Lemma 16.7.1. Let $R$ be a Noetherian ring. Let $\Lambda$ be an $R$-algebra. Let $\pi \in R$ and assume that $\text{Ann}_\Lambda (\pi ) = \text{Ann}_\Lambda (\pi ^2)$. Let $A \to \Lambda$ be an $R$-algebra map with $A$ of finite presentation. Assume

1. the image of $\pi$ is strictly standard in $A$ over $R$, and

2. there exists a section $\rho : A/\pi ^4 A \to R/\pi ^4 R$ which is compatible with the map to $\Lambda /\pi ^4 \Lambda$.

Then we can find $R$-algebra maps $A \to B \to \Lambda$ with $B$ of finite presentation such that $\mathfrak a B \subset H_{B/R}$ where $\mathfrak a = \text{Ann}_ R(\text{Ann}_ R(\pi ^2)/\text{Ann}_ R(\pi ))$.

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

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

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

(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. $\square$

Comment #2752 by Anonymous on

After introducing the $h_i$ it should read

$f_j = f_j(x_1-h_1,\dots,x_n-h_n) = \pi^2 \sum_{k=1}^c \left( \sum_{i=1}^n r_{ji} s_{ik} \right) v_k + \pi^3 \sum_{i=1}^n r_{ji} w_i \text{ mod } \pi^4 = \pi^3 v_j + \pi^3 \sum_{i=1}^n r_{ji} w_i \text{ mod } \pi^4$

Comment #2753 by Anonymous on

And right after that, one has to mod out all $f_1,\dots,f_m$ in $B$.

Comment #2864 by on

Made these improvements to the latex. Thanks. See change here.

Comment #3998 by Kestutis Cesnavicius on

In the equation followed by "(Kronecker delta)", I think one should have $s_{ik}$ instead of $c_{ik}$.

Comment #4164 by Kestutis Cesnavicius on

The part of the proof that refers to both Lemmas 00ST and 00TF is effectively a fibral smoothness criterion based on counting the number of variables and relations. It may be worthwhile to formulate this criterion as an explicit lemma in the section of 00TF, especially since the same criterion has already been used in the "Lifting Lemma" earlier in the present chapter.

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