hcju/stacksqa huggingface dataset:
Choose a diagram $$ \xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] } $$ as in Lemma \ref{lemma-embed-map-Noetherian-complete-local-rings}. Let $t_1, \ldots, t_d$ be a regular system of parameters for $R$ with $t_1 = p$ in case the characteristic of $k$ is $p > 0$. As $B$ and $\overline{S} = S \otimes_A k$ are regular we see that $\Ker(\overline{S} \to B)$ is generated by elements $\overline{x}_1, \ldots, \overline{x}_r$ which form part of a regular system of parameters of $\overline{S}$, see Algebra, Lemma \ref{algebra-lemma-regular-quotient-regular}. Lift these elements to $x_1, \ldots, x_r \in S$. Then $t_1, \ldots, t_d, x_1, \ldots, x_r$ is part of a regular system of parameters for $S$. Hence $S/(x_1, \ldots, x_r)$ is a power series ring over a field (if the characteristic of $k$ is zero) or a power series ring over a Cohen ring (if the characteristic of $k$ is $p > 0$), see Lemma \ref{lemma-quotient-power-series-ring-over-Cohen}. Moreover, it is still the case that $R \to S/(x_1, \ldots, x_r)$ maps $t_1, \ldots, t_d$ to a part of a regular system of parameters of $S/(x_1, \ldots, x_r)$. In other words, we may replace $S$ by $S/(x_1, \ldots, x_r)$ and assume we have a diagram $$ \xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] } $$ as in Lemma \ref{lemma-embed-map-Noetherian-complete-local-rings} with moreover $\overline{S} = B$. In this case $R \to S$ is formally smooth in the $\mathfrak m_S$-adic topology by Proposition \ref{proposition-fs-flat-fibre-fs}. Hence the base change $C = S \otimes_R A$ is formally smooth over $A$ in the $\mathfrak m_C$-adic topology by Lemma \ref{lemma-base-change-fs}.
Flatness is preserved under any base change, see Algebra, Lemma \ref{algebra-lemma-flat-base-change}. Consider a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p \subset R$. The residue field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak p')$ is finitely generated as $R'$ is of finite type over $R$. Hence the fibre ring $$ (\Lambda \otimes_R R') \otimes_{R'} \kappa(\mathfrak p') = \Lambda \otimes_R \kappa(\mathfrak p) \otimes_{\kappa(\mathfrak p)}  \kappa(\mathfrak p') $$ is Noetherian by Algebra, Lemma \ref{algebra-lemma-Noetherian-field-extension} and the assumption on the fibre rings of $R \to \Lambda$. Geometric regularity of the fibres is preserved by Algebra, Lemma \ref{algebra-lemma-geometrically-regular}.
Let $\mathfrak p \subset A$ be a prime. Let $\kappa(\mathfrak p) \subset k$ be a finite purely inseparable extension. We have to show that $C \otimes_A k$ is regular. By Lemma \ref{lemma-regular-base-change} we may assume that $A = k$ and we reduce to proving that $C$ is regular. The assumption is that $B$ is regular and that $B \to C$ is flat with regular fibres. Then $C$ is regular by Algebra, Lemma \ref{algebra-lemma-flat-over-regular-with-regular-fibre}. Some details omitted.
Note that $\Lambda$ is flat over $R$ by Algebra, Lemmas \ref{algebra-lemma-colimit-flat} and \ref{algebra-lemma-smooth-syntomic}. Let $\kappa(\mathfrak p) \subset k$ be a finite purely inseparable extension. Note that $$ \Lambda \otimes_R \kappa(\mathfrak p) \otimes_{\kappa(\mathfrak p)} k = \Lambda \otimes_R k = \colim A_i \otimes_R k $$ is a colimit of smooth $k$-algebras, see Algebra, Lemma \ref{algebra-lemma-base-change-smooth}. Since each local ring of a smooth $k$-algebra is regular by Algebra, Lemma \ref{algebra-lemma-characterize-smooth-over-field} we conclude that all local rings of $\Lambda \otimes_R k$ are regular by Algebra, Lemma \ref{algebra-lemma-colimit-regular}. This proves the lemma.
