Timeline for Proposition 5.13 (ii) in Scholze's Perfectoid Spaces
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Mar 28, 2021 at 20:30 | vote | accept | Kenny Lau | ||
| Mar 21, 2021 at 21:20 | answer | added | Peter Scholze | timeline score: 9 | |
| Mar 21, 2021 at 13:14 | comment | added | ali | if you are comfortable with french in illusie thesis all this things defined and discussed with great details | |
| Mar 21, 2021 at 13:13 | comment | added | ali | sorry i didn't see the $S\circ$ in your comment I was talking about the case that base is an actual ring. also what do you mean by pointwise?I mean tensor each term with $S$(in general you put $\oplus_{i+j=n}M_i\otimes N_j$ in degree $n$ of $M_\circ\otimes N_\circ$.).in the case of $L_{S/R}$ the last step when you tensor with $S$ over $S_\circ$ does not really change the complex up to an quasi-isomorphism because $S_\circ$ and $S$ are quisi isomorphic to each other.(so you can forget about it) | |
| Mar 21, 2021 at 12:20 | history | edited | Kenny Lau | CC BY-SA 4.0 |
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| Mar 21, 2021 at 12:19 | comment | added | Kenny Lau | If $\otimes_{S_\bullet}$ is defined pointwise and $\Bbb L_{S/R} = \Omega^1_{S_\bullet/R} \otimes_{S_\bullet} S$ and $S$ is concentrated at degree $0$ then does that mean $\Bbb L_{S/R}$ is concentrated at degree $0$? | |
| Mar 21, 2021 at 11:57 | history | edited | Kenny Lau | CC BY-SA 4.0 |
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| Mar 21, 2021 at 11:52 | comment | added | ali | first, it is not $\otimes^L$ it is $\otimes$ so yes it is pointwise,in this particular case $R_\Phi$ and $S$ and hence each free $S$ algebra are Tor independent so there is no difference. also for your last question as you said $S_\Phi,R_\Phi$ as ring are just $R,S$ so obviously $L_{S_\phi/R_\phi}=L_{R/S}$! | |
| Mar 21, 2021 at 9:22 | comment | added | Kenny Lau | @ali I apologize for the basic question, but is $\otimes_{S^\bullet}$ also defined pointwise? | |
| Mar 21, 2021 at 9:11 | comment | added | ali | module differential of a complex is just obtained by the complex of the module of differentials of each term of that complex. so the fact about rings by definition gives the fact about complexes | |
| Mar 21, 2021 at 7:52 | comment | added | Kenny Lau | @ali Re $\Omega_{S\otimes_{R} A/A}=\Omega_{S/R}\otimes A$: Wikipedia has a similar fact, but that are for rings; here $S^\bullet$ is a complex, so I don't know if I can still use it. | |
| Mar 21, 2021 at 7:50 | comment | added | Kenny Lau | @ali I've thought more about the last statement and I have edited to add my thoughts. | |
| Mar 21, 2021 at 7:50 | history | edited | Kenny Lau | CC BY-SA 4.0 |
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| Mar 21, 2021 at 6:59 | comment | added | ali | do you know that $\Omega_{S\otimes_{R} A/A}=\Omega_{S/R}\otimes A$? and your last statement does not make sense because you say yourself that $R_{(\Phi)} \otimes_R^{\Bbb L} \Bbb L_{S/R} \cong \Bbb L_{S_{(\Phi)}/R_{(\Phi)}}$ unless in your context $R$ is prefect and $R=R_{\Phi}$. | |
| Mar 21, 2021 at 0:26 | history | asked | Kenny Lau | CC BY-SA 4.0 |