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\input{preamble}
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\begin{document}
\title{More on Morphisms}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
In this chapter we continue our study of properties of morphisms of schemes.
A fundamental reference is \cite{EGA}.
\section{Thickenings}
\label{section-thickenings}
\noindent
The following terminology may not be completely standard, but it is convenient.
\begin{definition}
\label{definition-thickening}
Thickenings.
\begin{enumerate}
\item We say a scheme $X'$ is a {\it thickening} of a scheme $X$ if
$X$ is a closed subscheme of $X'$ and the underlying topological spaces
are equal.
\item We say a scheme $X'$ is a {\it first order thickening} of a scheme $X$ if
$X$ is a closed subscheme of $X'$ and the quasi-coherent sheaf of ideals
$\mathcal{I} \subset \mathcal{O}_{X'}$ defining $X$ has square zero.
\item Given two thickenings $X \subset X'$ and $Y \subset Y'$ a
{\it morphism of thickenings} is a morphism $f' : X' \to Y'$ such that
$f'(X) \subset Y$, i.e., such that $f'|_X$ factors through the closed
subscheme $Y$. In this situation we set $f = f'|_X : X \to Y$ and we say
that $(f, f') : (X \subset X') \to (Y \subset Y')$ is a morphism of
thickenings.
\item Let $S$ be a scheme. We similarly define {\it thickenings over $S$}, and
{\it morphisms of thickenings over $S$}. This means that the schemes
$X, X', Y, Y'$ above are schemes over $S$, and that the morphisms
$X \to X'$, $Y \to Y'$ and $f' : X' \to Y'$ are morphisms over $S$.
\end{enumerate}
\end{definition}
\noindent
Finite order thickenings. Let $i_X : X \to X'$ be a thickening.
Any local section of the kernel
$\mathcal{I} = \Ker(i_X^\sharp)$ is locally nilpotent.
Let us say that $X \subset X'$ is a {\it finite order thickening}
if the ideal sheaf $\mathcal{I}$ is ``globally'' nilpotent, i.e.,
if there exists an $n \geq 0$ such that $\mathcal{I}^{n + 1} = 0$.
Technically the class of finite order thickenings $X \subset X'$
is much easier to handle than the general case.
Namely, in this case we have a filtration
$$
0 \subset \mathcal{I}^n \subset \mathcal{I}^{n - 1} \subset
\ldots \subset \mathcal{I} \subset \mathcal{O}_{X'}
$$
and we see that $X'$ is filtered by closed subspaces
$$
X = X_0 \subset X_1 \subset \ldots \subset X_{n - 1} \subset X_{n + 1} = X'
$$
such that each pair $X_i \subset X_{i + 1}$ is a first order thickening
over $S$. Using simple induction arguments many results proved for first order
thickenings can be rephrased as results on finite order thickenings.
\medskip\noindent
First order thickening are described as follows (see
Modules, Lemma \ref{modules-lemma-double-structure-gives-derivation}).
\begin{lemma}
\label{lemma-first-order-thickening}
Let $X$ be a scheme over a base $S$. Consider a short exact sequence
$$
0 \to \mathcal{I} \to \mathcal{A} \to \mathcal{O}_X \to 0
$$
of sheaves on $X$ where $\mathcal{A}$ is a sheaf of
$f^{-1}\mathcal{O}_S$-algebras,
$\mathcal{A} \to \mathcal{O}_X$ is a surjection
of sheaves of $f^{-1}\mathcal{O}_S$-algebras, and $\mathcal{I}$ is its kernel.
If
\begin{enumerate}
\item $\mathcal{I}$ is an ideal of square zero in $\mathcal{A}$, and
\item $\mathcal{I}$ is quasi-coherent as an $\mathcal{O}_X$-module
\end{enumerate}
then $X' = (X, \mathcal{A})$ is a scheme and $X \to X'$ is a first
order thickening over $S$. Moreover, any first order thickening over
$S$ is of this form.
\end{lemma}
\begin{proof}
It is clear that $X'$ is a locally ringed space. Let $U = \Spec(B)$
be an affine open of $X$. Set $A = \Gamma(U, \mathcal{A})$. Note that
since $H^1(U, \mathcal{I}) = 0$ (see Cohomology of Schemes, Lemma
\ref{coherent-lemma-quasi-coherent-affine-cohomology-zero})
the map $A \to B$ is surjective. By assumption the kernel
$I = \mathcal{I}(U)$ is an ideal of square zero in the ring $A$.
By
Schemes, Lemma \ref{schemes-lemma-morphism-into-affine}
there is a canonical morphism of locally ringed spaces
$$
(U, \mathcal{A}|_U) \longrightarrow \Spec(A)
$$
coming from the map $B \to \Gamma(U, \mathcal{A})$. Since this morphism
fits into the commutative diagram
$$
\xymatrix{
(U, \mathcal{O}_X|_U) \ar[d] \ar[r] & \Spec(B) \ar[d] \\
(U, \mathcal{A}|_U) \ar[r] & \Spec(A)
}
$$
we see that it is a homeomorphism on underlying topological spaces.
Thus to see that it is an isomorphism, it suffices to check it induces
an isomorphism on the local rings.
For $u \in U$ corresponding to the prime $\mathfrak p \subset A$
we obtain a commutative diagram of short exact sequences
$$
\xymatrix{
0 \ar[r] &
I_{\mathfrak p} \ar[r] \ar[d] &
A_{\mathfrak p} \ar[r] \ar[d] &
B_{\mathfrak p} \ar[r] \ar[d] & 0 \\
0 \ar[r] &
\mathcal{I}_u \ar[r] &
\mathcal{A}_u \ar[r] &
\mathcal{O}_{X, u} \ar[r] & 0.
}
$$
The left and right vertical arrows are isomorphisms because
$\mathcal{I}$ and $\mathcal{O}_X$ are quasi-coherent sheaves.
Hence also the middle map is an isomorphism. Hence every point
of $X' = (X, \mathcal{A})$ has an affine neighbourhood and $X'$ is a
scheme as desired.
\end{proof}
\begin{lemma}
\label{lemma-thickening-affine-scheme}
Any thickening of an affine scheme is affine.
\end{lemma}
\begin{proof}
This is a special case of
Limits, Proposition \ref{limits-proposition-affine}.
\end{proof}
\begin{proof}[Proof for a finite order thickening]
Suppose that $X \subset X'$ is a finite order thickening with $X$ affine. Then
we may use Serre's criterion to prove $X'$ is affine. More precisely, we will
use Cohomology of Schemes, Lemma
\ref{coherent-lemma-quasi-compact-h1-zero-covering}. Let $\mathcal{F}$ be a
quasi-coherent $\mathcal{O}_{X'}$-module. It suffices to show that
$H^1(X', \mathcal{F}) = 0$. Denote $i : X \to X'$ the given closed immersion
and denote
$\mathcal{I} = \Ker(i^\sharp : \mathcal{O}_{X'} \to i_*\mathcal{O}_X)$.
By our discussion of finite order thickenings (following
Definition \ref{definition-thickening}) there exists an $n \geq 0$
and a filtration
$$
0 = \mathcal{F}_{n + 1} \subset \mathcal{F}_n \subset
\mathcal{F}_{n - 1} \subset \ldots \subset
\mathcal{F}_0 = \mathcal{F}
$$
by quasi-coherent submodules such that $\mathcal{F}_a/\mathcal{F}_{a + 1}$ is
annihilated by $\mathcal{I}$. Namely, we can take
$\mathcal{F}_a = \mathcal{I}^a\mathcal{F}$. Then
$\mathcal{F}_a/\mathcal{F}_{a + 1} = i_*\mathcal{G}_a$ for some quasi-coherent
$\mathcal{O}_X$-module $\mathcal{G}_a$, see Morphisms, Lemma
\ref{morphisms-lemma-i-star-equivalence}. We obtain
$$
H^1(X', \mathcal{F}_a/\mathcal{F}_{a + 1}) =
H^1(X', i_*\mathcal{G}_a) = H^1(X, \mathcal{G}_a) = 0
$$
The second equality comes from Cohomology of Schemes, Lemma
\ref{coherent-lemma-relative-affine-cohomology}
and the last equality from Cohomology of Schemes, Lemma
\ref{coherent-lemma-quasi-coherent-affine-cohomology-zero}.
Thus $\mathcal{F}$ has a finite filtration whose successive quotients
have vanishing first cohomology and it follows by a simple
induction argument that $H^1(X', \mathcal{F}) = 0$.
\end{proof}
\begin{lemma}
\label{lemma-base-change-thickening}
Let $S \subset S'$ be a thickening of schemes. Let $X' \to S'$ be a morphism
and set $X = S \times_{S'} X'$. Then $(X \subset X') \to (S \subset S')$
is a morphism of thickenings. If $S \subset S'$ is a first
(resp.\ finite order) thickening, then $X \subset X'$ is a first
(resp.\ finite order) thickening.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-composition-thickening}
If $S \subset S'$ and $S' \subset S''$ are thickenings, then so is
$S \subset S''$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-descending-property-thickening}
The property of being a thickening is fpqc local.
Similarly for first order thickenings.
\end{lemma}
\begin{proof}
The statement means the following: Let $X \to X'$ be a morphism
of schemes and let $\{g_i : X'_i \to X'\}$
be an fpqc covering such that the base change $X_i \to X'_i$
is a thickening for all $i$. Then $X \to X'$ is a thickening.
Since the morphisms $g_i$ are jointly surjective we conclude
that $X \to X'$ is surjective. By
Descent, Lemma \ref{descent-lemma-descending-property-closed-immersion}
we conclude that $X \to X'$ is a closed immersion.
Thus $X \to X'$ is a thickening. We omit the proof in the
case of first order thickenings.
\end{proof}
\section{Morphisms of thickenings}
\label{section-morphisms-thickenings}
\noindent
If $(f, f') : (X \subset X') \to (Y \subset Y')$ is a morphism
of thickenings of schemes, then often properties of the morphism
$f$ are inherited by $f'$. There are several variants.
\begin{lemma}
\label{lemma-thicken-property-morphisms}
Let $(f, f') : (X \subset X') \to (S \subset S')$ be a morphism
of thickenings. Then
\begin{enumerate}
\item $f$ is an affine morphism if and only if $f'$ is an affine morphism,
\item $f$ is a surjective morphism if and only if $f'$ is a surjective morphism,
\item $f$ is quasi-compact if and only if $f'$ quasi-compact,
\item $f$ is universally closed if and only if $f'$ is universally closed,
\item $f$ is integral if and only if $f'$ is integral,
\item $f$ is (quasi-)separated if and only if $f'$ is (quasi-)separated,
\item $f$ is universally injective if and only if $f'$ is universally injective,
\item $f$ is universally open if and only if $f'$ is universally open,
\item $f$ is quasi-affine if and only if $f'$ is quasi-affine, and
\item add more here.