If $k \to K$ is regular, then $K$ is geometrically reduced over $k$, hence $K$ is separable over $k$ by Algebra, Proposition \ref{algebra-proposition-characterize-separable-field-extensions}. Conversely, if $K/k$ is separable, then $K$ is a colimit of smooth $k$-algebras, see Algebra, Lemma \ref{algebra-lemma-colimit-syntomic} hence is regular by Lemma \ref{lemma-colimit-smooth-regular}.
By Algebra, Lemma \ref{algebra-lemma-flat-permanence} we see that $A \to B$ is flat. Let $\mathfrak p \subset A$ be a prime. The ring map $B \otimes_A \kappa(\mathfrak p) \to C \otimes_A \kappa(\mathfrak p)$ is flat and surjective on spectra. Hence $B \otimes_A \kappa(\mathfrak p)$ is geometrically regular by Algebra, Lemma \ref{algebra-lemma-geometrically-regular-descent}.
By Algebra, Lemma \ref{algebra-lemma-completion-complete} the map $A \to A^\wedge$ induces isomorphisms $A/\mathfrak m^n = A^\wedge/(\mathfrak m^\wedge)^n$ for $n \geq 1$. By Algebra, Lemma \ref{algebra-lemma-pushdown-module} this implies that $$ \text{length}_A(A/\mathfrak m^n) = \text{length}_{A^\wedge}(A^\wedge/(\mathfrak m^\wedge)^n) $$ for all $n \geq 1$. Thus $d(A) = d(A^\wedge)$ and we conclude by Algebra, Proposition \ref{algebra-proposition-dimension}. An alternative proof is to use Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-equals-total}.
If $A^\wedge$ is regular, then $A$ is regular by Algebra, Lemma \ref{algebra-lemma-flat-under-regular}. Assume $A$ is regular. Let $\mathfrak m$ be the maximal ideal of $A$. Then $\dim_{\kappa(\mathfrak m)} \mathfrak m/\mathfrak m^2 = \dim(A) = \dim(A^\wedge)$ (Lemma \ref{lemma-completion-dimension}). On the other hand, $\mathfrak mA^\wedge$ is the maximal ideal of $A^\wedge$ and hence $\mathfrak m_{A^\wedge}$ is generated by at most $\dim(A^\wedge)$ elements. Thus $A^\wedge$ is regular. (You can also use Algebra, Lemma \ref{algebra-lemma-flat-over-regular-with-regular-fibre}.)
As $A \to A^\wedge$ is faithfully flat we have (1) by Algebra, Lemma \ref{algebra-lemma-descent-reduced}. For (2) see Algebra, Example \ref{algebra-example-bad-dvr-char-p} (there are also examples in characteristic zero, see Algebra, Remark \ref{algebra-remark-resolution-dim-1}). For (3) see Algebra, Lemmas \ref{algebra-lemma-local-nagata-domain-analytically-unramified} and \ref{algebra-lemma-analytically-unramified-easy}.
Consider the commutative diagram $$ \xymatrix{ A^\wedge \ar[r] & B^\wedge \\ A \ar[r] \ar[u] & B \ar[u] } $$ The vertical arrows are faithfully flat. Assume that $A^\wedge \to B^\wedge$ is flat. Then $A \to B^\wedge$ is flat. Hence $B$ is flat over $A$ by Algebra, Lemma \ref{algebra-lemma-flatness-descends-more-general}. \medskip\noindent Assume that $A \to B$ is flat. Then $A \to B^\wedge$ is flat. Hence $B^\wedge/\mathfrak m_A^n B^\wedge$ is flat over $A/\mathfrak m_A^n$ for all $n \geq 1$. Note that $\mathfrak m_A^n A^\wedge$ is the $n$th power of the maximal ideal $\mathfrak m_A^\wedge$ of $A^\wedge$ and $A/\mathfrak m_A^n = A^\wedge/(\mathfrak m_A^\wedge)^n$. Thus we see that $B^\wedge$ is flat over $A^\wedge$ by applying Algebra, Lemma \ref{algebra-lemma-flat-module-powers} (with $R = A^\wedge$, $I = \mathfrak m_A^\wedge$, $S = B^\wedge$, $M = S$).
