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formal-defos.tex
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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Formal Deformation Theory}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
This chapter develops formal deformation theory in a form applicable
later in the stacks project, closely following Rim \cite[Exposee VI]{SGA7-I}
and Schlessinger \cite{Sch}. We strongly encourage the reader new to
this topic to read the paper by Schlessinger first, as it is sufficiently
general for most applications, and Schlessinger's results are indeed
used in most papers that use this kind of formal deformation theory.
\medskip\noindent
Let $\Lambda$ be a complete Noetherian local ring with residue field $k$,
and let $\mathcal{C}_\Lambda$ denote the category of Artinian local
$\Lambda$-algebras with residue field $k$. Given a functor
$F : \mathcal{C}_\Lambda \to \textit{Sets}$ such that $F(k)$
is a one element set, Schlessinger's paper introduced conditions
(H1)-(H4) such that:
\begin{enumerate}
\item $F$ has a ``hull'' if and only if (H1)-(H3) hold.
\item $F$ is prorepresentable if and only (H1)-(H4) hold.
\end{enumerate}
The purpose of this chapter is to generalize these results in two ways
exactly as is done in Rim's paper:
\begin{enumerate}
\item[(A)] The functor $F$ is replaced by a category $\mathcal{F}$ cofibered
in groupoids over $\mathcal{C}_\Lambda$, see
Section \ref{section-CLambda}.
\item[(B)] We let $\Lambda$ be a Noetherian ring and $\Lambda \to k$
a finite ring map to a field. The category $\mathcal{C}_\Lambda$ is
the category of Artinian local $\Lambda$-algebras $A$ endowed with a
given identification $A/\mathfrak m_A = k$.
\end{enumerate}
The analogue of the condition that $F(k)$ is a one element set is that
$\mathcal{F}(k)$ is the trivial groupoid. If $\mathcal{F}$ satisfies this
condition then we say it is a {\it predeformation category}, but in general
we do not make this assumption. Rim's paper \cite[Exposee VI]{SGA7-I} is the
original source for the results in this document. We also mention the useful
paper \cite{Vistoli}, which discusses deformation theory with groupoids but
in less generality than we do here.
\medskip\noindent
An important role is played by the ``completion''
$\widehat{\mathcal{C}}_\Lambda$ of the category $\mathcal{C}_\Lambda$.
An object of $\widehat{\mathcal{C}}_\Lambda$ is a Noetherian complete
local $\Lambda$-algebra $R$ whose residue field is identified with $k$, see
Section \ref{section-category-completion-CLambda}.
On the one hand $\mathcal{C}_\Lambda \subset \widehat{\mathcal{C}}_\Lambda$
is a strictly full subcategory and on the other hand
$\widehat{\mathcal{C}}_\Lambda$ is a full subcategory of the category
of pro-objects of $\mathcal{C}_\Lambda$. A functor
$\mathcal{C}_\Lambda \to \textit{Sets}$ is {\it prorepresentable}
if it is isomorphic to the restriction of a representable functor
$\underline{R} = \Mor_{\widehat{\mathcal{C}}_\Lambda}(R, -)$
to $\mathcal{C}_\Lambda$ where
$R \in \Ob(\widehat{\mathcal{C}}_\Lambda)$.
\medskip\noindent
{\it Categories cofibred in groupoids} are dual to categories fibred in
groupoids; we introduced them in Section \ref{section-preliminary}.
A {\it smooth} morphism of categories cofibred in groupoids over
$\mathcal{C}_\Lambda$ is one that satisfies the infinitesimal lifting
criterion for objects, see
Section \ref{section-smooth-morphisms}.
This is analogous to the definition of a formally smooth ring map, see
Algebra, Definition \ref{algebra-definition-formally-smooth}
and is exactly dual to the notion in
Criteria for Representability, Section \ref{criteria-section-formally-smooth}.
This is an important notion as we eventually want to prove that certain
kinds of categories cofibred in groupoids have a smooth prorepresentable
presentation, much like the characterization of algebraic stacks in
Algebraic Stacks, Sections \ref{algebraic-section-stack-to-presentation} and
\ref{algebraic-section-smooth-groupoid-gives-algebraic-stack}.
A {\it versal formal object} of a category $\mathcal{F}$ cofibred
in groupoids over $\mathcal{C}_\Lambda$ is an object
$\xi \in \widehat{\mathcal{F}}(R)$ of the completion such that the
associated morphism $\underline{\xi} : \underline{R} \to \mathcal{F}$
is smooth.
\medskip\noindent
In
Section \ref{section-schlessinger-conditions},
we define conditions (S1) and (S2) on $\mathcal{F}$ generalizing
Schlessinger's (H1) and (H2). The analogue of Schlessinger's
(H3)---the condition that $\mathcal{F}$ has finite dimensional
tangent space---is not given a name.
A key step in the development of the theory is the existence of
versal formal objects for predeformation categories satisfying
(S1), (S2) and (H3), see
Lemma \ref{lemma-versal-object-existence}.
Schlessinger's notion of a {\it hull} for a functor
$F : \mathcal{C}_\Lambda \to \textit{Sets}$
is, in our terminology, a versal formal object $\xi \in \widehat{F}(R)$
such that the induced map of tangent spaces
$d\underline{\xi} : T\underline{R} \to TF$ is an isomorphism.
In the literature a hull is often called a ``miniversal'' object.
We do not do so, and here is why. It can happen that a functor has a
versal formal object without having a hull. Moreover, we show in
Section \ref{section-minimal-versal}
that if a predeformation category has a versal formal object, then
it always has a {\it minimal} one (as defined in
Definition \ref{definition-minimal-versal})
which is unique up to isomorphism, see
Lemma \ref{lemma-minimal-versal}.
But it can happen that the minimal versal formal object does not
induce an isomorphism on tangent spaces! (See
Examples \ref{example-do-not-get-S2} and
\ref{example-smooth-continued}.)
\medskip\noindent
Keeping in mind the differences pointed out above,
Theorem \ref{theorem-miniversal-object-existence}
is the direct generalization of (1) above: it recovers Schlessinger's
result in the case that $\mathcal{F}$ is a functor and it characterizes
minimal versal formal objects, in the presence of conditions
(S1) and (S2), in terms of the map $d\underline{\xi} : T\underline{R} \to TF$
on tangent spaces.
\medskip\noindent
In Section \ref{section-RS-condition},
we define Rim's condition (RS) on $\mathcal{F}$ generalizing
Schlessinger's (H4). A {\it deformation category} is defined as a
predeformation category satisfying (RS).
