hcju/stacksqa huggingface dataset:
We will show that the criterion of Lemma \ref{lemma-J-1} applies. Let $\mathfrak p \in \text{Reg}(X)$ be a prime of height $r$. Pick $f_1, \ldots, f_r \in \mathfrak p$ which map to generators of $\mathfrak pR_\mathfrak p$. Since $\mathfrak p \in \text{Reg}(X)$ we see that $f_1, \ldots, f_r$ maps to a regular sequence in $R_\mathfrak p$, see Algebra, Lemma \ref{algebra-lemma-regular-ring-CM}. Thus by Algebra, Lemma \ref{algebra-lemma-regular-sequence-in-neighbourhood} we see that after replacing $R$ by $R_g$ for some $g \in R$, $g \not \in \mathfrak p$ the sequence $f_1, \ldots, f_r$ is a regular sequence in $R$. After another replacement we may also assume $f_1, \ldots, f_r$ generate $\mathfrak p$. Next, let $\mathfrak p \subset \mathfrak q$ be a prime ideal such that $(R/\mathfrak p)_\mathfrak q$ is a regular local ring. By the assumption of the lemma there exists a non-empty open subset of $V(\mathfrak p)$ consisting of such primes, hence it suffices to prove $R_\mathfrak q$ is regular. Note that $f_1, \ldots, f_r$ is a regular sequence in $R_\mathfrak q$ such that $R_\mathfrak q/(f_1, \ldots, f_r)R_\mathfrak q$ is regular. Hence $R_\mathfrak q$ is regular by Algebra, Lemma \ref{algebra-lemma-regular-mod-x}.
After replacing $S$ by $S_g$ for some nonzero $g \in S$ we may assume that $S$ is a regular ring. By generic flatness we may assume that also $R \to S$ is faithfully flat, see Algebra, Lemma \ref{algebra-lemma-generic-flatness-Noetherian}. Then $R$ is regular by Algebra, Lemma \ref{algebra-lemma-descent-regular}.
We may replace $R$ by a principal localization and assume $R$ is a regular ring. By Algebra, Lemma \ref{algebra-lemma-smooth-at-generic-point} the ring map $R \to S$ is smooth at $(0)$. Hence after replacing $S$ by a principal localization we may assume that $S$ is smooth over $R$. Then $S$ is regular too, see Algebra, Lemma \ref{algebra-lemma-regular-goes-up}.
It is clear that we have the implications (1) $\Rightarrow$ (2) and (2) $\Rightarrow$ (4). Recall that a domain which is J-1 is J-0. Hence we also have the implications (1) $\Rightarrow$ (3) and (3) $\Rightarrow$ (4). \medskip\noindent Let $R \to S$ be a finite type ring map and let's try to show $S$ is J-1. By Lemma \ref{lemma-intersection-regular-with-closed} it suffices to prove that $S/\mathfrak q$ is J-0 for every prime $\mathfrak q$ of $S$. In this way we see (2) $\Rightarrow$ (1). \medskip\noindent Assume (4). We will show that (2) holds which will finish the proof. Let $R \to S$ be a finite type ring map with $S$ a domain. Let $\mathfrak p = \Ker(R \to S)$. Let $K$ be the fraction field of $S$. There exists a diagram of fields $$ \xymatrix{ K \ar[r] & K' \\ \kappa(\mathfrak p) \ar[u] \ar[r] & L \ar[u] } $$ where the horizontal arrows are finite purely inseparable field extensions and where $K'/L$ is separable, see Algebra, Lemma \ref{algebra-lemma-make-separably-generated}. Choose $R' \subset L$ as in (4) and let $S'$ be the image of the map $S \otimes_R R' \to K'$. Then $S'$ is a domain whose fraction field is $K'$, hence $S'$ is J-0 by Lemma \ref{lemma-J-0-goes-up} and our choice of $R'$. Then we apply Lemma \ref{lemma-J-0-goes-down} to see that $S$ is J-0 as desired.
