Timeline for Local property of split exact sequence
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 14, 2017 at 0:55 | vote | accept | Jian | ||
| Jun 13, 2017 at 16:09 | answer | added | Jeremy Rickard | timeline score: 9 | |
| S Jun 13, 2017 at 14:11 | review | First posts | |||
| Jun 13, 2017 at 14:45 | |||||
| Jun 13, 2017 at 14:11 | history | reopened |
Daniel Loughran Yemon Choi Mikhail Katz Stefan Kohl♦ Peter LeFanu Lumsdaine |
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| Jun 13, 2017 at 12:09 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Language editing.
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| Jun 13, 2017 at 3:01 | review | Reopen votes | |||
| S Jun 13, 2017 at 14:11 | |||||
| Jun 9, 2017 at 2:54 | vote | accept | Jian | ||
| Jun 9, 2017 at 2:54 | |||||
| Jun 8, 2017 at 10:20 | comment | added | js21 | @Jeremy Rickard: Very nice observation! Thank you. | |
| Jun 8, 2017 at 9:56 | comment | added | Jeremy Rickard | @js21 Actually, even if $M$ is just finitely presented, the natural map $\text{Ext}^1_A(M,N)\otimes_AB\to\text{Ext}^1_B(M\otimes_AB,N\otimes_AB)$ is injective, which is enough for this question. | |
| Jun 8, 2017 at 9:25 | comment | added | js21 | @Jeremy Rickard: You are right. The correct statement is that it is true (for $\mathrm{Ext}^1$) whenever $M$ is $(-2)$-pseudo-coherent. Using Lemma 10.72.1 from Stacks Project $087M$, this amounts to show that $\mathrm{Ext}^1_A(M , N \otimes_A B) = \mathrm{Ext}^1_A(M , N ) \otimes_A B$. By Lazard's theorem, $B$ is a cofiltered colimit of finite free $A$-modules. One concludes using the finiteness assumtion on $M$ | |
| Jun 8, 2017 at 8:09 | comment | added | Jeremy Rickard | @js21 I think you need some extra finiteness condition, such as $A$ Noetherian, for the statement about $\text{Ext}^*$? | |
| Jun 8, 2017 at 8:01 | review | Reopen votes | |||
| Jun 8, 2017 at 10:19 | |||||
| Jun 8, 2017 at 6:36 | comment | added | Niels | I don't really understand why this question should be off-topic. As the mistaken answer below shows, it is not so easy to come up with a counterexample, is it ? | |
| Jun 7, 2017 at 12:29 | history | closed |
js21 Jeremy Rickard Henry.L abx user21574 |
Not suitable for this site | |
| Jun 7, 2017 at 11:50 | answer | added | Fernando Muro | timeline score: -2 | |
| Jun 7, 2017 at 11:06 | review | Close votes | |||
| Jun 7, 2017 at 12:29 | |||||
| Jun 7, 2017 at 10:53 | comment | added | js21 | If $M$ is a finitely generated $A$-module, and $A \rightarrow B$ is a flat morphism of rings, then $\mathrm{Ext}^*_B(M \otimes_A B, N \otimes_A B) = \mathrm{Ext}^*_A(M , N ) \otimes_A B$ : apply this with $B$ equal to various localizations of $A$ to get your result, when the last term of the exact sequence is finitely generated. Without the finite generation hypothesis, your assertion is false. | |
| Jun 7, 2017 at 10:00 | history | asked | Jian | CC BY-SA 3.0 |