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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