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smoothing.tex
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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Smoothing Ring Maps}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
The main result of this chapter is the following:
$$
\fbox{A regular map of Noetherian rings is a filtered colimit
of smooth ones.}
$$
This theorem is due to Popescu, see \cite{popescu-letter}.
A readable exposition of Popescu's proof was given by Richard Swan,
see \cite{swan} who used notes by Andr\'e and a paper of Ogoma, see
\cite{Ogoma}.
\medskip\noindent
Our exposition follows Swan's, but we first prove an intermediate result
which lets us work in a slightly simpler situation. Here is an overview.
We first solve the following ``lifting problem'': A flat infinitesimal
deformation of a filtered colimit of smooth algebras is a filtered colimit
of smooth algebras. This result essentially says that it suffices to prove
the main theorem for maps between reduced Noetherian rings. Next we prove
two very clever lemmas called the ``lifting lemma'' and the
``desingularization lemma''. We show that these lemmas combined
reduce the main theorem to proving a Noetherian, geometrically regular
algebra $\Lambda$ over a field $k$ is a filtered colimit of smooth $k$-algebras.
Next, we discuss the necessary local tricks that go into the
Popescu-Ogoma-Swan-Andr\'e proof. Finally, in the last three sections
we give the proof.
\medskip\noindent
We end this introduction with some pointers to references.
Let $A$ be a henselian Noetherian local ring.
We say $A$ has the {\it approximation property} if for any
$f_1, \ldots, f_m \in A[x_1, \ldots, x_n]$ the system of equations
$f_1 = 0, \ldots, f_m = 0$ has a solution in the completion
of $A$ if and only if it has a solution in $A$. This definition
is due to Artin.
Artin first proved the approximation property for analytic systems of
equations, see \cite{Artin-Analytic-Approximation}.
In \cite{Artin-Algebraic-Approximation} Artin proved the
approximation property for local rings
essentially of finite type over an excellent discrete valuation ring.
Artin conjectured (page 26 of \cite{Artin-Algebraic-Approximation})
that every excellent henselian local ring should have the
approximation property.
\medskip\noindent
At some point in time it became a conjecture
that every regular homomorphism of Noetherian rings is a
filtered colimit of smooth algebras (see for example
\cite{Raynaud-Rennes}, \cite{popescu-global}, \cite{Artin-power-series},
\cite{Artin-Denef}). We're not sure who this conjecture\footnote{The
question/conjecture as formulated in \cite{Artin-power-series},
\cite{Artin-Denef}, and \cite{popescu-global} is stronger and was shown
to be equivalent to the original version in \cite{Cipu}.}
is due to. The relationship with the approximation property is that if
$A \to A^\wedge$ is a colimit of smooth algebras with $A$ as above,
then the approximation
property holds (insert future reference here). Moreover, the main theorem
applies to the map $A \to A^\wedge$ if $A$ is an excellent local ring, as one
of the conditions of an excellent local ring is that the formal
fibres are geometrically regular. Note that excellent local rings
were defined by Grothendieck and their definition appeared in
print in 1965.
\medskip\noindent
In \cite{Artin-power-series} it was shown that
$R \to R^\wedge$ is a filtered colimit of smooth algebras for any
local ring $R$ essentially of finite type over a field.
In \cite{Rotthaus-Artin} it was shown that $R \to R^\wedge$
is a filtered colimit of smooth algebras for any local ring $R$
essentially of finite type over an excellent discrete valuation ring.
Finally, the main theorem was shown in
\cite{popescu-GND}, \cite{popescu-GNDA}, \cite{popescu-letter},
\cite{Ogoma}, and \cite{swan} as discussed above.
\medskip\noindent
Conversely, using some of the results above, in \cite{Rotthaus-excellent}
it was shown that any Noetherian local ring with the approximation property
is excellent.
\medskip\noindent
The paper \cite{Spivakovsky} provides an alternative approach to the
main theorem, but it seems hard to read (for example
\cite[Lemma 5.2]{Spivakovsky} appears to be an incorrectly reformulated
version of \cite[Lemma 3]{Elkik}). There is also a Bourbaki
lecture about this material, see \cite{Teissier}.
\section{Singular ideals}
\label{section-singular-ideal}
\noindent
Let $R \to A$ be a ring map. The singular ideal of $A$ over $R$
is the radical ideal in $A$ cutting out the singular locus of the
morphism $\Spec(A) \to \Spec(R)$. Here is a formal definition.
\begin{definition}
\label{definition-singular-ideal}
Let $R \to A$ be a ring map. The {\it singular ideal of $A$ over $R$},
denoted $H_{A/R}$ is the unique radical ideal $H_{A/R} \subset A$ with
$$
V(H_{A/R}) = \{\mathfrak q \in \Spec(A) \mid R \to A
\text{ not smooth at }\mathfrak q\}
$$
\end{definition}
\noindent
This makes sense because the set of primes where $R \to A$ is smooth
is open, see
Algebra, Definition \ref{algebra-definition-smooth-at-prime}.
In order to find an explicit set
of generators for the singular ideal we first prove the following lemma.
\begin{lemma}
\label{lemma-find-strictly-standard}
Let $R$ be a ring. Let $A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$.
Let $\mathfrak q \subset A$ be a prime ideal. Assume $R \to A$ is smooth
at $\mathfrak q$. Then there exists an $a \in A$, $a \not \in \mathfrak q$,
an integer $c$, $0 \leq c \leq \min(n, m)$, subsets
$U \subset \{1, \ldots, n\}$, $V \subset \{1, \ldots, m\}$
of cardinality $c$ such that
$$
a = a' \det(\partial f_j/\partial x_i)_{j \in V, i \in U}
$$
for some $a' \in A$ and
$$
a f_\ell \in (f_j, j \in V) + (f_1, \ldots, f_m)^2
$$
for all $\ell \in \{1, \ldots, m\}$.
\end{lemma}
\begin{proof}
Set $I = (f_1, \ldots, f_m)$ so that the naive cotangent
complex of $A$ over $R$ is homotopy equivalent to
$I/I^2 \to \bigoplus A\text{d}x_i$, see
Algebra, Lemma \ref{algebra-lemma-NL-homotopy}.
