Here is another fiendishly clever lemma.

**Proof.**
Choose a presentation

\[ A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m) \]

and $0 \leq c \leq \min (n, m)$ such that (16.2.3.3) holds for $\pi $ and such that

16.7.1.1
\begin{equation} \label{smoothing-equation-star} \pi f_{c + j} \in (f_1, \ldots , f_ c) + (f_1, \ldots , f_ m)^2 \end{equation}

for $j = 1, \ldots , m - c$. Say $\rho $ maps $x_ i$ to the class of $r_ i \in R$. Then we can replace $x_ i$ by $x_ i - r_ i$. Hence we may assume $\rho (x_ i) = 0$ in $R/\pi ^4 R$. This implies that $f_ j(0) \in \pi ^4R$ and that $A \to \Lambda $ maps $x_ i$ to $\pi ^4\lambda _ i$ for some $\lambda _ i \in \Lambda $. Write

\[ f_ j = f_ j(0) + \sum \nolimits _{i = 1, \ldots , n} r_{ji} x_ i + \text{h.o.t.} \]

This implies that the constant term of $\partial f_ j/\partial x_ i$ is $r_{ji}$. Apply $\rho $ to (16.2.3.3) for $\pi $ and we see that

\[ \pi = \sum \nolimits _{I \subset \{ 1, \ldots , n\} ,\ |I| = c} r_ I \det (r_{ji})_{j = 1, \ldots , c,\ i \in I} \bmod \pi ^4R \]

for some $r_ I \in R$. Thus we have

\[ u\pi = \sum \nolimits _{I \subset \{ 1, \ldots , n\} ,\ |I| = c} r_ I \det (r_{ji})_{j = 1, \ldots , c,\ i \in I} \]

for some $u \in 1 + \pi ^3R$. By Algebra, Lemma 10.14.5 this implies there exists a $n \times c$ matrix $(s_{ik})$ such that

\[ u\pi \delta _{jk} = \sum \nolimits _{i = 1, \ldots , n} r_{ji}c_{ik}\quad \text{for all } j, k = 1, \ldots , c \]

(Kronecker delta). We introduce auxiliary variables $v_1, \ldots , v_ c, w_1, \ldots , w_ n$ and we set

\[ h_ i = x_ i - \pi ^2 \sum \nolimits _{j = 1, \ldots c} s_{ij} v_ j - \pi ^3 w_ i \]

In the following we will use that

\[ R[x_1, \ldots , x_ n, v_1, \ldots , v_ c, w_1, \ldots , w_ n]/ (h_1, \ldots , h_ n) = R[v_1, \ldots , v_ c, w_1, \ldots , w_ n] \]

without further mention. In $R[x_1, \ldots , x_ n, v_1, \ldots , v_ c, w_1, \ldots , w_ n]/ (h_1, \ldots , h_ n)$ we have

\begin{align*} f_ j & = f_ j(x_1 - h_1, \ldots , x_ n - h_ n) \\ & = \pi ^2 \sum \nolimits _{k = 1}^ c \left(\sum \nolimits _{i = 1}^ n r_{ji} s_{ik}\right) v_ k + \pi ^3 \sum \nolimits _{i = 1}^ n r_{ji}w_ i \bmod \pi ^4 \\ & = \pi ^3 v_ j + \pi ^3 \sum \nolimits _{i = 1}^ n r_{ji}w_ i \bmod \pi ^4 \end{align*}

for $1 \leq j \leq c$. Hence we can choose elements $g_ j \in R[v_1, \ldots , v_ c, w_1, \ldots , w_ n]$ such that $g_ j = v_ j + \sum r_{ji}w_ i \bmod \pi $ and such that $f_ j = \pi ^3 g_ j$ in the $R$-algebra $R[x_1, \ldots , x_ n, v_1, \ldots , v_ c, w_1, \ldots , w_ n]/ (h_1, \ldots , h_ n)$. We set

\[ B = R[x_1, \ldots , x_ n, v_1, \ldots , v_ c, w_1, \ldots , w_ n]/ (f_1, \ldots , f_ m, h_1, \ldots , h_ n, g_1, \ldots , g_ c). \]

