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A quasi-affine variety is an open subset of an affine variety.

Open under Zariski topology? How does this make sense?

algebraically_speaking
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    Meaning "relatively open". So a quasi-affine is the intersection of an affine variety with an open subset of the ambient affine space. The usual example is $\mathbf{A}^2 - {(0, 0)}$. – Hoot Dec 31 '14 at 17:08
  • @Hoot- Would $x+y=0\cap [(0,1)\times(2,3)]$ be another example? – algebraically_speaking Dec 31 '14 at 17:10
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    @algebraically_speaking: No that set would not be quasi-affine: you have to intersect with a Zariski-open set, not a Euclidean-open set. The set $(0,1)\times (2,3)$ is not Zariski-open because its complement isn't the solution set of a polynomial.

    Hoot: He's referring to a piece of the line $y=-x$, intersected with an open rectangle.

     – pre-kidney Dec 31 '14 at 17:50

1 Answers1

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Yes, open under the Zariski topology. Here is how it makes sense:

An affine variety is the solution set of polynomials.

A quasi-affine variety is a solution set of polynomials minus another solution set.

Jo Liss
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pre-kidney
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    Yeah, I know that it has been five years since you wrote this answer and everything, but thanks for stating it so clearly, thereby allowing me to sort out something that's been bugging me the entire day. This need to be printed large in every textbook on the topic so that dummies like me can proceed without any ambiguities! :D – StormyTeacup Oct 13 '19 at 19:26
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    Great to hear the answer was helpful to you. Try this link https://golem.ph.utexas.edu/string/archives/000849.html for plain English explanations of other concepts in algebraic geometry (although that exposition uses more category theory than is strictly necessary, you can ignore some of those parts on a first reading) – pre-kidney Oct 13 '19 at 23:39