hcju/stacksqa huggingface dataset:
Set $X = \Spec(A)$, $X_0 = \Spec(A_0)$, $Y_0 = \Spec(B_0)$, $Z = \Spec(C)$, $Z_0 = \Spec(C_0)$. Furthermore, denote $U \subset X$, $U_0 \subset X_0$, $V_0 \subset Y_0$, $W \subset Z$, $W_0 \subset Z_0$ the complement of the vanishing set of $I$. Here is a picture to help visualize the situation: $$ \xymatrix{ Z \ar[dd] & Z_0 \ar[l] \ar[d] \\ & Y_0 \ar[d] \\ X & X_0 \ar[l] } \quad\quad\quad \xymatrix{ W \ar[dd] & W_0 \ar[l] \ar[d] \\ & V_0 \ar[d] \\ U & U_0 \ar[l] } $$ The conditions in the lemma guarantee that $$ \xymatrix{ W_0 \ar[r] \ar[d] & Z_0 \ar[d] \\ V_0 \ar[r] & Y_0 } $$ is an elementary distinguished square, see Derived Categories of Spaces, Definition \ref{spaces-perfect-definition-elementary-distinguished-square}. In addition we know that $W_0 \to U_0$ and $V_0 \to U_0$ are \'etale. The morphism $X_0 \subset X$ is a finite order thickening as $J$ is assumed nilpotent. By the topological invariance of the \'etale site we can find a unique \'etale morphism $V \to X$ of schemes with $V_0 = V \times_X X_0$ and we can lift the given morphism $W_0 \to V_0$ to a unique morphism $W \to V$ over $X$. See \'Etale Morphisms, Theorem \ref{etale-theorem-remarkable-equivalence}. Since $W_0 \to V_0$ is separated, the morphism $W \to V$ is separated too, see for example More on Morphisms, Lemma \ref{more-morphisms-lemma-deform-property}. By Pushouts of Spaces, Lemma \ref{spaces-pushouts-lemma-construct-elementary-distinguished-square} we can construct an elementary distinguished square $$ \xymatrix{ W \ar[r] \ar[d] & Z \ar[d] \\ V \ar[r] & Y } $$ in the category of algebraic spaces over $X$. Since the base change of an elementary distinguished square is an elementary distinguished square (Derived Categories of Spaces, Lemma \ref{spaces-perfect-lemma-make-more-elementary-distinguished-squares}) we see that $$ \xymatrix{ W_0 \ar[r] \ar[d] & Z_0 \ar[d] \\ V_0 \ar[r] & Y \times_X X_0 } $$ is an elementary distinguished square. It follows that there is a unique isomorphism $Y \times_X X_0 = Y_0$ compatible with the two squares involving these spaces because elementary distinguished squares are pushouts (Pushouts of Spaces, Lemma \ref{spaces-pushouts-lemma-elementary-distinguished-square-pushout}). It follows that $Y$ is affine by Limits of Spaces, Proposition \ref{spaces-limits-proposition-affine}. Write $Y = \Spec(B)$. It is clear that $B$ fits into the desired diagram and satisfies all the properties required of it.
Choose a presentation $B = A[x_1, \ldots, x_r]^\wedge/J$. By Lemma \ref{lemma-equivalent-with-artin} part (6) we can find $c \geq 0$ and $f_1, \ldots, f_r \in J$ such that $\det_{1 \leq i, j \leq r}(\partial f_j/\partial x_i)$ divides $a^c$ in $B$ and $a^c J \subset (f_1, \ldots, f_r) + J^2$. Hence Lemma \ref{lemma-approximate-presentation-rig-smooth} applies. This finishes the proof, but we'd like to point out that in this case the use of Lemma \ref{lemma-get-morphism-general-better} can be replaced by the much easier Lemma \ref{lemma-get-morphism-principal}.