The analogue to prorepresentable functors are the categories
cofibred in groupoids over $\mathcal{C}_\Lambda$ which have
a {\it presentation by a smooth prorepresentable groupoid in functors}
on $\mathcal{C}_\Lambda$, see
Definitions \ref{definition-groupoid-in-functors},
\ref{definition-prorepresentable-groupoid-in-functors}, and
\ref{definition-smooth-groupoid-in-functors}.
This notion of a presentation takes into account the groupoid structure
of the fibers of $\mathcal{F}$. In
Theorem \ref{theorem-presentation-deformation-groupoid}
we prove that $\mathcal{F}$ has a presentation by a smooth prorepresentable
groupoid in functors if and only if $\mathcal{F}$ has a finite dimensional
tangent space and finite dimensional infinitesimal automorphism space.
This is the generalization of (2) above: it reduces to Schlessinger's result
in the case that $\mathcal{F}$ is a functor.
There is a final
Section \ref{section-minimality}
where we discuss how to use minimal versal formal objects
to produce a (unique up to isomorphism) minimal presentation
by a smooth prorepresentable groupoid in functors.
\medskip\noindent
We also find the following conceptual explanation for Schlessinger's
conditions. If a predeformation category $\mathcal{F}$ satisfies (RS),
then the associated functor of isomorphism classes
$\overline{\mathcal{F}}: \mathcal{C}_\Lambda \to \textit{Sets}$
satisfies (H1) and (H2)
(Lemmas \ref{lemma-RS-implies-S1-S2} and
\ref{lemma-S1-S2-associated-functor}).
Conversely, if a functor
$F : \mathcal{C}_\Lambda \to \textit{Sets}$
arises naturally as the functor of isomorphism classes of
a category $\mathcal{F}$ cofibered in groupoids, then it seems to happen in
practice that an argument showing $F$ satisfies (H1) and (H2) will also show
$\mathcal{F}$ satisfies (RS) (see
Artin's Axioms, Section \ref{artin-section-examples}
for examples). Moreover, if $\mathcal{F}$ satisfies (RS), then condition
(H4) for $\overline{\mathcal{F}}$ has a simple interpretation in terms of
extending automorphisms of objects of $\mathcal{F}$
(Lemma \ref{lemma-RS-associated-functor}).
These observations suggest that (RS) should be regarded as the
fundamental deformation theoretic glueing condition.
\section{Notation and Conventions}
\label{section-notations-conventions}
\noindent
A ring is commutative with $1$. The maximal ideal of a local ring $A$
is denoted by $\mathfrak{m}_A$. The set of positive integers is denoted
by $\mathbf{N} = \{1, 2, 3, \dots\}$. If $U$ is an object of a
category $\mathcal{C}$, we denote by $\underline{U}$
the functor
$\Mor_\mathcal{C}(U, -): \mathcal{C} \to \textit{Sets}$, see
Remarks \ref{remarks-cofibered-groupoids} (\ref{item-definition-yoneda}).
Warning: this may conflict with the notation in other chapters where sometimes
use $\underline{U}$ to denote $h_U(-) = \Mor_\mathcal{C}(-, U)$.
\medskip\noindent
Throughout this chapter $\Lambda$ is a Noetherian ring and
$\Lambda \to k$ is a finite ring map from $\Lambda$ to a field.
The kernel of this map is denoted $\mathfrak m_\Lambda$ and the
image $k' \subset k$. It turns out that $\mathfrak m_\Lambda$ is
a maximal ideal, $k' = \Lambda/\mathfrak m_\Lambda$ is a field, and
the extension $k' \subset k$ is finite. See discussion surrounding
(\ref{equation-k-prime}).
\section{The category $\mathcal{C}_\Lambda$}
\label{section-CLambda}
\noindent
Motivation. An important application of formal deformation theory is
to criteria for representability by algebraic spaces. Suppose given a
locally Noetherian base change $S$ and a functor
$F : (\Sch/S)_{fppf}^{opp} \to \textit{Sets}$.
Let $k$ be a finite type field over $S$, i.e., we are given a
finite type morphism $\Spec(k) \to S$.
One of Artin's criteria is that for any element $x \in F(\Spec(k))$
the predeformation functor associated to
the triple $(S, k, x)$ should be prorepresentable. By
Morphisms, Lemma \ref{morphisms-lemma-point-finite-type}
the condition that $k$ is of finite type over $S$ means that there exists
an affine open $\Spec(\Lambda) \subset S$ such that $k$
is a finite $\Lambda$-algebra. This motivates why we work throughout
this chapter with a base category as follows.
\begin{definition}
\label{definition-CLambda}
Let $\Lambda$ be a Noetherian ring and let $\Lambda \to k$ be a finite
ring map where $k$ is a field. We define {\it $\mathcal{C}_\Lambda$} to be
the category with
\begin{enumerate}
\item objects are pairs $(A, \varphi)$ where $A$ is an Artinian local
$\Lambda$-algebra and where $\varphi : A/\mathfrak m_A \to k$ is a
$\Lambda$-algebra isomorphism, and
\item morphisms $f : (B, \psi) \to (A, \varphi)$ are local $\Lambda$-algebra
homomorphisms such that $\varphi \circ (f \bmod \mathfrak m) = \psi$.
\end{enumerate}
We say we are in the {\it classical case} if $\Lambda$ is a Noetherian
complete local ring and $k$ is its residue field.
\end{definition}
\noindent
Note that if $\Lambda \to k$ is surjective and if $A$ is an Artinian local
$\Lambda$-algebra, then the identification $\varphi$, if it exists,
is unique. Moreover, in this case any $\Lambda$-algebra map $A \to B$ is
going to be compatible with the identifications. Hence in this case
$\mathcal{C}_\Lambda$ is just the category of local Artinian $\Lambda$-algebras
whose residue field ``is'' $k$. By abuse of notation we also denote objects of
$\mathcal{C}_\Lambda$ simply $A$ in the general case. Moreover, we will
often write $A/\mathfrak m = k$, i.e., we will pretend all rings in
$\mathcal{C}_\Lambda$ have residue field $k$ (since all ring maps in
$\mathcal{C}_\Lambda$ are compatible with the given identifications this
should never cause any problems).
Throughout the rest of this chapter the base ring $\Lambda$ and the
field $k$ are fixed. The category $\mathcal{C}_\Lambda$ will be the base
category for the cofibered categories considered below.
\begin{definition}
\label{definition-small-extension}
Let $f: B \to A$ be a ring map in $\mathcal{C}_\Lambda$. We say $f$
is a {\it small extension} if it is surjective and $\ker(f)$ is a nonzero
principal ideal which is annihilated by $\mathfrak{m}_B$.