We will use the formation of the naive cotangent complex commutes with
localization, see Algebra, Section \ref{algebra-section-netherlander},
especially Algebra, Lemma \ref{algebra-lemma-localize-NL}.
By Algebra, Definitions \ref{algebra-definition-smooth} and
\ref{algebra-definition-smooth-at-prime}
we see that $(I/I^2)_a \to \bigoplus A_a\text{d}x_i$
is a split injection for some $a \in A$, $a \not \in \mathfrak q$.
After renumbering $x_1, \ldots, x_n$ and $f_1, \ldots, f_m$ we may
assume that $f_1, \ldots, f_c$ form a basis for
the vector space $I/I^2 \otimes_A \kappa(\mathfrak q)$ and that
$\text{d}x_{c + 1}, \ldots, \text{d}x_n$ map to a basis of
$\Omega_{A/R} \otimes_A \kappa(\mathfrak q)$. Hence after replacing $a$
by $aa'$ for some $a' \in A$, $a' \not \in \mathfrak q$ we may assume
$f_1, \ldots, f_c$ form a basis for $(I/I^2)_a$ and that
$\text{d}x_{c + 1}, \ldots, \text{d}x_n$ map to a basis of
$(\Omega_{A/R})_a$. In this situation $a^N$ for some large integer
$N$ satisfies the conditions of the lemma (with $U = V = \{1, \ldots, c\}$).
\end{proof}
\noindent
We will use the notion of a {\it strictly standard} element in
$A$ over $R$. Our notion is slightly weaker than the
one in Swan's paper \cite{swan}. We also define an {\it elementary
standard} element to be one of the type we found in the lemma above.
We compare the different types of elements in
Lemma \ref{lemma-compare-standard}.
\begin{definition}
\label{definition-strictly-standard}
Let $R \to A$ be a ring map of finite presentation.
We say an element $a \in A$ is {\it elementary standard in $A$ over $R$}
if there exists a presentation
$A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$
and $0 \leq c \leq \min(n, m)$ such that
\begin{equation}
\label{equation-elementary-standard-one}
a = a' \det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}
\end{equation}
for some $a' \in A$ and
\begin{equation}
\label{equation-elementary-standard-two}
a f_{c + j} \in (f_1, \ldots, f_c) + (f_1, \ldots, f_m)^2
\end{equation}
for $j = 1, \ldots, m - c$. We say $a \in A$ is
{\it strictly standard in $A$ over $R$} if there exists a presentation
$A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$
and $0 \leq c \leq \min(n, m)$ such that
\begin{equation}
\label{equation-strictly-standard-one}
a = \sum\nolimits_{I \subset \{1, \ldots, n\},\ |I| = c}
a_I \det(\partial f_j/\partial x_i)_{j = 1, \ldots, c,\ i \in I}
\end{equation}
for some $a_I \in A$ and
\begin{equation}
\label{equation-strictly-standard-two}
a f_{c + j} \in (f_1, \ldots, f_c) + (f_1, \ldots, f_m)^2
\end{equation}
for $j = 1, \ldots, m - c$.
\end{definition}
\noindent
The following lemma is useful to find implications of
(\ref{equation-strictly-standard-one}).
\begin{lemma}
\label{lemma-parse-equation-strictly-standard-one}
Let $R$ be a ring. Let $A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$
and write $I = (f_1, \ldots, f_m)$. Let $a \in A$. Then
(\ref{equation-strictly-standard-one}) implies
there exists an $A$-linear map
$\psi : \bigoplus\nolimits_{i = 1, \ldots, n} A \text{d}x_i \to A^{\oplus c}$
such that the composition
$$
A^{\oplus c} \xrightarrow{(f_1, \ldots, f_c)}
I/I^2 \xrightarrow{f \mapsto \text{d}f}
\bigoplus\nolimits_{i = 1, \ldots, n} A \text{d}x_i
\xrightarrow{\psi}
A^{\oplus c}
$$
is multiplication by $a$. Conversely, if such a $\psi$ exists, then
$a^c$ satisfies (\ref{equation-strictly-standard-one}).
\end{lemma}
\begin{proof}
This is a special case of
Algebra, Lemma \ref{algebra-lemma-matrix-left-inverse}.
\end{proof}
\begin{lemma}[Elkik]
\label{lemma-elkik}
Let $R \to A$ be a ring map of finite presentation.
The singular ideal $H_{A/R}$ is the radical of the ideal
generated by strictly standard elements in $A$ over $R$
and also the radical of the ideal generated by elementary
standard elements in $A$ over $R$.
\end{lemma}
\begin{proof}
Assume $a$ is strictly standard in $A$ over $R$. We claim that
$A_a$ is smooth over $R$, which proves that $a \in H_{A/R}$. Namely,
let $A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$, $c$, and $a' \in A$
be as in Definition \ref{definition-strictly-standard}.
Write $I = (f_1, \ldots, f_m)$ so that the naive cotangent
complex of $A$ over $R$ is given by $I/I^2 \to \bigoplus A\text{d}x_i$.
Assumption (\ref{equation-strictly-standard-two})
implies that $(I/I^2)_a$ is generated by the classes of $f_1, \ldots, f_c$.
Assumption (\ref{equation-strictly-standard-one}) implies
that the differential $(I/I^2)_a \to \bigoplus A_a\text{d}x_i$
has a left inverse, see
Lemma \ref{lemma-parse-equation-strictly-standard-one}.
Hence $R \to A_a$ is smooth by definition and
Algebra, Lemma \ref{algebra-lemma-localize-NL}.
\medskip\noindent
Let $H_e, H_s \subset A$ be the radical of the ideal generated by
elementary, resp.\ strictly standard elements of $A$ over $R$.
By definition and what we just proved we have
$H_e \subset H_s \subset H_{A/R}$. The inclusion $H_{A/R} \subset H_e$
follows from Lemma \ref{lemma-find-strictly-standard}.
\end{proof}
\begin{example}
\label{example-not-quasi-compact}
The set of points where a finitely presented ring map is smooth
needn't be a quasi-compact open. For example, let
$R = k[x, y_1, y_2, y_3, \ldots]/(xy_i)$ and $A = R/(x)$.
