Does somebody know an example for an 1-dim., Noeth., local domain $D$ which is unibranched (that is, its integral closure $D'$ is local) but not analytically irreducible (that is, its $\mathfrak{m}$-adic completion $\widehat{D}$ is not a domain where $\mathfrak{m}$ is the maximal ideal of $D$)?
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$\begingroup$ 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. $\endgroup$– Jason StarrCommented Aug 16, 2013 at 17:30
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1$\begingroup$ 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. $\endgroup$– YoungsuCommented Aug 16, 2013 at 20:05
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$\begingroup$ @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. $\endgroup$– Jason StarrCommented Aug 16, 2013 at 20:20
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$\begingroup$ 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 $D'$) $\endgroup$– RRrCommented Aug 19, 2013 at 8:24
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$\begingroup$ I found an example (of Akizuki) here http://arxiv.org/pdf/alg-geom/9503017v1.pdf $\endgroup$– RRrCommented Aug 28, 2013 at 14:50
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