\end{definition}
\noindent
By the following lemma we can often reduce arguments involving surjective ring
maps in $\mathcal{C}_\Lambda$ to the case of small extensions.
\begin{lemma}
\label{lemma-factor-small-extension}
Let $f: B \to A$ be a surjective ring map in $\mathcal{C}_\Lambda$.
Then $f$ can be factored as a composition of small extensions.
\end{lemma}
\begin{proof}
Let $I$ be the kernel of $f$. The maximal ideal $\mathfrak{m}_B$ is
nilpotent since $B$ is Artinian, say $\mathfrak{m}_B^n = 0$. Hence we get a
factorization
$$
B = B/I\mathfrak{m}_B^{n-1} \to B/I\mathfrak{m}_B^{n-2} \to
\cdots \to B/I \cong A
$$
of $f$ into a composition of surjective maps whose kernels are annihilated by
the maximal ideal. Thus it suffices to prove the lemma when $f$ itself is such
a map, i.e.\ when $I$ is annihilated by $\mathfrak{m}_B$. In this case
$I$ is a $k$-vector space, which has finite dimension, see
Algebra, Lemma \ref{algebra-lemma-artinian-finite-length}.
Take a basis $x_1, \dots, x_n$ of $I$ as a $k$-vector space to get a
factorization
$$
B \to B/(x_1) \to \cdots \to B/(x_1, \dots, x_n) \cong A
$$
of $f$ into a composition of small extensions.
\end{proof}
\noindent
The next lemma says that we can compute the length of a module over a local
$\Lambda$-algebra with residue field $k$ in terms of the length over
$\Lambda$. To explain the notation in the statement, let $k' \subset k$
be the image of our fixed finite ring map $\Lambda \to k$. Note
that $k/k'$ is a finite extension of rings. Hence $k'$ is a field
and $k'/k$ is a finite extension, see
Algebra, Lemma \ref{algebra-lemma-integral-under-field}.
Moreover, as $\Lambda \to k'$ is surjective we see that its kernel
is a maximal ideal $\mathfrak m_\Lambda$. Thus
\begin{equation}
\label{equation-k-prime}
[k : k'] = [k : \Lambda/\mathfrak m_\Lambda] < \infty
\end{equation}
and in the classical case we have $k = k'$. The notation
$k' = \Lambda/\mathfrak m_\Lambda$ will be fixed throughout this chapter.
\begin{lemma}
\label{lemma-length}
Let $A$ be a local $\Lambda$-algebra with residue field $k$.
Let $M$ be an $A$-module. Then
$[k : k'] \text{length}_A(M) = \text{length}_\Lambda(M)$.
In the classical case we have
$\text{length}_A(M) = \text{length}_\Lambda(M)$.
\end{lemma}
\begin{proof}
If $M$ is a simple $A$-module then $M \cong k$ as an $A$-module, see
Algebra, Lemma \ref{algebra-lemma-characterize-length-1}.
In this case $\text{length}_A(M) = 1$ and
$\text{length}_\Lambda(M) = [k' : k]$, see
Algebra, Lemma \ref{algebra-lemma-dimension-is-length}.
If $\text{length}_A(M)$ is finite, then the result follows on
choosing a filtration of $M$ by $A$-submodules with simple quotients
using additivity, see
Algebra, Lemma \ref{algebra-lemma-length-additive}.
If $\text{length}_A(M)$ is infinite, the result follows from the obvious
inequality $\text{length}_A(M) \leq \text{length}_\Lambda(M)$.
\end{proof}
\begin{lemma}
\label{lemma-surjective}
Let $A \to B$ be a ring map in $\mathcal{C}_\Lambda$.
The following are equivalent
\begin{enumerate}
\item $f$ is surjective,
\item $\mathfrak m_A/\mathfrak m_A^2 \to \mathfrak m_B/\mathfrak m_B^2$
is surjective, and
\item $\mathfrak m_A/(\mathfrak m_\Lambda A + \mathfrak m_A^2)
\to \mathfrak m_B/(\mathfrak m_\Lambda B + \mathfrak m_B^2)$ is surjective.
\end{enumerate}
\end{lemma}
\begin{proof}
For any ring map $f : A \to B$ in $\mathcal{C}_\Lambda$ we have
$f(\mathfrak m_A) \subset \mathfrak m_B$ for example because
$\mathfrak m_A$, $\mathfrak m_B$ is the set of nilpotent elements of
$A$, $B$. Suppose $f$ is surjective. Let $y \in \mathfrak m_B$.
Choose $x \in A$ with $f(x) = y$. Since $f$ induces an isomorphism
$A/\mathfrak m_A \to B/\mathfrak m_B$ we see that $x \in \mathfrak m_A$.
Hence the induced map
$\mathfrak m_A/\mathfrak m_A^2 \to \mathfrak m_B/\mathfrak m_B^2$
is surjective. In this way we see that (1) implies (2).
\medskip\noindent
It is clear that (2) implies (3). The map $A \to B$ gives rise
to a canonical commutative diagram
$$
\xymatrix{
\mathfrak m_\Lambda/\mathfrak m_\Lambda^2 \otimes_{k'} k \ar[r] \ar[d] &
\mathfrak m_A/\mathfrak m_A^2 \ar[r] \ar[d] &
\mathfrak m_A/(\mathfrak m_\Lambda A + \mathfrak m_A^2) \ar[r] \ar[d] & 0 \\
\mathfrak m_\Lambda/\mathfrak m_\Lambda^2 \otimes_{k'} k \ar[r] &
\mathfrak m_B/\mathfrak m_B^2 \ar[r] &
\mathfrak m_B/(\mathfrak m_\Lambda B + \mathfrak m_B^2) \ar[r] & 0
}
$$
with exact rows. Hence if (3) holds, then so does (2).
\medskip\noindent
Assume (2). To show that $A \to B$ is surjective it suffices by
Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK})
to show that $A/\mathfrak m_A \to B/\mathfrak m_AB$ is surjective.
(Note that $\mathfrak m_A$ is a nilpotent ideal.)
As $k = A/\mathfrak m_A = B/\mathfrak m_B$ it suffices to show that
$\mathfrak m_AB \to \mathfrak m_B$ is surjective. Applying
Nakayama's lemma once more we see that it suffices to see that
$\mathfrak m_AB/\mathfrak m_A\mathfrak m_B \to \mathfrak m_B/\mathfrak m_B^2$
is surjective which is what we assumed.
