subvariety if and only if $Y = Z - Z_a $ where $ Z,Z_a \subset X$ are closed projective sets.

Well obviously one side is direct. I have troubles with proving that if $ Y \subset X $ is a quasiprojective subvariety , then I can write $Y$ in that form.

I have the following : $ X = Z_1 \cap U_1 $ and $ Y = Z_2 \cap U_2 $ where $Z_i$ are closed projective sets and $U_i$ are open projective sets. And I also know that $ Y \subset X $ . Well... how can I write $Y$ in the form $Y = Z-Z_a $ where $Z, Z_a \subset X $ and closed projective sets? Please help me )=!

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  • $\begingroup$ Why do you think it is true? $\endgroup$ Dec 30, 2012 at 20:09
  • $\begingroup$ It appears in the book of Shafarevich and is used a lot of times $\endgroup$
    – Kuru
    Dec 30, 2012 at 20:11
  • $\begingroup$ The book does not prove it? $\endgroup$ Dec 30, 2012 at 20:18
  • $\begingroup$ No , look here sccs.swarthmore.edu/users/09/hyeok/tempupload/3540548122.pdf is the page 46 $\endgroup$
    – Kuru
    Dec 30, 2012 at 20:24
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    $\begingroup$ It says $Y=Z−Z_1$ with $Z,Z_1 \subset X$ closed subsets. I think it means that $Z,Z_1$ are closed subsets of $X$, not closed subsets of the ambient projective space, in which case the assertion is trivially true. $\endgroup$ Dec 30, 2012 at 20:50

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The correct statement is given in comments by Makoto.

In general it is not possible to write $Y=Z\setminus Z_a$ as you described. Indeed, take $Y=X$. Then $Z=X$. So $Z$ is never projective if $X$ is not projective.

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