Then the smooth locus of $R \to A$ is
$\bigcup D(y_i)$ which is not quasi-compact.
\end{example}
\begin{lemma}
\label{lemma-strictly-standard-base-change}
Let $R \to A$ be a ring map of finite presentation.
Let $R \to R'$ be a ring map. If $a \in A$ is elementary,
resp.\ strictly standard in $A$ over $R$, then $a \otimes 1$
is elementary, resp.\ strictly standard in $A \otimes_R R'$ over $R'$.
\end{lemma}
\begin{proof}
If $A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ is a presentation
of $A$ over $R$, then
$A \otimes_R R' = R'[x_1, \ldots, x_n]/(f'_1, \ldots, f'_m)$
is a presentation of $A \otimes_R R'$ over $R'$. Here $f'_j$ is
the image of $f_j$ in $R'[x_1, \ldots, x_n]$.
Hence the result follows from the definitions.
\end{proof}
\begin{lemma}
\label{lemma-final-solve}
Let $R \to A \to \Lambda$ be ring maps with $A$ of finite presentation
over $R$. Assume that $H_{A/R} \Lambda = \Lambda$. Then there exists
a factorization $A \to B \to \Lambda$ with $B$ smooth over $R$.
\end{lemma}
\begin{proof}
Choose $f_1, \ldots, f_r \in H_{A/R}$ and
$\lambda_1, \ldots, \lambda_r \in \Lambda$ such that
$\sum f_i\lambda_i = 1$ in $\Lambda$. Set
$B = A[x_1, \ldots, x_r]/(f_1x_1 + \ldots + f_rx_r - 1)$
and define $B \to \Lambda$ by mapping $x_i$ to $\lambda_i$.
To check that $B$ is smooth over $R$ use that $A_{f_i}$ is smooth
over $R$ by definition of $H_{A/R}$ and that $B_{f_i}$ is smooth
over $A_{f_i}$. Details omitted.
\end{proof}
\section{Presentations of algebras}
\label{section-presentations}
\noindent
Some of the results in this section are due to Elkik. Note that the algebra
$C$ in the following lemma is a symmetric algebra over $A$. Moreover, if
$R$ is Noetherian, then $C$ is of finite presentation over $R$.
\begin{lemma}
\label{lemma-improve-presentation}
Let $R$ be a ring and let $A$ be a finitely presented $R$-algebra.
There exists finite type $R$-algebra map $A \to C$ which has a
retraction with the following two properties
\begin{enumerate}
\item for each $a \in A$ such that $R \to A_a$ is a local complete
intersection (More on Algebra, Definition
\ref{more-algebra-definition-local-complete-intersection})
the ring $C_a$ is smooth over $A_a$ and has a presentation
$C_a = R[y_1, \ldots, y_m]/J$ such that $J/J^2$ is free over $C_a$, and
\item for each $a \in A$ such that $A_a$ is smooth over $R$ the
module $\Omega_{C_a/R}$ is free over $C_a$.
\end{enumerate}
\end{lemma}
\begin{proof}
Choose a presentation $A = R[x_1, \ldots, x_n]/I$ and write
$I = (f_1, \ldots, f_m)$. Define the $A$-module $K$ by the short exact sequence
$$
0 \to K \to A^{\oplus m} \to I/I^2 \to 0
$$
where the $j$th basis vector $e_j$ in the middle is mapped to the class of
$f_j$ on the right. Set
$$
C = \text{Sym}^*_A(I/I^2).
$$
The retraction is just the projection onto the degree $0$ part of $C$.
We have a surjection $R[x_1, \ldots, x_n, y_1, \ldots, y_m] \to C$
which maps $y_j$ to the class of $f_j$ in $I/I^2$. The kernel $J$ of this
map is generated by the elements $f_1, \ldots, f_m$ and by elements
$\sum h_j y_j$ with $h_j \in R[x_1, \ldots, x_n]$ such that
$\sum h_j e_j$ defines an element of $K$. By
Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL}
applied to $R \to A \to C$ and the presentations above and
More on Algebra, Lemma
\ref{more-algebra-lemma-cotangent-complex-symmetric-algebra}
there is a short exact sequence
\begin{equation}
\label{equation-sequence}
I/I^2 \otimes_A C \to J/J^2 \to K \otimes_A C \to 0
\end{equation}
of $C$-modules. Let $h \in R[x_1, \ldots, x_n]$ be an element
with image $a \in A$. We will use as presentations for the localized rings
$$
A_a = R[x_0, x_1, \ldots, x_n]/I'
\quad\text{and}\quad
C_a = R[x_0, x_1, \ldots, x_n, y_1, \ldots, y_m]/J'
$$
where $I' = (hx_0 - 1, I)$ and $J' = (hx_0 - 1, J)$. Hence
$I'/(I')^2 = A_a \oplus (I/I^2)_a$ as $A_a$-modules and
$J'/(J')^2 = C_a \oplus (J/J^2)_a$ as $C_a$-modules.
Thus we obtain
\begin{equation}
\label{equation-sequence-localized}
C_a \oplus I/I^2 \otimes_A C_a \to
C_a \oplus (J/J^2)_a \to
K \otimes_A C_a \to 0
\end{equation}
as the sequence of
Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL}
corresponding to $R \to A_a \to C_a$ and the presentations above.
\medskip\noindent
Next, assume that $a \in A$ is such that $A_a$ is a local complete
intersection over $R$. Then $(I/I^2)_a$ is finite projective over $A_a$, see
More on Algebra, Lemma
\ref{more-algebra-lemma-quasi-regular-ideal-finite-projective}.
Hence we see $K_a \oplus (I/I^2)_a \cong A_a^{\oplus m}$ is free.
In particular $K_a$ is finite projective too.
By More on Algebra, Lemma \ref{more-algebra-lemma-transitive-lci-at-end}
the sequence (\ref{equation-sequence-localized}) is exact on the left.