\end{enumerate}
\end{lemma}
\begin{proof}
Observe that $S \to S'$ and $X \to X'$ are universal homeomorphisms
(see for example
Morphisms, Lemma \ref{morphisms-lemma-reduction-universal-homeomorphism}).
This immediately implies parts (2), (3), (4), (7), and (8).
Part (1) follows from Lemma \ref{lemma-thickening-affine-scheme}
which tells us that there is a 1-to-1 correspondence between
affine opens of $S$ and $S'$ and between affine opens of $X$ and $X'$.
Part (9) follows from
Limits, Lemma \ref{limits-lemma-thickening-quasi-affine}
and the remark just made about affine opens of $S$ and $S'$.
Part (5) follows from (1) and (4) by
Morphisms, Lemma \ref{morphisms-lemma-integral-universally-closed}.
Finally, note that
$$
S \times_X S = S \times_{X'} S \to S \times_{X'} S' \to S' \times_{X'} S'
$$
is a thickening (the two arrows are thickenings by
Lemma \ref{lemma-base-change-thickening}).
Hence applying (3) and (4) to the morphism
$(S \subset S') \to (S \times_X S \to S' \times_{X'} S')$
we obtain (6).
\end{proof}
\begin{lemma}
\label{lemma-thicken-property-relatively-ample}
Let $(f, f') : (X \subset X') \to (S \subset S')$ be a morphism
of thickenings. Let $\mathcal{L}'$ be an invertible sheaf on $X'$
and denote $\mathcal{L}$ the restriction to $X$.
Then $\mathcal{L}'$ is $f'$-ample if and only if
$\mathcal{L}$ is $f$-ample.
\end{lemma}
\begin{proof}
Recall that being relatively ample is a condition for each
affine open in the base, see
Morphisms, Definition \ref{morphisms-definition-relatively-ample}.
By Lemma \ref{lemma-thickening-affine-scheme}
there is a 1-to-1 correspondence between
affine opens of $S$ and $S'$.
Thus we may assume $S$ and $S'$ are affine
and we reduce to proving that
$\mathcal{L}'$ is ample if and only if
$\mathcal{L}$ is ample.
This is Limits, Lemma \ref{limits-lemma-ample-on-reduction}.
\end{proof}
\begin{lemma}
\label{lemma-thicken-property-morphisms-cartesian}
Let $(f, f') : (X \subset X') \to (S \subset S')$ be a morphism
of thickenings such that $X = S \times_{S'} X'$. If $S \subset S'$
is a finite order thickening, then
\begin{enumerate}
\item $f$ is a closed immersion if and only if $f'$ is a closed immersion,
\item $f$ is locally of finite type if and only if $f'$ is
locally of finite type,
\item $f$ is locally quasi-finite if and only if $f'$ is locally
quasi-finite,
\item $f$ is locally of finite type of relative dimension $d$ if and
only if $f'$ is locally of finite type of relative dimension $d$,
\item $\Omega_{X/S} = 0$ if and only if $\Omega_{X'/S'} = 0$,
\item $f$ is unramified if and only if $f'$ is unramified,
\item $f$ is proper if and only if $f'$ is proper,
\item $f$ is finite if and only if $f'$ is finite,
\item $f$ is a monomorphism if and only if $f'$ is a monomorphism,
\item $f$ is an immersion if and only if $f'$ is an immersion, and
\item add more here.
\end{enumerate}
\end{lemma}
\begin{proof}
The properties $\mathcal{P}$ listed in the lemma are all stable
under base change, hence if $f'$ has property $\mathcal{P}$, then so
does $f$. See
Schemes, Lemmas \ref{schemes-lemma-base-change-immersion} and
\ref{schemes-lemma-base-change-monomorphism}
and
Morphisms, Lemmas
\ref{morphisms-lemma-base-change-finite-type},
\ref{morphisms-lemma-base-change-quasi-finite},
\ref{morphisms-lemma-base-change-relative-dimension-d},
\ref{morphisms-lemma-base-change-differentials},
\ref{morphisms-lemma-base-change-unramified},
\ref{morphisms-lemma-base-change-proper}, and
\ref{morphisms-lemma-base-change-finite}.
\medskip\noindent
The interesting direction in each case is therefore to assume
that $f$ has the property and deduce that $f'$ has it too.
By induction on the order of the thickening we may
assume that $S \subset S'$ is a first order thickening, see
discussion immediately following
Definition \ref{definition-thickening}.
\medskip\noindent
Most of the proofs will use a reduction to the affine case. Let
$U' \subset S'$ be an affine open and let $V' \subset X'$ be an affine open
lying over $U'$. Let $U' = \Spec(A')$ and denote $I \subset A'$ be the ideal
defining the closed subscheme $U' \cap S$. Say $V' = \Spec(B')$.
Then $V' \cap X = \Spec(B'/IB')$. Setting $A = A'/I$ and
$B = B'/IB'$ we get a commutative diagram
$$
\xymatrix{
0 \ar[r] &
IB' \ar[r] &
B' \ar[r] &
B \ar[r] & 0 \\
0 \ar[r] &
IA' \ar[r] \ar[u] &
A' \ar[r] \ar[u] &
A \ar[r] \ar[u] & 0
}
$$
with exact rows and $I^2 = 0$.
\medskip\noindent
The translation of (1) into algebra: If $A \to B$ is surjective,
then $A' \to B'$ is surjective. This follows from
Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}).
\medskip\noindent
The translation of (2) into algebra: If $A \to B$ is a finite type ring
map, then $A' \to B'$ is a finite type ring map. This follows from
Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK})
applied to a map $A'[x_1, \ldots, x_n] \to B'$ such that
$A[x_1, \ldots, x_n] \to B$ is surjective.
\medskip\noindent
Proof of (3). Follows from (2) and that quasi-finiteness of a morphism
which is locally of finite type can be checked on fibres, see
Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-point-characterize}.
\medskip\noindent
Proof of (4). Follows from (2) and that the additional property of ``being of
relative dimension $d$'' can be checked on fibres (by definition, see
Morphisms, Definition \ref{morphisms-definition-relative-dimension-d}.
\medskip\noindent
The translation of (5) into algebra: If $\Omega_{B/A} = 0$, then
$\Omega_{B'/A'} = 0$. By
Algebra, Lemma \ref{algebra-lemma-differentials-base-change}
we have $0 = \Omega_{B/A} = \Omega_{B'/A'}/I\Omega_{B'/A'}$.
Hence $\Omega_{B'/A'} = 0$ by
Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}).
\medskip\noindent
The translation of (6) into algebra: If $A \to B$ is unramified
map, then $A' \to B'$ is unramified. Since $A \to B$ is of finite
type we see that $A' \to B'$ is of finite type by (2) above.
Since $A \to B$ is unramified we have $\Omega_{B/A} = 0$. By
part (5) we have $\Omega_{B'/A'} = 0$. Thus $A' \to B'$ is unramified.
\medskip\noindent
Proof of (7). Follows by combining (2) with
results of Lemma \ref{lemma-thicken-property-morphisms}
and the fact that proper equals quasi-compact $+$
separated $+$ locally of finite type $+$ universally closed.
\medskip\noindent
Proof of (8). Follows by combining (2) with
results of Lemma \ref{lemma-thicken-property-morphisms}
and using the fact that finite equals integral $+$ locally
of finite type (Morphisms, Lemma \ref{morphisms-lemma-finite-integral}).
\medskip\noindent
Proof of (9). As $f$ is a monomorphism we have $X = X \times_S X$.
We may apply the results proved so far to the morphism of thickenings
$(X \subset X') \to (X \times_S X \subset X' \times_{S'} X')$.
We conclude $X' \to X' \times_{S'} X'$ is a closed immersion by (1).
In fact, it is a first order thickening as the ideal defining the
closed immersion
$X' \to X' \times_{S'} X'$ is contained in the pullback of the ideal
$\mathcal{I} \subset \mathcal{O}_{S'}$ cutting out $S$ in $S'$.
Indeed, $X = X \times_S X = (X' \times_{S'} X') \times_{S'} S$
is contained in $X'$. Hence by
Morphisms, Lemma \ref{morphisms-lemma-differentials-diagonal}
it suffices to show that
$\Omega_{X'/S'} = 0$ which follows from (5)
and the corresponding statement for $X/S$.
\medskip\noindent
Proof of (10). If $f : X \to S$ is an immersion, then it factors as
$X \to U \to S$ where $U \to S$ is an open immersion and $X \to U$ is a
closed immersion. Let $U' \subset S'$ be the open subscheme whose
underlying topological space is the same as $U$. Then $X' \to S'$
factors through $U'$ and we conclude that $X' \to U'$ is a closed
immersion by part (1). This finishes the proof.
\end{proof}
\noindent
The following lemma is a variant on the preceding one. Rather than assume
that the thickenings involved are finite order (which allows us to transfer
the property of being locally of finite type from $f$ to $f'$),
we instead take as given that each of $f$ and $f'$ is locally of
finite type.
\begin{lemma}
\label{lemma-properties-that-extend-over-thickenings}
Let $(f, f') : (X \subset X') \to (Y \to Y')$ be a morphism
of thickenings. Assume $f$ and $f'$ are locally of finite type
and $X = Y \times_{Y'} X'$. Then
\begin{enumerate}
\item $f$ is locally quasi-finite if and only if $f'$ is locally quasi-finite,
\item $f$ is finite if and only if $f'$ is finite,
\item $f$ is a closed immersion if and only if $f'$ is a closed immersion,
\item $\Omega_{X/Y} = 0$ if and only if $\Omega_{X'/Y'} = 0$,
\item $f$ is unramified if and only if $f'$ is unramified,
\item $f$ is a monomorphism if and only if $f'$ is a monomorphism,
\item $f$ is an immersion if and only if $f'$ is an immersion,
\item $f$ is proper if and only if $f'$ is proper, and
\item add more here.
\end{enumerate}
\end{lemma}
\begin{proof}
The properties $\mathcal{P}$ listed in the lemma are all stable
under base change, hence if $f'$ has property $\mathcal{P}$, then so
does $f$. See
Schemes, Lemmas \ref{schemes-lemma-base-change-immersion} and
\ref{schemes-lemma-base-change-monomorphism}
and
Morphisms, Lemmas
\ref{morphisms-lemma-base-change-quasi-finite},
\ref{morphisms-lemma-base-change-relative-dimension-d},
\ref{morphisms-lemma-base-change-differentials},
\ref{morphisms-lemma-base-change-unramified},
\ref{morphisms-lemma-base-change-proper}, and
\ref{morphisms-lemma-base-change-finite}.