It suffices to prove this when $R$ is a local ring with maximal ideal $\mathfrak m$ and residue field $\kappa$. In this case it suffices to prove that $f \not \in \mathfrak m^2$, see Algebra, Lemma \ref{algebra-lemma-regular-ring-CM}. However, if $f \in \mathfrak m^2$ then $D(f) \in \mathfrak m$ by the Leibniz rule, a contradiction.
It suffices to prove that $f_1, \ldots, f_m$ are $\kappa$-linearly independent in $\mathfrak m/\mathfrak m^2$, see Algebra, Lemma \ref{algebra-lemma-regular-ring-CM}. However, if there is a nontrivial linear relation the we get $\sum a_i f_i \in \mathfrak m^2$ for some $a_i \in R$ but not all $a_i \in \mathfrak m$. Observe that $D_i(\mathfrak m^2) \subset \mathfrak m$ and $D_i(a_j f_j) \equiv a_j D_i(f_j) \bmod \mathfrak m$ by the Leibniz rule for derivations. Hence this would imply $$ \sum a_j D_i(f_j) \in \mathfrak m $$ which would contradict the assumption on the determinant.
Let $R$ be a Noetherian complete local ring such that there exists a finite type ring map $R \to B$. Of course we may replace $R$ by its image in $B$, hence we may assume $R$ is a domain of characteristic $p > 0$ (as well as Noetherian complete local). By Algebra, Lemma \ref{algebra-lemma-complete-local-Noetherian-domain-finite-over-regular} we can write $R$ as a finite extension of $k[[x_1, \ldots, x_n]]$ for some field $k$ and integer $n$. Hence we may replace $R$ by $k[[x_1, \ldots, x_n]]$. Next, we use Algebra, Lemma \ref{algebra-lemma-Noether-normalization-over-a-domain} to factor $R \to B$ as $$ R \subset R[y_1, \ldots, y_d] \subset B' \subset B $$ with $B'$ finite over $R[y_1, \ldots, y_d]$ and $B'_g \cong B_g$ for some nonzero $g \in R$. Note that $f' = g^{pN} f \in B'$ for some large integer $N$. It is clear that $f'$ is not a $p$th power in the fraction field of $B'$. If we can find a derivation $D' : B' \to B'$ with $D'(f') \not = 0$, then Lemma \ref{lemma-derivation-extends} guarantees that $D = g^MD'$ extends to $B$ for some $M > 0$. Then $D(f) = g^ND'(f) = g^MD'(g^{-pN}f') = g^{M - pN}D'(f')$ is nonzero. Thus it suffices to prove the lemma in case $B$ is a finite extension of $A = k[[x_1, \ldots, x_n]][y_1, \ldots, y_m]$. \medskip\noindent Assume $B$ is a finite extension of $A = k[[x_1, \ldots, x_n]][y_1, \ldots, y_m]$. Denote $L$ the fraction field of $B$. Note that $\text{d}f$ is not zero in $\Omega_{L/\mathbf{F}_p}$, see Algebra, Lemma \ref{algebra-lemma-derivative-zero-pth-power}. We apply Lemma \ref{lemma-power-series-ring-subfields} to find a subfield $k' \subset k$ of finite index such that with $A' = k'[[x_1^p, \ldots, x_n^p]][y_1^p, \ldots, y_m^p]$ the element $\text{d}f$ does not map to zero in $\Omega_{L/K'}$ where $K'$ is the fraction field of $A'$. Thus we can choose a $K'$-derivation $D' : L \to L$ with $D'(f) \not = 0$. Since $A' \subset A$ and $A \subset B$ are finite by construction we see that $A' \subset B$ is finite. Choose $b_1, \ldots, b_t \in B$ which generate $B$ as an $A'$-module. Then $D'(b_i) = f_i/g_i$ for some $f_i, g_i \in B$ with $g_i \not = 0$. Setting $D = g_1 \ldots g_t D'$ we win.
