Timeline for Proposition 5.13 (ii) in Scholze's Perfectoid Spaces
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 28, 2021 at 20:30 | vote | accept | Kenny Lau | ||
| Mar 22, 2021 at 8:19 | comment | added | Peter Scholze | You're welcome! For this point, first note that "K-flatness" is only relevant for complexes unbounded to the right, so not here, where flat resolution are good enough. But actually, by the usual formalism of derived functors, acyclic resolutions (for the functor $M\mapsto R'\otimes_R^L M$) are enough, and the given resolution is $R'\otimes_R -$-acyclic, as all terms are free over $S$ and $R'\otimes_R^L S$ sits in degree $0$. | |
| Mar 22, 2021 at 1:23 | comment | added | Kenny Lau | $\newcommand{\OSR}{\Omega^1_{S_\bullet/R}}\newcommand{\ot}{\otimes}$Thank you for your answer. I understand that the two simplicial $R'$-modules, $R'\ot_R(\OSR\ot_{S_\bullet}S)$ and $\Omega^1_{S'_\bullet/R'}\ot_{S'_\bullet}S'$, agree. I also accept that $\Bbb L_{S/R}=\OSR\ot_{S_\bullet}S$. What I am confused about is why $R'\ot_R^L\Bbb L_{S/R}=R'\ot_R(\OSR\ot_{S_\bullet}S)$. According to Definition 15.58.15 in Stacks, one needs a K-flat resolution to compute $\ot_R^L$, and it is not obvious to me why either $R'$ or $\OSR\ot_{S_\bullet}S$ is K-flat. | |
| Mar 21, 2021 at 21:20 | history | answered | Peter Scholze | CC BY-SA 4.0 |