Hence in each case we need only to prove that if $f$ has
the desired property, so does $f'$.
\medskip\noindent
A morphism is locally quasi-finite if and only if it is locally
of finite type and the scheme theoretic fibres are discrete spaces, see
Morphisms, Lemma \ref{morphisms-lemma-locally-quasi-finite-fibres}.
Since the underlying topological space is unchanged by
passing to a thickening, we see that $f'$ is locally quasi-finite if
(and only if) $f$ is. This proves (1).
\medskip\noindent
Case (2) follows from case (5) of Lemma \ref{lemma-thicken-property-morphisms}
and the fact that the finite morphisms are precisely
the integral morphisms that are locally of finite type
(Morphisms, Lemma \ref{morphisms-lemma-finite-integral}).
\medskip\noindent
Case (3). This follows immediately from
Morphisms, Lemma \ref{morphisms-lemma-check-closed-infinitesimally}.
\medskip\noindent
Case (4) follows from the following algebra statement: Let $A$ be a ring and
let $I \subset A$ be a locally nilpotent ideal. Let $B$ be a finite type
$A$-algebra. If $\Omega_{(B/IB)/(A/I)} = 0$, then $\Omega_{B/A} = 0$.
Namely, the assumption means that $I\Omega_{B/A} = 0$, see
Algebra, Lemma \ref{algebra-lemma-differentials-base-change}.
On the other hand $\Omega_{B/A}$ is a finite $B$-module, see
Algebra, Lemma \ref{algebra-lemma-differentials-finitely-generated}.
Hence the vanishing of $\Omega_{B/A}$ follows from Nakayama's
lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) and the fact
that $IB$ is contained in the radical of $B$.
\medskip\noindent
Case (5) follows immediately from (4) and
Morphisms, Lemma \ref{morphisms-lemma-unramified-omega-zero}.
\medskip\noindent
Proof of (6). As $f$ is a monomorphism we have $X = X \times_Y X$.
We may apply the results proved so far to the morphism of thickenings
$(X \subset X') \to (X \times_Y X \subset X' \times_{Y'} X')$.
We conclude $\Delta_{X'/Y'} : X' \to X' \times_{Y'} X'$
is a closed immersion by (3). In fact $\Delta_{X'/Y'}$ is a bijection on
underlying sets, hence $\Delta_{X'/Y'}$ is a thickening. On the other hand
$\Delta_{X'/Y'}$ is locally of finite presentation by
Morphisms, Lemma \ref{morphisms-lemma-diagonal-morphism-finite-type}.
In other words, $\Delta_{X'/Y'}(X')$ is cut out by
a quasi-coherent sheaf of ideals
$\mathcal{J} \subset \mathcal{O}_{X' \times_{Y'} X'}$ of finite type.
Since $\Omega_{X'/Y'} = 0$ by (5) we see that
the conormal sheaf of $X' \to X' \times_{Y'} X'$ is zero by
Morphisms, Lemma \ref{morphisms-lemma-differentials-diagonal}.
In other words, $\mathcal{J}/\mathcal{J}^2 = 0$.
This implies $\Delta_{X'/Y'}$ is an isomorphism, for example
by Algebra, Lemma \ref{algebra-lemma-ideal-is-squared-union-connected}.
\medskip\noindent
Proof of (7). If $f : X \to Y$ is an immersion, then it factors as
$X \to V \to Y$ where $V \to Y$ is an open immersion and $X \to V$ is a
closed immersion. Let $V' \subset Y'$ be the open subscheme whose
underlying topological space is the same as $V$. Then $X' \to V'$
factors through $V'$ and we conclude that $X' \to V'$ is a closed
immersion by part (3).
\medskip\noindent
Case (8) follows from Lemma \ref{lemma-thicken-property-morphisms}
and the definition of proper morphisms as being the quasi-compact,
universally closed, and separated morphisms that are locally of finite type.
\end{proof}
\section{Picard groups of thickenings}
\label{section-picard-group-thickening}
\noindent
Some material on Picard groups of thickenings.
\begin{lemma}
\label{lemma-picard-group-first-order-thickening}
Let $X \subset X'$ be a first order thickening
with ideal sheaf $\mathcal{I}$. Then there is a canonical
exact sequence
$$
\xymatrix{
0 \ar[r] &
H^0(X, \mathcal{I}) \ar[r] &
H^0(X', \mathcal{O}_{X'}^*) \ar[r] &
H^0(X, \mathcal{O}^*_X) \ar `r[d] `d[l] `l[llld] `d[dll] [dll] \\
& H^1(X, \mathcal{I}) \ar[r] &
\text{Pic}(X') \ar[r] &
\text{Pic}(X) \ar `r[d] `d[l] `l[llld] `d[dll] [dll] \\
& H^2(X, \mathcal{I}) \ar[r] & \ldots \ar[r] & \ldots
}
$$
of abelian groups.
\end{lemma}
\begin{proof}
This is the long exact cohomology sequence associated to the
short exact sequence of sheaves of abelian groups
$$
0 \to \mathcal{I} \to \mathcal{O}_{X'}^* \to \mathcal{O}_X^* \to 0
$$
where the first map sends a local section $f$ of $\mathcal{I}$
to the invertible section $1 + f$ of $\mathcal{O}_{X'}$.
We also use the identification of the Picard group of a
ringed space with the first cohomology group of the sheaf
of invertible functions, see
Cohomology, Lemma \ref{cohomology-lemma-h1-invertible}.
\end{proof}
\begin{lemma}
\label{lemma-torsion-pic-thickening}
Let $X \subset X'$ be a thickening. Let $n$ be an integer
invertible in $\mathcal{O}_X$. Then the map
$\text{Pic}(X')[n] \to \text{Pic}(X)[n]$ is bijective.
\end{lemma}
\begin{proof}[Proof for a finite order thickening]
By the general principle explained following
Definition \ref{definition-thickening}
this reduces to the case of a first order thickening.
Then may use Lemma \ref{lemma-picard-group-first-order-thickening}
to see that it suffices to show that
$H^1(X, \mathcal{I})[n]$, $H^1(X, \mathcal{I})/n$, and
$H^2(X, \mathcal{I})[n]$ are zero.
This follows as multiplication by $n$ on $\mathcal{I}$
is an isomorphism as it is an $\mathcal{O}_X$-module.
\end{proof}
\begin{proof}[Proof in general]
Let $\mathcal{I} \subset \mathcal{O}_{X'}$ be the quasi-coherent ideal
sheaf cutting out $X$. Then we have a short exact sequence of
abelian groups
$$
0 \to (1 + \mathcal{I})^* \to \mathcal{O}_{X'}^* \to \mathcal{O}_X^* \to 0
$$
We obtain a long exact cohomology sequence as in the statement of
Lemma \ref{lemma-picard-group-first-order-thickening}
with $H^i(X, \mathcal{I})$ replaced by $H^i(X, (1 + \mathcal{I})^*)$.
Thus it suffices to show that raising to the $n$th power is an
isomorphism $(1 + \mathcal{I})^* \to (1 + \mathcal{I})^*$.
Taking sections over affine opens this follows from
Algebra, Lemma \ref{algebra-lemma-lift-nth-roots}.
\end{proof}
\section{First order infinitesimal neighbourhood}
\label{section-first-order-infinitesimal-neighbourhood}
\noindent
A natural construction of first order thickenings is the following.
Suppose that $i : Z \to X$ be an immersion of schemes. Choose an
open subscheme $U \subset X$ such that $i$ identifies $Z$ with a closed
subscheme $Z \subset U$. Let $\mathcal{I} \subset \mathcal{O}_U$ be the
quasi-coherent sheaf of ideals defining $Z$ in $U$. Then we can consider
the closed subscheme $Z' \subset U$ defined by the quasi-coherent sheaf
of ideals $\mathcal{I}^2$.
\begin{definition}
\label{definition-first-order-infinitesimal-neighbourhood}
Let $i : Z \to X$ be an immersion of schemes. The
{\it first order infinitesimal neighbourhood} of $Z$ in $X$ is
the first order thickening $Z \subset Z'$ over $X$ described above.
\end{definition}
\noindent
This thickening has the following universal property (which will assuage
any fears that the construction above depends on the choice of the open
$U$).
\begin{lemma}
\label{lemma-first-order-infinitesimal-neighbourhood}
Let $i : Z \to X$ be an immersion of schemes. The first order infinitesimal
neighbourhood $Z'$ of $Z$ in $X$ has the following universal property:
Given any commutative diagram
$$
\xymatrix{
Z \ar[d]_i & T \ar[l]^a \ar[d] \\
X & T' \ar[l]_b
}
$$
where $T \subset T'$ is a first order thickening over $X$, there exists
a unique morphism $(a', a) : (T \subset T') \to (Z \subset Z')$ of
thickenings over $X$.
\end{lemma}
\begin{proof}
Let $U \subset X$ be the open used in the construction of $Z'$, i.e., an
open such that $Z$ is identified with a closed subscheme of $U$ cut out by
the quasi-coherent sheaf of ideals $\mathcal{I}$.
Since $|T| = |T'|$ we see that $b(T') \subset U$. Hence we can
think of $b$ as a morphism into $U$. Let $\mathcal{J} \subset \mathcal{O}_{T'}$
be the ideal cutting out $T$. Since $b(T) \subset Z$ by the diagram above
we see that $b^\sharp(b^{-1}\mathcal{I}) \subset \mathcal{J}$. As
$T'$ is a first order thickening of $T$ we see that $\mathcal{J}^2 = 0$
hence $b^\sharp(b^{-1}(\mathcal{I}^2)) = 0$. By
Schemes, Lemma \ref{schemes-lemma-characterize-closed-subspace}
this implies that $b$ factors through $Z'$. Denote $a' : T' \to Z'$
this factorization and everything is clear.
\end{proof}
\begin{lemma}
\label{lemma-infinitesimal-neighbourhood-conormal}
Let $i : Z \to X$ be an immersion of schemes. Let $Z \subset Z'$ be
the first order infinitesimal neighbourhood of $Z$ in $X$.
Then the diagram
$$
\xymatrix{
Z \ar[r] \ar[d] & Z' \ar[d] \\
Z \ar[r] & X
}
$$
induces a map of conormal sheaves $\mathcal{C}_{Z/X} \to \mathcal{C}_{Z/Z'}$ by
Morphisms, Lemma \ref{morphisms-lemma-conormal-functorial}.
This map is an isomorphism.
\end{lemma}
\begin{proof}
This is clear from the construction of $Z'$ above.
