Timeline for Are there (enough) injectives in condensed abelian groups?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Mar 4, 2022 at 22:07 | history | edited | Matthieu Romagny | CC BY-SA 4.0 |
added "is"
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| Apr 1, 2020 at 15:45 | vote | accept | Maxime Ramzi | ||
| Apr 1, 2020 at 15:45 | comment | added | Maxime Ramzi | Ah ok, I had missed that one, great ! Thanks a lot ! | |
| Apr 1, 2020 at 15:37 | comment | added | Peter Scholze | See Proposition 2.1 in math.uni-bonn.de/people/scholze/Analytic.pdf | |
| Apr 1, 2020 at 14:01 | comment | added | Maxime Ramzi | to use the map to the solidification (which is itself a product of $\mathbb Z$'s), but I'm not sure that map is injective (my scribblings don't get me anywhere, I seem to need to rely on $\mathbb Z^{presheaf}[S]$ being a seprated presheaf, which isn't clear to me). Could you explain that ? | |
| Apr 1, 2020 at 14:01 | comment | added | Maxime Ramzi | Thanks for answering ! The result (and the proof) seem to indicate that my intuition that set-theoretic issues aren't silly here isn't completely off. There's just a point in your argument that I'm not entirely sure about : I'm not entirely sure how you embed $\mathbb Z[S]$ into a product of $\mathbb Z$'s - I don't understand how you guarantee that you get an injection from sufficiently many maps $S\to \mathbb Z$ (surely you can get an injection $S\to \prod_A\mathbb Z$, but I'm not sure how you get to $\mathbb Z[S]$). One possible way would be | |
| Apr 1, 2020 at 11:02 | history | answered | Peter Scholze | CC BY-SA 4.0 |