Timeline for non constant regular function derivative is zero
Current License: CC BY-SA 4.0
14 events
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| Sep 22, 2020 at 6:41 | review | Close votes | |||
| Sep 27, 2020 at 3:02 | |||||
| Sep 22, 2020 at 5:06 | comment | added | Sunny | It seems like only possible counter example to the above claim are of the same flavour i. e. Take an algebaric extension of rational so that it's module of differential is zero and attach finite many variable. Is it true these are the only possible way the above claim could fail. | |
| Sep 22, 2020 at 4:59 | comment | added | abx | Answer to your edit: no. Take $k=\mathbb{Q}$, $R=\overline{\mathbb{Q}}[x]$. | |
| Sep 22, 2020 at 4:27 | history | edited | Sunny | CC BY-SA 4.0 |
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| Sep 21, 2020 at 17:01 | comment | added | Sunny | Yes I agree with @Achinger. May be I should edit the question. | |
| Sep 21, 2020 at 16:20 | comment | added | Piotr Achinger | I think the correct statement should be that $H^0_{\rm dR}(R/k)$ is a finite $k$-algebra (and hence a product of finite field extensions). | |
| Sep 21, 2020 at 15:44 | comment | added | Donu Arapura | OK, building on Mohan's comment, take $R=\mathbb{Q}(i)[x]$. A bit more precision perhaps? | |
| Sep 21, 2020 at 15:31 | history | edited | Sunny | CC BY-SA 4.0 |
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| Sep 21, 2020 at 15:30 | comment | added | Sunny | Ok well you are right. But in some sense these are kind of trivial counter example which I don't want. May be I should be precise about my question and should say dimension of R is greater than 0. | |
| Sep 21, 2020 at 15:19 | comment | added | Mohan | Take $k=\mathbb{Q}$ and $R$ be any finite extension . | |
| Sep 21, 2020 at 12:36 | comment | added | Sunny | Sure we can assume that as this is also needed in manifold case as well. | |
| Sep 21, 2020 at 12:34 | comment | added | Sam Gunningham | Probably you want to assume $Spec(R)$ is connected? | |
| Sep 21, 2020 at 11:57 | history | edited | Sunny | CC BY-SA 4.0 |
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| Sep 21, 2020 at 11:49 | history | asked | Sunny | CC BY-SA 4.0 |