\end{proof}
\section{Formally unramified morphisms}
\label{section-formally-unramified}
\noindent
Recall that a ring map $R \to A$ is called {\it formally unramified}
(see Algebra, Definition \ref{algebra-definition-formally-unramified})
if for every commutative solid diagram
$$
\xymatrix{
A \ar[r] \ar@{-->}[rd] & B/I \\
R \ar[r] \ar[u] & B \ar[u]
}
$$
where $I \subset B$ is an ideal of square zero, at most one dotted
arrow exists which makes the diagram commute. This motivates
the following analogue for morphisms of schemes.
\begin{definition}
\label{definition-formally-unramified}
Let $f : X \to S$ be a morphism of schemes.
We say $f$ is {\it formally unramified} if given any solid commutative diagram
$$
\xymatrix{
X \ar[d]_f & T \ar[d]^i \ar[l] \\
S & T' \ar[l] \ar@{-->}[lu]
}
$$
where $T \subset T'$ is a first order thickening of affine schemes over $S$
there exists at most one dotted arrow making the diagram commute.
\end{definition}
\noindent
We first prove some formal lemmas, i.e., lemmas which can be proved by
drawing the corresponding diagrams.
\begin{lemma}
\label{lemma-formally-unramified-not-affine}
If $f : X \to S$ is a formally unramified morphism, then given
any solid commutative diagram
$$
\xymatrix{
X \ar[d]_f & T \ar[d]^i \ar[l] \\
S & T' \ar[l] \ar@{-->}[lu]
}
$$
where $T \subset T'$ is a first order thickening of schemes over $S$
there exists at most one dotted arrow making the diagram commute.
In other words, in
Definition \ref{definition-formally-unramified}
the condition that $T$ be affine may be dropped.
\end{lemma}
\begin{proof}
This is true because a morphism is determined by its restrictions
to affine opens.
\end{proof}
\begin{lemma}
\label{lemma-composition-formally-unramified}
A composition of formally unramified morphisms is formally unramified.
\end{lemma}
\begin{proof}
This is formal.
\end{proof}
\begin{lemma}
\label{lemma-base-change-formally-unramified}
A base change of a formally unramified morphism is formally unramified.
\end{lemma}
\begin{proof}
This is formal.
\end{proof}
\begin{lemma}
\label{lemma-formally-unramified-on-opens}
Let $f : X \to S$ be a morphism of schemes.
Let $U \subset X$ and $V \subset S$ be open such that
$f(U) \subset V$. If $f$ is formally unramified, so is $f|_U : U \to V$.
\end{lemma}
\begin{proof}
Consider a solid diagram
$$
\xymatrix{
U \ar[d]_{f|_U} & T \ar[d]^i \ar[l]^a \\
V & T' \ar[l] \ar@{-->}[lu]
}
$$
as in Definition \ref{definition-formally-unramified}. If $f$ is formally
ramified, then there exists at most one
$S$-morphism $a' : T' \to X$ such that $a'|_T = a$.
Hence clearly there exists at most one such morphism into $U$.
\end{proof}
\begin{lemma}
\label{lemma-affine-formally-unramified}
Let $f : X \to S$ be a morphism of schemes.
Assume $X$ and $S$ are affine.
Then $f$ is formally unramified if and only if
$\mathcal{O}_S(S) \to \mathcal{O}_X(X)$ is a formally unramified
ring map.
\end{lemma}
\begin{proof}
This is immediate from the definitions
(Definition \ref{definition-formally-unramified} and
Algebra, Definition \ref{algebra-definition-formally-unramified})
by the equivalence of categories of rings and affine schemes,
see
Schemes, Lemma \ref{schemes-lemma-category-affine-schemes}.
\end{proof}
\noindent
Here is a characterization in terms of the sheaf of differentials.
\begin{lemma}
\label{lemma-formally-unramified-differentials}
Let $f : X \to S$ be a morphism of schemes.
Then $f$ is formally unramified if and only if $\Omega_{X/S} = 0$.
\end{lemma}
\begin{proof}
We give two proofs.
\medskip\noindent
First proof. It suffices to show that $\Omega_{X/S}$ is zero on the members of
an affine open covering of $X$. Choose an affine open $U \subset X$
with $f(U) \subset V$ where $V \subset S$ is an affine open of $S$. By
Lemma \ref{lemma-formally-unramified-on-opens}
the restriction $f_U : U \to V$ is formally unramified. By
Morphisms, Lemma \ref{morphisms-lemma-differentials-affine}
we see that $\Omega_{X/S}|_U$ is the quasi-coherent sheaf associated to
the $\mathcal{O}_X(U)$-module $\Omega_{\mathcal{O}_X(U)/\mathcal{O}_S(V)}$. By
Lemma \ref{lemma-affine-formally-unramified}
we see that $\mathcal{O}_S(V) \to \mathcal{O}_X(U)$ is a formally unramified
ring map. Hence by
Algebra, Lemma \ref{algebra-lemma-characterize-formally-unramified}
we conclude that $\Omega_{X/S}|_U = 0$ as desired.
\medskip\noindent
Second proof. We recall some of the arguments of the proof of
Morphisms, Lemma \ref{morphisms-lemma-differentials-affine}.
Let $W \subset X \times_S X$ be an open such that
$\Delta : X \to X \times_S X$ induces a closed immersion into $W$.
Let $\mathcal{J} \subset \mathcal{O}_W$ be the ideal sheaf of this
closed immersion. Let $X' \subset W$ be the closed subscheme
defined by the quasi-coherent sheaf of ideals $\mathcal{J}^2$.
Consider the two morphisms $p_1, p_2 : X' \to X$ induced by
the two projections $X \times_S X \to X$.
Note that $p_1$ and $p_2$ agree when composed with $\Delta : X \to X'$
and that $X \to X'$ is a closed immersion defined by a an ideal
whose square is zero. Moreover there is a short exact sequence
$$
0 \to \mathcal{J}/\mathcal{J}^2 \to \mathcal{O}_{X'} \to \mathcal{O}_X \to 0
$$
and $\Omega_{X/S} = \mathcal{J}/\mathcal{J}^2$. Moreover,
$\mathcal{J}/\mathcal{J}^2$ is generated by the local
sections $p_1^\sharp(f) - p_2^\sharp(f)$ for $f$ a local section of
$\mathcal{O}_X$.
\medskip\noindent
Suppose that $f : X \to S$ is formally unramified.
By assumption this means that $p_1 = p_2$ when restricted to any
affine open $T' \subset X'$. Hence $p_1 = p_2$. By what was said above
we conclude that $\Omega_{X/S} = \mathcal{J}/\mathcal{J}^2 = 0$.
\medskip\noindent
Conversely, suppose that $\Omega_{X/S} = 0$. Then $X' = X$. Take any pair
of morphisms $f'_1, f'_2 : T' \to X$ fitting as dotted arrows in
the diagram of
Definition \ref{definition-formally-unramified}.
This gives a morphism $(f'_1, f'_2) : T' \to X \times_S X$.
Since $f'_1|_T = f'_2|_T$ and $|T| =|T'|$ we see that the image of $T'$
under $(f'_1, f'_2)$ is contained in the open $W$ chosen above. Since
$(f'_1, f'_2)(T) \subset \Delta(X)$ and since $T$ is defined by an ideal
of square zero in $T'$ we see that $(f'_1, f'_2)$ factors through $X'$.
As $X' = X$ we conclude $f_1' = f'_2$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-unramified-formally-unramified}
\begin{slogan}
Unramified morphisms are the same as formally unramified morphism that
are locally of finite type.
\end{slogan}
Let $f : X \to S$ be a morphism of schemes.
The following are equivalent:
\begin{enumerate}
\item The morphism $f$ is unramified (resp.\ G-unramified), and
\item the morphism $f$ is locally of finite type (resp.\ locally of finite
presentation) and formally unramified.
\end{enumerate}
\end{lemma}
\begin{proof}
Use Lemma \ref{lemma-formally-unramified-differentials} and
Morphisms, Lemma \ref{morphisms-lemma-unramified-omega-zero}.
\end{proof}
\section{Universal first order thickenings}
\label{section-universal-thickening}
\noindent
Let $h : Z \to X$ be a morphism of schemes. A {\it universal first order
thickening} of $Z$ over $X$ is a first order thickening $Z \subset Z'$
over $X$ such that given any first order thickening $T \subset T'$
over $X$ and a solid commutative diagram
$$
\xymatrix{
& Z \ar[ld] & & T \ar[rd] \ar[ll]^a \\
Z' \ar[rrd] & & & & T' \ar@{..>}[llll]_{a'} \ar[lld]^b \\
& & X
}
$$
there exists a unique dotted arrow making the diagram commute.
Note that in this situation $(a, a') : (T \subset T') \to (Z \subset Z')$
is a morphism of thickenings over $X$. Thus if a universal first order
thickening exists, then it is unique up to unique isomorphism.
In general a universal first order thickening
does not exist, but if $h$ is formally unramified then it does.
\begin{lemma}
\label{lemma-universal-thickening}
Let $h : Z \to X$ be a formally unramified morphism of schemes.
There exists a universal first order thickening $Z \subset Z'$ of
$Z$ over $X$.
\end{lemma}
\begin{proof}
During this proof we will say $Z \subset Z'$ is a universal first order
thickening of $Z$ over $X$ if it satisfies the condition of the lemma.
We will construct the universal first order thickening $Z \subset Z'$ over $X$
by glueing, starting with the affine case which is
Algebra, Lemma \ref{algebra-lemma-universal-thickening}.
We begin with some general remarks.
\medskip\noindent
If a universal first order thickening of $Z$ over $X$ exists, then it is unique
up to unique isomorphism. Moreover, suppose that $V \subset Z$ and
$U \subset X$ are open subschemes such that $h(V) \subset U$. Let
$Z \subset Z'$ be a universal first order thickening of $Z$ over $X$.
Let $V' \subset Z'$ be the open subscheme such that $V = Z \cap V'$.
Then we claim that $V \subset V'$ is the universal first order thickening of
$V$ over $U$. Namely, suppose given any diagram
$$
\xymatrix{
V \ar[d]_h & T \ar[l]^a \ar[d] \\
U & T' \ar[l]_b
}
$$
where $T \subset T'$ is a first order thickening over $U$. By the universal
property of $Z'$ we obtain $(a, a') : (T \subset T') \to (Z \subset Z')$.
But since we have equality $|T| = |T'|$ of underlying topological spaces
we see that $a'(T') \subset V'$. Hence we may think of $(a, a')$
as a morphism of thickenings $(a, a') : (T \subset T') \to (V \subset V')$
over $U$. Uniqueness is clear also. In a completely similar manner one proves
that if $h(Z) \subset U$ and $Z \subset Z'$ is a universal first order
thickening over $U$, then $Z \subset Z'$ is a universal first order thickening
over $X$.