Hence
$$
J'/(J')^2 \cong
C_a \oplus I/I^2 \otimes_A C_a \oplus K \otimes_A C_a \cong
C_a^{\oplus m + 1}
$$
This proves (1). Finally, suppose that in addition $A_a$ is smooth over
$R$. Then the same presentation shows that $\Omega_{C_a/R}$
is the cokernel of the map
$$
J'/(J')^2 \longrightarrow
\bigoplus\nolimits_i C_a\text{d}x_i \oplus \bigoplus\nolimits_j C_a\text{d}y_j
$$
The summand $C_a$ of $J'/(J')^2$ in the decomposition above
corresponds to $hx_0 - 1$ and hence maps
isomorphically to the summand $C_a\text{d}x_0$. The summand
$I/I^2 \otimes_A C_a$ of $J'/(J')^2$ maps injectively to
$\bigoplus_{i = 1, \ldots, n} C_a\text{d}x_i$
with quotient $\Omega_{A_a/R} \otimes_{A_a} C_a$. The summand
$K \otimes_A C_a$ maps injectively to
$\bigoplus_{j \geq 1} C_a\text{d}y_j$ with quotient isomorphic to
$I/I^2 \otimes_A C_a$. Thus the cokernel of the last displayed
map is the module
$I/I^2 \otimes_A C_a \oplus \Omega_{A_a/R} \otimes_{A_a} C_a$.
Since $(I/I^2)_a \oplus \Omega_{A_a/R}$ is
free (from the definition of smooth ring maps) we see that (2) holds.
\end{proof}
\noindent
The following proposition was proved for smooth ring maps over henselian
pairs by Elkik in \cite{Elkik}. For smooth ring maps it can be found in
\cite{Arabia}, where it is also proven that ring maps between smooth
algebras can be lifted.
\begin{proposition}
\label{proposition-lift-smooth}
\begin{slogan}
Smooth and syntomic algebras lift along surjections
\end{slogan}
Let $R \to R_0$ be a surjective ring map with kernel $I$.
\begin{enumerate}
\item If $R_0 \to A_0$ is a syntomic ring map, then there exists a syntomic
ring map $R \to A$ such that $A/IA \cong A_0$.
\item If $R_0 \to A_0$ is a smooth ring map, then there exists a smooth
ring map $R \to A$ such that $A/IA \cong A_0$.
\end{enumerate}
\end{proposition}
\begin{proof}
Assume $R_0 \to A_0$ syntomic, in particular a local complete intersection
(More on Algebra, Lemma \ref{more-algebra-lemma-syntomic-lci}).
Choose a presentation $A_0 = R_0[x_1, \ldots, x_n]/J_0$. Set
$C_0 = \text{Sym}^*_{A_0}(J_0/J_0^2)$. Note that $J_0/J_0^2$ is a finite
projective $A_0$-module (Algebra, Lemma
\ref{algebra-lemma-syntomic-presentation-ideal-mod-squares}).
By Lemma \ref{lemma-improve-presentation} the ring map
$A_0 \to C_0$ is smooth and we can find a presentation
$C_0 = R_0[y_1, \ldots, y_m]/K_0$ with $K_0/K_0^2$ free over $C_0$.
By Algebra, Lemma \ref{algebra-lemma-huber} we can assume
$C_0 = R_0[y_1, \ldots, y_m]/(\overline{f}_1, \ldots, \overline{f}_c)$
where $\overline{f}_1, \ldots, \overline{f}_c$ maps to a basis of
$K_0/K_0^2$ over $C_0$. Choose
$f_1, \ldots, f_c \in R[y_1, \ldots, y_c]$ lifting
$\overline{f}_1, \ldots, \overline{f}_c$ and set
$$
C = R[y_1, \ldots, y_m]/(f_1, \ldots, f_c)
$$
By construction $C_0 = C/IC$. By Algebra, Lemma
\ref{algebra-lemma-localize-relative-complete-intersection}
we can after replacing $C$ by $C_g$ assume that $C$ is a relative
global complete intersection over $R$.
We conclude that there exists a finite projective $A_0$-module
$P_0$ such that $C_0 = \text{Sym}^*_{A_0}(P_0)$
is isomorphic to $C/IC$ for some syntomic $R$-algebra $C$.
\medskip\noindent
Choose an integer $n$ and a direct sum decomposition
$A_0^{\oplus n} = P_0 \oplus Q_0$.
By More on Algebra, Lemma \ref{more-algebra-lemma-lift-projective-module}
we can find an \'etale ring map $C \to C'$ which induces
an isomorphism $C/IC \to C'/IC'$ and a finite projective
$C'$-module $Q$ such that $Q/IQ$ is isomorphic to
$Q_0 \otimes_{A_0} C/IC$.
Then $D = \text{Sym}_{C'}^*(Q)$ is a smooth $C'$-algebra (see
More on Algebra, Lemma \ref{more-algebra-lemma-symmetric-algebra-smooth}).
Picture
$$
\xymatrix{
R \ar[d] \ar[rr] & &
C \ar[r] \ar[d] &
C' \ar[r] \ar[d] &
D \ar[d] \\
R/I \ar[r] &
A_0 \ar[r] &
C/IC \ar[r]^{\cong} &
C'/IC' \ar[r] &
D/ID
}
$$
Observe that our choice of $Q$ gives
\begin{align*}
D/ID & =
\text{Sym}_{C/IC}^*(Q_0 \otimes_{A_0} C/IC) \\
& =
\text{Sym}_{A_0}^*(Q_0) \otimes_{A_0} C/IC \\
& =
\text{Sym}_{A_0}^*(Q_0) \otimes_{A_0}
\text{Sym}_{A_0}^*(P_0) \\
& =
\text{Sym}_{A_0}^*(Q_0 \oplus P_0) \\
& =
\text{Sym}_{A_0}^*(A_0^{\oplus n}) \\
& =
A_0[x_1, \ldots, x_n]
\end{align*}
Choose $f_1, \ldots, f_n \in D$ which map to $x_1, \ldots, x_n$
in $D/ID = A_0[x_1, \ldots, x_n]$. Set $A = D/(f_1, \ldots, f_n)$.
