Timeline for Dense Affine Subvarieties of Algebraic Varieties
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| May 2, 2013 at 14:42 | vote | accept | Peter Crooks | ||
| May 1, 2013 at 23:36 | comment | added | Will Sawin | What about disjoint union of infinitely many affines? I guess that's not really a variety. | |
| May 1, 2013 at 23:17 | history | edited | Steven Landsburg | CC BY-SA 3.0 |
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| May 1, 2013 at 23:03 | history | edited | Steven Landsburg | CC BY-SA 3.0 |
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| May 1, 2013 at 20:36 | comment | added | Sam Gunningham | @Dan: Sorry, didn't see your comment in time. | |
| May 1, 2013 at 20:35 | comment | added | Sam Gunningham | 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. | |
| May 1, 2013 at 20:33 | comment | added | Dan Petersen | 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 $X'$. | |
| May 1, 2013 at 20:02 | comment | added | Peter Crooks | 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. | |
| May 1, 2013 at 19:53 | comment | added | Angelo | This must be done a little more carefully: the union is not necessarily affine, or even a subvariety. | |
| May 1, 2013 at 19:38 | vote | accept | Peter Crooks | ||
| May 1, 2013 at 20:00 | |||||
| May 1, 2013 at 19:33 | history | answered | Steven Landsburg | CC BY-SA 3.0 |