\end{proof}
\noindent
If $A \to B$ is a ring map in $\mathcal{C}_\Lambda$, then the map
$\mathfrak m_A/(\mathfrak m_\Lambda A + \mathfrak m_A^2)
\to \mathfrak m_B/(\mathfrak m_\Lambda B + \mathfrak m_B^2)$
is the map on relative cotangent spaces. Here is a formal definition.
\begin{definition}
\label{definition-tangent-space-ring}
Let $R \to S$ be a local homomorphism of local rings. The
{\it relative cotangent space}\footnote{Caution: We will see later
that in our general setting the tangent
space of an object $A \in \mathcal{C}_\Lambda$ over $\Lambda$ should
not be defined simply as the $k$-linear dual of the relative
cotangent space. In fact, the correct definition of the relative
cotangent space is
$\Omega_{S/R} \otimes_S S/\mathfrak m_S$.} of $R$ over $S$ is the
$S/\mathfrak m_S$-vector space
$\mathfrak m_S/(\mathfrak m_R S + \mathfrak m_S^2)$.
\end{definition}
\noindent
If $f_1: A_1 \to A$ and $f_2: A_2 \to A$ are two ring maps, then the fiber
product $A_1 \times_A A_2$ is the subring of $A_1 \times A_2$ consisting of
elements whose two projections to $A$ are equal. Throughout this chapter we
will be considering conditions involving such a fiber product when $f_1$
and $f_2$ are in $\mathcal{C}_\Lambda$. It isn't always the case that the
fibre product is an object of $\mathcal{C}_\Lambda$.
\begin{example}
\label{example-fibre-product}
Let $p$ be a prime number and let $n \in \mathbf{N}$.
Let $\Lambda = \mathbf{F}_p(t_1, t_2, \ldots, t_n)$ and let
$k = \mathbf{F}_p(x_1, \ldots, x_n)$ with map $\Lambda \to k$ given
by $t_i \mapsto x_i^p$. Let $A = k[\epsilon] = k[x]/(x^2)$.
Then $A$ is an object of $\mathcal{C}_\Lambda$. Suppose that
$D : k \to k$ is a derivation of $k$ over $\Lambda$, for example
$D = \partial/\partial x_i$. Then the map
$$
f_D : k \longrightarrow k[\epsilon], \quad
a \mapsto a + D(a)\epsilon
$$
is a morphism of $\mathcal{C}_\Lambda$. Set $A_1 = A_2 = k$ and set
$f_1 = f_{\partial/\partial x_1}$ and $f_2(a) = a$. Then
$A_1 \times_A A_2 = \{a \in k \mid \partial/\partial x_1(a) = 0\}$
which does not surject onto $k$. Hence the fibre product isn't
an object of $\mathcal{C}_\Lambda$.
\end{example}
\noindent
It turns out that this problem can only occur if the residue field
extension $k' \subset k$ (\ref{equation-k-prime}) is inseparable
and neither $f_1$ nor $f_2$ is surjective.
\begin{lemma}
\label{lemma-fiber-product-CLambda}
Let $f_1 : A_1 \to A$ and $f_2 : A_2 \to A$ be ring maps in
$\mathcal{C}_\Lambda$. Then:
\begin{enumerate}
\item If $f_1$ or $f_2$ is surjective, then
$A_1 \times_A A_2$ is in $\mathcal{C}_\Lambda$.
\item If $f_2$ is a small extension, then so is
$A_1 \times_A A_2 \to A_1$.
\item If the field extension $k' \subset k$ is separable, then
$A_1 \times_A A_2$ is in $\mathcal{C}_\Lambda$.
\end{enumerate}
\end{lemma}
\begin{proof}
The ring $A_1 \times_A A_2$ is a $\Lambda$-algebra via the map
$\Lambda \to A_1 \times_A A_2$ induced by the maps
$\Lambda \to A_1$ and $\Lambda \to A_2$. It is a local ring with unique
maximal ideal
$$
\mathfrak m_{A_1} \times_{\mathfrak m_A} \mathfrak m_{A_2} =
\text{Ker}(A_1 \times_A A_2 \longrightarrow k)
$$
A ring is Artinian if and only if it has finite length as a module
over itself, see
Algebra, Lemma \ref{algebra-lemma-artinian-finite-length}.
Since $A_1$ and $A_2$ are Artinian, Lemma \ref{lemma-length} implies
$\text{length}_\Lambda(A_1)$ and $\text{length}_\Lambda(A_2)$,
and hence $\text{length}_\Lambda(A_1 \times A_2)$, are all finite. As
$A_1 \times_A A_2 \subset A_1 \times A_2$ is a $\Lambda$-submodule, this
implies
$\text{length}_{A_1 \times_A A_2}(A_1 \times_A A_2) \leq
\text{length}_\Lambda(A_1 \times_A A_2)$ is finite. So $A_1
\times_A A_2$ is Artinian. Thus the only thing that is keeping
$A_1 \times_A A_2$ from being an object of $\mathcal{C}_\Lambda$ is
the possibility that its residue field maps to a proper subfield of $k$
via the map $A_1 \times_A A_2 \to A \to A/\mathfrak m_A = k$ above.
\medskip\noindent
Proof of (1). If $f_2$ is surjective, then the projection
$A_1 \times_A A_2 \to A_1$ is surjective. Hence the composition
$A_1 \times_A A_2 \to A_1 \to A_1/\mathfrak m_{A_1} = k$ is surjective
and we conclude that $A_1 \times_A A_2$ is an object of $\mathcal{C}_\Lambda$.
\medskip\noindent
Proof of (2). If $f_2$ is a small extension then $A_2 \to A$ and
$A_1 \times_A A_2 \to A_1$ are both surjective with the same kernel.
Hence the kernel of $A_1 \times_A A_2 \to A_1$ is a $1$-dimensional
$k$-vector space and we see that $A_1 \times_A A_2 \to A_1$ is a small
extension.
\medskip\noindent
Proof of (3). Choose $\overline{x} \in k$ such that
$k = k'(\overline{x})$ (see
Fields, Lemma \ref{fields-lemma-primitive-element}).
Let $P'(T) \in k'[T]$ be the minimal polynomial of $\overline{x}$ over $k'$.
Since $k/k'$ is separable we see that
$\text{d}P/\text{d}T(\overline{x}) \not = 0$.
