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Tag 07FE

Chapter 16: Smoothing Ring Maps > Section 16.10: Separable residue fields

Lemma 16.10.3. Let $k \to A \to \Lambda \supset \mathfrak q$ be as in Situation 16.9.1 where

  1. $k$ is a field,
  2. $\Lambda$ is Noetherian,
  3. $\mathfrak q$ is minimal over $\mathfrak h_A$,
  4. $\Lambda_\mathfrak q$ is a regular local ring, and
  5. the field extension $k \subset \kappa(\mathfrak q)$ is separable.

Then $k \to A \to \Lambda \supset \mathfrak q$ can be resolved.

Proof. Set $d = \dim \Lambda_\mathfrak q$. Set $R = k[x_1, \ldots, x_d]$. Choose $n > 0$ such that $\mathfrak q^n\Lambda_\mathfrak q \subset \mathfrak h_A\Lambda_\mathfrak q$ which is possible as $\mathfrak q$ is minimal over $\mathfrak h_A$. Choose generators $a_1, \ldots, a_r$ of $H_{A/R}$. Set $$ B = A[x_1, \ldots, x_d, z_{ij}]/(x_i^n - \sum z_{ij}a_j) $$ Each $B_{a_j}$ is smooth over $R$ it is a polynomial algebra over $A_{a_j}[x_1, \ldots, x_d]$ and $A_{a_j}$ is smooth over $k$. Hence $B_{x_i}$ is smooth over $R$. Let $B \to C$ be the $R$-algebra map constructed in Lemma 16.3.1 which comes with a $R$-algebra retraction $C \to B$. In particular a map $C \to \Lambda$ fitting into the diagram above. By construction $C_{x_i}$ is a smooth $R$-algebra with $\Omega_{C_{x_i}/R}$ free. Hence we can find $c > 0$ such that $x_i^c$ is strictly standard in $C/R$, see Lemma 16.3.7. Now choose $\pi_1, \ldots, \pi_d \in \Lambda$ as in Lemma 16.10.2 where $n = n$, $e = 8c$, $\mathfrak q = \mathfrak q$ and $I = \mathfrak h_A$. Write $\pi_i^n = \sum \lambda_{ij} a_j$ for some $\pi_{ij} \in \Lambda$. There is a map $B \to \Lambda$ given by $x_i \mapsto \pi_i$ and $z_{ij} \mapsto \lambda_{ij}$. Set $R = k[x_1, \ldots, x_d]$. Diagram $$ \xymatrix{ R \ar[r] & B \ar[rd] \\ k \ar[u] \ar[r] & A \ar[u] \ar[r] & \Lambda } $$ Now we apply Lemma 16.9.2 to $R \to C \to \Lambda \supset \mathfrak q$ and the sequence of elements $x_1^c, \ldots, x_d^c$ of $R$. Assumption (2) is clear. Assumption (1) holds for $R$ by inspection and for $\Lambda$ by our choice of $\pi_1, \ldots, \pi_d$. (Note that if $\text{Ann}_\Lambda(\pi) = \text{Ann}_\Lambda(\pi^2)$, then we have $\text{Ann}_\Lambda(\pi) = \text{Ann}_\Lambda(\pi^c)$ for all $c > 0$.) Thus it suffices to resolve $$ R/(x_1^e, \ldots, x_d^e) \to C/(x_1^e, \ldots, x_d^e) \to \Lambda/(\pi_1^e, \ldots, \pi_d^e) \supset \mathfrak q/(\pi_1^e, \ldots, \pi_d^e) $$ for $e = 8c$. By Lemma 16.9.4 it suffices to resolve this after localizing at $\mathfrak q$. But since $x_1, \ldots, x_d$ map to a regular sequence in $\Lambda_\mathfrak q$ we see that $R \to \Lambda$ is flat, see Algebra, Lemma 10.127.2. Hence $$ R/(x_1^e, \ldots, x_d^e) \to \Lambda_\mathfrak q/(\pi_1^e, \ldots, \pi_d^e) $$ is a flat ring map of Artinian local rings. Moreover, this map induces a separable field extension on residue fields by assumption. Thus this map is a filtered colimit of smooth algebras by Algebra, Lemma 10.152.11 and Proposition 16.5.3. Existence of the desired solution follows from Algebra, Lemma 10.126.4. $\square$

    The code snippet corresponding to this tag is a part of the file smoothing.tex and is located in lines 2324–2336 (see updates for more information).

