## 16.7 The desingularization lemma

Here is another fiendishly clever lemma.

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 10.15.5 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 10.136.11 and 10.137.17 (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 10.51.4. 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. $\square$

Lemma 16.7.2. Let $R$ be a Noetherian ring. Let $\Lambda$ be an $R$-algebra. Let $\pi \in R$ and assume that $\text{Ann}_ R(\pi ) = \text{Ann}_ R(\pi ^2)$ and $\text{Ann}_\Lambda (\pi ) = \text{Ann}_\Lambda (\pi ^2)$. Let $A \to \Lambda$ and $D \to \Lambda$ be $R$-algebra maps with $A$ and $D$ of finite presentation. Assume

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

2. there exists an $R$-algebra map $A/\pi ^4 A \to D/\pi ^4 D$ compatible with the maps to $\Lambda /\pi ^4 \Lambda$.

Then we can find an $R$-algebra map $B \to \Lambda$ with $B$ of finite presentation and $R$-algebra maps $A \to B$ and $D \to B$ compatible with the maps to $\Lambda$ such that $H_{D/R}B \subset H_{B/D}$ and $H_{D/R}B \subset H_{B/R}$.

Proof. We apply Lemma 16.7.1 to

$D \longrightarrow A \otimes _ R D \longrightarrow \Lambda$

and the image of $\pi$ in $D$. By Lemma 16.2.7 we see that $\pi$ is strictly standard in $A \otimes _ R D$ over $D$. As our section $\rho : (A \otimes _ R D)/\pi ^4 (A \otimes _ R D) \to D/\pi ^4 D$ we take the map induced by the map in (2). Thus Lemma 16.7.1 applies and we obtain a factorization $A \otimes _ R D \to B \to \Lambda$ with $B$ of finite presentation and $\mathfrak a B \subset H_{B/D}$ where

$\mathfrak a = \text{Ann}_ D(\text{Ann}_ D(\pi ^2)/\text{Ann}_ D(\pi )).$

For any prime $\mathfrak q$ of $D$ such that $D_\mathfrak q$ is flat over $R$ we have $\text{Ann}_{D_\mathfrak q}(\pi ^2)/\text{Ann}_{D_\mathfrak q}(\pi ) = 0$ because annihilators of elements commutes with flat base change and we assumed $\text{Ann}_ R(\pi ) = \text{Ann}_ R(\pi ^2)$. Because $D$ is Noetherian we see that $\text{Ann}_ D(\pi ^2)/\text{Ann}_ D(\pi )$ is a finite $D$-module, hence formation of its annihilator commutes with localization. Thus we see that $\mathfrak a \not\subset \mathfrak q$. Hence we see that $D \to B$ is smooth at any prime of $B$ lying over $\mathfrak q$. Since any prime of $D$ where $R \to D$ is smooth is one where $D_\mathfrak q$ is flat over $R$ we conclude that $H_{D/R}B \subset H_{B/D}$. The final inclusion $H_{D/R}B \subset H_{B/R}$ follows because compositions of smooth ring maps are smooth (Algebra, Lemma 10.137.14). $\square$

Lemma 16.7.3. Let $R$ be a Noetherian ring. Let $\Lambda$ be an $R$-algebra. Let $\pi \in R$ and assume that $\text{Ann}_ R(\pi ) = \text{Ann}_ R(\pi ^2)$ and $\text{Ann}_\Lambda (\pi ) = \text{Ann}_\Lambda (\pi ^2)$. Let $A \to \Lambda$ be an $R$-algebra map with $A$ of finite presentation and assume $\pi$ is strictly standard in $A$ over $R$. Let

$A/\pi ^8A \to \bar C \to \Lambda /\pi ^8\Lambda$

be a factorization with $\bar C$ of finite presentation. Then we can find a factorization $A \to B \to \Lambda$ with $B$ of finite presentation such that $R_\pi \to B_\pi$ is smooth and such that

$H_{\bar C/(R/\pi ^8 R)} \cdot \Lambda /\pi ^8\Lambda \subset \sqrt{H_{B/R} \Lambda } \bmod \pi ^8\Lambda .$

Proof. Apply Lemma 16.6.1 to get $R \to D \to \Lambda$ with a factorization $\bar C/\pi ^4\bar C \to D/\pi ^4 D \to \Lambda /\pi ^4\Lambda$ such that $R \to D$ is smooth at any prime not containing $\pi$ and at any prime lying over a prime of $\bar C/\pi ^4\bar C$ where $R/\pi ^8 R \to \bar C$ is smooth. By Lemma 16.7.2 we can find a finitely presented $R$-algebra $B$ and factorizations $A \to B \to \Lambda$ and $D \to B \to \Lambda$ such that $H_{D/R}B \subset H_{B/R}$. We omit the verification that this is a solution to the problem posed by the lemma. $\square$

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