By Algebra, Lemma \ref{algebra-lemma-finite-after-completion} we see that $B^\wedge$ is the $\mathfrak m_A$-adic completion of $B$ and that $A^\wedge \to B^\wedge$ is finite. Since $A \to B$ is flat we have $\text{Tor}_1^A(B, \kappa(\mathfrak m_A)) = 0$. Hence we see that $B^\wedge$ is flat over $A^\wedge$ by Lemma \ref{lemma-flat-after-completion}. Thus $B^\wedge$ is a free $A^\wedge$-module by Algebra, Lemma \ref{algebra-lemma-finite-flat-local}. Since $A^\wedge \to B^\wedge$ induces an isomorphism $\kappa(\mathfrak m_A) = A^\wedge/\mathfrak m_A A^\wedge \to B^\wedge/\mathfrak m_A B^\wedge = B^\wedge/\mathfrak m_B B^\wedge = \kappa(\mathfrak m_B)$ by our assumptions (and Algebra, Lemma \ref{algebra-lemma-hathat-finitely-generated}), we see that $B^\wedge$ is free of rank $1$. Thus $A^\wedge \to B^\wedge$ is an isomorphism.
Since $A_\mathfrak p \to B_\mathfrak q$ is faithfully flat we see that $B_\mathfrak q$ Noetherian implies that $A_\mathfrak p$ is Noetherian, see Algebra, Lemma \ref{algebra-lemma-descent-Noetherian}. Conversely, if $A_\mathfrak p$ is Noetherian, then $B_\mathfrak q$ is Noetherian as it is a localization of a finite type $A_\mathfrak p$-algebra.
Namely, because $A_{\mathfrak p} \to B_{\mathfrak q}$ is flat we have going down, and hence the inequality $\dim(A_{\mathfrak p}) \leq \dim(B_{\mathfrak q})$, see Algebra, Lemma \ref{algebra-lemma-dimension-going-up}. On the other hand, suppose that $\mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset \mathfrak q_n$ is a chain of primes in $B_{\mathfrak q}$. Then the corresponding sequence of primes $\mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_n$ (with $\mathfrak p_i = \mathfrak q_i \cap A_{\mathfrak p}$) is chain also (i.e., no equalities in the sequence) as an \'etale ring map is quasi-finite (see Algebra, Lemma \ref{algebra-lemma-etale-quasi-finite}) and a quasi-finite ring map induces a map of spectra with discrete fibres (by definition). This means that $\dim(A_{\mathfrak p}) \geq \dim(B_{\mathfrak q})$ as desired.
By Lemma \ref{lemma-Noetherian-etale-extension} we may assume both $A_\mathfrak p$ and $B_\mathfrak q$ are Noetherian in order to prove the equivalence. Let $x_1, \ldots, x_t \in \mathfrak pA_\mathfrak p$ be a minimal set of generators. As $A_\mathfrak p \to B_\mathfrak q$ is faithfully flat we see that the images $y_1, \ldots, y_t$ in $B_\mathfrak q$ form a minimal system of generators for $\mathfrak pB_\mathfrak q = \mathfrak q B_\mathfrak q$ (Algebra, Lemma \ref{algebra-lemma-etale-at-prime}). Regularity of $A_\mathfrak p$ by definition means $t = \dim(A_\mathfrak p)$ and similarly for $B_\mathfrak q$. Hence the lemma follows from the equality $\dim(A_\mathfrak p) = \dim(B_\mathfrak q)$ of Lemma \ref{lemma-dimension-etale-extension}.
The statement on the local rings follows from Lemmas \ref{lemma-dimension-etale-extension} and \ref{lemma-regular-etale-extension} and Algebra, Lemma \ref{algebra-lemma-characterize-dvr}. It follows that $B$ is a Noetherian normal ring of dimension $1$. By Algebra, Lemma \ref{algebra-lemma-characterize-reduced-ring-normal} we conclude that $B$ is a finite product of normal domains of dimension $1$. These are Dedekind domains by Algebra, Lemma \ref{algebra-lemma-characterize-Dedekind}.
