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The Stacks project

[IV, Theorem 18.6.6 and Proposition 18.8.8, EGA]

Lemma 15.45.3. Let R be a local ring. The following are equivalent

  1. R is Noetherian,

  2. R^ h is Noetherian, and

  3. R^{sh} is Noetherian.

In this case we have

  1. (R^ h)^\wedge and (R^{sh})^\wedge are Noetherian complete local rings,

  2. R^\wedge \to (R^ h)^\wedge is an isomorphism,

  3. R^ h \to (R^ h)^\wedge and R^{sh} \to (R^{sh})^\wedge are flat,

  4. R^\wedge \to (R^{sh})^\wedge is formally smooth in the \mathfrak m_{(R^{sh})^\wedge }-adic topology,

  5. (R^\wedge )^{sh} = R^\wedge \otimes _{R^ h} R^{sh}, and

  6. ((R^\wedge )^{sh})^\wedge = (R^{sh})^\wedge .

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 on

A reference is EGA IV_4, Theorem 18.6.6 and Proposition 18.8.8.


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