\medskip\noindent
Before we glue affine pieces let us show that the lemma holds if
$Z$ and $X$ are affine. Say $X = \Spec(R)$ and $Z = \Spec(S)$. By
Algebra, Lemma \ref{algebra-lemma-universal-thickening}
there exists a first order thickening $Z \subset Z'$ over $X$
which has the universal property of the lemma for diagrams
$$
\xymatrix{
Z \ar[d]_h & T \ar[l]^a \ar[d] \\
X & T' \ar[l]_b
}
$$
where $T, T'$ are affine. Given a general diagram we can choose an affine
open covering $T' = \bigcup T'_i$ and we obtain morphisms
$a'_i : T'_i \to Z'$ over $X$ such that $a'_i|_{T_i} = a|_{T_i}$.
By uniqueness we see that $a'_i$ and $a'_j$ agree on any affine open
of $T'_i \cap T'_j$. Hence the morphisms $a'_i$ glue to a global morphism
$a' : T' \to Z'$ over $X$ as desired. Thus the lemma holds if $X$ and $Z$
are affine.
\medskip\noindent
Choose an affine open covering $Z = \bigcup Z_i$ such that each $Z_i$
maps into an affine open $U_i$ of $X$. By
Lemma \ref{lemma-formally-unramified-on-opens}
the morphisms $Z_i \to U_i$ are formally unramified.
Hence by the affine case we obtain universal first order thickenings
$Z_i \subset Z_i'$ over $U_i$. By the general remarks above
$Z_i \subset Z_i'$ is also a universal first order thickening of
$Z_i$ over $X$. Let $Z'_{i, j} \subset Z'_i$ be the open subscheme
such that $Z_i \cap Z_j = Z'_{i, j} \cap Z_i$. By the general remarks
we see that both $Z'_{i, j}$ and $Z'_{j, i}$ are universal first
order thickenings of $Z_i \cap Z_j$ over $X$. Thus, by
the first of our general remarks, we see that there is a canonical isomorphism
$\varphi_{ij} : Z'_{i, j} \to Z'_{j, i}$ inducing the identity on
$Z_i \cap Z_j$. We claim that these morphisms satisfy the cocycle condition of
Schemes, Section \ref{schemes-section-glueing-schemes}.
(Verification omitted. Hint: Use that $Z'_{i, j} \cap Z'_{i, k}$ is the
universal first order thickening of $Z_i \cap Z_j \cap Z_k$ which determines
it up to unique isomorphism by what was said above.)
Hence we can use the results of
Schemes, Section \ref{schemes-section-glueing-schemes}
to get a first order thickening $Z \subset Z'$ over $X$ which the property
that the open subscheme $Z'_i \subset Z'$ with $Z_i = Z'_i \cap Z$
is a universal first order thickening of $Z_i$ over $X$.
\medskip\noindent
It turns out that this implies formally that $Z'$ is a universal first order
thickening of $Z$ over $X$. Namely, we have the universal property for any
diagram
$$
\xymatrix{
Z \ar[d]_h & T \ar[l]^a \ar[d] \\
X & T' \ar[l]_b
}
$$
where $a(T)$ is contained in some $Z_i$. Given a general diagram we can
choose an open covering $T' = \bigcup T'_i$ such that $a(T_i) \subset Z_i$.
We obtain morphisms $a'_i : T'_i \to Z'$ over $X$ such that
$a'_i|_{T_i} = a|_{T_i}$. We see that $a'_i$ and $a'_j$ necessarily agree
on $T'_i \cap T'_j$ since both $a'_i|_{T'_i \cap T'_j}$ and
$a'_j|_{T'_i \cap T'_j}$ are solutions of the problem of mapping into the
universal first order thickening $Z'_i \cap Z'_j$ of $Z_i \cap Z_j$ over $X$.
Hence the morphisms $a'_i$ glue to a global morphism
$a' : T' \to Z'$ over $X$ as desired. This finishes the proof.
\end{proof}
\begin{definition}
\label{definition-universal-thickening}
Let $h : Z \to X$ be a formally unramified morphism of schemes.
\begin{enumerate}
\item The {\it universal first order thickening} of $Z$ over $X$
is the thickening $Z \subset Z'$ constructed in
Lemma \ref{lemma-universal-thickening}.
\item The {\it conormal sheaf of $Z$ over $X$} is the conormal sheaf
of $Z$ in its universal first order thickening $Z'$ over $X$.
\end{enumerate}
We often denote the conormal sheaf $\mathcal{C}_{Z/X}$ in this situation.
\end{definition}
\noindent
Thus we see that there is a short exact sequence of sheaves
$$
0 \to \mathcal{C}_{Z/X} \to \mathcal{O}_{Z'} \to \mathcal{O}_Z \to 0
$$
on $Z$.
The following lemma proves that there is no conflict between this definition
and the definition in case $Z \to X$ is an immersion.
\begin{lemma}
\label{lemma-immersion-universal-thickening}
Let $i : Z \to X$ be an immersion of schemes. Then
\begin{enumerate}
\item $i$ is formally unramified,
\item the universal first order thickening of $Z$ over $X$ is the first order
infinitesimal neighbourhood of $Z$ in $X$ of
Definition \ref{definition-first-order-infinitesimal-neighbourhood}, and
\item the conormal sheaf of $i$ in the sense of
Morphisms, Definition \ref{morphisms-definition-conormal-sheaf}
agrees with the conormal sheaf of $i$ in the sense of
Definition \ref{definition-universal-thickening}.
\end{enumerate}
\end{lemma}
\begin{proof}
By
Morphisms, Lemmas \ref{morphisms-lemma-open-immersion-unramified} and
\ref{morphisms-lemma-closed-immersion-unramified}
an immersion is unramified, hence formally unramified by
Lemma \ref{lemma-unramified-formally-unramified}.
The other assertions follow by combining
Lemmas \ref{lemma-first-order-infinitesimal-neighbourhood} and
\ref{lemma-infinitesimal-neighbourhood-conormal}
and the definitions.
\end{proof}
\begin{lemma}
\label{lemma-universal-thickening-unramified}
Let $Z \to X$ be a formally unramified morphism of schemes.
Then the universal first order thickening $Z'$ is formally
unramified over $X$.
\end{lemma}
\begin{proof}
There are two proofs. The first is to show that $\Omega_{Z'/X} = 0$
by working affine locally and applying
Algebra, Lemma \ref{algebra-lemma-differentials-universal-thickening}.
Then
Lemma \ref{lemma-formally-unramified-differentials}
implies what we want.
The second is a direct argument as follows.
\medskip\noindent
Let $T \subset T'$ be a first order thickening. Let
$$
\xymatrix{
Z' \ar[d] & T \ar[l]^c \ar[d] \\
X & T' \ar[l] \ar[lu]^{a, b}
}
$$
be a commutative diagram. Consider two morphisms $a, b : T' \to Z'$
fitting into the diagram. Set $T_0 = c^{-1}(Z) \subset T$ and
$T'_a = a^{-1}(Z)$ (scheme theoretically).
Since $Z'$ is a first order thickening of $Z$, we see that $T'$
is a first order thickening of $T'_a$. Moreover, since $c = a|_T$ we see that
$T_0 = T \cap T'_a$ (scheme theoretically). As $T'$ is a first order
thickening of $T$ it follows that $T'_a$
is a first order thickening of $T_0$. Now $a|_{T'_a}$ and $b|_{T'_a}$
are morphisms of $T'_a$ into $Z'$ over $X$ which agree on $T_0$ as
morphisms into $Z$. Hence by the universal property of $Z'$ we conclude that
$a|_{T'_a} = b|_{T'_a}$. Thus $a$ and $b$ are morphism from
the first order thickening $T'$ of $T'_a$ whose restrictions to
$T'_a$ agree as morphisms into $Z$. Thus using the universal property of
$Z'$ once more we conclude that $a = b$. In other words, the defining
property of a formally unramified morphism holds for $Z' \to X$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-universal-thickening-functorial}
Consider a commutative diagram of schemes
$$
\xymatrix{
Z \ar[r]_h \ar[d]_f & X \ar[d]^g \\
W \ar[r]^{h'} & Y
}
$$
with $h$ and $h'$ formally unramified. Let $Z \subset Z'$ be the universal
first order thickening of $Z$ over $X$. Let $W \subset W'$ be the universal
first order thickening of $W$ over $Y$. There exists a canonical morphism
$(f, f') : (Z, Z') \to (W, W')$ of thickenings over $Y$ which fits into
the following commutative diagram
$$
\xymatrix{
& & & Z' \ar[ld] \ar[d]^{f'} \\
Z \ar[rr] \ar[d]_f \ar[rrru] & & X \ar[d] & W' \ar[ld] \\
W \ar[rrru]|!{[rr];[rruu]}\hole \ar[rr] & & Y
}
$$
In particular the morphism $(f, f')$ of thickenings induces a morphism
of conormal sheaves $f^*\mathcal{C}_{W/Y} \to \mathcal{C}_{Z/X}$.
\end{lemma}
\begin{proof}
The first assertion is clear from the universal property of $W'$.
The induced map on conormal sheaves is the map of
Morphisms, Lemma \ref{morphisms-lemma-conormal-functorial}
applied to $(Z \subset Z') \to (W \subset W')$.
\end{proof}
\begin{lemma}
\label{lemma-universal-thickening-fibre-product}
Let
$$
\xymatrix{
Z \ar[r]_h \ar[d]_f & X \ar[d]^g \\
W \ar[r]^{h'} & Y
}
$$
be a fibre product diagram in the category of schemes with
$h'$ formally unramified. Then $h$ is formally unramified and if
$W \subset W'$ is the universal first order thickening of $W$ over $Y$,
then $Z = X \times_Y W \subset X \times_Y W'$ is the universal
first order thickening of $Z$ over $X$. In particular the canonical map
$f^*\mathcal{C}_{W/Y} \to \mathcal{C}_{Z/X}$ of
Lemma \ref{lemma-universal-thickening-functorial}
is surjective.
\end{lemma}
\begin{proof}
The morphism $h$ is formally unramified by
Lemma \ref{lemma-base-change-formally-unramified}.
It is clear that $X \times_Y W'$ is a first order thickening.
It is straightforward to check that it has the universal property
because $W'$ has the universal property (by mapping properties of
fibre products). See
Morphisms, Lemma \ref{morphisms-lemma-conormal-functorial-flat}
for why this implies that the map of conormal sheaves is surjective.
