Timeline for Completion of a local ring of an arithmetic surface
Current License: CC BY-SA 4.0
11 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Jun 24, 2019 at 4:28 | comment | added | Angelo | The local rings that you describe are all regular, so any arithmetic surface that is not regular can not have this form. In general you get complete locar rings of the form $R[[S,T]]/(ST-\pi^n)$ for some $n \ge 0$. | |
Jun 23, 2019 at 20:39 | history | edited | nowhere dense | CC BY-SA 4.0 |
added 214 characters in body
|
Dec 27, 2018 at 21:27 | history | edited | nowhere dense | CC BY-SA 4.0 |
edited title
|
Dec 21, 2018 at 12:10 | history | edited | nowhere dense | CC BY-SA 4.0 |
added 133 characters in body
|
Dec 21, 2018 at 10:52 | history | edited | nowhere dense | CC BY-SA 4.0 |
deleted 31 characters in body
|
Dec 20, 2018 at 21:16 | history | edited | YCor |
edited tags
|
|
Dec 20, 2018 at 21:11 | history | edited | nowhere dense | CC BY-SA 4.0 |
added 126 characters in body
|
Dec 20, 2018 at 21:09 | comment | added | nowhere dense | @JasonStarr oh I see. So I think in order to have a chance to be true we should at least ask for $k$ to be algebraically closed (or maybe just $x$ a rational point). I will add this hypothesis to the question as it is still useful with that for me. | |
Dec 20, 2018 at 21:05 | comment | added | Jason Starr | Welcome new contributor. That is false. For instance, the residue field of $\mathcal{O}_{X,x}$ may be a nontrivial extension of the residue field of $R$. | |
Dec 20, 2018 at 21:05 | review | First posts | |||
Dec 20, 2018 at 21:08 | |||||
Dec 20, 2018 at 21:02 | history | asked | nowhere dense | CC BY-SA 4.0 |