Proof.
Since R \to R^ h \to R^{sh} are faithfully flat (Lemma 15.45.1), we see that R^ h or R^{sh} being Noetherian implies that R is Noetherian, see Algebra, Lemma 10.164.1. 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 15.45.1. Hence (R^ h)^\wedge and (R^{sh})^\wedge are Noetherian by Algebra, Lemma 10.160.3. This proves (a).
Note that (b) is immediate from Lemma 15.45.1. In particular we see that (R^ h)^\wedge is flat over R, see Algebra, Lemma 10.97.3.
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 10.39.6 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 15.45.1 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 10.164.1 that R^ h and R^{sh} are Noetherian.
Part (d) follows from Lemma 15.45.2 and Lemma 15.37.4.
Part (e) follows from Algebra, Lemma 10.155.13 and the fact that R^\wedge is henselian by Algebra, Lemma 10.153.9.
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 10.155.10. 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