\end{proof}
\begin{lemma}
\label{lemma-universal-thickening-fibre-product-flat}
Let
$$
\xymatrix{
Z \ar[r]_h \ar[d]_f & X \ar[d]^g \\
W \ar[r]^{h'} & Y
}
$$
be a fibre product diagram in the category of schemes with
$h'$ formally unramified and $g$ flat. In this case the corresponding
map $Z' \to W'$ of universal first order thickenings is flat, and
$f^*\mathcal{C}_{W/Y} \to \mathcal{C}_{Z/X}$ is an isomorphism.
\end{lemma}
\begin{proof}
Flatness is preserved under base change, see
Morphisms, Lemma \ref{morphisms-lemma-base-change-flat}.
Hence the first statement follows from the description of
$W'$ in Lemma \ref{lemma-universal-thickening-fibre-product}.
It is clear that $X \times_Y W'$ is a first order thickening.
It is straightforward to check that it has the universal property
because $W'$ has the universal property (by mapping properties of
fibre products). See
Morphisms, Lemma \ref{morphisms-lemma-conormal-functorial-flat}
for why this implies that the map of conormal sheaves is an isomorphism.
\end{proof}
\begin{lemma}
\label{lemma-universal-thickening-localize}
Taking the universal first order thickenings commutes with taking opens.
More precisely, let $h : Z \to X$ be a formally unramified morphism of schemes.
Let $V \subset Z$, $U \subset X$ be opens such that $h(V) \subset U$.
Let $Z'$ be the universal first order thickening of $Z$ over $X$.
Then $h|_V : V \to U$ is formally unramified and the universal first
order thickening of $V$ over $U$ is the open subscheme $V' \subset Z'$
such that $V = Z \cap V'$. In particular,
$\mathcal{C}_{Z/X}|_V = \mathcal{C}_{V/U}$.
\end{lemma}
\begin{proof}
The first statement is
Lemma \ref{lemma-formally-unramified-on-opens}.
The compatibility of universal thickenings can be deduced from the proof of
Lemma \ref{lemma-universal-thickening},
or from
Algebra, Lemma \ref{algebra-lemma-universal-thickening-localize}
or deduced from
Lemma \ref{lemma-universal-thickening-fibre-product-flat}.
\end{proof}
\begin{lemma}
\label{lemma-differentials-universally-unramified}
Let $h : Z \to X$ be a formally unramified morphism of schemes over $S$.
Let $Z \subset Z'$ be the universal first order thickening of $Z$
over $X$ with structure morphism $h' : Z' \to X$. The canonical map
$$
c_{h'} : (h')^*\Omega_{X/S} \longrightarrow \Omega_{Z'/S}
$$
induces an isomorphism
$h^*\Omega_{X/S} \to \Omega_{Z'/S} \otimes \mathcal{O}_Z$.
\end{lemma}
\begin{proof}
The map $c_{h'}$ is the map defined in
Morphisms, Lemma \ref{morphisms-lemma-functoriality-differentials}.
If $i : Z \to Z'$ is the given closed immersion, then
$i^*c_{h'}$ is a map
$h^*\Omega_{X/S} \to \Omega_{Z'/S} \otimes \mathcal{O}_Z$.
Checking that it is an isomorphism reduces to the affine case
by localization, see
Lemma \ref{lemma-universal-thickening-localize}
and
Morphisms, Lemma \ref{morphisms-lemma-differentials-restrict-open}.
In this case the result is
Algebra, Lemma \ref{algebra-lemma-differentials-universal-thickening}.
\end{proof}
\begin{lemma}
\label{lemma-universally-unramified-differentials-sequence}
Let $h : Z \to X$ be a formally unramified morphism of schemes over $S$.
There is a canonical exact sequence
$$
\mathcal{C}_{Z/X} \to h^*\Omega_{X/S} \to \Omega_{Z/S} \to 0.
$$
The first arrow is induced by $\text{d}_{Z'/S}$ where
$Z'$ is the universal first order neighbourhood of $Z$ over $X$.
\end{lemma}
\begin{proof}
We know that there is a canonical exact sequence
$$
\mathcal{C}_{Z/Z'} \to
\Omega_{Z'/S} \otimes \mathcal{O}_Z \to
\Omega_{Z/S} \to 0.
$$
see
Morphisms, Lemma \ref{morphisms-lemma-differentials-relative-immersion}.
Hence the result follows on applying
Lemma \ref{lemma-differentials-universally-unramified}.
\end{proof}
\begin{lemma}
\label{lemma-two-unramified-morphisms}
Let
$$
\xymatrix{
Z \ar[r]_i \ar[rd]_j & X \ar[d] \\
& Y
}
$$
be a commutative diagram of schemes where $i$ and $j$ are formally
unramified. Then there is a canonical exact sequence
$$
\mathcal{C}_{Z/Y} \to
\mathcal{C}_{Z/X} \to
i^*\Omega_{X/Y} \to 0
$$
where the first arrow comes from
Lemma \ref{lemma-universal-thickening-functorial}
and the second from
Lemma \ref{lemma-universally-unramified-differentials-sequence}.
\end{lemma}
\begin{proof}
Denote $Z \to Z'$ the universal first order thickening of $Z$ over $X$.
Denote $Z \to Z''$ the universal first order thickening of $Z$ over $Y$.
By
Lemma \ref{lemma-universally-unramified-differentials-sequence}
here is a canonical morphism $Z' \to Z''$ so that we have a commutative
diagram
$$
\xymatrix{
Z \ar[r]_{i'} \ar[rd]_{j'} & Z' \ar[r] \ar[d] & X \ar[d] \\
& Z'' \ar[r] & Y
}
$$
Apply
Morphisms, Lemma \ref{morphisms-lemma-two-immersions}
to the left triangle to get an exact sequence
$$
\mathcal{C}_{Z/Z''} \to
\mathcal{C}_{Z/Z'} \to
(i')^*\Omega_{Z'/Z''} \to 0
$$
As $Z''$ is formally unramified over $Y$ (see
Lemma \ref{lemma-universal-thickening-unramified})
we have
$\Omega_{Z'/Z''} = \Omega_{Z/Y}$ (by combining
Lemma \ref{lemma-formally-unramified-differentials}
and
Morphisms, Lemma \ref{morphisms-lemma-triangle-differentials}).
Then we have $(i')^*\Omega_{Z'/Y} = i^*\Omega_{X/Y}$ by
Lemma \ref{lemma-differentials-universally-unramified}.
\end{proof}
\begin{lemma}
\label{lemma-transitivity-conormal}
Let $Z \to Y \to X$ be formally unramified morphisms of schemes.
\begin{enumerate}
\item If $Z \subset Z'$ is the universal first order thickening of $Z$
over $X$ and $Y \subset Y'$ is the universal first order thickening of $Y$
over $X$, then there is a morphism $Z' \to Y'$ and $Y \times_{Y'} Z'$ is
the universal first order thickening of $Z$ over $Y$.
\item There is a canonical exact sequence
$$
i^*\mathcal{C}_{Y/X} \to
\mathcal{C}_{Z/X} \to
\mathcal{C}_{Z/Y} \to 0
$$
where the maps come from
Lemma \ref{lemma-universal-thickening-functorial}
and $i : Z \to Y$ is the first morphism.
\end{enumerate}
\end{lemma}
\begin{proof}
The map $h : Z' \to Y'$ in (1) comes from
Lemma \ref{lemma-universal-thickening-functorial}.
The assertion that $Y \times_{Y'} Z'$ is the universal first order
thickening of $Z$ over $Y$ is clear from the universal properties
of $Z'$ and $Y'$. By
Morphisms, Lemma \ref{morphisms-lemma-transitivity-conormal}
we have an exact sequence
$$
(i')^*\mathcal{C}_{Y \times_{Y'} Z'/Z'} \to
\mathcal{C}_{Z/Z'} \to
\mathcal{C}_{Z/Y \times_{Y'} Z'} \to 0
$$
where $i' : Z \to Y \times_{Y'} Z'$ is the given morphism. By
Morphisms, Lemma \ref{morphisms-lemma-conormal-functorial-flat}
there exists a surjection
$h^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{Y \times_{Y'} Z'/Z'}$.
Combined with the equalities
$\mathcal{C}_{Y/Y'} = \mathcal{C}_{Y/X}$,
$\mathcal{C}_{Z/Z'} = \mathcal{C}_{Z/X}$, and
$\mathcal{C}_{Z/Y \times_{Y'} Z'} = \mathcal{C}_{Z/Y}$
this proves the lemma.
\end{proof}
\section{Formally \'etale morphisms}
\label{section-formally-etale}
\noindent
Recall that a ring map $R \to A$ is called {\it formally \'etale}
(see Algebra, Definition \ref{algebra-definition-formally-etale})
if for every commutative solid diagram
$$
\xymatrix{
A \ar[r] \ar@{-->}[rd] & B/I \\
R \ar[r] \ar[u] & B \ar[u]
}
$$
where $I \subset B$ is an ideal of square zero, there exists
exactly one dotted arrow which makes the diagram commute. This motivates
the following analogue for morphisms of schemes.
\begin{definition}
\label{definition-formally-etale}
Let $f : X \to S$ be a morphism of schemes.
We say $f$ is {\it formally \'etale} if given any solid commutative diagram
$$
\xymatrix{
X \ar[d]_f & T \ar[d]^i \ar[l] \\
S & T' \ar[l] \ar@{-->}[lu]
}
$$
where $T \subset T'$ is a first order thickening of affine schemes over $S$
there exists exactly one dotted arrow making the diagram commute.
\end{definition}
\noindent
It is clear that a formally \'etale morphism is formally unramified.
Hence if $f : X \to S$ is formally \'etale, then $\Omega_{X/S}$ is zero, see
Lemma \ref{lemma-formally-unramified-differentials}.
\begin{lemma}
\label{lemma-formally-etale-not-affine}
If $f : X \to S$ is a formally \'etale morphism, then given
any solid commutative diagram
$$
\xymatrix{
X \ar[d]_f & T \ar[d]^i \ar[l] \\
S & T' \ar[l] \ar@{-->}[lu]
}
$$
where $T \subset T'$ is a first order thickening of schemes over $S$
there exists exactly one dotted arrow making the diagram commute.
In other words, in
Definition \ref{definition-formally-etale}
the condition that $T$ be affine may be dropped.
\end{lemma}
\begin{proof}
Let $T' = \bigcup T'_i$ be an affine open covering, and let
$T_i = T \cap T'_i$. Then we get morphisms $a'_i : T'_i \to X$ fitting
into the diagram. By uniqueness we see that $a'_i$ and $a'_j$ agree on
any affine open subscheme of $T'_i \cap T'_j$. Hence $a'_i$ and
$a'_j$ agree on $T'_i \cap T'_j$. Thus we see that the morphisms $a'_i$
glue to a global morphism $a' : T' \to X$. The uniqueness of
$a'$ we have seen in
Lemma \ref{lemma-formally-unramified-not-affine}.