Choose a monic $P \in \Lambda[T]$ which maps to $P'$ under the surjective map
$\Lambda[T] \to k'[T]$. Because $A, A_1, A_2$ are henselian, see
Algebra, Lemma \ref{algebra-lemma-local-dimension-zero-henselian},
we can find $x, x_1, x_2 \in A, A_1, A_2$ with $P(x) = 0, P(x_1) = 0,
P(x_2) = 0$ and such that the image of $x, x_1, x_2$ in $k$ is $\overline{x}$.
Then $(x_1, x_2) \in A_1 \times_A A_2$ because $x_1, x_2$
map to $x \in A$ by uniqueness, see
Algebra, Lemma \ref{algebra-lemma-uniqueness}.
Hence the residue field of
$A_1 \times_A A_2$ contains a generator of $k$ over $k'$ and we win.
\end{proof}
\noindent
Next we define essential surjections in $\mathcal{C}_\Lambda$. A necessary
and sufficient condition for a surjection in $\mathcal{C}_\Lambda$ to be
essential is given in Lemma \ref{lemma-essential-surjection}.
\begin{definition}
\label{definition-essential-surjection}
Let $f: B \to A$ be a ring map in $\mathcal{C}_\Lambda$. We say $f$
is an {\it essential surjection} if it has the following properties:
\begin{enumerate}
\item $f$ is surjective.
\item If $g: C \to B$ is a ring map in $\mathcal{C}_\Lambda$ such that
$f \circ g$ is surjective, then $g$ is surjective.
\end{enumerate}
\end{definition}
\noindent
Using Lemma \ref{lemma-surjective}, we can characterize
essential surjections in $\mathcal{C}_\Lambda$ as follows.
\begin{lemma}
\label{lemma-essential-surjection-mod-squares}
Let $f: B \to A$ be a ring map in $\mathcal{C}_\Lambda$.
The following are equivalent
\begin{enumerate}
\item $f$ is an essential surjection,
\item the map $B/\mathfrak m_B^2 \to A/\mathfrak m_A^2$ is an essential
surjection, and
\item the map
$B/(\mathfrak m_\Lambda B + \mathfrak m_B^2) \to
A/(\mathfrak m_\Lambda A + \mathfrak m_A^2)$ is an essential surjection.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (3). Let $C \to B$ be a ring map in $\mathcal{C}_\Lambda$ such
that $C \to A$ is surjective. Then
$C \to A/(\mathfrak m_\Lambda A + \mathfrak m_A^2)$ is surjective
too. We conclude that $C \to B/(\mathfrak m_\Lambda B + \mathfrak m_B^2)$
is surjective by our assumption. Hence $C \to B$ is surjective by applying
Lemma \ref{lemma-surjective} (2 times).
\medskip\noindent
Assume (1). Let $C \to B/(\mathfrak m_\Lambda B + \mathfrak m_B^2)$
be a morphism of $\mathcal{C}_\Lambda$ such that
$C \to A/(\mathfrak m_\Lambda A + \mathfrak m_A^2)$ is surjective. Set
$C' = C \times_{B/(\mathfrak m_\Lambda B + \mathfrak m_B^2)} B$
which is an object of $\mathcal{C}_\Lambda$ by
Lemma \ref{lemma-fiber-product-CLambda}.
Note that $C' \to A/(\mathfrak m_\Lambda A + \mathfrak m_A^2)$
is still surjective, hence $C' \to A$ is surjective by
Lemma \ref{lemma-surjective}.
Thus $C' \to B$ is surjective by our assumption. This implies
that $C' \to B/(\mathfrak m_\Lambda B + \mathfrak m_B^2)$ is
surjective, which implies by the construction of $C'$ that
$C \to B/(\mathfrak m_\Lambda B + \mathfrak m_B^2)$ is surjective.
\medskip\noindent
In the first paragraph we proved (3) $\Rightarrow$ (1) and in the second
paragraph we proved (1) $\Rightarrow$ (3). The equivalence of
(2) and (3) is a special case of the equivalence of (1) and (3), hence
we are done.
\end{proof}
\noindent
To analyze essential surjections in $\mathcal{C}_\Lambda$ a bit more
we introduce some notation. Suppose that $A$ is an object
of $\mathcal{C}_\Lambda$. There is a canonical exact sequence
\begin{equation}
\label{equation-sequence}
\mathfrak m_A/\mathfrak m_A^2 \xrightarrow{\text{d}_A}
\Omega_{A/\Lambda} \otimes_A k \to
\Omega_{k/\Lambda} \to 0
\end{equation}
see
Algebra, Lemma \ref{algebra-lemma-differential-seq}.
Note that $\Omega_{k/\Lambda} = \Omega_{k/k'}$ with $k'$ as
in (\ref{equation-k-prime}). Let $H_1(L_{k/\Lambda})$
be the first homology module of the naive cotangent complex of $k$
over $\Lambda$, see
Algebra, Definition \ref{algebra-definition-naive-cotangent-complex}.
Then we can extend (\ref{equation-sequence})
to the exact sequence
\begin{equation}
\label{equation-sequence-extended}
H_1(L_{k/\Lambda}) \to
\mathfrak m_A/\mathfrak m_A^2 \xrightarrow{\text{d}_A}
\Omega_{A/\Lambda} \otimes_A k \to
\Omega_{k/\Lambda} \to 0,
\end{equation}
see
Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL}.
If $B \to A$ is a ring map in $\mathcal{C}_\Lambda$ then we obtain a
commutative diagram
\begin{equation}
\label{equation-sequence-functorial}
\vcenter{
\xymatrix{
H_1(L_{k/\Lambda}) \ar[r] \ar@{=}[d] &
\mathfrak m_B/\mathfrak m_B^2 \ar[r]_{\text{d}_B} \ar[d] &
\Omega_{B/\Lambda} \otimes_B k \ar[r] \ar[d] &
\Omega_{k/\Lambda} \ar[r] \ar@{=}[d] & 0 \\
H_1(L_{k/\Lambda}) \ar[r] &
\mathfrak m_A/\mathfrak m_A^2 \ar[r]^{\text{d}_A} &
\Omega_{A/\Lambda} \otimes_A k \ar[r] &
\Omega_{k/\Lambda} \ar[r] & 0
}
}
\end{equation}
with exact rows.
\begin{lemma}
\label{lemma-H1-separable-case}
There is a canonical map
$$
\mathfrak m_\Lambda/\mathfrak m_\Lambda^2 \longrightarrow H_1(L_{k/\Lambda}).
$$
If $k' \subset k$ is separable (for example if the characteristic
of $k$ is zero), then this map induces an isomorphism
$\mathfrak m_\Lambda/\mathfrak m_\Lambda^2 \otimes_{k'} k = H_1(L_{k/\Lambda})$.