By Algebra, Lemma \ref{algebra-lemma-complete-local-Noetherian-domain-finite-over-regular} we can find a regular subring $A_0 \subset A$ with $A$ finite over $A_0$. The induced extension $K/K_0$ of fraction fields is finite. If $K/K_0$ is separable, then we are done by Lemma \ref{lemma-J-0-goes-up}. If not, then $A_0$ and $A$ have characteristic $p > 0$. For any subextension $K/M/K_0$ there exists a finite subextension $A_0 \subset B \subset A$ whose fraction field is $M$. Hence, arguing by induction on $[K : K_0]$ we may assume there exists $A_0 \subset B \subset A$ such that $B$ is J-0 and $K/M$ has no nontrivial subextensions. In this case, if $K/M$ is separable, then we see that $A$ is J-0 by Lemma \ref{lemma-J-0-goes-up}. If not, then $K = M[z]/(z^p - b)$ for some $b \in B$ which is not a $p$th power in $M$. By Lemma \ref{lemma-find-D} we can find a derivation $D : B \to B$ with $D(f) \not = 0$. Applying Lemma \ref{lemma-degree-p-extension-regular} we see that $A_\mathfrak p$ is regular for any prime $\mathfrak p$ of $A$ lying over a regular prime of $B$ and not containing $D(f)$. As $B$ is J-0 we conclude $A$ is too.
Denote $B[\epsilon] = B[x]/(x^2)$ the ring of dual numbers over $B$. Consider the ring map $\psi : A \to B[\epsilon]$, $a \mapsto a + \epsilon D(a)$. Consider the commutative diagram $$ \xymatrix{ B \ar[r]_1 & B \\ A \ar[u] \ar[r]^\psi & B[\epsilon] \ar[u] } $$ By Lemma \ref{lemma-lift-continuous} and the assumption of formal smoothness of $B/A$ we find a map $\varphi : B \to B[\epsilon]$ fitting into the diagram. Write $\varphi(b) = b + \epsilon D'(b)$. Then $D' : B \to B$ is the desired extension.
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 \ref{algebra-lemma-completion-at-quasi-finite-prime} 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 \ref{algebra-lemma-geometrically-regular}.
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 \ref{algebra-lemma-finite-free-given-residue-field-extension}. By Algebra, Lemma \ref{algebra-lemma-completion-finite-extension} 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 \ref{lemma-G-ring-goes-up-quasi-finite}.
Let $L \supset K$ 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 \ref{lemma-completion-regular}), 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 $$ A_\mathfrak p^\wedge \otimes_A M = B_\mathfrak r^\wedge \otimes_B M $$ see Algebra, Lemma \ref{algebra-lemma-completion-finite-extension}. 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 \ref{lemma-find-D} 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 \ref{lemma-derivation-extends} $D$ 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 \ref{lemma-degree-p-extension-regular} and we win.
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 $$ \xymatrix{ R_\mathfrak m^\wedge \ar[r] & (R_\mathfrak m^\wedge)_{\mathfrak p'} \ar[r] & (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge \\ R_\mathfrak m \ar[u] \ar[r] & R_\mathfrak p \ar[u] \ar[r] & R_\mathfrak p^\wedge \ar[u] } $$ By assumption the ring map $R_\mathfrak m \to R_\mathfrak m^\wedge$ is regular. By Proposition \ref{proposition-Noetherian-complete-G-ring} $(R_\mathfrak m^\wedge)_{\mathfrak p'} \to (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge$ 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 \ref{lemma-regular-composition}. 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 \ref{lemma-regular-permanence} to see that $R_\mathfrak p \to R_\mathfrak p^\wedge$ is regular.
We will use the criterion of Lemma \ref{lemma-check-G-ring-maximal-ideals}. 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 \ref{lemma-fibres-henselization}. Using that $(R^h)^\wedge = R^\wedge$ (Lemma \ref{lemma-henselization-noetherian}) 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 \ref{algebra-lemma-geometrically-regular-over-separable-algebraic} that $(R^h)^\wedge \otimes_{R^h} \kappa(\mathfrak q_i)$ is geometrically regular over $\kappa(\mathfrak q_i)$. \medskip\noindent 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 \ref{lemma-fibres-henselization}. 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 \ref{lemma-henselization-noetherian}. Hence $R^\wedge \to (R^{sh})^\wedge$ is regular by Proposition \ref{proposition-fs-regular}. We conclude that $(R^{sh})^\wedge \otimes_{R^{sh}} \kappa(\mathfrak r_i)$ is regular over $\kappa(\mathfrak p)$ by Lemma \ref{lemma-regular-composition} 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 \ref{algebra-lemma-geometrically-regular-over-separable-algebraic} that $(R^{sh})^\wedge \otimes_{R^{sh}} \kappa(\mathfrak r_i)$ is geometrically regular over $\kappa(\mathfrak r_i)$.
