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Let us call a Noetherian local ring $A$ unibranch if it is a domain and the normalization map is finite and induces a bijection on spectra.

My question is as follows: is this property preserved when one completes at the maximal ideal of $A$? What about being geometrically unibranch?

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    $\begingroup$ If the completion is a domain, then it's unibranch. An exercise in Bourbaki's Commutative Algebra: books.google.com/… $\endgroup$
    – Graham Leuschke
    Jun 14, 2011 at 0:52
  • $\begingroup$ In that exercise "unibranch" means something much weaker: the normalization has only one point over the closed point. $\endgroup$
    – Tom Goodwillie
    Jun 14, 2011 at 1:28
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    $\begingroup$ 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. $\endgroup$
    – Tom Goodwillie
    Jun 14, 2011 at 1:46
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    $\begingroup$ 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. $\endgroup$
    – Laurent Moret-Bailly
    Jun 14, 2011 at 13:16
  • $\begingroup$ Thank you both for your answers. Tom's counter-example is instructive. $\endgroup$
    – user1594
    Jun 14, 2011 at 15:28

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