Timeline for Example for 1-dim, Noeth., local domain which is unibranched but not analytically irreducible
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 28, 2013 at 14:50 | comment | added | RRr | I found an example (of Akizuki) here http://arxiv.org/pdf/alg-geom/9503017v1.pdf | |
| Aug 19, 2013 at 8:24 | comment | added | RRr | Thank you for the hints and yes Youngsu, that is what I wanted to ask. Also it is known that under the assumptions above, $D$ is analytically irreducible if and only if $D$ is unibranched and $D'$ is finitely generated as $D$-module if and only if $D$ is unibranched and the $\mathfrak{m}'$-adic topologyon $D'$ induces the $\mathfrak{m}$-adic topology (where $\mathfrak{m}'$ is the maximal ideal of $D'$) | |
| Aug 16, 2013 at 20:20 | comment | added | Jason Starr | @Youngsu: Thank you for your clarification. Since $\widehat{D}$ is $D$-flat, the inclusion $D\hookrightarrow D'$ induces an inclusion $\widehat{D}\hookrightarrow \widehat{D}\otimes_D D'$. So also, for any example, the induced map $\widehat{D}\otimes_D D' \to \widehat{D'}$ must be noninjective. | |
| Aug 16, 2013 at 20:05 | comment | added | Youngsu | I think the original post asks if there is an example where $\widehat{D}$ is not a domain not $\widehat{D'}$. In addition to your answer, if $\widehat{D}$ is reduced, then there is a 1-1 correspondence between the maximal ideals of $D'$ and the minimal primes of $\widehat{D}$. Since $D'$ is local, there is unique minimal prime of $\widehat{D}$. Since $\widehat{D}$ is reduced, it is a domain. Therefore, if there is an example, $\widehat{D}$ needs to be non reduced. | |
| Aug 16, 2013 at 17:30 | comment | added | Jason Starr | I am not sure that I understand. By the Krull-Akizuki theorem, the integral closure $D'$ of $D$ is still Noetherian. Thus, if $D'$ is local, then it is a Noetherian, local domain that is also integrally closed. Since $D\subset D'$ is an integral extension, then it satisfies the "incomparability property" so that the dimension of the local ring $D'$ is at most $1$. As a Noetherian, normal, local domain of dimension $1$, $D'$ is regular. Thus the completion of $D'$ is also a regular local ring, hence the completion is a domain. | |
| Aug 16, 2013 at 9:36 | review | First posts | |||
| Aug 16, 2013 at 9:43 | |||||
| Aug 16, 2013 at 9:16 | history | asked | RRr | CC BY-SA 3.0 |