Timeline for Wanted: example of a non-algebraic singularity
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 28, 2011 at 20:41 | answer | added | Joël | timeline score: 17 | |
| Sep 27, 2011 at 23:52 | vote | accept | Anton Geraschenko | ||
| Sep 26, 2011 at 18:24 | answer | added | Laurent Moret-Bailly | timeline score: 36 | |
| Sep 25, 2011 at 0:12 | comment | added | Joël | but I have not been able to do it. An other way would be to prove that a formal singularity of this type is at least analytic, and then to apply the theorem quoted in Ulrich's comment. I thought that it was a consequence of Artin's approximation theorem that any formal singularity was analytic, but I wasn't able to get sure that it was true and I have to go for dinner. I am very curious though. | |
| Sep 25, 2011 at 0:07 | comment | added | Joël | @Anton: I like your question, and I am surprised it has not been fully answered yet. There must be no active singularity theorists on math overflow at this moment. So for what has been written so far in comment, it seems that there will be no example of a non-algebraic formal singularity of the form $\mathbb{C}[[x,y]]/(f(X,Y))$ but I can't prove it (I am really a great beginner in that theory, I should say). I have two ideas about how one could prove it: One way would be to use Theorem 3.8 of Artin quoted in my comment, and to prove that a singularity of this form is necessarily isolated... | |
| Sep 24, 2011 at 2:59 | answer | added | Jorge Vitório Pereira | timeline score: 11 | |
| Sep 23, 2011 at 23:58 | comment | added | Anton Geraschenko | @Joël: That's very interesting; thank you for the reference! (archive.numdam.org/article/PMIHES_1969__36__23_0.pdf) The result is over an arbitrary field, too. I'd like to say that this means that the singularity at the generic point of the singular locus must be algebraic, but of course localization does not commute with completion. | |
| Sep 23, 2011 at 23:32 | comment | added | Joël | If the singularity is isolated (that is if $\mathbb{ℂ}[[x,y]]/f(x,y)$ minus its closed point is formally smooth over $\mathbb{C}$), then it is algebraic. This is Theorem 3.8 in Artin's paper on "Algebraic approximation of structures over Henselian local rings", at IHES, hence easily available on numdam.org. It uses Artin's approximation theorem plus works of Hironaka | |
| Sep 23, 2011 at 21:55 | history | edited | Anton Geraschenko |
shameless bump
|
|
| Aug 29, 2011 at 15:58 | comment | added | Anton Geraschenko | @S. Carnahan: that information is automatically included, since the topology is induced by the maximal ideal. | |
| Aug 29, 2011 at 15:33 | comment | added | S. Carnahan♦ | Do you want the singularity type to include the datum of the topology on the completed local ring? | |
| Aug 29, 2011 at 14:48 | comment | added | Michael Bächtold | Not an answer to your question, but maybe you find something relevant in Angelos answer here: mathoverflow.net/questions/51530/… | |
| Aug 29, 2011 at 12:41 | comment | added | naf | Regarding your expectation, any analytic plane curve singularity is in fact algebraic. See, for example, Corollary 7.7.3 of the book by Casas-Alvero "Singularities of plane curves", London Mathematical Society Lecture Note Series, 276. I'm not sure though whether a similar result holds for formal planar singularities but it seems possible to me. | |
| Aug 29, 2011 at 6:46 | comment | added | Anton Geraschenko | One idea I had is to take a non-algebraic variety $X$, and consider the completed local ring at the cone point of a cone on $X$. The problem is that making a cone on $X$ involves choosing a very ample line bundle on $X$. But as soon as $X$ is projective, it is algebraic by GAGA. | |
| Aug 29, 2011 at 6:36 | history | asked | Anton Geraschenko | CC BY-SA 3.0 |