Taut ring homomorphisms are defined in Formal Spaces, Definition \ref{formal-spaces-definition-taut}. Adic ring homomorphisms are defined in Formal Spaces, Definition \ref{formal-spaces-definition-adic-homomorphism}. The lemma follows from a combination of Formal Spaces, Lemmas \ref{formal-spaces-lemma-quotient-restricted-power-series}, \ref{formal-spaces-lemma-quotient-restricted-power-series-admissible}, and \ref{formal-spaces-lemma-adic-homomorphism}. We omit the details. To be sure, there is no difference between the topological rings $A[x_1, \ldots, x_n]^\wedge$ and $A\{x_1, \ldots, x_r\}$, see Formal Spaces, Remark \ref{formal-spaces-remark-I-adic-completion-and-restricted-power-series}.
The statement makes sense by Lemma \ref{lemma-finite-type}. To see that it is true assume we have morphisms $B \to A$ and $B \to C$ in $\textit{WAdm}^{adic*}$ and that as a topological $B$-algebra we have $A = B\{x_1, \ldots, x_r\}/J$ for some closed ideal $J$. Then $A \widehat{\otimes}_B C$ is isomorphic to the quotient of $C\{x_1, \ldots, x_r\}/J'$ where $J'$ is the closure of $JC\{x_1, \ldots, x_r\}$. Some details omitted.
Consider a commutative diagram $$ \xymatrix{ B \ar[r] & (B')^\wedge \\ A \ar[r] \ar[u]^\varphi & (A')^\wedge \ar[u]_{\varphi'} } $$ as in Formal Spaces, Situation \ref{formal-spaces-situation-local-property}. Taking $\text{Spf}$ of this diagram we obtain $$ \xymatrix{ \text{Spf}(B) \ar[d] & \text{Spf}((B')^\wedge) \ar[l] \ar[d] \\ \text{Spf}(A) & \text{Spf}((A')^\wedge) \ar[l] } $$ of affine formal algebraic spaces whose horizontal arrows are representable by algebraic spaces and \'etale by Formal Spaces, Lemma \ref{formal-spaces-lemma-etale}. Hence we obtain a commutative diagram of affine schemes $$ \xymatrix{ \text{Spf}(B)_{red} \ar[d]^f & \text{Spf}((B')^\wedge)_{red} \ar[l]^g \ar[d]^{f'} \\ \text{Spf}(A)_{red} & \text{Spf}((A')^\wedge)_{red} \ar[l] } $$ whose horizontal arrows are \'etale by Formal Spaces, Lemma \ref{formal-spaces-lemma-reduction-smooth}. By Formal Spaces, Example \ref{formal-spaces-example-reduction-affine-formal-spectrum} and Lemma \ref{formal-spaces-lemma-etale-surjective} conditions (1), (2), and (3) of Formal Spaces, Situation \ref{formal-spaces-situation-local-property} translate into the following statements \begin{enumerate} \item if $f$ is locally of finite type, then $f'$ is locally of finite type, \item if $f'$ is locally of finite type and $g$ is surjective, then $f$ is locally of finite type, and \item if $T_i \to S$, $i = 1, \ldots, n$ are locally of finite type, then $\coprod_{i = 1, \ldots, n} T_i \to S$ is locally of finite type. \end{enumerate} Properties (1) and (2) follow from the fact that being locally of finite type is local on the source and target in the \'etale topology, see discussion in Morphisms of Spaces, Section \ref{spaces-morphisms-section-finite-type}. Property (3) is a straightforward consequence of the definition.
The statement makes sense by Lemma \ref{lemma-finite-type-red}. To see that it is true assume we have morphisms $B \to A$ and $B \to C$ in $\textit{WAdm}^{count}$ such that $B/\mathfrak b \to A/\mathfrak a$ is of finite type where $\mathfrak b \subset B$ and $\mathfrak a \subset A$ are the ideals of topologically nilpotent elements. Since $A$ and $B$ are weakly admissible, the ideals $\mathfrak a$ and $\mathfrak b$ are open. Let $\mathfrak c \subset C$ be the (open) ideal of topologically nilpotent elements. Then we find a surjection $A \widehat{\otimes}_B C \to A/\mathfrak a \otimes_{B/\mathfrak b} C/\mathfrak c$ whose kernel is a weak ideal of definition and hence consists of topologically nilpotent elements (please compare with the proof of Formal Spaces, Lemma \ref{formal-spaces-lemma-completed-tensor-product}). Since already $C/\mathfrak c \to A/\mathfrak a \otimes_{B/\mathfrak b} C/\mathfrak c$ is of finite type as a base change of $B/\mathfrak b \to A/\mathfrak a$ we conclude.