\end{proof}
\begin{lemma}
\label{lemma-composition-formally-etale}
A composition of formally \'etale morphisms is formally \'etale.
\end{lemma}
\begin{proof}
This is formal.
\end{proof}
\begin{lemma}
\label{lemma-base-change-formally-etale}
A base change of a formally \'etale morphism is formally \'etale.
\end{lemma}
\begin{proof}
This is formal.
\end{proof}
\begin{lemma}
\label{lemma-formally-etale-on-opens}
Let $f : X \to S$ be a morphism of schemes.
Let $U \subset X$ and $V \subset S$ be open subschemes such that
$f(U) \subset V$. If $f$ is formally \'etale, so is $f|_U : U \to V$.
\end{lemma}
\begin{proof}
Consider a solid diagram
$$
\xymatrix{
U \ar[d]_{f|_U} & T \ar[d]^i \ar[l]^a \\
V & T' \ar[l] \ar@{-->}[lu]
}
$$
as in Definition \ref{definition-formally-etale}. If $f$ is formally
ramified, then there exists exactly one $S$-morphism $a' : T' \to X$
such that $a'|_T = a$. Since $|T'| = |T|$ we conclude that $a'(T') \subset U$
which gives our unique morphism from $T'$ into $U$.
\end{proof}
\begin{lemma}
\label{lemma-characterize-formally-etale}
Let $f : X \to S$ be a morphism of schemes.
The following are equivalent:
\begin{enumerate}
\item $f$ is formally \'etale,
\item $f$ is formally unramified and the universal first order thickening
of $X$ over $S$ is equal to $X$,
\item $f$ is formally unramified and $\mathcal{C}_{X/S} = 0$, and
\item $\Omega_{X/S} = 0$ and $\mathcal{C}_{X/S} = 0$.
\end{enumerate}
\end{lemma}
\begin{proof}
Actually, the last assertion only make sense because $\Omega_{X/S} = 0$
implies that $\mathcal{C}_{X/S}$ is defined via
Lemma \ref{lemma-formally-unramified-differentials}
and
Definition \ref{definition-universal-thickening}.
This also makes it clear that (3) and (4) are equivalent.
\medskip\noindent
Either of the assumptions (1), (2), and (3) imply that $f$ is formally
unramified. Hence we may assume $f$ is formally unramified. The equivalence
of (1), (2), and (3) follow from the universal property of the universal
first order thickening $X'$ of $X$ over $S$ and the fact that
$X = X' \Leftrightarrow \mathcal{C}_{X/S} = 0$ since
after all by definition $\mathcal{C}_{X/S} = \mathcal{C}_{X/X'}$
is the ideal sheaf of $X$ in $X'$.
\end{proof}
\begin{lemma}
\label{lemma-unramified-flat-formally-etale}
An unramified flat morphism is formally \'etale.
\end{lemma}
\begin{proof}
Say $X \to S$ is unramified and flat. Then $\Delta : X \to X \times_S X$
is an open immersion, see
Morphisms, Lemma \ref{morphisms-lemma-diagonal-unramified-morphism}.
We have to show that $\mathcal{C}_{X/S}$ is zero.
Consider the two projections $p, q : X \times_S X \to X$.
As $f$ is formally unramified (see
Lemma \ref{lemma-unramified-formally-unramified}),
$q$ is formally unramified (see
Lemma \ref{lemma-base-change-formally-unramified}).
As $f$ is flat, $p$ is flat, see
Morphisms, Lemma \ref{morphisms-lemma-base-change-flat}.
Hence $p^*\mathcal{C}_{X/S} = \mathcal{C}_q$ by
Lemma \ref{lemma-universal-thickening-fibre-product-flat}
where $\mathcal{C}_q$ denotes the conormal sheaf of the formally
unramified morphism $q : X \times_S X \to X$.
But $\Delta(X) \subset X \times_S X$ is an open subscheme
which maps isomorphically to $X$ via $q$. Hence by
Lemma \ref{lemma-universal-thickening-localize}
we see that $\mathcal{C}_q|_{\Delta(X)} = \mathcal{C}_{X/X} = 0$.
In other words, the pullback of $\mathcal{C}_{X/S}$ to $X$ via
the identity morphism is zero, i.e., $\mathcal{C}_{X/S} = 0$.
\end{proof}
\begin{lemma}
\label{lemma-affine-formally-etale}
Let $f : X \to S$ be a morphism of schemes.
Assume $X$ and $S$ are affine.
Then $f$ is formally \'etale if and only if
$\mathcal{O}_S(S) \to \mathcal{O}_X(X)$ is a formally \'etale
ring map.
\end{lemma}
\begin{proof}
This is immediate from the definitions
(Definition \ref{definition-formally-etale} and
Algebra, Definition \ref{algebra-definition-formally-etale})
by the equivalence of categories of rings and affine schemes,
see
Schemes, Lemma \ref{schemes-lemma-category-affine-schemes}.
\end{proof}
\begin{lemma}
\label{lemma-etale-formally-etale}
Let $f : X \to S$ be a morphism of schemes.
The following are equivalent:
\begin{enumerate}
\item The morphism $f$ is \'etale, and
\item the morphism $f$ is locally of finite presentation and
formally \'etale.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume $f$ is \'etale.
An \'etale morphism is locally of finite presentation, flat and unramified, see
Morphisms, Section \ref{morphisms-section-etale}.
Hence $f$ is locally of finite presentation and formally \'etale, see
Lemma \ref{lemma-unramified-flat-formally-etale}.
\medskip\noindent
Conversely, suppose that $f$ is locally of finite presentation and
formally \'etale. Being \'etale is local in the Zariski topology on
$X$ and $S$, see
Morphisms, Lemma \ref{morphisms-lemma-etale-characterize}.
By
Lemma \ref{lemma-formally-etale-on-opens}
we can cover $X$ by affine opens $U$ which map into affine opens
$V$ such that $U \to V$ is formally \'etale (and of finite presentation, see
Morphisms,
Lemma \ref{morphisms-lemma-locally-finite-presentation-characterize}).
By
Lemma \ref{lemma-affine-formally-etale}
we see that the ring maps $\mathcal{O}(V) \to \mathcal{O}(U)$ are
formally \'etale (and of finite presentation).
We win by
Algebra, Lemma \ref{algebra-lemma-formally-etale-etale}.
(We will give another proof of this implication when we discuss
formally smooth morphisms.)
\end{proof}
\section{Infinitesimal deformations of maps}
\label{section-action-by-derivations}
\noindent
In this section we explain how a derivation can be used to
infinitesimally move a map. Throughout this section we use that
a sheaf on a thickening $X'$ of $X$ can be seen as a sheaf on $X$.
\begin{lemma}
\label{lemma-difference-derivation}
Let $S$ be a scheme.
Let $X \subset X'$ and $Y \subset Y'$ be two first order thickenings
over $S$. Let $(a, a'), (b, b') : (X \subset X') \to (Y \subset Y')$
be two morphisms of thickenings over $S$. Assume that
\begin{enumerate}
\item $a = b$, and
\item the two maps $a^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$
(Morphisms, Lemma \ref{morphisms-lemma-conormal-functorial})
are equal.
\end{enumerate}
Then the map $(a')^\sharp - (b')^\sharp$ factors as
$$
\mathcal{O}_{Y'} \to \mathcal{O}_Y \xrightarrow{D}
a_*\mathcal{C}_{X/X'} \to a_*\mathcal{O}_{X'}
$$
where $D$ is an $\mathcal{O}_S$-derivation.
\end{lemma}
\begin{proof}
Instead of working on $Y$ we work on $X$. The advantage is that the pullback
functor $a^{-1}$ is exact. Using (1) and (2) we obtain a commutative diagram
with exact rows
$$
\xymatrix{
0 \ar[r] &
\mathcal{C}_{X/X'} \ar[r] &
\mathcal{O}_{X'} \ar[r] &
\mathcal{O}_X \ar[r] & 0 \\
0 \ar[r] &
a^{-1}\mathcal{C}_{Y/Y'} \ar[r] \ar[u] &
a^{-1}\mathcal{O}_{Y'}
\ar[r] \ar@<1ex>[u]^{(a')^\sharp} \ar@<-1ex>[u]_{(b')^\sharp} &
a^{-1}\mathcal{O}_Y \ar[r] \ar[u] & 0
}
$$
Now it is a general fact that in such a situation the difference of the
$\mathcal{O}_S$-algebra maps $(a')^\sharp$ and $(b')^\sharp$ is an
$\mathcal{O}_S$-derivation from $a^{-1}\mathcal{O}_Y$ to $\mathcal{C}_{X/X'}$.
By adjointness of the functors $a^{-1}$ and $a_*$ this is the same
thing as an $\mathcal{O}_S$-derivation from
$\mathcal{O}_Y$ into $a_*\mathcal{C}_{X/X'}$. Some details omitted.
\end{proof}
\noindent
Note that in the situation of the lemma above we may write
$D$ as
\begin{equation}
\label{equation-D}
D = \text{d}_{Y/S} \circ \theta
\end{equation}
where $\theta$ is an $\mathcal{O}_Y$-linear map
$\theta : \Omega_{Y/S} \to a_*\mathcal{C}_{X/X'}$.
Of course, then by adjunction again we may view $\theta$ as an
$\mathcal{O}_X$-linear map
$\theta : a^*\Omega_{Y/S} \to \mathcal{C}_{X/X'}$.
\begin{lemma}
\label{lemma-action-by-derivations}
Let $S$ be a scheme.
Let $(a, a') : (X \subset X') \to (Y \subset Y')$
be a morphism of first order thickenings over $S$.
Let
$$
\theta : a^*\Omega_{Y/S} \to \mathcal{C}_{X/X'}
$$
be an $\mathcal{O}_X$-linear map. Then there exists a unique morphism of pairs
$(b, b') : (X \subset X') \to (Y \subset Y')$ such that
(1) and (2) of
Lemma \ref{lemma-difference-derivation}
hold and the derivation $D$ and $\theta$ are related by
Equation (\ref{equation-D}).
\end{lemma}
\begin{proof}
We simply set $b = a$ and we define $(b')^\sharp$ to be the map
$$
(a')^\sharp + D : a^{-1}\mathcal{O}_{Y'} \to \mathcal{O}_{X'}
$$
where $D$ is as in Equation (\ref{equation-D}). We omit the verification
that $(b')^\sharp$ is a map of sheaves of $\mathcal{O}_S$-algebras and
that (1) and (2) of
Lemma \ref{lemma-difference-derivation}
hold. Equation (\ref{equation-D}) holds by construction.
