# The Stacks Project

## Tag 07FE

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 $R_\mathfrak p \to \Lambda_\mathfrak q$ is flat, see Algebra, Lemma 10.127.2. Hence $$R_\mathfrak p/(x_1^e, \ldots, x_d^e) \to \Lambda_\mathfrak q/(\pi_1^e, \ldots, \pi_d^e)$$ 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 2328–2340 (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 $R_\mathfrak p \to \Lambda_\mathfrak q$
is flat, see Algebra, Lemma \ref{algebra-lemma-flat-over-regular}. Hence
$$R_\mathfrak p/(x_1^e, \ldots, x_d^e) \to \Lambda_\mathfrak q/(\pi_1^e, \ldots, \pi_d^e)$$
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}

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 $R_\mathfrak{p}/(x_1^e,\dots,x_d^e)$.

Comment #2868 by Johan (site) on October 4, 2017 a 6:23 pm UTC

Thanks! Fixed here.

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