The equivalence of (1), (2), and (3) follows from Lemma \ref{lemma-finite-type-red} and an application of Formal Spaces, Lemma \ref{formal-spaces-lemma-property-defines-property-morphisms}. Let $Y_j$ and $X_{ji}$ be as in (2). Then \begin{itemize} \item The families $\{Y_{j, red} \to Y_{red}\}$ and $\{X_{ji, red} \to X_{red}\}$ are \'etale coverings by affine schemes. This follows from the discussion in the proof of Formal Spaces, Lemma \ref{formal-spaces-lemma-reduction-formal-algebraic-space} or directly from Formal Spaces, Lemma \ref{formal-spaces-lemma-reduction-smooth}. \item If $X_{ji} \to Y_j$ corresponds to the morphism $B_j \to A_{ji}$ of $\textit{WAdm}^{count}$, then $X_{ji, red} \to Y_{j, red}$ corresponds to the ring map $B_j/\mathfrak b_j \to A_{ji}/\mathfrak a_{ji}$ where $\mathfrak b_j$ and $\mathfrak a_{ji}$ are the ideals of topologically nilpotent elements. This follows from Formal Spaces, Example \ref{formal-spaces-example-reduction-affine-formal-spectrum}. Hence $X_{ji, red} \to Y_{j, red}$ is locally of finite type if and only if $B_j \to A_{ji}$ satisfies the property defined in Lemma \ref{lemma-finite-type-red}. \end{itemize} The equivalence of (2) and (4) follows from these remarks because being locally of finite type is a property of morphisms of algebraic spaces which is \'etale local on source and target, see discussion in Morphisms of Spaces, Section \ref{spaces-morphisms-section-finite-type}.
Let us recall what the statement signifies. First,  $\textit{WAdm}^{Noeth}$ is the category whose objects are adic Noetherian topological rings and whose morphisms are continuous ring homomorphisms. Consider a commutative diagram $$ \xymatrix{ B \ar[r] & (B')^\wedge \\ A \ar[r] \ar[u]^\varphi & (A')^\wedge \ar[u]_{\varphi'} } $$ satisfying the following conditions: $A$ and $B$ are adic Noetherian topological rings, $A \to A'$ and $B \to B'$ are \'etale ring maps, $(A')^\wedge = \lim A'/I^nA'$ for some ideal of definition $I \subset A$, $(B')^\wedge = \lim B'/J^nB'$ for some ideal of definition $J \subset B$, and $\varphi : A \to B$ and $\varphi' : (A')^\wedge \to (B')^\wedge$ are continuous. Note that $(A')^\wedge$ and $(B')^\wedge$ are adic Noetherian topological rings by Formal Spaces, Lemma \ref{formal-spaces-lemma-completion-in-sub}. We have to show \begin{enumerate} \item $\varphi$ is flat $\Rightarrow \varphi'$ is flat, \item if $B \to B'$ faithfully flat, then $\varphi'$ is flat $\Rightarrow \varphi$ is flat, and \item if $A \to B_i$ is flat for $i = 1, \ldots, n$, then $A \to \prod_{i = 1, \ldots, n} B_i$ is flat. \end{enumerate} We will use without further mention that completions of Noetherian rings are flat (Algebra, Lemma \ref{algebra-lemma-completion-flat}). Since of course $A \to A'$ and $B \to B'$ are flat, we see in particular that the horizontal arrows in the diagram are flat. \medskip\noindent Proof of (1). If $\varphi$ is flat, then the composition $A \to (A')^\wedge \to (B')^\wedge$ is flat. Hence $A' \to (B')^\wedge$ is flat by More on Flatness, Lemma \ref{flat-lemma-etale-flat-up-down}. Hence we see that $(A')^\wedge \to (B')^\wedge$ is flat by applying More on Algebra, Lemma \ref{more-algebra-lemma-flat-after-completion} with $R = A'$, with ideal $I(A')$, and with $M = (B')^\wedge = M^\wedge$. \medskip\noindent Proof of (2). Assume $\varphi'$ is flat and $B \to B'$ is faithfully flat. Then the composition $A \to (A')^\wedge \to (B')^\wedge$ is flat. Also we see that $B \to (B')^\wedge$ is faithfully flat by Formal Spaces, Lemma \ref{formal-spaces-lemma-etale-surjective}. Hence by Algebra, Lemma \ref{algebra-lemma-flatness-descends-more-general} we find that $\varphi : A \to B$ is flat. \medskip\noindent Proof of (3). Omitted.
