Timeline for How to show integrally closed implies topologically unibranch
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| May 15, 2015 at 1:40 | comment | added | grghxy | You will enjoy item 6.1 in Chapter V of math.upenn.edu/~chai/624_08/mumford-oda_chap1-6.pdf (the AG book which Mumford had intended to write some day!). This shows that Mumford distinguished the notion of "analytically unibranch" (what I had in mind) and "topologically unibranch" (the notion defined in his earlier book), and for analytifications of varieties he shows at that link that the two are equivalent. Nonetheless, I think that "analytically unibranch" is a more natural and interesting notion (e.g., it intrinsic to $X^{\rm{an}}$, unlike the other). | |
| May 15, 2015 at 1:23 | comment | added | grghxy | "Topologically unibranch" is an analytic notion, so the definition should not use only complements of Zariski-closed sets (as opposed to analytic sets, or more general closed sets) in analytic open neighborhoods of $x$ in $X^{\rm{an}}$. The definition is intrinsic to complex-analytic spaces, and one gets this "correct" notion from my first CAS reference (or from p. 261 of Joe Harris' AG book, for curves). The correctness of Mumford's implication with a stronger meaning of the conclusion is what I sketched how to prove. (Above Prop. A.5 Mumford gives yet another definition!) | |
| May 14, 2015 at 19:59 | comment | added | Rene Schipperus | The definition is not "wrong" it is taken from Mumford's book and is the meaning of the quoted implication. But thanks for the references. Ill take a look at the Coherent sheaves book. | |
| May 14, 2015 at 18:42 | comment | added | grghxy | The definition you give for "topologically unibranch" is wrong (the definition involves prime germs at $x$ in the analytic space $X^{\rm{an}}$: see 4.1.3-4.1.4 in "Coherent Analytic Sheaves"). By Serre's homological criteria for normality and the identification of the completions of the local noetherian rings on $X$ and $X^{\rm{an}}$ at $x$, the space $X^{\rm{an}}$ is normal at $x$ since the completion is normal (by excellence of $X$!). By openness of the normal locus, we can conclude via the local structure of normal analytic spaces: see 7.4/2 and 9.2/3 in "Coherent Analytic Sheaves". | |
| May 14, 2015 at 12:08 | history | asked | Rene Schipperus | CC BY-SA 3.0 |