Let $X$ be a complex algebraic variety. Is it true that $X$ contains a dense affine subvariety?
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asked May 1, 2013 at 19:25
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$\begingroup$ How do you define "variety"? Integral separated scheme of finite type? $\endgroup$– S. Carnahan ♦Commented May 2, 2013 at 1:07
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1 Answer
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If $X$ is irreducible, then any open subset is dense, and since $X$ can be covered by open affines, any one of these will do. If $X$ isn't irreducible, take an affine dense subvariety of each component, and take their union.
Edited to add: As commenters have noted, you'll want to be sure the affine dense subvarieties you choose are disjoint from one another (in order to insure that their union is affine). But this is easy: For each of these subvarities, just throw away all its intersections with the other components, then take an affine open subvariety of what remains.
answered May 1, 2013 at 19:33
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2$\begingroup$ This must be done a little more carefully: the union is not necessarily affine, or even a subvariety. $\endgroup$– AngeloCommented May 1, 2013 at 19:53
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$\begingroup$ Good point! The irreducible case is straightforward, but the more general case is subtle. I suppose that if the irreducible components coincided with the connected components, then the above argument would work. $\endgroup$ Commented May 1, 2013 at 20:02
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3$\begingroup$ You can trivially reduce to the case when the connected components of $X$ are irreducible: just shrink $X$ to a smaller variety $X'$ by throwing out all points lying on more than one irreducible component. Then $X'$ is open and dense in $X$ and it suffices to find a dense affine in $X'$. $\endgroup$ Commented May 1, 2013 at 20:33
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1$\begingroup$ Can't you just take open affines in each irreducible component which do not intersect any other component? Then the union of such will be disjoint. $\endgroup$ Commented May 1, 2013 at 20:35
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$\begingroup$ @Dan: Sorry, didn't see your comment in time. $\endgroup$ Commented May 1, 2013 at 20:36