Timeline for Unibranch rings
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 14, 2011 at 15:28 | comment | added | user1594 | Thank you both for your answers. Tom's counter-example is instructive. | |
| Jun 14, 2011 at 13:16 | comment | added | Laurent Moret-Bailly | As far as I know, the accepted definition of "unibranch" is given above by Tom. References: EGA 0_I 6.5.10 (Springer edition), and Raynaud, LNM 169, chapter IX which implies that $A$ is unibranch iff its henselization is a domain. Whether the completion is then irreducible was an open problem when EGA IV_4 was written (EGA IV, 18.9.6 (ii)) but it is true if $A$ is excellent (18.9.1). With JT's definition, Tom's example does look like a counterexample. | |
| Jun 14, 2011 at 1:46 | comment | added | Tom Goodwillie | Isn't this a counterexample? Let $B$ be $\mathbb C[x,y]$ and let $A\subset B$ consist of those $f(x,y)$ such that $f(1-t^2,t-t^3)=f(1-t^2,t^3-t)$. The normalization is $B$. Localize at the origin. | |
| Jun 14, 2011 at 1:28 | comment | added | Tom Goodwillie | In that exercise "unibranch" means something much weaker: the normalization has only one point over the closed point. | |
| Jun 14, 2011 at 0:52 | comment | added | Graham Leuschke | If the completion is a domain, then it's unibranch. An exercise in Bourbaki's Commutative Algebra: books.google.com/… | |
| Jun 13, 2011 at 21:51 | history | asked | user1594 | CC BY-SA 3.0 |