The lemma is proven by the following steps in the given order. We will justify each of these steps below. \begin{enumerate} \item \label{item-power} Pick an integer $N > 0$ such that $\mathfrak q^N\Lambda_\mathfrak q \subset H_{A/k}\Lambda_\mathfrak q$. \item \label{item-generators} Pick generators $a_1, \ldots, a_t \in A$ of the ideal $H_{A/R}$. \item \label{item-dimension} Set $d = \dim(\Lambda_\mathfrak q)$. \item \label{item-standardizer} Set $B = A[x_1, \ldots, x_d, z_{ij}]/(x_i^{2N} - \sum z_{ij}a_j)$. \item \label{item-elkik} Consider $B$ as a $k[x_1, \ldots, x_d]$-algebra and let $B \to C$ be as in Lemma \ref{lemma-improve-presentation}. We also obtain a section $C \to B$. \item \label{item-strictly-standard} Choose $c > 0$ such that each $x_i^c$ is strictly standard in $C$ over $k[x_1, \ldots, x_d]$. \item \label{item-set-n} Set $n = N + dc$ and $e = 8c$. \item \label{item-set-E} Let $E \subset \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$ be the images of generators of $A$ as a $k$-algebra. \item \label{item-NP} Choose an integer $m$ and a $k$-algebra map $\varphi : k[y_1, \ldots, y_m] \to \Lambda$ and a factorization by local Artinian rings $$ k[y_1, \ldots, y_m]_\mathfrak p/\mathfrak p^n k[y_1, \ldots, y_m]_\mathfrak p \to D \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q $$ such that the first arrow is essentially smooth, the second is flat, $E$ is contained in $D$, with $\mathfrak p = \varphi^{-1}(\mathfrak q)$ the map $k[y_1, \ldots, y_m]_\mathfrak p \to \Lambda_\mathfrak q$ is flat, and $\mathfrak p \Lambda_\mathfrak q = \mathfrak q \Lambda_\mathfrak q$. \item \label{item-choose-pii} Choose $\pi_1, \ldots, \pi_d \in \mathfrak p$ which map to a regular system of parameters of $k[y_1, \ldots, y_m]_\mathfrak p$. \item \label{item-set-R} Let $R = k[y_1, \ldots, y_m, t_1, \ldots, t_m]$ and $\gamma_i = \pi_i t_i$. \item \label{item-modify-pii} If necessary modify the choice of $\pi_i$ such that for $i = 1, \ldots, d$ we have  $$ \text{Ann}_{R/(\gamma_1^e, \ldots, \gamma_{i - 1}^e)R}(\gamma_i) = \text{Ann}_{R/(\gamma_1^e, \ldots, \gamma_{i - 1}^e)R}(\gamma_i^2) $$ \item \label{item-choose-deltai} There exist $\delta_1, \ldots, \delta_d \in \Lambda$, $\delta_i \not \in \mathfrak q$ and a factorization $D \to D' \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$ with $D'$ local Artinian, $D \to D'$ essentially smooth, the map $D' \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$ flat such that, with $\pi_i' = \delta_i \pi_i$, we have for $i = 1, \ldots, d$ \begin{enumerate} \item $(\pi_i')^{2N} = \sum a_j\lambda_{ij}$ in $\Lambda$ where $\lambda_{ij} \bmod \mathfrak q^n\Lambda_\mathfrak q$ is an element of $D'$, \item $\text{Ann}_{\Lambda/({\pi'}_1^e, \ldots, {\pi'}_{i - 1}^e)}({\pi'}_i) = \text{Ann}_{\Lambda/({\pi'}_1^e, \ldots, {\pi'}_{i - 1}^e)}({\pi'}_i^2)$, \item $\delta_i \bmod \mathfrak q^n\Lambda_\mathfrak q$ is an element of $D'$. \end{enumerate} \item \label{item-map-B-Lambda} Define $B \to \Lambda$ by sending $x_i$ to $\pi'_i$ and $z_{ij}$ to $\lambda_{ij}$ found above. Define $C \to \Lambda$ by composing the map $B \to \Lambda$ with the retraction $C \to B$. \item \label{item-set-map-R} Map $R \to \Lambda$ by $\varphi$ on $k[y_1, \ldots, y_m]$ and by sending $t_i$ to $\delta_i$. Further introduce a map $$ k[x_1, \ldots, x_d] \longrightarrow R = k[y_1, \ldots, y_m, t_1, \ldots, t_d] $$ by sending $x_i$ to $\gamma_i = \pi_i t_i$. \item \label{item-first-solve} It suffices to resolve $$ R \to C \otimes_{k[x_1, \ldots, x_d]} R \to \Lambda \supset \mathfrak q $$ \item \label{item-set-I} Set $I = (\gamma_1^e, \ldots, \gamma_d^e) \subset R$. \item \label{item-second-resolve} It suffices to resolve $$ R/I \to C \otimes_{k[x_1, \ldots, x_d]} R/I \to \Lambda/I\Lambda \supset \mathfrak q/I\Lambda $$ \item \label{item-set-r} We denote $\mathfrak r \subset R = k[y_1, \ldots, y_m, t_1, \ldots, t_d]$ the inverse image of $\mathfrak q$. \item \label{item-third-resolve} It suffices to resolve $$ (R/I)_\mathfrak r \to C \otimes_{k[x_1, \ldots, x_d]} (R/I)_\mathfrak r \to \Lambda_\mathfrak q/I\Lambda_\mathfrak q \supset \mathfrak q\Lambda_\mathfrak q/I\Lambda_\mathfrak q $$ \item \label{item-fourth-resolve} Set $J = (\pi_1^e, \ldots, \pi_d^e)$ in $k[y_1, \ldots, y_m]$. \item \label{item-fifth-resolve} It suffices to resolve $$ (R/JR)_\mathfrak p \to C \otimes_{k[x_1, \ldots, x_d]} (R/JR)_\mathfrak p \to \Lambda_\mathfrak q/J\Lambda_\mathfrak q \supset \mathfrak q\Lambda_\mathfrak q/J\Lambda_\mathfrak q $$ \item \label{item-sixth-resolve} It suffices to resolve $$ (R/\mathfrak p^nR)_\mathfrak p \to C \otimes_{k[x_1, \ldots, x_d]} (R/\mathfrak p^nR)_\mathfrak p \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q \supset \mathfrak q\Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q $$ \item \label{item-seventh-resolve} It suffices to resolve $$ (R/\mathfrak p^nR)_\mathfrak p \to B \otimes_{k[x_1, \ldots, x_d]} (R/\mathfrak p^nR)_\mathfrak p \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q \supset \mathfrak q\Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q $$ \item \label{item-eighth-resolve} The ring $D'[t_1, \ldots, t_d]$ is given the structure of an $R_\mathfrak p/\mathfrak p^nR_\mathfrak p$-algebra by the given map $k[y_1, \ldots, y_m]_\mathfrak p/\mathfrak p^n k[y_1, \ldots, y_m]_\mathfrak p \to D'$ and by sending $t_i$ to $t_i$. It suffices to find a factorization $$ B \otimes_{k[x_1, \ldots, x_d]} (R/\mathfrak p^nR)_\mathfrak p \to D'[t_1, \ldots, t_d] \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q $$ where the second arrow sends $t_i$ to $\delta_i$ and induces the given homomorphism $D' \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$. \item \label{item-done} Such a factorization exists by our choice of $D'$ above. \end{enumerate} We now give the justification for each of the steps, except that we skip justifying the steps which just introduce notation. \medskip\noindent Ad (\ref{item-power}). This is possible as $\mathfrak q$ is minimal over $\mathfrak h_A = \sqrt{H_{A/k}\Lambda}$. \medskip\noindent Ad (\ref{item-strictly-standard}). Note that $A_{a_i}$ is smooth over $k$. Hence $B_{a_j}$, which is isomorphic to a polynomial algebra over $A_{a_j}[x_1, \ldots, x_d]$, is smooth over $k[x_1, \ldots, x_d]$. Thus $B_{x_i}$ is smooth over $k[x_1, \ldots, x_d]$. By Lemma \ref{lemma-improve-presentation} we see that $C_{x_i}$ is smooth over $k[x_1, \ldots, x_d]$ with finite free module of differentials. Hence some power of $x_i$ is strictly standard in $C$ over $k[x_1, \ldots, x_n]$ by Lemma \ref{lemma-compare-standard}. \medskip\noindent Ad (\ref{item-NP}). This follows by applying Lemma \ref{lemma-solution-modulo}. \medskip\noindent Ad (\ref{item-choose-pii}). Since $k[y_1, \ldots, y_m]_\mathfrak p \to \Lambda_\mathfrak q$ is flat and $\mathfrak p \Lambda_\mathfrak q = \mathfrak q \Lambda_\mathfrak q$ by construction we see that $\dim(k[y_1, \ldots, y_m]_\mathfrak p) = d$ by Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-equals-total}. Thus we can find $\pi_1, \ldots, \pi_d \in \Lambda$ which map to a regular system of parameters in $\Lambda_\mathfrak q$. \medskip\noindent Ad (\ref{item-modify-pii}). By Algebra, Lemma \ref{algebra-lemma-regular-ring-CM} any permutation of the sequence $\pi_1, \ldots, \pi_d$ is a regular sequence in $k[y_1, \ldots, y_m]_\mathfrak p$. Hence $\gamma_1 = \pi_1 t_1, \ldots, \gamma_d = \pi_d t_d$ is a regular sequence in $R_\mathfrak p = k[y_1, \ldots, y_m]_\mathfrak p[t_1, \ldots, t_d]$, see Algebra, Lemma \ref{algebra-lemma-regular-sequence-in-polynomial-ring}. Let $S = k[y_1, \ldots, y_m] \setminus \mathfrak p$ so that $R_\mathfrak p = S^{-1}R$. Note that $\pi_1, \ldots, \pi_d$ and $\gamma_1, \ldots, \gamma_d$ remain regular sequences if we multiply our $\pi_i$ by elements of $S$. Suppose that $$ \text{Ann}_{R/(\gamma_1^e, \ldots, \gamma_{i - 1}^e)R}(\gamma_i) = \text{Ann}_{R/(\gamma_1^e, \ldots, \gamma_{i - 1}^e)R}(\gamma_i^2) $$ holds for $i = 1, \ldots, t$ for some $t \in \{0, \ldots, d\}$. Note that $\gamma_1^e, \ldots, \gamma_t^e, \gamma_{t + 1}$ is a regular sequence in $S^{-1}R$ by Algebra, Lemma \ref{algebra-lemma-regular-sequence-powers}. Hence we see that $$ \text{Ann}_{S^{-1}R/(\gamma_1^e, \ldots, \gamma_{i - 1}^e)}(\gamma_i) = \text{Ann}_{S^{-1}R/(\gamma_1^e, \ldots, \gamma_{i - 1}^e)}(\gamma_i^2). $$ Thus we get $$ \text{Ann}_{R/(\gamma_1^e, \ldots, \gamma_t^e)R}(\gamma_{t + 1}) = \text{Ann}_{R/(\gamma_1^e, \ldots, \gamma_t^e)R}(\gamma_{t + 1}^2) $$ after replacing $\pi_{t + 1}$ by $s\pi_{t + 1}$ for some $s \in S$ by Lemma \ref{lemma-ogoma}. By induction on $t$ this produces the desired sequence. \medskip\noindent Ad (\ref{item-choose-deltai}). Let $S = \Lambda \setminus \mathfrak q$ so that $\Lambda_\mathfrak q = S^{-1}\Lambda$. Set $\bar \Lambda = \Lambda_\mathfrak q/\mathfrak q^n \Lambda_\mathfrak q$. Suppose that we have a $t \in \{0, \ldots, d\}$ and $\delta_1, \ldots, \delta_t \in S$ and a factorization $D \to D' \to \bar \Lambda$ as in (\ref{item-choose-deltai}) such that (a), (b), (c) hold for $i = 1, \ldots, t$. We have $\pi_{t + 1}^N \in H_{A/k}\Lambda_\mathfrak q$ as $\mathfrak q^N \Lambda_\mathfrak q \subset H_{A/k}\Lambda_\mathfrak q$ by (\ref{item-power}). Hence $\pi_{t + 1}^N \in H_{A/k} \bar\Lambda$. Hence $\pi_{t + 1}^N \in H_{A/k}D'$ as $D' \to \bar \Lambda$ is faithfully flat, see Algebra, Lemma \ref{algebra-lemma-faithfully-flat-universally-injective}. Recall that $H_{A/k} = (a_1, \ldots, a_t)$. Say $\pi_{t + 1}^N = \sum a_j d_j$ in $D'$ and choose $c_j \in \Lambda_\mathfrak q$ lifting $d_j \in D'$. Then $\pi_{t + 1}^N = \sum c_j a_j + \epsilon$ with $\epsilon \in \mathfrak q^n\Lambda_\mathfrak q \subset \mathfrak q^{n - N}H_{A/k}\Lambda_\mathfrak q$. Write $\epsilon = \sum a_j c'_j$ for some $c'_j \in \mathfrak q^{n - N}\Lambda_\mathfrak q$. Hence $\pi_{t + 1}^{2N} = \sum (\pi_{t + 1}^N c_j + \pi_{t + 1}^N c'_j) a_j$. Note that $\pi_{t + 1}^Nc'_j$ maps to zero in $\bar \Lambda$; this trivial but key observation will ensure later that (a) holds. Now we choose $s \in S$ such that there exist $\mu_{t + 1j} \in \Lambda$ such that on the one hand $\pi_{t + 1}^N c_j + \pi_{t + 1}^N c'_j = \mu_{t + 1j}/s^{2N}$ in $S^{-1}\Lambda$ and on the other $(s \pi_{t + 1})^{2N} = \sum \mu_{t + 1j}a_j$ in $\Lambda$ (minor detail omitted). We may further replace $s$ by a power and enlarge $D'$ such that $s$ maps to an element of $D'$. With these choices $\mu_{t + 1j}$ maps to $s^{2N}d_j$ which is an element of $D'$. Note that $\pi_1, \ldots, \pi_d$ are a regular sequence of parameters in $S^{-1}\Lambda$ by our choice of $\varphi$. Hence $\pi_1, \ldots, \pi_d$ forms a regular sequence in $\Lambda_\mathfrak q$ by Algebra, Lemma \ref{algebra-lemma-regular-ring-CM}. It follows that ${\pi'}_1^e, \ldots, {\pi'}_t^e, s\pi_{t + 1}$ is a regular sequence in $S^{-1}\Lambda$ by Algebra, Lemma \ref{algebra-lemma-regular-sequence-powers}. Thus we get $$ \text{Ann}_{S^{-1}\Lambda/({\pi'}_1^e, \ldots, {\pi'}_t^e)}(s\pi_{t + 1}) = \text{Ann}_{S^{-1}\Lambda/({\pi'}_1^e, \ldots, {\pi'}_t^e)}((s\pi_{t + 1})^2). $$ Hence we may apply Lemma \ref{lemma-ogoma} to find an $s' \in S$ such that $$ \text{Ann}_{\Lambda/({\pi'}_1^e, \ldots, {\pi'}_t^e)}((s')^qs\pi_{t + 1}) = \text{Ann}_{\Lambda/({\pi'}_1^e, \ldots, {\pi'}_t^e)}(((s')^qs\pi_{t + 1})^2). $$ for any $q > 0$. By Lemma \ref{lemma-enlarge-solution-modulo} we can choose $q$ and enlarge $D'$ such that $(s')^q$ maps to an element of $D'$. Setting $\delta_{t + 1} = (s')^qs$ and we conclude that (a), (b), (c) hold for $i = 1, \ldots, t + 1$. For (a) note that $\lambda_{t + 1j} = (s')^{2Nq}\mu_{t + 1j}$ works. By induction on $t$ we win. \medskip\noindent Ad (\ref{item-first-solve}). By construction the radical of $H_{(C \otimes_{k[x_1, \ldots, x_d]} R)/R} \Lambda$ contains $\mathfrak h_A$. Namely, the elements $a_j \in H_{A/k}$ map to elements of $H_{B/k[x_1, \ldots, x_n]}$, hence map to elements of $H_{C/k[x_1, \ldots, x_n]}$, hence $a_j \otimes 1$ map to elements of $H_{C \otimes_{k[x_1, \ldots, x_d]} R/R}$. Moreover, if we have a solution $C \otimes_{k[x_1, \ldots, x_n]} R \to T \to \Lambda$ of $$ R \to C \otimes_{k[x_1, \ldots, x_d]} R \to \Lambda \supset \mathfrak q $$ then $H_{T/R} \subset H_{T/k}$ as $R$ is smooth over $k$. Hence $T$ will also be a solution for the original situation $k \to A \to \Lambda \supset \mathfrak q$. \medskip\noindent Ad (\ref{item-second-resolve}). Follows on applying Lemma \ref{lemma-lift-solution} to $R \to C \otimes_{k[x_1, \ldots, x_d]} R \to \Lambda \supset \mathfrak q$ and the sequence of elements $\gamma_1^c, \ldots, \gamma_d^c$. We