    \begin{lemma}
    \label{lemma-resolve-special}
    Let $k \to A \to \Lambda \supset \mathfrak q$ be as in
    Situation \ref{situation-local} where
    \begin{enumerate}
    \item $k$ is a field,
    \item $\Lambda$ is Noetherian,
    \item $\mathfrak q$ is minimal over $\mathfrak h_A$,
    \item $\Lambda_\mathfrak q$ is a regular local ring, and
    \item the field extension $k \subset \kappa(\mathfrak q)$ is separable.
    \end{enumerate}
    Then $k \to A \to \Lambda \supset \mathfrak q$ can be resolved.
    \end{lemma}
    
    \begin{proof}
    Set $d = \dim \Lambda_\mathfrak q$. Set $R = k[x_1, \ldots, x_d]$.
    Choose $n > 0$ such that
    $\mathfrak q^n\Lambda_\mathfrak q \subset \mathfrak h_A\Lambda_\mathfrak q$
    which is possible as $\mathfrak q$ is minimal over $\mathfrak h_A$.
    Choose generators $a_1, \ldots, a_r$ of $H_{A/R}$. Set
    $$
    B = A[x_1, \ldots, x_d, z_{ij}]/(x_i^n - \sum z_{ij}a_j)
    $$
    Each $B_{a_j}$ is smooth over $R$ it is a polynomial
    algebra over $A_{a_j}[x_1, \ldots, x_d]$ and $A_{a_j}$ is smooth over $k$.
    Hence $B_{x_i}$ is smooth over $R$. Let $B \to C$ be the $R$-algebra
    map constructed in Lemma \ref{lemma-improve-presentation}
    which comes with a $R$-algebra retraction $C \to B$. In particular
    a map $C \to \Lambda$ fitting into the diagram above.
    By construction $C_{x_i}$ is a smooth $R$-algebra with
    $\Omega_{C_{x_i}/R}$ free. Hence we can find $c > 0$
    such that $x_i^c$ is strictly standard in $C/R$, see
    Lemma \ref{lemma-compare-standard}.
    Now choose $\pi_1, \ldots, \pi_d \in \Lambda$ as in
    Lemma \ref{lemma-find-sequence}
    where $n = n$, $e = 8c$, $\mathfrak q = \mathfrak q$ and $I = \mathfrak h_A$.
    Write $\pi_i^n = \sum \lambda_{ij} a_j$ for some $\pi_{ij} \in \Lambda$.
    There is a map $B \to \Lambda$ given by $x_i \mapsto \pi_i$
    and $z_{ij} \mapsto \lambda_{ij}$. Set $R = k[x_1, \ldots, x_d]$.
    Diagram
    $$
    \xymatrix{
    R \ar[r] & B \ar[rd] \\
    k \ar[u] \ar[r] & A \ar[u] \ar[r] & \Lambda
    }
    $$
    Now we apply
    Lemma \ref{lemma-lift-solution}
    to $R \to C \to \Lambda \supset \mathfrak q$
    and the sequence of elements $x_1^c, \ldots, x_d^c$ of $R$.
    Assumption (2) is clear. Assumption (1) holds for $R$
    by inspection and for $\Lambda$ by our choice of
    $\pi_1, \ldots, \pi_d$. (Note that if
    $\text{Ann}_\Lambda(\pi) = \text{Ann}_\Lambda(\pi^2)$, then we have
    $\text{Ann}_\Lambda(\pi) = \text{Ann}_\Lambda(\pi^c)$ for all $c > 0$.)
    Thus it suffices to resolve
    $$
    R/(x_1^e, \ldots, x_d^e) \to C/(x_1^e, \ldots, x_d^e) \to
    \Lambda/(\pi_1^e, \ldots, \pi_d^e) \supset
    \mathfrak q/(\pi_1^e, \ldots, \pi_d^e)
    $$
    for $e = 8c$. By
    Lemma \ref{lemma-delocalize-height-zero}
    it suffices to resolve this after localizing at $\mathfrak q$.
    But since $x_1, \ldots, x_d$ map to a regular sequence
    in $\Lambda_\mathfrak q$ we see that $R \to \Lambda$ is flat, see
    Algebra, Lemma \ref{algebra-lemma-flat-over-regular}. Hence
    $$
    R/(x_1^e, \ldots, x_d^e) \to 
    \Lambda_\mathfrak q/(\pi_1^e, \ldots, \pi_d^e)
    $$
    is a flat ring map of Artinian local rings.
    Moreover, this map induces a separable field extension
    on residue fields by assumption. Thus this map is a filtered colimit
    of smooth algebras by
    Algebra, Lemma \ref{algebra-lemma-colimit-syntomic}
    and Proposition \ref{proposition-lift}.
    Existence of the desired solution follows from
    Algebra, Lemma \ref{algebra-lemma-when-colimit}.
    \end{proof}

    Comments (1)

    Comment #2760 by Anonymous on August 5, 2017 a 2:46 pm UTC

    Towards the end of the proof it should say $R_\mathfrak{p} \rightarrow \Lambda_\mathfrak{q}$ is flat by Algebra, Lemma 10.127.2. And the following map should map from $R_\mathfrak{p}/(x_1^e,\dots,x_d^e)$.

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