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Let $R$ be a Noetherian domain, and let $\mathfrak{p}$ be a prime ideal; consider the completion $\hat R_{\mathfrak{p}}$ of $R$ at $\mathfrak{p}$ (the inverse limit of the system of quotients $R/\mathfrak{p}^n$). If $R$ is a PID, it is easy to see that $\hat R_{\mathfrak{p}}$ is a domain.

Someone asked in sci.math if $\hat R_{\mathfrak{p}}$ would always be a domain. I thought it would, but looking at Eisenbud's "Commutative Algebra", I found a reference to a theorem of Larfeldt and Lech that says that if $A$ is any finite-dimensional algebra over a field $k$, then there is a Noetherian local integral domain $R$ with maximal ideal $\mathfrak{m}$ such that $\hat{R_{\mathfrak{M}}}\cong A[[x_1,\ldots,x_n]]$ for some $n$; and so this completion will not be a domain if $A$ is not a domain. I would like to know an example directly, if possible.

Does someone know an easy example of a noetherian domain $R$ and a prime ideal $\mathfrak{p}$ such that $\hat R_{\mathfrak{p}}$ is not a domain? Thanks in advance.

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    $\begingroup$ Serre's homological conditions $R_i$ and $S_i$ are well-behaved under completion for excellent local rings, so properties characterized in this way such as reducedness ($R_0$ + $S_1$) and normality ($R_1$ + $S_2$) are preserved under completion for such rings. So those notions are well-behaved under analytification of algebraic schemes (over complex numbers, as well as over non-archimedean fields) since algebraic and analytic local rings are excellent and have the "same" completion. $\endgroup$
    – BCnrd
    Commented Mar 17, 2010 at 16:12
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    $\begingroup$ By the way, a local ring whose completion is a domain is called sometimes analytically irreducible, for obvious reasons. $\endgroup$
    – Karl Schwede
    Commented Jan 27, 2011 at 19:44
  • $\begingroup$ Hm, perhaps I should also mention that a curve singularity whose completion is a domain is called unibranch. $\endgroup$
    – Karl Schwede
    Commented Mar 20, 2013 at 12:51

4 Answers 4

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Let $R=\mathbb{C}[x,y]/(y^2-x^2(x-1))$. This is the nodal cubic in the plane. Look at the prime $\mathfrak{p}=(x,y)$, corresponding to the nodal point. The completion here is isomorphic to $\mathbb{C}[[x,y]]/(xy)$.

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It might be worth pointing out that you get an example of this by localizing at any point of a variety (scheme) which is irreducible but not (analytically/formally/étale) locally irreducible. In particular, any self-intersecting curve will give an example, just like the one above by Charles.

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    $\begingroup$ Yes, and related to this, one has the fact that being irreducible is not an etale local property (or an analytically local property), which is one reason why irreducible components of (complex or rigid) analytic varieties, or stacks, are somewhat involved to define. $\endgroup$
    – Emerton
    Commented Oct 4, 2010 at 4:07
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At least for algebraic local domains R, we have 1-1 correspondence between the minimal primes in the completion $\cap{R}$ and the maximal ideals in the integral closure of R in its quotient field. This will produce a lot of examples including curves near a point whose neighbourhood in the curve can't be covered by a single parameterization. In fact, various connections come to the fore here. For several ways to characterize the number of minimal primes in the completion of a local ring of an irreducible plane curve, one can look at S. S. Abhyankar's Chavounet prize winning paper "Historical Ramblings in Algebraic Geometry and related algebra".

Several connections are worth mentioning, eg. concepts such as Henselization and Hensel's lemma, Zariski's Main theorem (as a special case, it mentions that the completion of a normal algebraic local domain is again a normal domain), links associated to the singularities of algebraic curves, various reciprocity laws from Kummer to Artin in Algebraic Number Theory etc.

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There's another example (going back to Nagat) in here.

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