If $k = k'$ (for example in the classical case), then
$\mathfrak m_\Lambda/\mathfrak m_\Lambda^2 = H_1(L_{k/\Lambda})$.
The composition
$$
\mathfrak m_\Lambda/\mathfrak m_\Lambda^2 \longrightarrow
H_1(L_{k/\Lambda}) \longrightarrow \mathfrak m_A/\mathfrak m_A^2
$$
comes from the canonical map $\mathfrak m_\Lambda \to \mathfrak m_A$.
\end{lemma}
\begin{proof}
Note that $H_1(L_{k'/\Lambda}) = \mathfrak m_\Lambda/\mathfrak m_\Lambda^2$
as $\Lambda \to k'$ is surjective with kernel $\mathfrak m_\Lambda$.
The map arises from functoriality of the naive cotangent complex.
If $k' \subset k$ is separable, then $k' \to k$ is an \'etale ring map, see
Algebra, Lemma \ref{algebra-lemma-etale-over-field}.
Thus its naive cotangent complex has trivial homology groups, see
Algebra, Definition \ref{algebra-definition-etale}.
Then
Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL}
applied to the ring maps $\Lambda \to k' \to k$ implies that
$\mathfrak m_\Lambda/\mathfrak m_\Lambda^2 \otimes_{k'} k = H_1(L_{k/\Lambda})$.
We omit the proof of the final statement.
\end{proof}
\begin{lemma}
\label{lemma-essential-surjection}
Let $f: B \to A$ be a ring map in $\mathcal{C}_\Lambda$.
Notation as in (\ref{equation-sequence-functorial}).
\begin{enumerate}
\item The equivalent conditions of
Lemma \ref{lemma-essential-surjection-mod-squares}
characterizing when $f$ is surjective are also equivalent to
\begin{enumerate}
\item $\text{Im}(\text{d}_B) \to \text{Im}(\text{d}_A)$ is surjective, and
\item the map $\Omega_{B/\Lambda} \otimes_B k \to
\Omega_{A/\Lambda} \otimes_A k$ is surjective.
\end{enumerate}
\item The following are equivalent
\begin{enumerate}
\item $f$ is an essential surjection,
\item the map $\text{Im}(\text{d}_B) \to \text{Im}(\text{d}_A)$ is an
isomorphism, and
\item the map $\Omega_{B/\Lambda} \otimes_B k \to
\Omega_{A/\Lambda} \otimes_A k$ is an isomorphism.
\end{enumerate}
\item If $k/k'$ is separable, then $f$ is an essential surjection if
and only if the map
$\mathfrak m_B/(\mathfrak m_\Lambda B + \mathfrak m_B^2) \to
\mathfrak m_A/(\mathfrak m_\Lambda A + \mathfrak m_A^2)$
is an isomorphism.
\item If $f$ is a small extension, then $f$ is not essential if and only if
$f$ has a section $s: A \to B$ in $\mathcal{C}_\Lambda$
with $f \circ s = \text{id}_A$.
\end{enumerate}
\end{lemma}
\begin{proof}
Proof of (1). It follows from (\ref{equation-sequence-functorial})
that (1)(a) and (1)(b) are equivalent. Also, if
$A \to B$ is surjective, then (1)(a) and (1)(b) hold. Assume (1)(a).
Since the kernel of $\text{d}_A$ is the image of
$H_1(L_{k/\Lambda})$ which also maps to
$\mathfrak m_B/\mathfrak m_B^2$ we conclude that
$\mathfrak m_B/\mathfrak m_B^2 \to \mathfrak m_A/\mathfrak m_A^2$
is surjective. Hence $B \to A$ is surjective by
Lemma \ref{lemma-surjective}. This finishes the proof of (1).
\medskip\noindent
Proof of (2). The equivalence of (2)(b) and (2)(c) is immediate from
(\ref{equation-sequence-functorial}).
\medskip\noindent
Assume (2)(b). Let $g : C \to B$ be a ring map in $\mathcal{C}_\Lambda$
such that $f \circ g$ is surjective. We conclude that
$\mathfrak m_C/\mathfrak m_C^2 \to \mathfrak m_A/\mathfrak m_A^2$
is surjective by
Lemma \ref{lemma-surjective}.
Hence
$\text{Im}(\text{d}_C) \to \text{Im}(\text{d}_A)$ is surjective
and by the assumption we see that
$\text{Im}(\text{d}_C) \to \text{Im}(\text{d}_B)$ is surjective.
It follows that $C \to B$ is surjective by (1).
\medskip\noindent
Assume (2)(a). Then $f$ is surjective and we see that
$\Omega_{B/\Lambda} \otimes_B k \to \Omega_{A/\Lambda} \otimes_A k$
is surjective. Let $K$ be the kernel. Note that
$K = \text{d}_B(\text{Ker}(\mathfrak m_B/\mathfrak m_B^2 \to
\mathfrak m_A/\mathfrak m_A^2))$ by (\ref{equation-sequence-functorial}).
Choose a splitting
$$
\Omega_{B/\Lambda} \otimes_B k =
\Omega_{A/\Lambda} \otimes_A k \oplus K
$$
of $k$-vector space. The map $\text{d} : B \to \Omega_{B/\Lambda}$
induces via the projection onto $K$ a map $D : B \to K$. Set
$C = \{b \in B \mid D(b) = 0\}$. The Leibniz rule shows that this is
a $\Lambda$-subalgebra of $B$. Let $\overline{x} \in k$. Choose $x \in B$
mapping to $\overline{x}$. If $D(x) \not = 0$, then we can find an element
$y \in \mathfrak m_B$ such that $D(y) = D(x)$. Hence $x - y \in C$ is
an element which maps to $\overline{x}$. Thus $C \to k$ is surjective
and $C$ is an object of $\mathcal{C}_\Lambda$. Similarly, pick
$\omega \in \text{Im}(\text{d}_A)$. We can find $x \in \mathfrak m_B$
such that $\text{d}_B(x)$ maps to $\omega$ by (1). If $D(x) \not = 0$, then
we can find an element $y \in \mathfrak m_B$ which maps to zero
in $\mathfrak m_A/\mathfrak m_A^2$ such that $D(y) = D(x)$.
Hence $z = x - y$ is an element of $\mathfrak m_C$ whose
image $\text{d}_C(z) \in \Omega_{C/k} \otimes_C k$ maps to $\omega$.
Hence $\text{Im}(\text{d}_C) \to \text{Im}(\text{d}_A)$ is surjective.
