**Proof.**
Since $R \to R^ h \to R^{sh}$ are faithfully flat (Lemma 15.44.1), we see that $R^ h$ or $R^{sh}$ being Noetherian implies that $R$ is Noetherian, see Algebra, Lemma 10.158.1. In the rest of the proof we assume $R$ is Noetherian.

As $\mathfrak m \subset R$ is finitely generated it follows that $\mathfrak m^ h = \mathfrak m R^ h$ and $\mathfrak m^{sh} = \mathfrak mR^{sh}$ are finitely generated, see Lemma 15.44.1. Hence $(R^ h)^\wedge $ and $(R^{sh})^\wedge $ are Noetherian by Algebra, Lemma 10.154.3. This proves (a).

Note that (b) is immediate from Lemma 15.44.1. In particular we see that $(R^ h)^\wedge $ is flat over $R$, see Algebra, Lemma 10.96.3.

Next, we show that $R^ h \to (R^ h)^\wedge $ is flat. Write $R^ h = \mathop{\mathrm{colim}}\nolimits _ i R_ i$ as a directed colimit of localizations of étale $R$-algebras. By Algebra, Lemma 10.38.6 if $(R^ h)^\wedge $ is flat over each $R_ i$, then $R^ h \to (R^ h)^\wedge $ is flat. Note that $R^ h = R_ i^ h$ (by construction). Hence $R_ i^\wedge = (R^ h)^\wedge $ by part (b) is flat over $R_ i$ as desired. To finish the proof of (c) we show that $R^{sh} \to (R^{sh})^\wedge $ is flat. To do this, by a limit argument as above, it suffices to show that $(R^{sh})^\wedge $ is flat over $R$. Note that it follows from Lemma 15.44.1 that $(R^{sh})^\wedge $ is the completion of a free $R$-module. By Lemma 15.27.2 we see this is flat over $R$ as desired. This finishes the proof of (c).

At this point we know (c) is true and that $(R^ h)^\wedge $ and $(R^{sh})^\wedge $ are Noetherian. It follows from Algebra, Lemma 10.158.1 that $R^ h$ and $R^{sh}$ are Noetherian.

Part (d) follows from Lemma 15.44.2 and Lemma 15.36.4.

Part (e) follows from Algebra, Lemma 10.150.17 and the fact that $R^\wedge $ is henselian by Algebra, Lemma 10.148.9.

Proof of (f). Using (e) there is a map $R^{sh} \to (R^\wedge )^{sh}$ which induces a map $(R^{sh})^\wedge \to ((R^\wedge )^{sh})^\wedge $ upon completion. Using (e) there is a map $R^\wedge \to (R^{sh})^\wedge $. Since $(R^{sh})^\wedge $ is strictly henselian (see above) this map induces a map $(R^\wedge )^{sh} \to (R^{sh})^\wedge $ by Algebra, Lemma 10.150.12. Completing we obtain a map $((R^\wedge )^{sh})^\wedge \to (R^{sh})^\wedge $. We omit the verification that these two maps are mutually inverse.
$\square$

## Comments (2)

Comment #2730 by Johan on

Comment #2855 by Johan on