Timeline for answer to Does a scheme have a "separification"? by Will Sawin
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| when toggle format | what | by | license | comment | |
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| Jul 29, 2013 at 4:54 | comment | added | Will Sawin | Yes, this is of course the obvious flaw. But at least we can see that any counterexample must be pathological in some exciting new way, namely that it has infinitely many incompatible maps to separated schemes. | |
| Jul 29, 2013 at 4:40 | comment | added | S. Carnahan♦ | The basic reason is that open sets in the product (or inverse limit) topology tend to be too big to be affine. | |
| Jul 29, 2013 at 4:22 | comment | added | Eric Wofsey | mathoverflow.net/questions/65506/… | |
| Jul 29, 2013 at 4:21 | comment | added | Will Sawin | What's an example where they don't? | |
| Jul 29, 2013 at 4:19 | comment | added | Jonathan Wise | Inverse limits of arbitrary diagrams don't exist in the category of schemes. How does the argument deal with that? At least it seems to show that $X^s$ exists as a pro-scheme... | |
| Jul 29, 2013 at 4:18 | comment | added | Will Sawin | Thorny or nonexistent? | |
| Jul 29, 2013 at 4:10 | comment | added | S. Carnahan♦ | This is a good idea, but I'm not convinced that $X^s$ is a scheme. Infinite products and inverse limits of non-affine schemes can be rather thorny. | |
| Jul 29, 2013 at 4:09 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Jul 29, 2013 at 3:15 | history | answered | Will Sawin | CC BY-SA 3.0 |