# Example of a reduced ring whose completion is not reduced

Let $(R,\mathfrak{m})$ be a local Noetherian ring. Give an example such that R is reduced but $\mathfrak{m}$-adic completion of R is not reduced.

• Matsumura refers to [Y. Akizuki, Eigene Bemerkungen über primäre Integritätsberreiche mit Teilerkettensatz. Proc. Phys.–Math. Soc.Japan 17 (1935), 327–36.] for the first example of a one-dimensional Noetherian local integral domain with non-reduced completion. – Mariano Suárez-Álvarez Jul 13 '13 at 18:36
• I think that Nagata's "Local Rings" works out such an example in detail. – paul Monsky Jul 14 '13 at 0:08
• @paul Monsky: Yes, E.6.1 on page 209 (where he treats in detail just the case of equicharacteristic 2 "for simplicity"). – user36938 Jul 14 '13 at 1:28
• I find this example slightly troubling. Your ring $R$ is a dimension 1 Noetherian local ring, and thus Cohen-Macauly. So then its completion will be a Cohen-Macaulay Noetherian local ring in dimension 1, but such rings are necessarily reduced. Am I missing something obvious? Please advise. – user37270 Jul 17 '13 at 23:26
• Sorry for posting what should have been a comment on the previous answer as an answer itself; I just couldn't figure out how to post it as a comment. – user37270 Jul 17 '13 at 23:42

## 2 Answers

Perhaps the "simplest" kind of example would be a local noetherian domain that is finite flat over a discrete valuation ring, though any such example has to be in equicharacteristic $p > 0$ since otherwise excellence considerations for the discrete valuation ring of generic characteristic 0 would rule it out.

Such an example is lurking in Example 11 in section 3.6 of the book "Neron Models". That example (which omits a few details in its justification) provides a discrete valuation ring $A$ over $k = \mathbf{F}_p$ with residue field $k$ such that there is $a \in A$ with no $p$th root in $A$ but a $p$th root in $\widehat{A}$. This ring $A$ is built between $k(T)$ and $k(T,U)$ (and it is not essentially of finite type over $k$, as otherwise it would be excellent, contradicting what is below).

The ring $R = A[x]/(x^p-a)$ is a domain since it is $A$-flat and its localization to the fraction field $K$ of $A$ is the ring $K[x]/(x^p-a)$ that is a field (as $x^p - a$ is irreducible over $K$, due to the normality of $A$). The ring $R$ is local noetherian of dimension 1 since it is finite flat over $A$ with special fiber $k[x]/(x^p- \overline{a})$ that is artin local. But $A$-finiteness also implies that $\widehat{R} = \widehat{A} \otimes_A R = \widehat{A}[x]/(x^p-a)$, and by inspection this is non-reduced (since $a$ admits a $p$th root in $\widehat{A}$ by design).

• Can any one tell me what are the prerequisites to understand this example. – MAT Jul 15 '13 at 18:37
• Have you tried to read the Example for yourself to understand it? Your question about "prerequisites" is somewhat vague without saying anything more about what you have tried to do and where you ran into difficulty. (You certainly don't need to know about excellence to understand the Example on its own terms.) – user36938 Jul 16 '13 at 1:21

In the Stacks project you can find the standard Example Tag 00PB as well as the more interesting Example Tag 02JD due to Ferrand-Raynaud.