The statement makes sense as each of the properties $P$, $Q$, and $R$ is a local property of morphisms of $\textit{WAdm}^{Noeth}$. Let $\varphi : B \to A$ and $\psi : B \to C$ be morphisms of $\textit{WAdm}^{Noeth}$. If either $Q(\varphi)$ or $Q(\psi)$ then we see that $A \widehat{\otimes}_B C$ is Noetherian by Formal Spaces, Lemma \ref{formal-spaces-lemma-completed-tensor-product}. Since $R$ implies $Q$ (Lemma \ref{lemma-finite-type-finite-type-red}), we find that this holds in both cases (1) and (2). This is the first thing we have to check. It remains to show that $C \to A \widehat{\otimes}_B C$ is flat. \medskip\noindent Proof of (1). Fix ideals of definition $I \subset A$ and $J \subset B$. By Lemma \ref{lemma-Noetherian-finite-type-red} the ring map $B \to C$ is topologically of finite type. Hence $B \to C/J^n$ is of finite type for all $n \geq 1$. Hence $A \otimes_B C/J^n$ is Noetherian as a ring (because it is of finite type over $A$ and $A$ is Noetherian). Thus the $I$-adic completion $A \widehat{\otimes}_B C/J^n$ of $A \otimes_B C/J^n$ is flat over $C/J^n$ because $C/J^n \to A \otimes_B C/J^n$ is flat as a base change of $B \to A$ and because $A \otimes_B C/J^n \to A \widehat{\otimes}_B C/J^n$ is flat by Algebra, Lemma \ref{algebra-lemma-completion-flat} Observe that $A \widehat{\otimes}_B C/J^n = (A \widehat{\otimes}_B C)/J^n(A \widehat{\otimes}_B C)$; details omitted. We conclude that $M = A \widehat{\otimes}_B C$ is a $C$-module which is complete with respect to the $J$-adic topology such that $M/J^nM$ is flat over $C/J^n$ for all $n \geq 1$. This implies that $M$ is flat over $C$ by More on Algebra, Lemma \ref{more-algebra-lemma-limit-flat}. \medskip\noindent Proof of (2). In this case $B \to A$ is adic and hence we have just $A \widehat{\otimes}_B C = \lim A \otimes_B C/J^n$. The rings $A \otimes_B C/J^n$ are Noetherian by an application of  Formal Spaces, Lemma \ref{formal-spaces-lemma-completed-tensor-product} with $C$ replaced by $C/J^n$. We conclude in the same manner as before.
The equivalence of (1) and (2) is Definition \ref{definition-flat}. The equivalence of (2), (3), and (4) follows from the fact that being flat is a local property of arrows of $\text{WAdm}^{Noeth}$ by Lemma \ref{lemma-flat-axioms} and an application of the variant of Formal Spaces, Lemma \ref{formal-spaces-lemma-property-defines-property-morphisms} for morphisms between locally Noetherian algebraic spaces mentioned in Formal Spaces, Remark \ref{formal-spaces-remark-variant-Noetherian}.
Part (1) follows from a combination of Formal Spaces, Remark \ref{formal-spaces-remark-base-change-variant-variant-Noetherian}, Lemma \ref{lemma-base-change-flat-continuous} part (1), Lemma \ref{lemma-flat-morphisms}, and Lemma \ref{lemma-finite-type-red-morphisms}. \medskip\noindent Part (2) follows from a combination of Formal Spaces, Remark \ref{formal-spaces-remark-base-change-variant-Noetherian}, Lemma \ref{lemma-base-change-flat-continuous} part (2), Lemma \ref{lemma-flat-morphisms}, and Lemma \ref{lemma-finite-type-morphisms}.