By construction $R^h$ is a colimit of \'etale $R$-algebras, see Algebra, Lemma \ref{algebra-lemma-henselization}. Since \'etale ring maps are flat (Algebra, Lemma \ref{algebra-lemma-etale}) we see that $R^h$ is flat over $R$ by Algebra, Lemma \ref{algebra-lemma-colimit-flat}. As a flat local ring homomorphism is faithfully flat (Algebra, Lemma \ref{algebra-lemma-local-flat-ff}) we see that $R \to R^h$ is faithfully flat. The ring map $R^h \to R^{sh}$ is a colimit of finite \'etale ring maps, see proof of Algebra, Lemma \ref{algebra-lemma-strict-henselization}. Hence the same arguments as above show that $R^h \to R^{sh}$ is faithfully flat. \medskip\noindent Part (2) follows from Algebra, Lemmas \ref{algebra-lemma-henselization} and \ref{algebra-lemma-strict-henselization}. Part (3) follows from Algebra, Lemma \ref{algebra-lemma-local-artinian-basis-when-flat} because $R/\mathfrak m \to R^h/\mathfrak mR^h$ is an isomorphism and $R/\mathfrak m^n \to R^h/\mathfrak m^nR^h$ is flat as a base change of the flat ring map $R \to R^h$ (Algebra, Lemma \ref{algebra-lemma-flat-base-change}). Let $\kappa^{sep}$ be the residue field of $R^{sh}$ (it is a separable algebraic closure of $\kappa$). Choose $x_i \in R^{sh}$ mapping to a basis of $\kappa^{sep}$ as a $\kappa$-vector space. Then (4) follows from Algebra, Lemma \ref{algebra-lemma-local-artinian-basis-when-flat} in exactly the same way as above.
Part (1) follows from the fact that $R^h$ and $R^{sh}$ are directed colimits of \'etale algebras (by construction), that \'etale algebras are formally \'etale (Algebra, Lemma \ref{algebra-lemma-formally-etale-etale}), and that colimits of formally \'etale algebras are formally \'etale (Algebra, Lemma \ref{algebra-lemma-colimit-formally-etale}). Part (2) follows from the fact that a formally \'etale ring map is formally smooth and Lemma \ref{lemma-formally-smooth}.
Since $R \to R^h \to R^{sh}$ are faithfully flat (Lemma \ref{lemma-dumb-properties-henselization}), we see that $R^h$ or $R^{sh}$ being Noetherian implies that $R$ is Noetherian, see Algebra, Lemma \ref{algebra-lemma-descent-Noetherian}. In the rest of the proof we assume $R$ is Noetherian. \medskip\noindent 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 \ref{lemma-dumb-properties-henselization}. Hence $(R^h)^\wedge$ and $(R^{sh})^\wedge$ are Noetherian by Algebra, Lemma \ref{algebra-lemma-complete-local-ring-Noetherian}. This proves (a). \medskip\noindent Note that (b) is immediate from Lemma \ref{lemma-dumb-properties-henselization}. In particular we see that $(R^h)^\wedge$ is flat over $R$, see Algebra, Lemma \ref{algebra-lemma-completion-faithfully-flat}. \medskip\noindent Next, we show that $R^h \to (R^h)^\wedge$ is flat. Write $R^h = \colim_i R_i$ as a directed colimit of localizations of \'etale $R$-algebras. By Algebra, Lemma \ref{algebra-lemma-colimit-rings-flat} 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 \ref{lemma-dumb-properties-henselization} that $(R^{sh})^\wedge$ is the completion of a free $R$-module. By Lemma \ref{lemma-completed-direct-sum-flat} we see this is flat over $R$ as desired. This finishes the proof of (c). \medskip\noindent At this point we know (c) is true and that $(R^h)^\wedge$ and $(R^{sh})^\wedge$ are Noetherian. It follows from Algebra, Lemma \ref{algebra-lemma-descent-Noetherian} that $R^h$ and $R^{sh}$ are Noetherian. \medskip\noindent Part (d) follows from Lemma \ref{lemma-henselization-formally-smooth} and Lemma \ref{lemma-formally-smooth-completion}. \medskip\noindent Part (e) follows from Algebra, Lemma \ref{algebra-lemma-sh-from-h-map} and the fact that $R^\wedge$ is henselian by Algebra, Lemma \ref{algebra-lemma-complete-henselian}. \medskip\noindent 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 \ref{algebra-lemma-strictly-henselian-functorial}. Completing we obtain a map $((R^\wedge)^{sh})^\wedge \to (R^{sh})^\wedge$. We omit the verification that these two maps are mutually inverse.
