**Proof.**
In this proof we will use the remarks immediately following Definition 10.162.9. As $R \to R^\wedge $ is a faithfully flat local ring homomorphism it is injective and (1) follows.

Let $\mathfrak q$ be a minimal prime of $R$, and assume $R$ is analytically unramified. Then $\mathfrak q$ is an associated prime of $R$ (see Proposition 10.63.6). Hence there exists an $f \in R$ such that $\{ x \in R \mid fx = 0\} = \mathfrak q$. Note that $(R/\mathfrak q)^\wedge = R^\wedge /\mathfrak q^\wedge $, and that $\{ x \in R^\wedge \mid fx = 0\} = \mathfrak q^\wedge $, because completion is exact (Lemma 10.97.2). If $x \in R^\wedge $ is such that $x^2 \in \mathfrak q^\wedge $, then $fx^2 = 0$ hence $(fx)^2 = 0$ hence $fx = 0$ hence $x \in \mathfrak q^\wedge $. Thus $\mathfrak q$ is analytically unramified and (2) holds.

Assume $R$ is reduced with minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$, and each $\mathfrak q_ i$ is analytically unramified. Then $R \to R/\mathfrak q_1 \times \ldots \times R/\mathfrak q_ t$ is injective. Since completion is exact (see Lemma 10.97.2) we see that $R^\wedge \subset (R/\mathfrak q_1)^\wedge \times \ldots \times (R/\mathfrak q_ t)^\wedge $. Hence (3) is clear.

Assume $R$ is analytically unramified. Let $\mathfrak p_1, \ldots , \mathfrak p_ s$ be the minimal primes of $R^\wedge $. Then we see that

\[ Q(R^\wedge ) = R^\wedge _{\mathfrak p_1} \times \ldots \times R^\wedge _{\mathfrak p_ s} \]

with each $R^\wedge _{\mathfrak p_ i}$ a field as $R^\wedge $ is reduced (see Lemma 10.25.4). Hence the integral closure $S$ of $R^\wedge $ in $Q(R^\wedge )$ is equal to $S = S_1 \times \ldots \times S_ s$ with $S_ i$ the integral closure of $R^\wedge /\mathfrak p_ i$ in its fraction field. In particular $S$ is finite over $R^\wedge $. Denote $R'$ the integral closure of $R$ in $Q(R)$. As $R \to R^\wedge $ is flat we see that $R' \otimes _ R R^\wedge \subset Q(R) \otimes _ R R^\wedge \subset Q(R^\wedge )$. Moreover $R' \otimes _ R R^\wedge $ is integral over $R^\wedge $ (Lemma 10.36.13). Hence $R' \otimes _ R R^\wedge \subset S$ is a $R^\wedge $-submodule. As $R^\wedge $ is Noetherian it is a finite $R^\wedge $-module. Thus we may find $f_1, \ldots , f_ n \in R'$ such that $R' \otimes _ R R^\wedge $ is generated by the elements $f_ i \otimes 1$ as a $R^\wedge $-module. By faithful flatness we see that $R'$ is generated by $f_1, \ldots , f_ n$ as an $R$-module. This proves (4).

Part (5) is a special case of part (4).
$\square$

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