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15.50 G-rings

Let A be a Noetherian local ring. In Section 15.43 we have seen that some but not all properties of A are reflected in the completion A^\wedge of A. To study this further we introduce some terminology. For a prime \mathfrak q of A the fibre ring

A^\wedge \otimes _ A \kappa (\mathfrak q) = (A^\wedge )_\mathfrak q/\mathfrak q(A^\wedge )_\mathfrak q = (A/\mathfrak q)^\wedge \otimes _{A/q} \kappa (\mathfrak q)

is called a formal fibre of A. We think of the formal fibre as an algebra over \kappa (\mathfrak q). Thus A \to A^\wedge is a regular ring homomorphism if and only if all the formal fibres are geometrically regular algebras.

Definition 15.50.1. A ring R is called a G-ring if R is Noetherian and for every prime \mathfrak p of R the ring map R_\mathfrak p \to (R_\mathfrak p)^\wedge is regular.

By the discussion above we see that R is a G-ring if and only if every local ring R_\mathfrak p has geometrically regular formal fibres. Note that if \mathbf{Q} \subset R, then it suffices to check the formal fibres are regular. Another way to express the G-ring condition is described in the following lemma.

Lemma 15.50.2. Let R be a Noetherian ring. Then R is a G-ring if and only if for every pair of primes \mathfrak q \subset \mathfrak p \subset R the algebra

(R/\mathfrak q)_\mathfrak p^\wedge \otimes _{R/\mathfrak q} \kappa (\mathfrak q)

is geometrically regular over \kappa (\mathfrak q).

Proof. This follows from the fact that

R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q) = (R/\mathfrak q)_\mathfrak p^\wedge \otimes _{R/\mathfrak q} \kappa (\mathfrak q)

as algebras over \kappa (\mathfrak q). \square

Lemma 15.50.3. Let R \to R' be a finite type map of Noetherian rings and let

\xymatrix{ \mathfrak q' \ar[r] & \mathfrak p' \ar[r] & R' \\ \mathfrak q \ar[r] \ar@{-}[u] & \mathfrak p \ar[r] \ar@{-}[u] & R \ar[u] }

be primes. Assume R \to R' is quasi-finite at \mathfrak p'.

  1. If the formal fibre R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q) is geometrically regular over \kappa (\mathfrak q), then the formal fibre (R'_{\mathfrak p'})^\wedge \otimes _{R'} \kappa (\mathfrak q') is geometrically regular over \kappa (\mathfrak q').

  2. If the formal fibres of R_\mathfrak p are geometrically regular, then the formal fibres of R'_{\mathfrak p'} are geometrically regular.

  3. If R \to R' is quasi-finite and R is a G-ring, then R' is a G-ring.

Proof. It is clear that (1) \Rightarrow (2) \Rightarrow (3). Assume R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q) is geometrically regular over \kappa (\mathfrak q). By Algebra, Lemma 10.124.3 we see that

R_\mathfrak p^\wedge \otimes _ R R' = (R'_{\mathfrak p'})^\wedge \times B

for some R_\mathfrak p^\wedge -algebra B. Hence R'_{\mathfrak p'} \to (R'_{\mathfrak p'})^\wedge is a factor of a base change of the map R_\mathfrak p \to R_\mathfrak p^\wedge . It follows that (R'_{\mathfrak p'})^\wedge \otimes _{R'} \kappa (\mathfrak q') is a factor of

R_\mathfrak p^\wedge \otimes _ R R' \otimes _{R'} \kappa (\mathfrak q') = R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q) \otimes _{\kappa (\mathfrak q)} \kappa (\mathfrak q').

Thus the result follows as extension of base field preserves geometric regularity, see Algebra, Lemma 10.166.1. \square

Lemma 15.50.4. Let R be a Noetherian ring. Then R is a G-ring if and only if for every finite free ring map R \to S the formal fibres of S are regular rings.

Proof. Assume that for any finite free ring map R \to S the ring S has regular formal fibres. Let \mathfrak q \subset \mathfrak p \subset R be primes and let \kappa (\mathfrak q) \subset L be a finite purely inseparable extension. To show that R is a G-ring it suffices to show that

R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q) \otimes _{\kappa (\mathfrak q)} L

is a regular ring. Choose a finite free extension R \to R' such that \mathfrak q' = \mathfrak qR' is a prime and such that \kappa (\mathfrak q') is isomorphic to L over \kappa (\mathfrak q), see Algebra, Lemma 10.159.3. By Algebra, Lemma 10.97.8 we have

R_\mathfrak p^\wedge \otimes _ R R' = \prod (R'_{\mathfrak p_ i'})^\wedge

where \mathfrak p_ i' are the primes of R' lying over \mathfrak p. Thus we have

R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q) \otimes _{\kappa (\mathfrak q)} L = R_\mathfrak p^\wedge \otimes _ R R' \otimes _{R'} \kappa (\mathfrak q') = \prod (R'_{\mathfrak p_ i'})^\wedge \otimes _{R'_{\mathfrak p'_ i}} \kappa (\mathfrak q')

Our assumption is that the rings on the right are regular, hence the ring on the left is regular too. Thus R is a G-ring. The converse follows from Lemma 15.50.3. \square

Lemma 15.50.5. Let k be a field of characteristic p. Let A = k[[x_1, \ldots , x_ n]][y_1, \ldots , y_ n] and denote K the fraction field of A. Let \mathfrak p \subset A be a prime. Then A_\mathfrak p^\wedge \otimes _ A K is geometrically regular over K.

Proof. Let L/K be a finite purely inseparable field extension. We will show by induction on [L : K] that A_\mathfrak p^\wedge \otimes L is regular. The base case is L = K: as A is regular, A_\mathfrak p^\wedge is regular (Lemma 15.43.4), hence the localization A_\mathfrak p^\wedge \otimes K is regular. Let K \subset M \subset L be a subfield such that L is a degree p extension of M obtained by adjoining a pth root of an element f \in M. Let B be a finite A-subalgebra of M with fraction field M. Clearing denominators, we may and do assume f \in B. Set C = B[z]/(z^ p -f) and note that B \subset C is finite and that the fraction field of C is L. Since A \subset B \subset C are finite and L/M/K are purely inseparable we see that for every element of B or C some power of it lies in A. Hence there is a unique prime \mathfrak r \subset B, resp. \mathfrak q \subset C lying over \mathfrak p. Note that

A_\mathfrak p^\wedge \otimes _ A M = B_\mathfrak r^\wedge \otimes _ B M

see Algebra, Lemma 10.97.8. By induction we know that this ring is regular. In the same manner we have

A_\mathfrak p^\wedge \otimes _ A L = C_\mathfrak r^\wedge \otimes _ C L = B_\mathfrak r^\wedge \otimes _ B M[z]/(z^ p - f)

the last equality because the completion of C = B[z]/(z^ p - f) equals B_\mathfrak r^\wedge [z]/(z^ p -f). By Lemma 15.48.5 we know there exists a derivation D : B \to B such that D(f) \not= 0. In other words, g = D(f) is a unit in M! By Lemma 15.48.1 D extends to a derivation of B_\mathfrak r, B_\mathfrak r^\wedge and B_\mathfrak r^\wedge \otimes _ B M (successively extending through a localization, a completion, and a localization). Since it is an extension we end up with a derivation of B_\mathfrak r^\wedge \otimes _ B M which maps f to g and g is a unit of the ring B_\mathfrak r^\wedge \otimes _ B M. Hence A_\mathfrak p^\wedge \otimes _ A L is regular by Lemma 15.48.4 and we win. \square

Proof. Let A be a Noetherian complete local ring. By Lemma 15.50.2 it suffices to check that B = A/\mathfrak q has geometrically regular formal fibres over the minimal prime (0) of B. Thus we may assume that A is a domain and it suffices to check the condition for the formal fibres over the minimal prime (0) of A. Let K be the fraction field of A.

We can choose a subring A_0 \subset A which is a regular complete local ring such that A is finite over A_0, see Algebra, Lemma 10.160.11. Moreover, we may assume that A_0 is a power series ring over a field or a Cohen ring. By Lemma 15.50.3 we see that it suffices to prove the result for A_0.

Assume that A is a power series ring over a field or a Cohen ring. Since A is regular the localizations A_\mathfrak p are regular (see Algebra, Definition 10.110.7 and the discussion preceding it). Hence the completions A_\mathfrak p^\wedge are regular, see Lemma 15.43.4. Hence the fibre A_{\mathfrak p}^\wedge \otimes _ A K is, as a localization of A_\mathfrak p^\wedge , also regular. Thus we are done if the characteristic of K is 0. The positive characteristic case is the case A = k[[x_1, \ldots , x_ d]] which is a special case of Lemma 15.50.5. \square

Lemma 15.50.7. Let R be a Noetherian ring. Then R is a G-ring if and only if R_\mathfrak m has geometrically regular formal fibres for every maximal ideal \mathfrak m of R.

Proof. Assume R_\mathfrak m \to R_\mathfrak m^\wedge is regular for every maximal ideal \mathfrak m of R. Let \mathfrak p be a prime of R and choose a maximal ideal \mathfrak p \subset \mathfrak m. Since R_\mathfrak m \to R_\mathfrak m^\wedge is faithfully flat we can choose a prime \mathfrak p' in R_\mathfrak m^\wedge lying over \mathfrak pR_\mathfrak m. Consider the commutative diagram

\xymatrix{ R_\mathfrak m^\wedge \ar[r] & (R_\mathfrak m^\wedge )_{\mathfrak p'} \ar[r] & (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge \\ R_\mathfrak m \ar[u] \ar[r] & R_\mathfrak p \ar[u] \ar[r] & R_\mathfrak p^\wedge \ar[u] }

By assumption the ring map R_\mathfrak m \to R_\mathfrak m^\wedge is regular. By Proposition 15.50.6 (R_\mathfrak m^\wedge )_{\mathfrak p'} \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge is regular. The localization R_\mathfrak m^\wedge \to (R_\mathfrak m^\wedge )_{\mathfrak p'} is regular. Hence R_\mathfrak m \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge is regular by Lemma 15.41.4. Since it factors through the localization R_\mathfrak p, also the ring map R_\mathfrak p \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge is regular. Thus we may apply Lemma 15.41.7 to see that R_\mathfrak p \to R_\mathfrak p^\wedge is regular. \square

Lemma 15.50.8. Let R be a Noetherian local ring which is a G-ring. Then the henselization R^ h and the strict henselization R^{sh} are G-rings.

Proof. We will use the criterion of Lemma 15.50.7. Let \mathfrak q \subset R^ h be a prime and set \mathfrak p = R \cap \mathfrak q. Set \mathfrak q_1 = \mathfrak q and let \mathfrak q_2, \ldots , \mathfrak q_ t be the other primes of R^ h lying over \mathfrak p, so that R^ h \otimes _ R \kappa (\mathfrak p) = \prod \nolimits _{i = 1, \ldots , t} \kappa (\mathfrak q_ i), see Lemma 15.45.13. Using that (R^ h)^\wedge = R^\wedge (Lemma 15.45.3) we see

\prod \nolimits _{i = 1, \ldots , t} (R^ h)^\wedge \otimes _{R^ h} \kappa (\mathfrak q_ i) = (R^ h)^\wedge \otimes _{R^ h} (R^ h \otimes _ R \kappa (\mathfrak p)) = R^\wedge \otimes _ R \kappa (\mathfrak p)

Hence (R^ h)^\wedge \otimes _{R^ h} \kappa (\mathfrak q_ i) is geometrically regular over \kappa (\mathfrak p) by assumption. Since \kappa (\mathfrak q_ i) is separable algebraic over \kappa (\mathfrak p) it follows from Algebra, Lemma 10.166.6 that (R^ h)^\wedge \otimes _{R^ h} \kappa (\mathfrak q_ i) is geometrically regular over \kappa (\mathfrak q_ i).

Let \mathfrak r \subset R^{sh} be a prime and set \mathfrak p = R \cap \mathfrak r. Set \mathfrak r_1 = \mathfrak r and let \mathfrak r_2, \ldots , \mathfrak r_ s be the other primes of R^{sh} lying over \mathfrak p, so that R^{sh} \otimes _ R \kappa (\mathfrak p) = \prod \nolimits _{i = 1, \ldots , s} \kappa (\mathfrak r_ i), see Lemma 15.45.13. Then we see that

\prod \nolimits _{i = 1, \ldots , s} (R^{sh})^\wedge \otimes _{R^{sh}} \kappa (\mathfrak r_ i) = (R^{sh})^\wedge \otimes _{R^{sh}} (R^{sh} \otimes _ R \kappa (\mathfrak p)) = (R^{sh})^\wedge \otimes _ R \kappa (\mathfrak p)

Note that R^\wedge \to (R^{sh})^\wedge is formally smooth in the \mathfrak m_{(R^{sh})^\wedge }-adic topology, see Lemma 15.45.3. Hence R^\wedge \to (R^{sh})^\wedge is regular by Proposition 15.49.2. We conclude that (R^{sh})^\wedge \otimes _{R^{sh}} \kappa (\mathfrak r_ i) is regular over \kappa (\mathfrak p) by Lemma 15.41.4 as R^\wedge \otimes _ R \kappa (\mathfrak p) is regular over \kappa (\mathfrak p) by assumption. Since \kappa (\mathfrak r_ i) is separable algebraic over \kappa (\mathfrak p) it follows from Algebra, Lemma 10.166.6 that (R^{sh})^\wedge \otimes _{R^{sh}} \kappa (\mathfrak r_ i) is geometrically regular over \kappa (\mathfrak r_ i). \square

Lemma 15.50.9. Let p be a prime number. Let A be a Noetherian complete local domain with fraction field K of characteristic p. Let \mathfrak q \subset A[x] be a maximal ideal lying over the maximal ideal of A and let (0) \not= \mathfrak r \subset \mathfrak q be a prime lying over (0) \subset A. Then A[x]_\mathfrak q^\wedge \otimes _{A[x]} \kappa (\mathfrak r) is geometrically regular over \kappa (\mathfrak r).

Proof. Note that K \subset \kappa (\mathfrak r) is finite. Hence, given a finite purely inseparable extension L/\kappa (\mathfrak r) there exists a finite extension of Noetherian complete local domains A \subset B such that \kappa (\mathfrak r) \otimes _ A B surjects onto L. Namely, you take B \subset L a finite A-subalgebra whose field of fractions is L. Denote \mathfrak r' \subset B[x] the kernel of the map B[x] = A[x] \otimes _ A B \to \kappa (\mathfrak r) \otimes _ A B \to L so that \kappa (\mathfrak r') = L. Then

A[x]_\mathfrak q^\wedge \otimes _{A[x]} L = A[x]_\mathfrak q^\wedge \otimes _{A[x]} B[x] \otimes _{B[x]} \kappa (\mathfrak r') = \prod B[x]_{\mathfrak q_ i}^\wedge \otimes _{B[x]} \kappa (\mathfrak r')

where \mathfrak q_1, \ldots , \mathfrak q_ t are the primes of B[x] lying over \mathfrak q, see Algebra, Lemma 10.97.8. Thus we see that it suffices to prove the rings B[x]_{\mathfrak q_ i}^\wedge \otimes _{B[x]} \kappa (\mathfrak r') are regular. This reduces us to showing that A[x]_\mathfrak q^\wedge \otimes _{A[x]} \kappa (\mathfrak r) is regular in the special case that K = \kappa (\mathfrak r).

Assume K = \kappa (\mathfrak r). In this case we see that \mathfrak r K[x] is generated by x - f for some f \in K and

A[x]_\mathfrak q^\wedge \otimes _{A[x]} \kappa (\mathfrak r) = (A[x]_\mathfrak q^\wedge \otimes _ A K)/(x - f)

The derivation D = \text{d}/\text{d}x of A[x] extends to K[x] and maps x - f to a unit of K[x]. Moreover D extends to A[x]_\mathfrak q^\wedge \otimes _ A K by Lemma 15.48.1. As A \to A[x]_\mathfrak q^\wedge is formally smooth (see Lemmas 15.37.2 and 15.37.4) the ring A[x]_\mathfrak q^\wedge \otimes _ A K is regular by Proposition 15.49.2 (the arguments of the proof of that proposition simplify significantly in this particular case). We conclude by Lemma 15.48.2. \square

Proposition 15.50.10. Let R be a G-ring. If R \to S is essentially of finite type then S is a G-ring.

Proof. Since being a G-ring is a property of the local rings it is clear that a localization of a G-ring is a G-ring. Conversely, if every localization at a prime is a G-ring, then the ring is a G-ring. Thus it suffices to show that S_\mathfrak q is a G-ring for every finite type R-algebra S and every prime \mathfrak q of S. Writing S as a quotient of R[x_1, \ldots , x_ n] we see from Lemma 15.50.3 that it suffices to prove that R[x_1, \ldots , x_ n] is a G-ring. By induction on n it suffices to prove that R[x] is a G-ring. Let \mathfrak q \subset R[x] be a maximal ideal. By Lemma 15.50.7 it suffices to show that

R[x]_\mathfrak q \longrightarrow R[x]_\mathfrak q^\wedge

is regular. If \mathfrak q lies over \mathfrak p \subset R, then we may replace R by R_\mathfrak p. Hence we may assume that R is a Noetherian local G-ring with maximal ideal \mathfrak m and that \mathfrak q \subset R[x] lies over \mathfrak m. Note that there is a unique prime \mathfrak q' \subset R^\wedge [x] lying over \mathfrak q. Consider the diagram

\xymatrix{ R[x]_\mathfrak q^\wedge \ar[r] & (R^\wedge [x]_{\mathfrak q'})^\wedge \\ R[x]_\mathfrak q \ar[r] \ar[u] & R^\wedge [x]_{\mathfrak q'} \ar[u] }

Since R is a G-ring the lower horizontal arrow is regular (as a localization of a base change of the regular ring map R \to R^\wedge ). Suppose we can prove the right vertical arrow is regular. Then it follows that the composition R[x]_\mathfrak q \to (R^\wedge [x]_{\mathfrak q'})^\wedge is regular, and hence the left vertical arrow is regular by Lemma 15.41.7. Hence we see that we may assume R is a Noetherian complete local ring and \mathfrak q a prime lying over the maximal ideal of R.

Let R be a Noetherian complete local ring and let \mathfrak q \subset R[x] be a maximal ideal lying over the maximal ideal of R. Let \mathfrak r \subset \mathfrak q be a prime ideal. We want to show that R[x]_\mathfrak q^\wedge \otimes _{R[x]} \kappa (\mathfrak r) is a geometrically regular algebra over \kappa (\mathfrak r). Set \mathfrak p = R \cap \mathfrak r. Then we can replace R by R/\mathfrak p and \mathfrak q and \mathfrak r by their images in R/\mathfrak p[x], see Lemma 15.50.2. Hence we may assume that R is a domain and that \mathfrak r \cap R = (0).

By Algebra, Lemma 10.160.11 we can find R_0 \subset R which is regular and such that R is finite over R_0. Applying Lemma 15.50.3 we see that it suffices to prove R[x]_\mathfrak q^\wedge \otimes _{R[x]} \kappa (\mathfrak r) is geometrically regular over \kappa (r) when, in addition to the above, R is a regular complete local ring.

Now R is a regular complete local ring, we have \mathfrak q \subset \mathfrak r \subset R[x], we have (0) = R \cap \mathfrak r and \mathfrak q is a maximal ideal lying over the maximal ideal of R. Since R is regular the ring R[x] is regular (Algebra, Lemma 10.163.10). Hence the localization R[x]_\mathfrak q is regular. Hence the completions R[x]_\mathfrak q^\wedge are regular, see Lemma 15.43.4. Hence the fibre R[x]_{\mathfrak q}^\wedge \otimes _{R[x]} \kappa (\mathfrak r) is, as a localization of R[x]_\mathfrak q^\wedge , also regular. Thus we are done if the characteristic of the fraction field of R is 0.

If the characteristic of R is positive, then R = k[[x_1, \ldots , x_ n]]. In this case we split the argument in two subcases:

  1. The case \mathfrak r = (0). The result is a direct consequence of Lemma 15.50.5.

  2. The case \mathfrak r \not= (0). This is Lemma 15.50.9.

\square

Remark 15.50.11. Let R be a G-ring and let I \subset R be an ideal. In general it is not the case that the I-adic completion R^\wedge is a G-ring. An example was given by Nishimura in [Nishimura]. A generalization and, in some sense, clarification of this example can be found in the last section of [Dumitrescu].

Proposition 15.50.12. The following types of rings are G-rings:

  1. fields,

  2. Noetherian complete local rings,

  3. \mathbf{Z},

  4. Dedekind domains with fraction field of characteristic zero,

  5. finite type ring extensions of any of the above.

Proof. For fields, \mathbf{Z} and Dedekind domains of characteristic zero this follows immediately from the definition and the fact that the completion of a discrete valuation ring is a discrete valuation ring. A Noetherian complete local ring is a G-ring by Proposition 15.50.6. The statement on finite type overrings is Proposition 15.50.10. \square

Lemma 15.50.13. Let (A, \mathfrak m) be a henselian local ring. Then A is a filtered colimit of a system of henselian local G-rings with local transition maps.

Proof. Write A = \mathop{\mathrm{colim}}\nolimits A_ i as a filtered colimit of finite type \mathbf{Z}-algebras. Let \mathfrak p_ i be the prime ideal of A_ i lying under \mathfrak m. We may replace A_ i by the localization of A_ i at \mathfrak p_ i. Then A_ i is a Noetherian local G-ring (Proposition 15.50.12). By Lemma 15.12.5 we see that A = \mathop{\mathrm{colim}}\nolimits A_ i^ h. By Lemma 15.50.8 the rings A_ i^ h are G-rings. \square

Lemma 15.50.14.reference Let A be a G-ring. Let I \subset A be an ideal and let A^\wedge be the completion of A with respect to I. Then A \to A^\wedge is regular.

Proof. The ring map A \to A^\wedge is flat by Algebra, Lemma 10.97.2. The ring A^\wedge is Noetherian by Algebra, Lemma 10.97.6. Thus it suffices to check the third condition of Lemma 15.41.2. Let \mathfrak m' \subset A^\wedge be a maximal ideal lying over \mathfrak m \subset A. By Algebra, Lemma 10.96.6 we have IA^\wedge \subset \mathfrak m'. Since A^\wedge /IA^\wedge = A/I we see that I \subset \mathfrak m, \mathfrak m/I = \mathfrak m'/IA^\wedge , and A/\mathfrak m = A^\wedge /\mathfrak m'. Since A^\wedge /\mathfrak m' is a field, we conclude that \mathfrak m is a maximal ideal as well. Then A_\mathfrak m \to A^\wedge _{\mathfrak m'} is a flat local ring homomorphism of Noetherian local rings which identifies residue fields and such that \mathfrak m A^\wedge _{\mathfrak m'} = \mathfrak m'A^\wedge _{\mathfrak m'}. Thus it induces an isomorphism on complete local rings, see Lemma 15.43.9. Let (A_\mathfrak m)^\wedge be the completion of A_\mathfrak m with respect to its maximal ideal. The ring map

(A^\wedge )_{\mathfrak m'} \to ((A^\wedge )_{\mathfrak m'})^\wedge = (A_\mathfrak m)^\wedge

is faithfully flat (Algebra, Lemma 10.97.3). Thus we can apply Lemma 15.41.7 to the ring maps

A_\mathfrak m \to (A^\wedge )_{\mathfrak m'} \to (A_\mathfrak m)^\wedge

to conclude because A_\mathfrak m \to (A_\mathfrak m)^\wedge is regular as A is a G-ring. \square

Lemma 15.50.15.referenceslogan Let A be a G-ring. Let I \subset A be an ideal. Let (A^ h, I^ h) be the henselization of the pair (A, I), see Lemma 15.12.1. Then A^ h is a G-ring.

Proof. Let \mathfrak m^ h \subset A^ h be a maximal ideal. We have to show that the map from A^ h_{\mathfrak m^ h} to its completion has geometrically regular fibres, see Lemma 15.50.7. Let \mathfrak m be the inverse image of \mathfrak m^ h in A. Note that I^ h \subset \mathfrak m^ h and hence I \subset \mathfrak m as (A^ h, I^ h) is a henselian pair. Recall that A^ h is Noetherian, I^ h = IA^ h, and that A \to A^ h induces an isomorphism on I-adic completions, see Lemma 15.12.4. Then the local homomorphism of Noetherian local rings

A_\mathfrak m \to A^ h_{\mathfrak m^ h}

induces an isomorphism on completions at maximal ideals by Lemma 15.43.9 (details omitted). Let \mathfrak q^ h be a prime of A^ h_{\mathfrak m^ h} lying over \mathfrak q \subset A_\mathfrak m. Set \mathfrak q_1 = \mathfrak q^ h and let \mathfrak q_2, \ldots , \mathfrak q_ t be the other primes of A^ h lying over \mathfrak q, so that A^ h \otimes _ A \kappa (\mathfrak q) = \prod \nolimits _{i = 1, \ldots , t} \kappa (\mathfrak q_ i), see Lemma 15.45.12. Using that (A^ h)_{\mathfrak m^ h}^\wedge = (A_\mathfrak m)^\wedge as discussed above we see

\prod \nolimits _{i = 1, \ldots , t} (A^ h_{\mathfrak m^ h})^\wedge \otimes _{A^ h_{\mathfrak m^ h}} \kappa (\mathfrak q_ i) = (A^ h_{\mathfrak m^ h})^\wedge \otimes _{A^ h_{\mathfrak m^ h}} (A^ h_{\mathfrak m^ h} \otimes _{A_{\mathfrak m}} \kappa (\mathfrak q)) = (A_{\mathfrak m})^\wedge \otimes _{A_{\mathfrak m}} \kappa (\mathfrak q)

Hence, as one of the components, the ring

(A^ h_{\mathfrak m^ h})^\wedge \otimes _{A^ h_{\mathfrak m^ h}} \kappa (\mathfrak q^ h)

is geometrically regular over \kappa (\mathfrak q) by assumption on A. Since \kappa (\mathfrak q^ h) is separable algebraic over \kappa (\mathfrak q) it follows from Algebra, Lemma 10.166.6 that

(A^ h_{\mathfrak m^ h})^\wedge \otimes _{A^ h_{\mathfrak m^ h}} \kappa (\mathfrak q^ h)

is geometrically regular over \kappa (\mathfrak q^ h) as desired. \square


Comments (2)

Comment #10117 by Doug Liu on

The second in the first sentence "Let be a Noetherian local ring " shall be a typo.


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