note that since $x_i^c$ are strictly standard in $C$ over $k[x_1, \ldots, x_d]$ the elements $\gamma_i^c$ are strictly standard in $C \otimes_{k[x_1, \ldots, x_d]} R$ over $R$ by Lemma \ref{lemma-strictly-standard-base-change}. The other assumption of Lemma \ref{lemma-lift-solution} holds by steps (\ref{item-modify-pii}) and (\ref{item-choose-deltai}). \medskip\noindent Ad (\ref{item-third-resolve}). Apply Lemma \ref{lemma-delocalize-height-zero} to the situation in (\ref{item-second-resolve}). In the rest of the arguments the target ring is local Artinian, hence we are looking for a factorization by a smooth algebra $T$ over the source ring. \medskip\noindent Ad (\ref{item-fifth-resolve}). Suppose that $C \otimes_{k[x_1, \ldots, x_d]} (R/JR)_\mathfrak p \to T \to \Lambda_\mathfrak q/J\Lambda_\mathfrak q$ is a solution to $$ (R/JR)_\mathfrak p \to C \otimes_{k[x_1, \ldots, x_d]} (R/JR)_\mathfrak p \to \Lambda_\mathfrak q/J\Lambda_\mathfrak q \supset \mathfrak q\Lambda_\mathfrak q/J\Lambda_\mathfrak q $$ Then $C \otimes_{k[x_1, \ldots, x_d]} (R/I)_\mathfrak r \to T_\mathfrak r \to \Lambda_\mathfrak q/I\Lambda_\mathfrak q$ is a solution to the situation in (\ref{item-third-resolve}). \medskip\noindent Ad (\ref{item-sixth-resolve}). Our $n = N + dc$ is large enough so that $\mathfrak p^nk[y_1, \ldots, y_m]_\mathfrak p \subset J_\mathfrak p$ and $\mathfrak q^n \Lambda_\mathfrak q \subset J\Lambda_\mathfrak q$. Hence if we have a solution $C \otimes_{k[x_1, \ldots, x_d]} (R/\mathfrak p^nR)_\mathfrak p \to T \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$ of (\ref{item-fifth-resolve} then we can take $T/JT$ as the solution for (\ref{item-sixth-resolve}). \medskip\noindent Ad (\ref{item-seventh-resolve}). This is true because we have a section $C \to B$ in the category of $R$-algebras. \medskip\noindent Ad (\ref{item-eighth-resolve}). This is true because $D'$ is essentially smooth over the local Artinian ring $k[y_1, \ldots, y_m]_\mathfrak p/\mathfrak p^n k[y_1, \ldots, y_m]_\mathfrak p$ and $$ R_\mathfrak p/\mathfrak p^nR_\mathfrak p = k[y_1, \ldots, y_m]_\mathfrak p/ \mathfrak p^n k[y_1, \ldots, y_m]_\mathfrak p[t_1, \ldots, t_d]. $$ Hence $D'[t_1, \ldots, t_d]$ is a filtered colimit of smooth $R_\mathfrak p/\mathfrak p^nR_\mathfrak p$-algebras and $B \otimes_{k[x_1, \ldots, x_d]} (R_\mathfrak p/\mathfrak p^nR_\mathfrak p)$ factors through one of these. \medskip\noindent Ad (\ref{item-done}). The final twist of the proof is that we cannot just use the map $B \to D'$ which maps $x_i$ to the image of $\pi_i'$ in $D'$ and $z_{ij}$ to the image of $\lambda_{ij}$ in $D'$ because we need the diagram $$ \xymatrix{ B \ar[r] & D'[t_1, \ldots, t_d] \\ k[x_1, \ldots, x_d] \ar[r] \ar[u] & R_\mathfrak p/\mathfrak p^nR_\mathfrak p \ar[u] } $$ to commute and we need the composition $B \to D'[t_1, \ldots, t_d] \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$ to be the map of (\ref{item-map-B-Lambda}). This requires us to map $x_i$ to the image of $\pi_i t_i$ in $D'[t_1, \ldots, t_d]$. Hence we map $z_{ij}$ to the image of $\lambda_{ij} t_i^{2N} / \delta_i^{2N}$ in $D'[t_1, \ldots, t_d]$ and everything is clear.