The map $A \to B$ is clear. We define $B \to \Lambda $ by mapping $x_ i \to \pi ^4\lambda _ i$, $v_ i \mapsto 0$, and $w_ i \mapsto \pi \lambda _ i$. Then it is clear that the elements $f_ j$ and $h_ i$ are mapped to zero in $\Lambda $. Moreover, it is clear that $g_ i$ is mapped to an element $t$ of $\pi \Lambda $ such that $\pi ^3t = 0$ (as $f_ i = \pi ^3 g_ i$ modulo the ideal generated by the $h$'s). Hence our assumption that $\text{Ann}_\Lambda (\pi ) = \text{Ann}_\Lambda (\pi ^2)$ implies that $t = 0$. Thus we are done if we can prove the statement about smoothness.

Note that $B_\pi \cong A_\pi [v_1, \ldots , v_ c]$ because the equations $g_ i = 0$ are implied by $f_ i = 0$. Hence $B_\pi $ is smooth over $R$ as $A_\pi $ is smooth over $R$ by the assumption that $\pi $ is strictly standard in $A$ over $R$, see Lemma 16.2.5.

Set $B' = R[v_1, \ldots , v_ c, w_1, \ldots , w_ n]/(g_1, \ldots , g_ c)$. As $g_ i = v_ i + \sum r_{ji}w_ i \bmod \pi $ we see that $B'/\pi B' = R/\pi R[w_1, \ldots , w_ n]$. Hence $R \to B'$ is smooth of relative dimension $n$ at every point of $V(\pi )$ by Algebra, Lemmas 10.134.11 and 10.135.16 (the first lemma shows it is syntomic at those primes, in particular flat, whereupon the second lemma shows it is smooth).

Let $\mathfrak q \subset B$ be a prime with $\pi \in \mathfrak q$ and for some $r \in \mathfrak a$, $r \not\in \mathfrak q$. Denote $\mathfrak q' = B' \cap \mathfrak q$. We claim the surjection $B' \to B$ induces an isomorphism of local rings $(B')_{\mathfrak q'} \to B_\mathfrak q$. This will conclude the proof of the lemma. Note that $B_\mathfrak q$ is the quotient of $(B')_{\mathfrak q'}$ by the ideal generated by $f_{c + j}$, $j = 1, \ldots , m - c$. We observe two things: first the image of $f_{c + j}$ in $(B')_{\mathfrak q'}$ is divisible by $\pi ^2$ and second the image of $\pi f_{c + j}$ in $(B')_{\mathfrak q'}$ can be written as $\sum b_{j_1 j_2} f_{c + j_1}f_{c + j_2}$ by (16.7.1.1). Thus we see that the image of each $\pi f_{c + j}$ is contained in the ideal generated by the elements $\pi ^2 f_{c + j'}$. Hence $\pi f_{c + j} = 0$ in $(B')_{\mathfrak q'}$ as this is a Noetherian local ring, see Algebra, Lemma 10.50.4. As $R \to (B')_{\mathfrak q'}$ is flat we see that

\[ \left(\text{Ann}_ R(\pi ^2)/\text{Ann}_ R(\pi )\right) \otimes _ R (B')_{\mathfrak q'} = \text{Ann}_{(B')_{\mathfrak q'}}(\pi ^2)/\text{Ann}_{(B')_{\mathfrak q'}}(\pi ) \]

Because $r \in \mathfrak a$ is invertible in $(B')_{\mathfrak q'}$ we see that this module is zero. Hence we see that the image of $f_{c + j}$ is zero in $(B')_{\mathfrak q'}$ as desired.
$\square$

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