This is a sanity check whose proof should be trivial but isn't quite. We urge the reader to skip the proof. Assume $f$ is representable by algebraic spaces and flat in the sense of Bootstrap, Definition \ref{bootstrap-definition-property-transformation}. Consider a commutative diagram $$ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } $$ with $U$ and $V$ affine formal algebraic spaces, $U \to X$ and $V \to Y$ representable by algebraic spaces and \'etale. Then the morphism $U \to V$ corresponds to a taut map $B \to A$ of $\textit{WAdm}^{Noeth}$ by Formal Spaces, Lemma \ref{formal-spaces-lemma-representable-local-property}. Observe that this means $B \to A$ is adic (Formal Spaces, Lemma \ref{formal-spaces-lemma-adic-homomorphism}) and in particular for any ideal of definition $J \subset B$ the topology on $A$ is the $J$-adic topology and the diagrams $$ \xymatrix{ \Spec(A/J^nA) \ar[r] \ar[d] & \Spec(B/J^n) \ar[d] \\ U \ar[r] & V } $$ are cartesian. \medskip\noindent Let $T \to V$ is a morphism where $T$ is a scheme. Then \begin{align*} X \times_Y T \to T\text{ is flat} & \Rightarrow U \times_Y T \to T\text{ is flat} \\ & \Rightarrow U \times_V V \times_Y T \to T\text{ is flat} \\ & \Rightarrow U \times_V V \times_Y T \to V \times_Y T\text{ is flat} \\ & \Rightarrow U \times_V T \to T\text{ is flat} \end{align*} The first statement is the assumption on $f$. The first implication because $U \to X$ is \'etale and hence flat and compositions of flat morphisms of algebraic spaces are flat. The second impliciation because $U \times_Y T = U \times_V V \times_Y T$. The third implication by More on Flatness, Lemma \ref{flat-lemma-etale-flat-up-down}. The fourth implication because we can pullback by the morphism $T \to V \times_Y T$. We conclude that $U \to V$ is flat in the sense of Bootstrap, Definition \ref{bootstrap-definition-property-transformation}. In terms of the continuous ring map $B \to A$ this means the ring maps $B/J^n \to A/J^nA$ are flat (see diagram above). \medskip\noindent Finally, we can conclude that $B \to A$ is flat for example by More on Algebra, Lemma \ref{more-algebra-lemma-limit-flat}.
It is clear that (1) and (2) are equivalent and for the same reason that (3) and (4) are equivalent. Since $V(I)$ is independent of the choice of the ideal of definition $I$ of $A$, we see that (2) and (4) are equivalent. \medskip\noindent Assume the equivalent conditions (1) -- (4) hold. If $\dim(A/\mathfrak q) > 1$ we can choose a maximal ideal $\mathfrak q \subset \mathfrak m \subset A$ such that $\dim((A/\mathfrak q)_\mathfrak m) > 1$. Then $\Spec((A/\mathfrak q)_\mathfrak m) - V(I(A/\mathfrak q)_\mathfrak m)$ would be infinite by Algebra, Lemma \ref{algebra-lemma-Noetherian-local-domain-dim-2-infinite-opens}. This contradicts the fact that $\mathfrak q$ is closed in $\Spec(A) \setminus V(I)$. Hence we see that (6) holds. Trivially (6) implies (5). \medskip\noindent Conversely, assume (5) holds. Let $I \subset A$ be an ideal of definition. Since $A/\mathfrak q$ is complete with respect to $I(A/\mathfrak q)$ (for example by Algebra, Lemma \ref{algebra-lemma-completion-tensor}) we see that all closed points of $\Spec(A/\mathfrak q)$ are contained in $V(IA/\mathfrak q)$ by Algebra, Lemma \ref{algebra-lemma-radical-completion}. Since $\dim(A/\mathfrak q) = 1$ and since $I \not \subset \mathfrak q$ we conclude two things: (a) $V(IA/\mathfrak q)$ must contain all points distinct from the generic point of $\Spec(A/\mathfrak q)$, and (b) $V(IA/\mathfrak q)$ must be a (finite) discrete set. From (a) we see that $\mathfrak q$ is a closed point of $\Spec(A) \setminus V(I)$ and we conclude that (2) holds. \medskip\noindent Continuing to assume (5) we see that the finite discrete space $V(IA/\mathfrak q)$ must be a singleton by More on Algebra, Lemma \ref{more-algebra-lemma-irreducible-henselian-pair-connected} for example (and the fact that complete pairs are henselian pairs, see More on Algebra, Lemma \ref{more-algebra-lemma-complete-henselian}). Hence we see that (8) is true. Conversely, it is clear that (8) implies (5). \medskip\noindent At this point we know that (1) -- (6) and (8) are equivalent. We omit the verification that these are also equivalent to (7).