The ring maps $R \to R^h \to R^{sh}$ are faithfully flat. Hence one direction of the implications follows from Algebra, Lemma \ref{algebra-lemma-descent-reduced}. Conversely, assume $R$ is reduced. Since $R^h$ and $R^{sh}$ are filtered colimits of \'etale, hence smooth $R$-algebras, the result follows from Algebra, Lemma \ref{algebra-lemma-reduced-goes-up}.
Note that $nil(R)$ is the biggest ideal consisting of nilpotent elements such that the quotient $R/nil(R)$ is reduced. Note that $nil(R)R^h$ consists of nilpotent elements by Algebra, Lemma \ref{algebra-lemma-locally-nilpotent}. Also, note that $R^h/nil(R) R^h$ is the henselization of $R/nil(R)$ by Algebra, Lemma \ref{algebra-lemma-quotient-henselization}. Hence $R^h/nil(R)R^h$ is reduced by Lemma \ref{lemma-henselization-reduced}. We conclude that $nil(R) R^h = nil(R^h)$ as desired. Similarly for the strict henselization but using Algebra, Lemma \ref{algebra-lemma-quotient-strict-henselization}.
A preliminary remark is that a local ring is normal if and only if it is a normal domain (see Algebra, Definition \ref{algebra-definition-ring-normal}). The ring maps $R \to R^h \to R^{sh}$ are faithfully flat. Hence one direction of the implications follows from Algebra, Lemma \ref{algebra-lemma-descent-normal}. Conversely, assume $R$ is normal. Since $R^h$ and $R^{sh}$ are filtered colimits of \'etale hence smooth $R$-algebras, the result follows from Algebra, Lemmas \ref{algebra-lemma-normal-goes-up} and \ref{algebra-lemma-colimit-normal-ring}.
Since $R \to R^{sh}$ is faithfully flat (Lemma \ref{lemma-dumb-properties-henselization}) we see that $\dim(R^{sh}) \geq \dim(R)$ by going down, see Algebra, Lemma \ref{algebra-lemma-dimension-going-up}. For the converse, we write $R^{sh} = \colim R_i$ as a directed colimit of local rings $R_i$ each of which is a localization of an \'etale $R$-algebra. Now if $\mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset \mathfrak q_n$ is a chain of prime ideals in $R^{sh}$, then for some sufficiently large $i$ the sequence $$ R_i \cap \mathfrak q_0 \subset R_i \cap \mathfrak q_1 \subset \ldots \subset R_i \cap \mathfrak q_n $$ is a chain of primes in $R_i$. Thus we see that $\dim(R^{sh}) \leq \sup_i \dim(R_i)$. But by the result of Lemma \ref{lemma-dimension-etale-extension} we have $\dim(R_i) = \dim(R)$ for each $i$ and we win.
By Lemma \ref{lemma-henselization-noetherian} we know that $R^h$ and $R^{sh}$ are Noetherian, hence the lemma makes sense. Since we have $\text{depth}(R) = \text{depth}(R^h) = \text{depth}(R^{sh})$ and $\dim(R) = \dim(R^h) = \dim(R^{sh})$ by Lemmas \ref{lemma-henselization-depth} and \ref{lemma-henselization-dimension} we conclude.