Note that $A_0 = A/IA$. We claim that $R \to A$ is syntomic
in a neighbourhood of $V(IA)$. If the claim is true, then we can
find a $f \in A$ mapping to $1 \in A_0$ such that $A_f$ is syntomic
over $R$ and the proof of (1) is finished.
\medskip\noindent
Proof of the claim. Observe that $R \to D$ is syntomic as a composition
of the syntomic ring map $R \to C$, the \'etale ring map $C \to C'$ and the
smooth ring map $C' \to D$ (Algebra, Lemmas
\ref{algebra-lemma-composition-syntomic} and
\ref{algebra-lemma-smooth-syntomic}).
The question is local on $\Spec(D)$, hence we
may assume that $D$ is a relative global complete intersection
(Algebra, Lemma \ref{algebra-lemma-syntomic}).
Say $D = R[y_1, \ldots, y_m]/(g_1, \ldots, g_s)$.
Let $f'_1, \ldots, f'_n \in R[y_1, \ldots, y_m]$ be lifts of
$f_1, \ldots, f_n$. Then we can apply
Algebra, Lemma \ref{algebra-lemma-localize-relative-complete-intersection}
to get the claim.
\medskip\noindent
Proof of (2). Since a smooth ring map is syntomic, we can find
a syntomic ring map $R \to A$ such that $A_0 = A/IA$.
By assumption the fibres of $R \to A$ are smooth over primes in $V(I)$
hence $R \to A$ is smooth in an open neighbourhood of $V(IA)$
(Algebra, Lemma \ref{algebra-lemma-flat-fibre-smooth}).
Thus we can replace $A$ by a localization to obtain the result we want.
\end{proof}
\noindent
We know that any syntomic ring map $R \to A$ is locally a relative global
complete intersection, see
Algebra, Lemma \ref{algebra-lemma-syntomic}.
The next lemma says that a vector bundle over $\Spec(A)$ is
a relative global complete intersection.
\begin{lemma}
\label{lemma-syntomic-complete-intersection}
Let $R \to A$ be a syntomic ring map. Then there exists a smooth $R$-algebra
map $A \to C$ with a retraction such that $C$ is a global relative complete
intersection over $R$, i.e.,
$$
C \cong R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)
$$
flat over $R$ and all fibres of dimension $n - c$.
\end{lemma}
\begin{proof}
Apply Lemma \ref{lemma-improve-presentation} to get $A \to C$.
By Algebra, Lemma \ref{algebra-lemma-huber}
we can write $C = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$
with $f_i$ mapping to a basis of $J/J^2$.
The ring map $R \to C$ is syntomic (hence flat)
as it is a composition of a syntomic and a smooth ring map.
The dimension of the fibres is $n - c$ by
Algebra, Lemma \ref{algebra-lemma-lci}
(the fibres are local complete intersections, so the lemma applies).
\end{proof}
\begin{lemma}
\label{lemma-smooth-standard-smooth}
Let $R \to A$ be a smooth ring map. Then there exists a smooth $R$-algebra
map $A \to B$ with a retraction such that $B$ is standard smooth over
$R$, i.e.,
$$
B \cong R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)
$$
and $\det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}$
is invertible in $B$.
\end{lemma}
\begin{proof}
Apply Lemma \ref{lemma-syntomic-complete-intersection}
to get a smooth $R$-algebra map $A \to C$ with a retraction such that
$C = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$
is a relative global complete intersection over $R$. As $C$ is smooth
over $R$ we have a short exact sequence
$$
0 \to
\bigoplus\nolimits_{j = 1, \ldots, c} C f_j \to
\bigoplus\nolimits_{i = 1, \ldots, n} C\text{d}x_i \to
\Omega_{C/R} \to 0
$$
Since $\Omega_{C/R}$ is a projective $C$-module this sequence is split.
Choose a left inverse $t$ to the first map. Say
$t(\text{d}x_i) = \sum c_{ij} f_j$
so that $\sum_i \frac{\partial f_j}{\partial x_i} c_{i\ell} = \delta_{j\ell}$
(Kronecker delta). Let
$$
B' = C[y_1, \ldots, y_c] =
R[x_1, \ldots, x_n, y_1, \ldots, y_c]/(f_1, \ldots, f_c)
$$
The $R$-algebra map $C \to B'$ has a retraction given by mapping $y_j$ to zero.
We claim that the map
$$
R[z_1, \ldots, z_n] \longrightarrow B',\quad
z_i \longmapsto x_i - \sum\nolimits_j c_{ij} y_j
$$
is \'etale at every point in the image of $\Spec(C) \to \Spec(B')$.
In $\Omega_{B'/R[z_1, \ldots, z_n]}$ we have
$$
0 =
\text{d}f_j - \sum\nolimits_i \frac{\partial f_j}{\partial x_i} \text{d}z_i
\equiv
\sum\nolimits_{i, \ell}
\frac{\partial f_j}{\partial x_i} c_{i\ell} \text{d}y_\ell
\equiv
\text{d}y_j \bmod (y_1, \ldots, y_c)\Omega_{B'/R[z_1, \ldots, z_n]}
$$
Since $0 = \text{d}z_i = \text{d}x_i$ modulo
$\sum B'\text{d}y_j + (y_1, \ldots, y_c)\Omega_{B'/R[z_1, \ldots, z_n]}$
we conclude that
$$
\Omega_{B'/R[z_1, \ldots, z_n]}/
(y_1, \ldots, y_c)\Omega_{B'/R[z_1, \ldots, z_n]} = 0.
$$
As $\Omega_{B'/R[z_1, \ldots, z_n]}$ is a finite $B'$-module
by Nakayama's lemma there exists a $g \in 1 + (y_1, \ldots, y_c)$
that $(\Omega_{B'/R[z_1, \ldots, z_n]})_g = 0$. This proves that
$R[z_1, \ldots, z_n] \to B'_g$ is unramified, see
Algebra, Definition \ref{algebra-definition-unramified}.
