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Chapter 16: Smoothing Ring Maps > Section 16.7: The desingularization 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 \begin{equation} \tag{16.7.1.1} \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 10.14.4 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. $\square$

    The code snippet corresponding to this tag is a part of the file smoothing.tex and is located in lines 1649–1664 (see updates for more information).

    \begin{lemma}
    \label{lemma-desingularize}
    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
    \begin{enumerate}
    \item the image of $\pi$ is strictly standard in $A$ over $R$, and
    \item there exists a section $\rho : A/\pi^4 A \to R/\pi^4 R$
    which is compatible with the map to $\Lambda/\pi^4 \Lambda$.
    \end{enumerate}
    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
    $\mathfrak a = \text{Ann}_R(\text{Ann}_R(\pi^2)/\text{Ann}_R(\pi))$.
    \end{lemma}
    
    \begin{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
    (\ref{equation-strictly-standard-one}) holds for $\pi$ and such that
    \begin{equation}
    \label{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 (\ref{equation-strictly-standard-one})
    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 \ref{algebra-lemma-matrix-left-inverse}
    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.
    
    \medskip\noindent
    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 \ref{lemma-elkik}.
    
    \medskip\noindent
    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
    \ref{algebra-lemma-localize-relative-complete-intersection} and
    \ref{algebra-lemma-flat-fibre-smooth}
    (the first lemma shows it is syntomic at those primes, in particular
    flat, whereupon the second lemma shows it is smooth).
    
    \medskip\noindent
    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
    (\ref{equation-star}). 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 \ref{algebra-lemma-intersect-powers-ideal-module-zero}.
    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.
    \end{proof}

    Comments (3)

    Comment #2752 by Anonymous on August 2, 2017 a 8:39 am UTC

    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 August 2, 2017 a 8:59 am UTC

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

    Comment #2864 by Johan (site) on October 4, 2017 a 5:54 pm UTC

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

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