We conclude that $C \to A$ is surjective by (1). Hence $C \to B$ is
surjective by assumption. Hence $D = 0$, i.e., $K = 0$, i.e., (2)(c) holds.
This finishes the proof of (2).
\medskip\noindent
Proof of (3). If $k'/k$ is separable, then
$H_1(L_{k/\Lambda}) =
\mathfrak m_\Lambda/\mathfrak m_\Lambda^2 \otimes_{k'} k$, see
Lemma \ref{lemma-H1-separable-case}.
Hence $\text{Im}(\text{d}_A) =
\mathfrak m_A/(\mathfrak m_\Lambda A + \mathfrak m_A^2)$
and similarly for $B$. Thus (3) follows from (2).
\medskip\noindent
Proof of (4). A section $s$ of $f$ is not surjective (by definition a
small extension has nontrivial kernel), hence $f$ is not essentially
surjective. Conversely, assume $f$ is a small surjection but not an
essential surjection. Choose a ring map $C \to B$ in $\mathcal{C}_\Lambda$
which is not surjective, such that $C \to A$ is surjective. Let
$C' \subset B$ be the image of $C \to B$. Then $C' \not = B$ but
$C'$ surjects onto $A$. Since $f : B \to A$ is a small extension,
$\text{length}_C(B) = \text{length}_C(A) + 1$. Thus
$\text{length}_C(C') \leq \text{length}_C(A)$ since
$C'$ is a proper subring of $B$. But $C' \to A$ is surjective, so in
fact we must have $\text{length}_C(C') = \text{length}_C(A)$ and
$C' \to A$ is an isomorphism which gives us our section.
\end{proof}
\begin{example}
\label{example-essential-surjection}
Let $\Lambda = k[[x]]$ be the power series ring in $1$ variable over $k$.
Set $A = k$ and $B = \Lambda/(x^2)$. Then $B \to A$ is an essential
surjection by
Lemma \ref{lemma-essential-surjection}
because it is a small extension and the map $B \to A$ does not have a
right inverse (in the category $\mathcal{C}_\Lambda$). But the map
$$
k \cong \mathfrak m_B/\mathfrak m_B^2
\longrightarrow
\mathfrak m_A/\mathfrak m_A^2 = 0
$$
is not an isomorphism. Thus in
Lemma \ref{lemma-essential-surjection} (3)
it is necessary to consider the map of relative cotangent spaces
$\mathfrak m_B/(\mathfrak m_\Lambda B + \mathfrak m_B^2) \to
\mathfrak m_A/(\mathfrak m_\Lambda A + \mathfrak m_A^2)$.
\end{example}
\section{The category $\widehat{\mathcal{C}}_\Lambda$}
\label{section-category-completion-CLambda}
\noindent
The following ``completion'' of the category $\mathcal{C}_\Lambda$ will serve
as the base category of the completion of a category cofibered in groupoids
over $\mathcal{C}_\Lambda$
(Section \ref{section-formal-objects}).
\begin{definition}
\label{definition-completion-CLambda}
Let $\Lambda$ be a Noetherian ring and let $\Lambda \to k$ be a finite
ring map where $k$ is a field. We define {\it $\widehat{\mathcal{C}}_\Lambda$}
to be the category with
\begin{enumerate}
\item objects are pairs $(R, \varphi)$ where $R$ is a Noetherian complete
local $\Lambda$-algebra and where $\varphi : R/\mathfrak m_R \to k$ is a
$\Lambda$-algebra isomorphism, and
\item morphisms $f : (S, \psi) \to (R, \varphi)$ are local $\Lambda$-algebra
homomorphisms such that $\varphi \circ (f \bmod \mathfrak m) = \psi$.
\end{enumerate}
\end{definition}
\noindent
As in the discussion following
Definition \ref{definition-CLambda}
we will usually denote an object of $\widehat{\mathcal{C}}_\Lambda$
simply $R$, with the identification $R/\mathfrak m_R = k$ understood.
In this section we discuss some basic properties of objects and morphisms
of the category $\widehat{\mathcal{C}}_\Lambda$ paralleling our discussion of
the category $\mathcal{C}_\Lambda$ in the previous section.
\medskip\noindent
Our first observation is that any object $A \in \mathcal{C}_\Lambda$
is an object of $\widehat{\mathcal{C}}_\Lambda$ as an Artinian local
ring is always Noetherian and complete with respect to its maximal ideal
(which is after all a nilpotent ideal). Moreover, it is clear from the
definitions that
$\mathcal{C}_\Lambda \subset \widehat{\mathcal{C}}_\Lambda$
is the strictly full subcategory consisting of all Artinian rings.
As it turns out, conversely every object of
$\widehat{\mathcal{C}}_\Lambda$ is a limit of objects of
$\mathcal{C}_\Lambda$.
\medskip\noindent
Suppose that $R$ is an object of $\widehat{\mathcal{C}}_\Lambda$.
Consider the rings $R_n = R/\mathfrak m_R^n$ for $n \in \mathbf{N}$.
These are Noetherian local rings with a unique nilpotent prime ideal, hence
Artinian, see
Algebra, Proposition \ref{algebra-proposition-dimension-zero-ring}.
The ring maps
$$
\ldots \to R_{n + 1} \to R_n \to \ldots \to R_2 \to R_1 = k
$$
are all surjective. Completeness of $R$ by definition means
that $R = \lim R_n$. If $f : R \to S$ is a ring map in
$\widehat{\mathcal{C}}_\Lambda$ then we obtain a system of ring maps
$f_n : R_n \to S_n$ whose limit is the given map.
\begin{lemma}
\label{lemma-surjective-cotangent-space}
Let $f: R \to S$ be a ring map in $\widehat{\mathcal{C}}_\Lambda$.
The following are equivalent
\begin{enumerate}
\item $f$ is surjective,
\item the map
$\mathfrak m_R/\mathfrak m_R^2 \to \mathfrak m_S/\mathfrak m_S^2$
is surjective, and
\item the map
$\mathfrak m_R/(\mathfrak m_\Lambda R + \mathfrak m_R^2) \to
\mathfrak m_S/(\mathfrak m_\Lambda S + \mathfrak m_S^2)$
is surjective.
\end{enumerate}
\end{lemma}
\begin{proof}
Note that for $n \geq 2$ we have the equality of relative cotangent spaces
$$
\mathfrak m_R/(\mathfrak m_\Lambda R + \mathfrak m_R^2)
=
\mathfrak m_{R_n}/(\mathfrak m_\Lambda R_n + \mathfrak m_{R_n}^2)
$$
and similarly for $S$. Hence by
Lemma \ref{lemma-surjective}
we see that $R_n \to S_n$ is surjective for all $n$.