For any ring map $R \to k$ where $k$ is a field we obtain an
unramified ring map $k[z_1, \ldots, z_n] \to (B'_g) \otimes_R k$
between smooth $k$-algebras of dimension $n$. It follows that
$k[z_1, \ldots, z_n] \to (B'_g) \otimes_R k$ is flat by
Algebra, Lemmas \ref{algebra-lemma-CM-over-regular-flat} and
\ref{algebra-lemma-characterize-smooth-kbar}. By the crit\`ere
de platitude par fibre
(Algebra, Lemma \ref{algebra-lemma-criterion-flatness-fibre})
we conclude that $R[z_1, \ldots, z_n] \to B'_g$ is flat.
Finally, Algebra, Lemma \ref{algebra-lemma-characterize-etale}
implies that $R[z_1, \ldots, z_n] \to B'_g$ is \'etale.
Set $B = B'_g$. Note that $C \to B$ is smooth and has a retraction,
so also $A \to B$ is smooth and has a retraction.
Moreover, $R[z_1, \ldots, z_n] \to B$ is \'etale.
By Algebra, Lemma \ref{algebra-lemma-etale-standard-smooth}
we can write
$$
B = R[z_1, \ldots, z_n, w_1, \ldots, w_c]/(g_1, \ldots, g_c)
$$
with $\det(\partial g_j/\partial w_i)$ invertible in $B$.
This proves the lemma.
\end{proof}
\begin{lemma}
\label{lemma-colimit-standard-smooth}
Let $R \to \Lambda$ be a ring map. If $\Lambda$ is a filtered colimit of
smooth $R$-algebras, then $\Lambda$ is a filtered colimit of standard
smooth $R$-algebras.
\end{lemma}
\begin{proof}
Let $A \to \Lambda$ be an $R$-algebra map with $A$
of finite presentation over $R$. According to
Algebra, Lemma \ref{algebra-lemma-when-colimit}
we have to factor this map through a standard smooth algebra, and
we know we can factor it as $A \to B \to \Lambda$ with $B$ smooth
over $R$. Choose an $R$-algebra map $B \to C$ with a retraction
$C \to B$ such that $C$ is standard smooth over $R$, see
Lemma \ref{lemma-smooth-standard-smooth}.
Then the desired factorization is $A \to B \to C \to B \to \Lambda$.
\end{proof}
\begin{lemma}
\label{lemma-standard-smooth-include-generators}
Let $R \to A$ be a standard smooth ring map.
Let $E \subset A$ be a finite subset of order $|E| = n$.
Then there exists a presentation
$A = R[x_1, \ldots, x_{n + m}]/(f_1, \ldots, f_c)$ with $c \geq n$,
with $\det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}$
invertible in $A$, and such that $E$ is the set of congruence classes of
$x_1, \ldots, x_n$.
\end{lemma}
\begin{proof}
Choose a presentation $A = R[y_1, \ldots, y_m]/(g_1, \ldots, g_d)$
such that the image of
$\det(\partial g_j/\partial y_i)_{i, j = 1, \ldots, d}$
is invertible in $A$. Choose an enumerations $E = \{a_1, \ldots, a_n\}$
and choose $h_i \in R[y_1, \ldots, y_m]$ whose image in $A$ is $a_i$.
Consider the presentation
$$
A = R[x_1, \ldots, x_n, y_1, \ldots, y_m]/
(x_1 - h_1, \ldots, x_n - h_n, g_1, \ldots, g_d)
$$
and set $c = n + d$.
\end{proof}
\begin{lemma}
\label{lemma-compare-standard}
Let $R \to A$ be a ring map of finite presentation.
Let $a \in A$. Consider the following conditions on $a$:
\begin{enumerate}
\item $A_a$ is smooth over $R$,
\item $A_a$ is smooth over $R$ and $\Omega_{A_a/R}$ is stably free,
\item $A_a$ is smooth over $R$ and $\Omega_{A_a/R}$ is free,
\item $A_a$ is standard smooth over $R$,
\item $a$ is strictly standard in $A$ over $R$,
\item $a$ is elementary standard in $A$ over $R$.
\end{enumerate}
Then we have
\begin{enumerate}
\item[(a)] (4) $\Rightarrow$ (3) $\Rightarrow$ (2) $\Rightarrow$ (1),
\item[(b)] (6) $\Rightarrow$ (5),
\item[(c)] (6) $\Rightarrow$ (4),
\item[(d)] (5) $\Rightarrow$ (2),
\item[(e)] (2) $\Rightarrow$ the elements $a^e$, $e \geq e_0$ are
strictly standard in $A$ over $R$,
\item[(f)] (4) $\Rightarrow$ the elements $a^e$, $e \geq e_0$ are
elementary standard in $A$ over $R$.
\end{enumerate}
\end{lemma}
\begin{proof}
Part (a) is clear from the definitions and
Algebra, Lemma \ref{algebra-lemma-standard-smooth}.
Part (b) is clear from Definition \ref{definition-strictly-standard}.
\medskip\noindent
Proof of (c). Choose a presentation
$A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ such that
(\ref{equation-elementary-standard-one}) and
(\ref{equation-elementary-standard-two}) hold.
Choose $h \in R[x_1, \ldots, x_n]$ mapping to $a$. Then
$$
A_a = R[x_0, x_1, \ldots, x_n]/(x_0h - 1, f_1, \ldots, f_m).
$$
Write $J = (x_0h - 1, f_1, \ldots, f_m)$.
By (\ref{equation-elementary-standard-two}) we see that the $A_a$-module
$J/J^2$ is generated by $x_0h - 1, f_1, \ldots, f_c$
over $A_a$. Hence, as in the proof of Algebra, Lemma \ref{algebra-lemma-huber},
we can choose a $g \in 1 + J$ such that
$$
A_a = R[x_0, \ldots, x_n, x_{n + 1}]/
(x_0h - 1, f_1, \ldots, f_m, gx_{n + 1} - 1).
$$
At this point (\ref{equation-elementary-standard-one})
implies that $R \to A_a$ is standard smooth (use the coordinates
$x_0, x_1, \ldots, x_c, x_{n + 1}$ to take derivatives).
\medskip\noindent
Proof of (d). Choose a presentation
$A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ such that
(\ref{equation-strictly-standard-one}) and
(\ref{equation-strictly-standard-two}) hold.
