8
$\begingroup$

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"?

Improve this question
$\endgroup$
9
  • 10
    $\begingroup$ This is homework $\endgroup$
    – Dinakar Muthiah
    Commented Mar 18, 2010 at 19:33
  • 3
    $\begingroup$ 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.) $\endgroup$
    – Ilya Grigoriev
    Commented Mar 18, 2010 at 19:44
  • 1
    $\begingroup$ @llya: this is true and it is equivalent to say any algebraic variety is semi-separated. $\endgroup$
    – Shizhuo Zhang
    Commented Mar 18, 2010 at 19:50
  • 2
    $\begingroup$ Muthiah was right, it's easy: just take the diagonal inside the product of the two open affines etc. $\endgroup$
    – Qfwfq
    Commented Mar 18, 2010 at 20:22
  • 2
    $\begingroup$ 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). $\endgroup$
    – Zoran Skoda
    Commented Mar 24, 2010 at 19:53

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .