The Stacks project

Proof. This follows from the fact that

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

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

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

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

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

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

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

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. $\square$

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 $p$th 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

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

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. $\square$

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 $\mathfrak p'$ if $R_\mathfrak m^\wedge $ lying over $\mathfrak pR_\mathfrak m$. Consider the commutative diagram

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. $\square$

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

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

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 $\kappa (\mathfrak r_ i)$. $\square$

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

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

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. $\square$

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

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

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.

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. $\square$

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. $\square$

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

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

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

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

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

Hence, as one of the components, the ring

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

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