quasiprojective subvariety

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 )=!

• Why do you think it is true? Dec 30, 2012 at 20:09
• It appears in the book of Shafarevich and is used a lot of times
– Kuru
Dec 30, 2012 at 20:11
• The book does not prove it? Dec 30, 2012 at 20:18
• No , look here sccs.swarthmore.edu/users/09/hyeok/tempupload/3540548122.pdf is the page 46
– Kuru
Dec 30, 2012 at 20:24
• 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. Dec 30, 2012 at 20:50

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.