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Suppose $R$ is a domain, finitely generated over an algebraically closed field, and $\mathfrak{m}\subset R$ is a maximal ideal. Is $\underleftarrow{\lim} R/\mathfrak{m}^n$ necessarily a domain?

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This phenomenon of posing a question in the title and the posing its negation in the body is very curious to me. Anyway, it is possible for the completion of a domain to be a non-domain.

Let $R = k[x, y]/(y^2 - x^3 - x^2)$ where $k$ is a field of characteristic not equal to $2$ and let $\mathfrak{m} = (x, y)$. In the completion $\sqrt{1 + x}$ exists (e.g. by Hensel's lemma, or more explicitly by the binomial theorem) and consequently the defining polynomial admits a factorization

$$(y - x \sqrt{1 + x})(y + x \sqrt{1 + x})$$

so the completion has a zero divisor. This corresponds geometrically to the fact that $y^2 = x^3 + x^2$ has a singularity at the origin (the curve looks locally like an intersection of a pair of lines and this can be detected by completing with respect to $\mathfrak{m}$).

For a number-theoretic example let $R = \mathbb{Z}[x]/(x^2 + 3)$ and let $\mathfrak{m} = (2, x + 1)$. In the completion $\sqrt{1 - 4}$ exists in $\mathbb{Z}_2$ (again by either Hensel's lemma or the binomial theorem), so the same argument as above produces a zero divisor. Morally speaking this is for the same reason as above: $\text{Spec } R$ is singular at $\mathfrak{m}$.

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Hmm, is this true for all singular points? (As in, is the completion of a finite dimension local ring an integral domain if and only if the local ring is regular?) –  only Aug 25 '12 at 1:00
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@only: I don't think so. Consider $k[x, y]/(y^2 - x^3)$ at $\mathfrak{m} = (x, y)$. It seems to me that the completion is still a domain (it just fails to be integrally closed). –  Qiaochu Yuan Aug 25 '12 at 1:03
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The phenomenon occurs because people don't want to sound repetitive. –  Kevin Aug 25 '12 at 1:15

The two examples given by Qiaochu are singular points, one gives rise to a domain when passing to the completion but not the other. What makes the difference ?

Consider the normalization morphism of $\mathrm{Spec}(R)$. In both cases the normalization is the affine line parametrized by $t:=y/x$. For the first curve, there are two points in the affine line lying over the singular point $x=y=0$: they are given by $t=1$ and $t=-1$. In the second case, there is only one point $t=0$ lying over the singular point. This actually characterizes the integrality:

Let $A$ be a noetherian integral domain. Suppose $A$ is excellent (e.g. $A$ is a localization of a finitely generated algebra over either a field or $\mathbb Z$). Let $B$ be the integral closure of $A$. Let $m$ be a maximal ideal of $A$. Then the $m$-adic completion $\hat{A}$ is a domain if and only if there is only one maximal ideal of $B$ lying over $m$.

Proof. By the excellent hypothesis, $B$ is finite over $A$. Let $m_1, \dots, m_n$ be the maximal ideals of $B$ lying over $A$. A second fact resulting from the excellent hypothesis is that $B$ is excellent, and the normalization commutes with the completion. In particular, the ($m_i$-adic) completion of $B_{m_i}$ is integrally closed. It is an integral domain because it is integrally closed and connected (it is a local ring).

Let $K=\mathrm{Frac}(A)$. Then $B\subseteq K$. By the flatness of $A\to \hat{A}$, we have $$\hat{A}\subseteq \hat{A}\otimes_A B\subseteq \hat{A}\otimes_A K\subseteq \mathrm{Frac}(\hat{A})$$ where the last Frac denotes the total ring of fractions. We conclude that $\hat{A}$ is a domain if and only if $\hat{A}\otimes_A B$ is a domain.

Finally, as a general fact, there is a canonical isomorphism $$ \hat{A}\otimes_A B \simeq \hat{B}=\prod_{1\le i\le n} \widehat{(B_{m_i})}$$ where $\hat{B}$ denotes the $mB$-adic completion of $B$. Therefore, $\hat{A}\otimes_A B$ (hence $\hat{A}$) is a domain if and only if $n=1$.

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