MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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"?

share|cite|improve this question
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. – 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
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.