We choose a presentation $$ \bar C = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m) $$ We also denote $I = (f_1, \ldots, f_m)$ and $\bar I$ the image of $I$ in $R/\pi^2R[x_1, \ldots, x_n]$. Since $R$ is Noetherian, so is $\bar C$. Hence the smooth locus of $R/\pi^2 R \to \bar C$ is quasi-compact, see Topology, Lemma \ref{topology-lemma-Noetherian}. Applying Lemma \ref{lemma-find-strictly-standard} we may choose a finite list of elements $a_1, \ldots, a_r \in R[x_1, \ldots, x_n]$ such that \begin{enumerate} \item the union of the open subspaces $\Spec(\bar C_{a_k}) \subset \Spec(\bar C)$ cover the smooth locus of $R/\pi^2 R \to \bar C$, and \item for each $k = 1, \ldots, r$ there exists a finite subset $E_k \subset \{1, \ldots, m\}$ such that $(\bar I/\bar I^2)_{a_k}$ is freely generated by the classes of $f_j$, $j \in E_k$. \end{enumerate} Set $I_k = (f_j, j \in E_k) \subset I$ and denote $\bar I_k$ the image of $I_k$ in $R/\pi^2R[x_1, \ldots, x_n]$. By (2) and Nakayama's lemma we see that $(\bar I/\bar I_k)_{a_k}$ is annihilated by $1 + b'_k$ for some $b'_k \in \bar I_{a_k}$. Suppose $b'_k$ is the image of $b_k/(a_k)^N$ for some $b_k \in I$ and some integer $N$. After replacing $a_k$ by $a_kb_k$ we get \begin{enumerate} \item[(3)] $(\bar I_k)_{a_k} = (\bar I)_{a_k}$. \end{enumerate} Thus, after possibly replacing $a_k$ by a high power, we may write \begin{enumerate} \item[(4)] $a_k f_\ell = \sum\nolimits_{j \in E_k} h_{k, \ell}^jf_j + \pi^2 g_{k, \ell}$ \end{enumerate} for any $\ell \in \{1, \ldots, m\}$ and some $h_{i, \ell}^j, g_{i, \ell} \in R[x_1, \ldots, x_n]$. If $\ell \in E_k$ we choose $h_{k, \ell}^j = a_k\delta_{\ell, j}$ (Kronecker delta) and $g_{k, \ell} = 0$. Set $$ D = R[x_1, \ldots, x_n, z_1, \ldots, z_m]/ (f_j - \pi z_j, p_{k, \ell}). $$ Here $j \in \{1, \ldots, m\}$, $k \in \{1, \ldots, r\}$, $\ell \in \{1, \ldots, m\}$, and $$ p_{k, \ell} = a_k z_\ell - \sum\nolimits_{j \in E_k} h_{k, \ell}^j z_j - \pi g_{k, \ell}. $$ Note that for $\ell \in E_k$ we have $p_{k, \ell} = 0$ by our choices above. \medskip\noindent The map $R \to D$ is the given one. Say $\bar C \to \Lambda/\pi^2\Lambda$ maps $x_i$ to the class of $\lambda_i$ modulo $\pi^2$. For an element $f \in R[x_1, \ldots, x_n]$ we denote $f(\lambda) \in \Lambda$ the result of substituting $\lambda_i$ for $x_i$. Then we know that $f_j(\lambda) = \pi^2 \mu_j$ for some $\mu_j \in \Lambda$. Define $D \to \Lambda$ by the rules $x_i \mapsto \lambda_i$ and $z_j \mapsto \pi\mu_j$. This is well defined because \begin{align*} p_{k, \ell} & \mapsto a_k(\lambda) \pi \mu_\ell - \sum\nolimits_{j \in E_k} h_{k, \ell}^j(\lambda) \pi \mu_j - \pi g_{k, \ell}(\lambda) \\ & = \pi\left(a_k(\lambda) \mu_\ell - \sum\nolimits_{j \in E_k} h_{k, \ell}^j(\lambda) \mu_j - g_{k, \ell}(\lambda)\right) \end{align*} Substituting $x_i = \lambda_i$ in (4) above we see that the expression inside the brackets is annihilated by $\pi^2$, hence it is annihilated by $\pi$ as we have assumed $\text{Ann}_\Lambda(\pi) = \text{Ann}_\Lambda(\pi^2)$. The map $\bar C \to D/\pi D$ is determined by $x_i \mapsto x_i$ (clearly well defined). Thus we are done if we can prove (b), (c), and (d). \medskip\noindent Using (4) we obtain the following key equality \begin{align*} \pi p_{k, \ell} & = \pi a_k z_\ell - \sum\nolimits_{j \in E_k} \pi h_{k, \ell}^jz_j - \pi^2 g_{k, \ell} \\ & = - a_k (f_\ell - \pi z_\ell) + a_k f_\ell + \sum\nolimits_{j \in E_k} h_{k, \ell}^j (f_j - \pi z_j) - \sum\nolimits_{j \in E_k} h_{k, \ell}^j f_j - \pi^2 g_{k, \ell} \\ & = -a_k(f_\ell - \pi z_\ell) + \sum\nolimits_{j \in E_k} h_{k, \ell}^j(f_j - \pi z_j) \end{align*} The end result is an element of the ideal generated by $f_j - \pi z_j$. In particular, we see that $D[1/\pi]$ is isomorphic to $R[1/\pi][x_1, \ldots, x_n, z_1, \ldots, z_m]/(f_j - \pi z_j)$ which is