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I would like to see a clear, rigorous and elementary proof of the following statement:

Let X be a (not necessary quasi-projective, separated) algebraic variety over the complex numbers, and let U,V be two affine open subsets of X. Then the intersection of U and V is affine.

Does the proof change if one substitutes "scheme" for "variety"?

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This is homework –  Dinakar Muthiah Mar 18 '10 at 19:33
    
Nice question! Form Ravi's old notes (link follows): "Another nice property of varieties: the intersection of any two affine opens is another affine open. I don't foresee using this, so I won't prove it, but you can find a proof in Mumford (p. 55) or Hartshorne (Exercise II.4.4)." I might look there or think about it. math.stanford.edu/~vakil/725/class12.pdf –  Ilya Grigoriev Mar 18 '10 at 19:36
    
Duh! I was confused... For schemes, this property is the most important consequence of separatedness. For non-separated schemes, it'll be wildly unture (there is a weaker notion of quasiseparatedness which means that intersections of open affines are finite unions of open affines, but there are schemes that don't even have this property.) –  Ilya Grigoriev Mar 18 '10 at 19:44
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Muthiah was right, it's easy: just take the diagonal inside the product of the two open affines etc. –  Qfwfq Mar 18 '10 at 20:22
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The asnwer is in textbooks as quoted. But some comments. Thomason and Trobaugh have introduced a notion of semi-separatedness, which means that this scheme has an affine cover with affine double intersections. Of course, this is weaker than separatedness. In noncommutative geometry almost all interesting schemes are not semi-separated (principal example: quantum flag varieties). –  Zoran Skoda Mar 24 '10 at 19:53
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