Write $I = (f_1, \ldots, f_m)$.
We already know that $A_a$ is smooth over $R$, see
Lemma \ref{lemma-elkik}. By
Lemma \ref{lemma-parse-equation-strictly-standard-one}
we see that $(I/I^2)_a$ is free on $f_1, \ldots, f_c$
and maps isomorphically to a direct summand of
$\bigoplus A_a \text{d}x_i$. Since
$\Omega_{A_a/R} = (\Omega_{A/R})_a$
is the cokernel of the map
$(I/I^2)_a \to \bigoplus A_a \text{d}x_i$
we conclude that it is stably free.
\medskip\noindent
Proof of (e). Choose a presentation
$A = R[x_1, \ldots, x_n]/I$ with $I$ finitely generated.
By assumption we have a short exact sequence
$$
0 \to (I/I^2)_a \to \bigoplus\nolimits_{i = 1, \ldots, n} A_a\text{d}x_i \to
\Omega_{A_a/R} \to 0
$$
which is split exact. Hence we see that
$(I/I^2)_a \oplus \Omega_{A_a/R}$ is a free $A_a$-module.
Since $\Omega_{A_a/R}$ is stably free we see that $(I/I^2)_a$
is stably free as well. Thus replacing the presentation chosen
above by $A = R[x_1, \ldots, x_n, x_{n + 1}, \ldots, x_{n + r}]/J$ with
$J = (I, x_{n + 1}, \ldots, x_{n + r})$ for some $r$ we get that $(J/J^2)_a$
is (finite) free. Choose $f_1, \ldots, f_c \in J$ which map to a basis of
$(J/J^2)_a$. Extend this to a list of generators
$f_1, \ldots, f_m \in J$. Consider the presentation
$A = R[x_1, \ldots, x_{n + r}]/(f_1, \ldots, f_m)$.
Then (\ref{equation-strictly-standard-two}) holds for $a^e$
for all sufficiently large $e$ by construction. Moreover, since
$(J/J^2)_a \to \bigoplus\nolimits_{i = 1, \ldots, n + r} A_a\text{d}x_i$
is a split injection we can find an $A_a$-linear left inverse.
Writing this left inverse in terms of the basis $f_1, \ldots, f_c$
and clearing denominators we find a linear map
$\psi_0 : A^{\oplus n + r} \to A^{\oplus c}$ such that
$$
A^{\oplus c} \xrightarrow{(f_1, \ldots, f_c)}
J/J^2 \xrightarrow{f \mapsto \text{d}f}
\bigoplus\nolimits_{i = 1, \ldots, n + r} A \text{d}x_i
\xrightarrow{\psi_0}
A^{\oplus c}
$$
is multiplication by $a^{e_0}$ for some $e_0 \geq 1$. By
Lemma \ref{lemma-parse-equation-strictly-standard-one}
we see (\ref{equation-strictly-standard-one})
holds for all $a^{ce_0}$ and hence for $a^e$ for all $e$ with $e \geq ce_0$.
\medskip\noindent
Proof of (f). Choose a presentation
$A_a = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ such that
$\det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}$
is invertible in $A_a$. We may assume that for some
$m < n$ the classes of the elements $x_1, \ldots, x_m$
correspond $a_i/1$ where $a_1, \ldots, a_m \in A$ are generators of $A$
over $R$, see Lemma \ref{lemma-standard-smooth-include-generators}.
After replacing $x_i$ by $a^Nx_i$ for $m < i \leq n$
we may assume the class of $x_i$ is $a_i/1 \in A_a$ for some
$a_i \in A$. Consider the ring map
$$
\Psi : R[x_1, \ldots, x_n] \longrightarrow A,\quad
x_i \longmapsto a_i.
$$
This is a surjective ring map. By replacing $f_j$ by $a^Nf_j$ we may
assume that $f_j \in R[x_1, \ldots, x_n]$ and that
$\Psi(f_j) = 0$ (since after all $f_j(a_1/1, \ldots, a_n/1) = 0$
in $A_a$). Let $J = \Ker(\Psi)$. Then $A = R[x_1, \ldots, x_n]/J$
is a presentation and $f_1, \ldots, f_c \in J$ are elements such that
$(J/J^2)_a$ is freely generated by $f_1, \ldots, f_c$ and such
that $\det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}$
maps to an invertible element of $A_a$. It follows that
(\ref{equation-elementary-standard-one}) and
(\ref{equation-elementary-standard-two})
hold for $a^e$ and all large enough $e$ as desired.
\end{proof}
\section{Intermezzo: N\'eron desingularization}
\label{section-neron}
\noindent
We interrupt the attack on the general case of Popescu's theorem to
an easier but already very interesting case, namely, when
$R \to \Lambda$ is a homomorphism of discrete valuation rings.
This is discussed in
\cite[Section 4]{Artin-Algebraic-Approximation}.
\begin{situation}
\label{situation-neron}
Here $R \subset \Lambda$ is an extension of discrete valuation rings
with ramification index $1$ (More on Algebra, Definition
\ref{more-algebra-definition-extension-discrete-valuation-rings}).
We assume given a factorization
$$
R \to A \xrightarrow{\varphi} \Lambda
$$
with $R \to A$ flat and of finite type. Let $\mathfrak q = \Ker(\varphi)$
and $\mathfrak p = \varphi^{-1}(\mathfrak m_\Lambda)$.
\end{situation}
\noindent
In Situation \ref{situation-neron} let $\pi \in R$ be a uniformizer.
Recall that flatness of $A$ over $R$ signifies that $\pi$
is a nonzerodivisor on $A$
(More on Algebra, Lemma
\ref{more-algebra-lemma-valuation-ring-torsion-free-flat}).
By our assumption on $R \subset \Lambda$ we see that $\pi$ maps
to a uniformizer of $\Lambda$. Since $\pi \in \mathfrak p$ we can consider
N\'eron's affine blowup algebra (see
Algebra, Section \ref{algebra-section-blow-up})
$$
\varphi' :
A' = A[\textstyle{\frac{\mathfrak p}{\pi}}]
\longrightarrow
\Lambda
$$
which comes endowed with an induced map to $\Lambda$ sending
$a/\pi^n$, $a \in \mathfrak p^n$ to $\pi^{-n}\varphi(a)$ in $\Lambda$.