isomorphic to $R[1/\pi][x_1, \ldots, x_n]$ hence smooth over $R$. This proves (b). \medskip\noindent For fixed $k \in \{1, \ldots, r\}$ consider the ring $$ D_k = R[x_1, \ldots, x_n, z_1, \ldots, z_m]/ (f_j - \pi z_j, j \in E_k, p_{k, \ell}) $$ The number of equations is $m = |E_k| + (m - |E_k|)$ as $p_{k, \ell}$ is zero if $\ell \in E_k$. Also, note that \begin{align*} (D_k/\pi D_k)_{a_k} & = R/\pi R[x_1, \ldots, x_n, 1/a_k, z_1, \ldots, z_m]/ (f_j, j \in E_k, p_{k, \ell}) \\ & = (\bar C/\pi \bar C)_{a_k}[z_1, \ldots, z_m]/ (a_kz_\ell - \sum\nolimits_{j \in E_k} h_{k, \ell}^j z_j) \\ & \cong (\bar C/\pi \bar C)_{a_k}[z_j, j \in E_k] \end{align*} In particular $(D_k/\pi D_k)_{a_k}$ is smooth over $(\bar C/\pi \bar C)_{a_k}$. By our choice of $a_k$ we have that $(\bar C/\pi \bar C)_{a_k}$ is smooth over $R/\pi R$ of relative dimension $n - |E_k|$, see (2). Hence for a prime $\mathfrak q_k \subset D_k$ containing $\pi$ and lying over $\Spec(\bar C_{a_k})$ the fibre ring of $R \to D_k$ is smooth at $\mathfrak q_k$ of dimension $n$. Thus $R \to D_k$ is syntomic at $\mathfrak q_k$ by our count of the number of equations above, see Algebra, Lemma \ref{algebra-lemma-localize-relative-complete-intersection}. Hence $R \to D_k$ is smooth at $\mathfrak q_k$, see Algebra, Lemma \ref{algebra-lemma-flat-fibre-smooth}. \medskip\noindent To finish the proof, let $\mathfrak q \subset D$ be a prime containing $\pi$ lying over a prime where $R/\pi^2 R \to \bar C$ is smooth. Then $a_k \not \in \mathfrak q$ for some $k$ by (1). We will show that the surjection $D_k \to D$ induces an isomorphism on local rings at $\mathfrak q$. Since we know that the ring maps $\bar C/\pi \bar C \to D_k/\pi D_k$ and $R \to D_k$ are smooth at the corresponding prime $\mathfrak q_k$ by the preceding paragraph this will prove (c) and (d) and thus finish the proof. \medskip\noindent First, note that for any $\ell$ the equation $\pi p_{k, \ell} = -a_k(f_\ell - \pi z_\ell) + \sum_{j \in E_k} h_{k, \ell}^j (f_j - \pi z_j)$ proved above shows that $f_\ell - \pi z_\ell$ maps to zero in $(D_k)_{a_k}$ and in particular in $(D_k)_{\mathfrak q_k}$. The relations (4) imply that $a_k f_\ell = \sum_{j \in E_k} h_{k, \ell}^j f_j$ in $I/I^2$. Since $(\bar I_k/\bar I_k^2)_{a_k}$ is free on $f_j$, $j \in E_k$ we see that $$ a_{k'} h_{k, \ell}^j - \sum\nolimits_{j' \in E_{k'}} h_{k', \ell}^{j'} h_{k, j'}^j $$ is zero in $\bar C_{a_k}$ for every $k, k', \ell$ and $j \in E_k$. Hence we can find a large integer $N$ such that $$ a_k^N\left( a_{k'} h_{k, \ell}^j - \sum\nolimits_{j' \in E_{k'}} h_{k', \ell}^{j'} h_{k, j'}^j \right) $$ is in $I_k + \pi^2R[x_1, \ldots, x_n]$. Computing modulo $\pi$ we have \begin{align*} & a_kp_{k', \ell} - a_{k'}p_{k, \ell} + \sum h_{k', \ell}^{j'} p_{k, j'} \\ & = - a_k \sum h_{k', \ell}^{j'} z_{j'} + a_{k'} \sum h_{k, \ell}^j z_j + \sum h_{k', \ell}^{j'} a_k z_{j'} - \sum \sum h_{k', \ell}^{j'} h_{k, j'}^j z_j \\ & = \sum \left( a_{k'} h_{k, \ell}^j - \sum h_{k', \ell}^{j'} h_{k, j'}^j \right) z_j \end{align*} with Einstein summation convention. Combining with the above we see $a_k^{N + 1} p_{k', \ell}$ is contained in the ideal generated by $I_k$ and $\pi$ in $R[x_1, \ldots, x_n, z_1, \ldots, z_m]$. Thus $p_{k', \ell}$ maps into $\pi (D_k)_{a_k}$. On the other hand, the equation $$ \pi p_{k', \ell} = -a_{k'} (f_\ell - \pi z_\ell) + \sum\nolimits_{j' \in E_{k'}} h_{k', \ell}^{j'}(f_{j'} - \pi z_{j'}) $$ shows that $\pi p_{k', \ell}$ is zero in $(D_k)_{a_k}$. Since we have assumed that $\text{Ann}_R(\pi) = \text{Ann}_R(\pi^2)$ and since $(D_k)_{\mathfrak q_k}$ is smooth hence flat over $R$ we see that $\text{Ann}_{(D_k)_{\mathfrak q_k}}(\pi) = \text{Ann}_{(D_k)_{\mathfrak q_k}}(\pi^2)$. We conclude that $p_{k', \ell}$ maps to zero as well, hence $D_{\mathfrak q} = (D_k)_{\mathfrak q_k}$ and we win.