Assume (1). Recall that $B/\mathfrak q$ is a Noetherian local ring of dimension $1$ whose topology is the adic topology coming from the maximal ideal. Since $\varphi$ is adic, we see that $A \to B/\mathfrak q$ is adic. Hence $\varphi(\mathfrak a)$ is a nonzero ideal in $B/\mathfrak q$. Hence $B/\mathfrak q + \varphi(\mathfrak a)$ has finite length. Hence $B/\mathfrak q + \varphi(\mathfrak a)$ is finite as an $A/\mathfrak a$-module by our assumption. Thus $B/\mathfrak q$ is finite over $A$ by Algebra, Lemma \ref{algebra-lemma-finite-over-complete-ring}. Thus (3) holds. \medskip\noindent Assume (3). Then $\Spec(B/\mathfrak q) \to \Spec(A/\mathfrak p)$ is surjective by Algebra, Lemma \ref{algebra-lemma-integral-overring-surjective}. This implies (2). \medskip\noindent Assume (2). Denote $\kappa'$ the residue field of $A/\mathfrak p$. By Lemma \ref{lemma-rig-closed-point-relative-residue-field} (and Lemma \ref{lemma-finite-type-finite-type-red}) the extension $\kappa/\kappa'$ is finitely generated as an algebra. By the Hilbert Nullstellensatz (Algebra, Lemma \ref{algebra-lemma-field-finite-type-over-domain}) we see that $\kappa/\kappa'$ is a finite extension. Hence we see that (1) holds.
By Lemma \ref{lemma-rig-closed-point-relative-residue-field} (combined with Lemma \ref{lemma-finite-type-finite-type-red}) the residue field $\kappa$ of $B/\mathfrak q$ is of finite type over $A/\mathfrak a$. Since $A/\mathfrak a$ is Jacobson, we see that $\kappa$ is finite over $A/\mathfrak a$ by Algebra, Lemma \ref{algebra-lemma-silly-jacobson}. We conclude by Lemma \ref{lemma-rig-closed-point-relative}.
The implication (2) $\Rightarrow$ (1) is immediate. Assume (1). To prove the existence of $\mathfrak q$ we may replace $A$ by $A/\mathfrak p$ and $B$ by $B/\mathfrak p B$ (some details omitted). Thus we may assume $(A, \mathfrak m, \kappa)$ is a local complete $1$-dimensional Noetherian ring, $\mathfrak m = \mathfrak a$, and $\mathfrak p = (0)$. Condition (1) just says that $B_0 = B \otimes_A \kappa = B/\mathfrak m B = B/\mathfrak a B$ is nonzero. Note that $B_0$ is of finite type over $\kappa$. Hence we can use induction on $\dim(B_0)$. If $\dim(B_0) = 0$, then any minimal prime $\mathfrak q \subset B$ will do (flatness of $A \to B$ insures that $\mathfrak q$ will lie over $\mathfrak p = (0)$). If $\dim(B_0) > 0$ then we can find an element $b \in B$ which maps to an element $b_0 \in B_0$ which is a nonzerodivisor and a nonunit, see Algebra, Lemma \ref{algebra-lemma-dim-not-zero-exists-nonzerodivisor-nonunit}. By Algebra, Lemma \ref{algebra-lemma-grothendieck} the ring $B' = B/bB$ is flat over $A$. Since $B'_0 = B' \otimes_A \kappa = B_0/(b_0)$ is not zero, we may apply the induction hypothesis to $B'$ and conclude. The final statement of the lemma is clear from Lemma \ref{lemma-rig-closed-point-relative}.
