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

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).