Now let $K_n$ be the kernel of $R_n \to S_n$. Then the sequences
$$
0 \to K_n \to R_n \to S_n \to 0
$$
form an exact sequence of directed inverse systems. The system $(K_n)$ is
Mittag-Leffler since each $K_n$ is Artinian. Hence by
Algebra, Lemma \ref{algebra-lemma-ML-exact-sequence}
taking limits preserves exactness. So
$\lim R_n \to \lim S_n$ is surjective, i.e., $f$ is surjective.
\end{proof}
\begin{lemma}
\label{lemma-CLambdahat-pushouts}
The category $\widehat{\mathcal{C}}_\Lambda$ admits pushouts.
\end{lemma}
\begin{proof}
Let $R \to S_1$ and $R \to S_2$ be morphisms of
$\widehat{\mathcal{C}}_\Lambda$. Consider the ring
$C = S_1 \otimes_R S_2$.
This ring has a finitely generated maximal ideal
$\mathfrak m = \mathfrak m_{S_1} \otimes S_2 +
S_1 \otimes \mathfrak m_{S_2}$ with residue field $k$.
Set $C^\wedge$ equal to the completion of $C$ with respect to $\mathfrak m$.
Then $C^\wedge$ is a Noetherian ring complete with respect to
the maximal ideal $\mathfrak m^\wedge = \mathfrak mC^\wedge$
whose residue field is identified with $k$, see
Algebra, Lemma \ref{algebra-lemma-completion-Noetherian}.
Hence $C^\wedge$ is an object of $\widehat{\mathcal{C}}_\Lambda$.
Then $S_1 \to C^\wedge$ and $S_2 \to C^\wedge$ turn $C^\wedge$
into a pushout over $R$ in $\widehat{\mathcal{C}}_\Lambda$ (details omitted).
\end{proof}
\noindent
We will not need the following lemma.
\begin{lemma}
\label{lemma-CLambdahat-coproducts}
The category $\widehat{\mathcal{C}}_\Lambda$ admits coproducts
of pairs of objects.
\end{lemma}
\begin{proof}
Let $R$ and $S$ be objects of $\widehat{\mathcal{C}}_\Lambda$.
Consider the ring $C = R \otimes_\Lambda S$. There is a canonical
surjective map $C \to R \otimes_\Lambda S \to k \otimes_\Lambda k \to k$
where the last map is the multiplication map. The kernel of
$C \to k$ is a maximal ideal $\mathfrak m$. Note that $\mathfrak m$
is generated by $\mathfrak m_R C$, $\mathfrak m_S C$ and finitely many
elements of $C$ which map to generators of the kernel of
$k \otimes_\Lambda k \to k$. Hence $\mathfrak m$ is a finitely
generated ideal. Set
$C^\wedge$ equal to the completion of $C$ with respect to $\mathfrak m$.
Then $C^\wedge$ is a Noetherian ring complete with respect to
the maximal ideal $\mathfrak m^\wedge = \mathfrak mC^\wedge$
with residue field $k$, see
Algebra, Lemma \ref{algebra-lemma-completion-Noetherian}.
Hence $C^\wedge$ is an object of $\widehat{\mathcal{C}}_\Lambda$.
Then $R \to C^\wedge$ and $S \to C^\wedge$ turn $C^\wedge$
into a coproduct in $\widehat{\mathcal{C}}_\Lambda$ (details omitted).
\end{proof}
\noindent
An empty coproduct in a category is an initial object of the category.
In the classical case $\widehat{\mathcal{C}}_\Lambda$ has an initial
object, namely $\Lambda$ itself. More generally, if $k' = k$, then
the completion $\Lambda^\wedge$ of $\Lambda$ with respect to
$\mathfrak m_\Lambda$ is an initial object. More generally still, if
$k' \subset k$ is separable, then $\widehat{\mathcal{C}}_\Lambda$ has an
initial object too. Namely, choose a monic polynomial $P \in \Lambda[T]$
such that $k \cong k'[T]/(P')$ where $p' \in k'[T]$ is the image
of $P$. Then $R = \Lambda^\wedge[T]/(P)$ is an initial object, see proof of
Lemma \ref{lemma-fiber-product-CLambda}.
\medskip\noindent
If $R$ is an initial object as above, then we have
$\mathcal{C}_\Lambda = \mathcal{C}_R$ and
$\widehat{\mathcal{C}}_\Lambda = \widehat{\mathcal{C}}_R$ which effectively
brings the whole discussion in this chapter back to the classical case.
But, if $k' \subset k$ is inseparable, then an initial object does not
exist.
\begin{lemma}
\label{lemma-derivations-finite}
Let $S$ be an object of $\widehat{\mathcal{C}}_\Lambda$.
Then $\dim_k \text{Der}_\Lambda(S, k) < \infty$.
\end{lemma}
\begin{proof}
Let $x_1, \ldots, x_n \in \mathfrak m_S$ map to a $k$-basis
for the relative cotangent space
$\mathfrak m_S/(\mathfrak m_\Lambda S + \mathfrak m_S^2)$.
Choose $y_1, \ldots, y_m \in S$ whose images in $k$ generate $k$
over $k'$. We claim that $\dim_k \text{Der}_\Lambda(S, k) \leq n + m$.
To see this it suffices to prove that if $D(x_i) = 0$ and
$D(y_j) = 0$, then $D = 0$. Let $a \in S$. We can find a
polynomial $P = \sum \lambda_J y^J$ with $\lambda_J \in \Lambda$
whose image in $k$ is the same as the image of $a$ in $k$.
Then we see that $D(a - P) = D(a) - D(P) = D(a)$ by our assumption
that $D(y_j) = 0$ for all $j$. Thus we may assume $a \in \mathfrak m_S$.
Write $a = \sum a_i x_i$ with $a_i \in S$. By the Leibniz rule
$$
D(a) = \sum x_iD(a_i) + \sum a_iD(x_i) = \sum x_iD(a_i)
$$
as we assumed $D(x_i) = 0$. We have $\sum x_iD(a_i) = 0$
as multiplication by $x_i$ is zero on $k$.
\end{proof}
\begin{lemma}
\label{lemma-derivations-surjective}
Let $f : R \to S$ be a morphism of $\widehat{\mathcal{C}}_\Lambda$.
If $\text{Der}_\Lambda(S, k) \to \text{Der}_\Lambda(R, k)$ is injective,
then $f$ is surjective.
\end{lemma}
\begin{proof}