Please see Lemma \ref{lemma-rig-closed-point-in-localization} for the description of $\mathfrak q'$. The only assertion the lemma makes is that for a suitable choice of $f$ the prime ideal $\mathfrak p'$ has the property $\dim((A_f)^\wedge/\mathfrak p') = 1$. By Lemma \ref{lemma-rig-closed-point-relative} this in turn just means that the residue field $\kappa$ of $B/\mathfrak q = (B_f)^\wedge/\mathfrak q'$ is finite over $(A_f)^\wedge/\mathfrak a' = (A/\mathfrak a)_f$. By Lemma \ref{lemma-rig-closed-point-relative-residue-field} we know that $A/\mathfrak a \to \kappa$ is a finite type algebra homomorphism. By the Hilbert Nullstellensatz in the form of Algebra, Lemma \ref{algebra-lemma-field-finite-type-over-domain} we can find an $f \in A$ which maps to a unit in $\kappa$ such that $\kappa$ is finite over $A_f$. This finishes the proof.
By Lemma \ref{lemma-rig-closed-point-relative} the ring map $A/\mathfrak p \to A\{x_1, \ldots, x_n\}/\mathfrak q$ is finite. For every $m \geq 1$ the module $\mathfrak q^m/\mathfrak q^{m + 1}$ is finite over $A$ as it is a finite $A\{x_1, \ldots, x_n\}/\mathfrak q$-module. Hence $A\{x_1, \ldots x_n\}/\mathfrak q^m$ is a finite $A$-module. Hence $A[x_1, \ldots, x_n] \to A\{x_1, \ldots, x_n\}/\mathfrak q^m$ is surjective (as the image is dense and an $A$-submodule). It follows in a straightforward manner that $A[x_1, \ldots, x_n]/(\mathfrak q')^m \to A\{x_1, \ldots, x_n\}/\mathfrak q^m$ is an isomorphism for all $m$. From this the lemma easily follows. Hint: Pick a topologically nilpotent $g \in A$ which is not contained in $\mathfrak p$. Then the map of completions is the map $$ \lim_m \left(A[x_1, \ldots, x_n]/(\mathfrak q')^m\right)_g \longrightarrow \left(A\{x_1, \ldots, x_n\}/\mathfrak q^m\right)_g $$ Some details omitted.
Let $\kappa$ be the residue field of the $1$-dimensional complete local ring $A/\mathfrak p$. Since $A/I \to B/IB$ is \'etale, we see that $B \otimes_A \kappa$ is a finite product of finite separable extensions of $\kappa$, see Algebra, Lemma \ref{algebra-lemma-etale-over-field}. One of these is the residue field of $B/\mathfrak q$. By Algebra, Lemma \ref{algebra-lemma-finite-over-complete-ring} we see that $B/\mathfrak p B$ is a finite $A/\mathfrak p$-algebra. It is also flat. Combining the above we see that $A/\mathfrak p  \to B /\mathfrak p B$ is finite \'etale, see Algebra, Lemma \ref{algebra-lemma-characterize-etale}. Hence $B/\mathfrak p B$ is reduced, which implies the statement of the lemma (details omitted).
We will use some fact on regular ring maps the reader can find in More on Algebra, Section \ref{more-algebra-section-regular}. Since $A/\mathfrak p$ is a complete local Noetherian ring it is excellent (More on Algebra, Proposition \ref{more-algebra-proposition-ubiquity-excellent}). Hence $A/\mathfrak p[x_1, \ldots, x_n]$ is excellent (by the same reference). Hence $A/\mathfrak p[x_1, \ldots, x_n] \to A/\mathfrak p\{x_1, \ldots, x_n\}$ is a regular ring homomorphism by More on Algebra, Lemma \ref{more-algebra-lemma-map-G-ring-to-completion-regular}. Of course $A/\mathfrak p \to A/\mathfrak p[x_1, \ldots, x_n]$ is smooth and hence regular. Since the composition of regular ring maps is regular the proof is complete.