We will denote $\mathfrak q' \subset \mathfrak p' \subset A'$
the corresponding prime ideals of $A'$. Observe that the isomorphism
class of $A'$ does not depend on our choice of uniformizer.
Repeating the construction we obtain a sequence
$$
A \to A' \to A'' \to \ldots \to \Lambda
$$
\begin{lemma}
\label{lemma-neron-functorial}
In Situation \ref{situation-neron} N\'eron's blowup is functorial
in the following sense
\begin{enumerate}
\item if $a \in A$, $a \not \in \mathfrak p$, then N\'eron's blowup
of $A_a$ is $A'_a$, and
\item if $B \to A$ is a surjection of flat finite type $R$-algebras
with kernel $I$, then $A'$ is the quotient of $B'/IB'$ by its
$\pi$-power torsion.
\end{enumerate}
\end{lemma}
\begin{proof}
Both (1) and (2) are special cases of
Algebra, Lemma \ref{algebra-lemma-blowup-base-change}.
In fact, whenever we have $A_1 \to A_2 \to \Lambda$ such that
$\mathfrak p_1 A_2 = \mathfrak p_2$, we have that $A_2'$ is
the quotient of $A_1' \otimes_{A_1} A_2$ by its $\pi$-power torsion.
\end{proof}
\begin{lemma}
\label{lemma-neron-blowup-smooth}
In Situation \ref{situation-neron} assume that $R \to A$ is smooth
at $\mathfrak p$ and that $R/\pi R \subset \Lambda/\pi \Lambda$
is a separable field extension. Then $R \to A'$ is smooth at
$\mathfrak p'$ and there is a short exact sequence
$$
0 \to
\Omega_{A/R} \otimes_A A'_{\mathfrak p'} \to
\Omega_{A'/R, \mathfrak p'} \to
(A'/\pi A')_{\mathfrak p'}^{\oplus c} \to 0
$$
where $c = \dim((A/\pi A)_\mathfrak p)$.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-neron-functorial} we may replace $A$ by a localization
at an element not in $\mathfrak p$; we will use this without further mention.
Write $\kappa = R/\pi R$. Since smoothness is stable under base change
(Algebra, Lemma \ref{algebra-lemma-base-change-smooth})
we see that $A/\pi A$ is smooth over $\kappa$ at $\mathfrak p$.
Hence $(A/\pi A)_\mathfrak p$ is a regular local ring
(Algebra, Lemma \ref{algebra-lemma-characterize-smooth-over-field}).
Choose $g_1, \ldots, g_c \in \mathfrak p$ which map to
a regular system of parameters in $(A/\pi A)_\mathfrak p$.
Then we see that $\mathfrak p = (\pi, g_1, \ldots, g_c)$
after possibly replacing $A$ by a localization.
Note that $\pi, g_1, \ldots, g_c$ is a regular sequence
in $A_\mathfrak p$ (first $\pi$ is a nonzerodivisor and
then Algebra, Lemma \ref{algebra-lemma-regular-ring-CM}
for the rest of the sequence).
After replacing $A$ by a localization we may assume that
$\pi, g_1, \ldots, g_c$ is a regular sequence in $A$
(Algebra, Lemma \ref{algebra-lemma-regular-sequence-in-neighbourhood}).
It follows that
$$
A' = A[y_1, \ldots, y_c]/(\pi y_1 - g_1, \ldots, \pi y_c - g_c) =
A[y_1, \ldots, y_c]/I
$$
by More on Algebra, Lemma \ref{more-algebra-lemma-blowup-regular-sequence}.
In the following we will use the definition of smoothness using the
naive cotangent complex (Algebra, Definition \ref{algebra-definition-smooth})
and the criterion of Algebra, Lemma \ref{algebra-lemma-smooth-at-point}
without further mention.
The exact sequence of Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL}
for $R \to A[y_1, \ldots, y_c] \to A'$ looks like this
$$
0 \to H_1(\NL_{A'/R}) \to I/I^2 \to
\Omega_{A/R} \otimes_A A' \oplus
\bigoplus\nolimits_{i = 1, \ldots, c} A' \text{d}y_i \to
\Omega_{A'/R} \to 0
$$
where the class of $\pi y_i - g_i$ in $I/I^2$ is mapped
to $- \text{d}g_i + \pi \text{d}y_i$ in the next term.
Here we have used Algebra, Lemma \ref{algebra-lemma-NL-surjection}
to compute $\NL_{A'/A[y_1, \ldots, y_c]}$ and we have used that
$R \to A[y_1, \ldots, y_c]$ is smooth, so
$H_1(\NL_{A[y_1, \ldots, y_c]/R}) = 0$ and
$\Omega_{A[y_1, \ldots, y_c]/R}$ is a finite projective (a fortiori flat)
$A[y_1, \ldots, y_c]$-module which is in fact the direct sum
of $\Omega_{A/R} \otimes_A A[y_1, \ldots, y_c]$ and a free
module with basis $\text{d}y_i$. To finish the proof it
suffices to show that $\text{d}g_1, \ldots, \text{d}g_c$
forms part of a basis for the finite free module $\Omega_{A/R, \mathfrak p}$.
Namely, this will show $(I/I^2)_\mathfrak p$ is free on $\pi y_i - g_i$,
the localization at $\mathfrak p$ of the middle map in the sequence is
injective, so $H_1(\NL_{A'/R})_\mathfrak p = 0$, and
that the cokernel $\Omega_{A'/R, \mathfrak p}$ is finite free.
To do this it suffices to show
that the images of $\text{d}g_i$ are $\kappa(\mathfrak p)$-linearly
independent in
$\Omega_{A/R, \mathfrak p}/\pi = \Omega_{(A/\pi A)/\kappa, \mathfrak p}$
(equality by Algebra, Lemma \ref{algebra-lemma-differentials-base-change}).
Since $\kappa \subset \kappa(\mathfrak p) \subset \Lambda/\pi \Lambda$