Note that $K \subset \kappa(\mathfrak r)$ is finite. Hence, given a finite purely inseparable extension $\kappa(\mathfrak r) \subset L$ 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 \ref{algebra-lemma-completion-finite-extension}. 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 $K = \kappa(\mathfrak r)$. \medskip\noindent 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 \ref{lemma-derivation-extends}. As $A \to A[x]_\mathfrak q^\wedge$ is formally smooth (see Lemmas \ref{lemma-formally-smooth} and \ref{lemma-formally-smooth-completion}) the ring $A[x]_\mathfrak q^\wedge \otimes_A K$ is regular by Proposition \ref{proposition-fs-regular} (the arguments of the proof of that proposition simplify significantly in this particular case). We conclude by Lemma \ref{lemma-quotient-regular}.
Write $A = \colim A_i$ 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 \ref{proposition-ubiquity-G-ring}). By Lemma \ref{lemma-henselization-colimit} we see that $A = \colim A_i^h$. By Lemma \ref{lemma-henselization-G-ring} the rings $A_i^h$ are G-rings.
The ring map $A \to A^\wedge$ is flat by Algebra, Lemma \ref{algebra-lemma-completion-flat}. The ring $A^\wedge$ is Noetherian by Algebra, Lemma \ref{algebra-lemma-completion-Noetherian-Noetherian}. Thus it suffices to check the third condition of Lemma \ref{lemma-regular-local}. Let $\mathfrak m' \subset A^\wedge$ be a maximal ideal lying over $\mathfrak m \subset A$. By Algebra, Lemma \ref{algebra-lemma-radical-completion} 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 \ref{lemma-flat-unramified}. 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 \ref{algebra-lemma-completion-faithfully-flat}). Thus we can apply Lemma \ref{lemma-regular-permanence} 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.
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 \ref{lemma-check-G-ring-maximal-ideals}. 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 \ref{lemma-henselization-Noetherian-pair}. 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 \ref{lemma-flat-unramified} (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 \ref{lemma-filtered-colimit-etale-noetherian-fibres}. 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 \ref{algebra-lemma-geometrically-regular-over-separable-algebraic} 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.
Let $\mathfrak p \subset R$ be a prime. Then the fibre over $\mathfrak p$ is the ring $\Lambda \otimes_R \kappa(\mathfrak p)$ whose spectrum maps bijectively onto the subset of $\Spec(\Lambda)$ consisting of primes $\mathfrak q$ lying over $\mathfrak p$, see Algebra, Remark \ref{algebra-remark-fundamental-diagram}. For such a prime $\mathfrak q$ choose a maximal ideal $\mathfrak q \subset \mathfrak m'$ and set $\mathfrak m = R \cap \mathfrak m'$. Then $\mathfrak p \subset \mathfrak m$ and we have $$ (\Lambda \otimes_R \kappa(\mathfrak p))_\mathfrak q \cong (\Lambda_{\mathfrak m'} \otimes_{R_\mathfrak m} \kappa(\mathfrak p))_\mathfrak q $$ as $\kappa(\mathfrak q)$-algebras. Thus (1), (2), and (3) are equivalent because by (B) we can check property $P$ on local rings.
It is clear that (1) $\Rightarrow$ (2) $\Rightarrow$ (3). Assume $P$ holds for $\kappa(\mathfrak q) \to R_\mathfrak p^\wedge \otimes_R \kappa(\mathfrak q)$. By Algebra, Lemma \ref{algebra-lemma-completion-at-quasi-finite-prime} 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 from the assumptions on $P$.