By Lemma \ref{lemma-henselization-noetherian} we know that $R^h$ and $R^{sh}$ are Noetherian, hence the lemma makes sense. Let $\mathfrak m$ be the maximal ideal of $R$. Let $x_1, \ldots, x_t \in \mathfrak m$ be a minimal system of generators of $\mathfrak m$, i.e., such that the images in $\mathfrak m/\mathfrak m^2$ form a basis over $\kappa = R/\mathfrak m$. Because $R \to R^h$ and $R \to R^{sh}$ are faithfully flat, it follows that the images $x_1^h, \ldots, x_t^h$ in $R^h$, resp.\  $x_1^{sh}, \ldots, x_t^{sh}$ in $R^{sh}$ are a minimal system of generators for $\mathfrak m^h = \mathfrak mR^h$, resp.\ $\mathfrak m^{sh} = \mathfrak mR^{sh}$. Regularity of $R$ by definition means $t = \dim(R)$ and similarly for $R^h$ and $R^{sh}$. Hence the lemma follows from the equality of dimensions $\dim(R) = \dim(R^h) = \dim(R^{sh})$ of Lemma \ref{lemma-henselization-dimension}
Write $B$ as a filtered colimit $B = \colim B_i$ with $A \to B_i$ \'etale. Then on the one hand $B \otimes_A \kappa(\mathfrak p) = \colim B_i \otimes_A \kappa(\mathfrak p)$ is a filtered colimit of \'etale $\kappa(\mathfrak p)$-algebras, and on the other hand it is Noetherian. An \'etale $\kappa(\mathfrak p)$-algebra is a finite product of finite separable field extensions (Algebra, Lemma \ref{algebra-lemma-etale-over-field}). Hence there are no nontrivial specializations between the primes (which are all maximal and minimal primes) of the algebras $B_i \otimes_A \kappa(\mathfrak p)$ and hence there are no nontrivial specializations between the primes of $B \otimes_A \kappa(\mathfrak p)$. Thus $B \otimes_A \kappa(\mathfrak p)$ is reduced and has finitely many primes which all minimal. Thus it is a finite product of fields (use Algebra, Lemma \ref{algebra-lemma-total-ring-fractions-no-embedded-points} or Algebra, Proposition \ref{algebra-proposition-dimension-zero-ring}). Each of these fields is a colimit of finite separable extensions and hence the final statement of the lemma follows.
This can be deduced from the more general Lemma \ref{lemma-filtered-colimit-etale-noetherian-fibres} using that the henselization and strict henselization are Noetherian (as we've seen above). But we also give a direct proof as follows. \medskip\noindent We will use without further mention the results of Lemmas \ref{lemma-dumb-properties-henselization} and \ref{lemma-henselization-noetherian}. Note that $R^h/\mathfrak pR^h$, resp.\ $R^{sh}/\mathfrak pR^{sh}$ is the henselization, resp.\ strict henselization of $R/\mathfrak p$, see Algebra, Lemma \ref{algebra-lemma-quotient-henselization} resp.\ Algebra, Lemma \ref{algebra-lemma-quotient-strict-henselization}. Hence we may replace $R$ by $R/\mathfrak p$ and assume that $R$ is a Noetherian local domain and that $\mathfrak p = (0)$. Since $R^h$, resp.\ $R^{sh}$ is Noetherian, it has finitely many minimal primes $\mathfrak q_1, \ldots, \mathfrak q_t$, resp.\ $\mathfrak r_1, \ldots, \mathfrak r_s$. Since $R \to R^h$, resp.\ $R \to R^{sh}$ is flat these are exactly the primes lying over $\mathfrak p = (0)$ (by going down). Finally, as $R$ is a domain, we see that $R^h$, resp.\ $R^{sh}$ is reduced, see Lemma \ref{lemma-henselization-reduced}. Thus we see that $R^h \otimes_R \kappa(\mathfrak p)$ resp.\ $R^{sh} \otimes_R \kappa(\mathfrak p)$ is a reduced Noetherian ring with finitely many primes, all of which are minimal (and hence maximal). Thus these rings are Artinian and are products of their localizations at maximal ideals, each necessarily a field (see Algebra, Proposition \ref{algebra-proposition-dimension-zero-ring} and Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}). \medskip\noindent The final statement follows from the fact that $R \to R^h$, resp.\ $R \to R^{sh}$ is a colimit of \'etale ring maps and hence the induced residue field extensions are colimits of finite separable extensions, see Algebra, Lemma \ref{algebra-lemma-etale-at-prime}.
Proof of (1). Choose a $p$-basis $\{x_i\}$ for $K$ over $\mathbf{F}_p$. Suppose that $\eta = \sum_{i \in I'} y_i \text{d}x_i$ maps to zero in $\Omega_{K/K_\alpha}$ for every $\alpha \in A$. Here the index set $I'$ is finite. By Lemma \ref{lemma-p-basis} this means that for every $\alpha$ there exists a relation $$ \sum\nolimits_E a_{E, \alpha} x^E,\quad a_{E, \alpha} \in K_\alpha $$ where $E$ runs over multi-indices $E = (e_i)_{i \in I'}$ with $0 \leq e_i < p$. On the other hand, Lemma \ref{lemma-p-basis} guarantees there is no such relation $\sum a_E x^E = 0$ with $a_E \in K^p$. This is a contradiction by Lemma \ref{lemma-intersection-subfields-subspace}. \medskip\noindent Proof of (2). Suppose that we have a tower $K \subset M \subset L$ of finite extensions of fields. Set $M_\alpha = M^p K_\alpha$ and $L_\alpha = L^p K_\alpha = L^p M_\alpha$. Then we can first prove that $M^p = \bigcap_{\alpha \in A} M_\alpha$, and after that prove that $L^p = \bigcap_{\alpha \in A} L_\alpha$. Hence it suffices to prove (2) for primitive field extensions having no nontrivial subfields. First, assume that $L = K(\theta)$ is separable over $K$. Then $L$ is generated by $\theta^p$ over $K$, hence we may assume that $\theta \in L^p$. In this case we see that $$ L^p = K^p \oplus K^p\theta \oplus \ldots K^p\theta^{d - 1} \quad\text{and}\quad L^pK_\alpha = K_\alpha \oplus K_\alpha \theta \oplus \ldots K_\alpha\theta^{d - 1} $$ where $d = [L : K]$. Thus the conclusion is clear in this case. The other case is where $L = K(\theta)$ with $\theta^p = t \in K$, $t \not \in K^p$. In this case we have $$ L^p = K^p \oplus K^pt \oplus \ldots K^pt^{p - 1} \quad\text{and}\quad L^pK_\alpha = K_\alpha \oplus K_\alpha t \oplus \ldots K_\alpha t^{p - 1} $$ Again the result is clear.
Write $A = k[[x_1, \ldots, x_n]][y_1, \ldots, y_m]$ and $A' = k'[[x_1^p, \ldots, x_n^p]][y_1^p, \ldots, y_m^p]$. We also denote $K'$ the fraction field of $A'$. The ring extension $k'[[x_1^p, \ldots, x_d^p]] \subset k[[x_1, \ldots, x_d]]$ is finite by Algebra, Lemma \ref{algebra-lemma-finite-after-completion} which implies that $A \to A'$ is finite. For $f \in A$ we see that $f^p \in A'$. Hence $K^p \subset K'$. Any element of $K'$ can be written as $a/b^p$ with $a \in A'$ and $b \in A$ nonzero. Suppose that $f/g^p \in K$, $f, g \in A$, $g \not = 0$ is contained in $K'$ for every choice of $k'$. Fix a choice of $k'$ for the moment. By the above we see $f/g^p = a/b^p$ for some $a \in A'$ and some nonzero $b \in A$. Hence $b^p f \in A'$. For any $A'$-derivation $D : A \to A$ we see that $0 = D(b^pf) = b^p D(f)$ hence $D(f) = 0$ as $A$ is a domain. Taking $D = \partial_{x_i}$ and $D = \partial_{y_j}$ we conclude that $f \in k[[x_1^p, \ldots, x_n^p]][y_1^p, \ldots, y_d^p]$. Applying a $k'$-derivation $\theta : k \to k$ we similarly conclude that all coefficients of $f$ are in $k'$, i.e., $f \in A'$. Since it is clear that $A^p = \bigcap\nolimits_{k'} A'$ where $k'$ ranges over all subfields as in the lemma we win.