\end{proof}
\begin{remark}
\label{remark-action-by-derivations}
Assumptions and notation as in Lemma \ref{lemma-action-by-derivations}.
The action of a local section $\theta$ on $a'$ is sometimes indicated by
$\theta \cdot a'$. Note that this means nothing else than the fact
that $(a')^\sharp$ and $(\theta \cdot a')^\sharp$ differ by a derivation
$D$ which is related to $\theta$ by Equation (\ref{equation-D}).
\end{remark}
\begin{lemma}
\label{lemma-sheaf}
Let $S$ be a scheme.
Let $X \subset X'$ and $Y \subset Y'$ be first order thickenings
over $S$. Assume given a morphism $a : X \to Y$ and a map
$A : a^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$ of
$\mathcal{O}_X$-modules. For an open subscheme $U' \subset X'$
consider morphisms $a' : U' \to Y'$ such that
\begin{enumerate}
\item $a'$ is a morphism over $S$,
\item $a'|_U = a|_U$, and
\item the induced map
$a^*\mathcal{C}_{Y/Y'}|_U \to \mathcal{C}_{X/X'}|_U$
is the restriction of $A$ to $U$.
\end{enumerate}
Here $U = X \cap U'$. Then the rule
\begin{equation}
\label{equation-sheaf}
U' \mapsto
\{a' : U' \to Y'\text{ such that (1), (2), (3) hold.}\}
\end{equation}
defines a sheaf of sets on $X'$.
\end{lemma}
\begin{proof}
Denote $\mathcal{F}$ the rule of the lemma.
The restriction mapping $\mathcal{F}(U') \to \mathcal{F}(V')$ for
$V' \subset U' \subset X'$
of $\mathcal{F}$ is really the restriction map $a' \mapsto a'|_{V'}$.
With this definition in place it is clear that $\mathcal{F}$ is a
sheaf since morphisms are defined locally.
\end{proof}
\noindent
In the following lemma we identify sheaves on $X$ and any thickening
of $X$.
\begin{lemma}
\label{lemma-action-sheaf}
Same notation and assumptions as in Lemma \ref{lemma-sheaf}.
There is an action of the sheaf
$$
\SheafHom_{\mathcal{O}_X}(a^*\Omega_{Y/S}, \mathcal{C}_{X/X'})
$$
on the sheaf (\ref{equation-sheaf}). Moreover, the action
is simply transitive for any open $U' \subset X'$ over which the sheaf
(\ref{equation-sheaf}) has a section.
\end{lemma}
\begin{proof}
This is a combination of
Lemmas \ref{lemma-difference-derivation},
\ref{lemma-action-by-derivations},
and \ref{lemma-sheaf}.
\end{proof}
\begin{remark}
\label{remark-special-case}
A special case of
Lemmas \ref{lemma-difference-derivation},
\ref{lemma-action-by-derivations},
\ref{lemma-sheaf}, and
\ref{lemma-action-sheaf}
is where $Y = Y'$. In this case the map $A$ is always zero.
The sheaf of
Lemma \ref{lemma-sheaf}
is just given by the rule
$$
U' \mapsto
\{a' : U' \to Y\text{ over }S\text{ with } a'|_U = a|_U\}
$$
and we act on this by the sheaf
$\SheafHom_{\mathcal{O}_X}(a^*\Omega_{Y/S}, \mathcal{C}_{X/X'})$.
\end{remark}
\begin{remark}
\label{remark-another-special-case}
Another special case of
Lemmas \ref{lemma-difference-derivation},
\ref{lemma-action-by-derivations},
\ref{lemma-sheaf}, and
\ref{lemma-action-sheaf}
is where $S$ itself is a thickening $Z \subset Z' = S$
and $Y = Z \times_{Z'} Y'$. Picture
$$
\xymatrix{
(X \subset X') \ar@{..>}[rr]_{(a, ?)} \ar[rd]_{(g, g')} & &
(Y \subset Y') \ar[ld]^{(h, h')} \\
& (Z \subset Z')
}
$$
In this case the map $A : a^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$
is determined by $a$: the map
$h^*\mathcal{C}_{Z/Z'} \to \mathcal{C}_{Y/Y'}$ is surjective (because
we assumed $Y = Z \times_{Z'} Y'$),
hence the pullback $g^*\mathcal{C}_{Z/Z'} = a^*h^*\mathcal{C}_{Z/Z'} \to
a^*\mathcal{C}_{Y/Y'}$ is surjective, and the composition
$g^*\mathcal{C}_{Z/Z'} \to a^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$
has to be the canonical map induced by $g'$. Thus the sheaf of
Lemma \ref{lemma-sheaf}
is just given by the rule
$$
U' \mapsto
\{a' : U' \to Y'\text{ over }Z'\text{ with } a'|_U = a|_U\}
$$
and we act on this by the sheaf
$\SheafHom_{\mathcal{O}_X}(a^*\Omega_{Y/Z}, \mathcal{C}_{X/X'})$.
\end{remark}
\begin{lemma}
\label{lemma-omega-deformation}
Let $S$ be a scheme. Let $X \subset X'$ be a first order thickening over
$S$. Let $Y$ be a scheme over $S$. Let
$a', b' : X' \to Y$ be two morphisms over $S$ with
$a = a'|_X = b'|_X$. This gives rise to a commutative diagram
$$
\xymatrix{
X \ar[r] \ar[d]_a & X' \ar[d]^{(b', a')} \\
Y \ar[r]^-{\Delta_{Y/S}} & Y \times_S Y
}
$$
Since the horizontal arrows are immersions with conormal sheaves
$\mathcal{C}_{X/X'}$ and $\Omega_{Y/S}$, by
Morphisms, Lemma \ref{morphisms-lemma-conormal-functorial},
we obtain a map $\theta : a^*\Omega_{Y/S} \to \mathcal{C}_{X/X'}$.
Then this $\theta$ and the derivation $D$ of
Lemma \ref{lemma-difference-derivation}
are related by Equation (\ref{equation-D}).
\end{lemma}
\begin{proof}
Omitted. Hint: The equality may be checked on affine opens where it
comes from the following computation. If $f$ is a local section of
$\mathcal{O}_Y$, then $1 \otimes f - f \otimes 1$ is a local section
of $\mathcal{C}_{Y/(Y \times_S Y)}$ corresponding to $\text{d}_{Y/S}(f)$.
It is mapped to the local section $(a')^\sharp(f) - (b')^\sharp(f) = D(f)$
of $\mathcal{C}_{X/X'}$. In other words, $\theta(\text{d}_{Y/S}(f)) = D(f)$.
\end{proof}
\noindent
For later purposes we need a result that roughly states that the
construction of
Lemma \ref{lemma-action-by-derivations}
is compatible with \'etale localization.
\begin{lemma}
\label{lemma-sheaf-differentials-etale-localization}
Let
$$
\xymatrix{
X_1 \ar[d] & X_2 \ar[l]^f \ar[d] \\
S_1 & S_2 \ar[l]
}
$$
be a commutative diagram of schemes with $X_2 \to X_1$ and $S_2 \to S_1$
\'etale. Then the map $c_f : f^*\Omega_{X_1/S_1} \to \Omega_{X_2/S_2}$ of
Morphisms, Lemma \ref{morphisms-lemma-functoriality-differentials}
is an isomorphism.
\end{lemma}
\begin{proof}
We recall that an \'etale morphism $U \to V$ is a smooth morphism
with $\Omega_{U/V} = 0$. Using this we see that
Morphisms, Lemma \ref{morphisms-lemma-triangle-differentials}
implies $\Omega_{X_2/S_2} = \Omega_{X_2/S_1}$ and
Morphisms, Lemma \ref{morphisms-lemma-triangle-differentials-smooth}
implies that the map $f^*\Omega_{X_1/S_1} \to \Omega_{X_2/S_1}$
(for the morphism $f$ seen as a morphism over $S_1$)
is an isomorphism. Hence the lemma follows.
\end{proof}
\begin{lemma}
\label{lemma-action-by-derivations-etale-localization}
Consider a commutative diagram of first order thickenings
$$
\vcenter{
\xymatrix{
(T_2 \subset T_2') \ar[d]_{(h, h')} \ar[rr]_{(a_2, a_2')} & &
(X_2 \subset X_2') \ar[d]^{(f, f')} \\
(T_1 \subset T_1') \ar[rr]^{(a_1, a_1')} & &
(X_1 \subset X_1')
}
}
\quad
\begin{matrix}
\text{and a commutative} \\
\text{diagram of schemes}
\end{matrix}
\quad
\vcenter{
\xymatrix{
X_2' \ar[r] \ar[d] & S_2 \ar[d] \\
X_1' \ar[r] & S_1
}
}
$$
with $X_2 \to X_1$ and $S_2 \to S_1$ \'etale.
For any $\mathcal{O}_{T_1}$-linear map
$\theta_1 : a_1^*\Omega_{X_1/S_1} \to \mathcal{C}_{T_1/T'_1}$ let
$\theta_2$ be the composition
$$
\xymatrix{
a_2^*\Omega_{X_2/S_2} \ar@{=}[r] &
h^*a_1^*\Omega_{X_1/S_1} \ar[r]^-{h^*\theta_1} &
h^*\mathcal{C}_{T_1/T'_1} \ar[r] &
\mathcal{C}_{T_2/T'_2}
}
$$
(equality sign is explained in the proof). Then the diagram
$$
\xymatrix{
T_2' \ar[rr]_{\theta_2 \cdot a_2'} \ar[d] & & X'_2 \ar[d] \\
T_1' \ar[rr]^{\theta_1 \cdot a_1'} & & X'_1
}
$$
commutes where the actions $\theta_2 \cdot a_2'$ and $\theta_1 \cdot a_1'$
are as in Remark \ref{remark-action-by-derivations}.
\end{lemma}
\begin{proof}
The equality sign comes from the identification
$f^*\Omega_{X_1/S_1} = \Omega_{X_2/S_2}$ of
Lemma \ref{lemma-sheaf-differentials-etale-localization}.
Namely, using this we have
$a_2^*\Omega_{X_2/S_2} = a_2^*f^*\Omega_{X_1/S_1} =
h^*a_1^*\Omega_{X_1/S_1}$ because $f \circ a_2 = a_1 \circ h$.
Having said this, the commutativity of the diagram may be checked
on affine opens. Hence we may assume the schemes in the initial
big diagram are affine. Thus we obtain commutative diagrams
$$
\vcenter{
\xymatrix{
(B'_2, I_2) & & (A'_2, J_2) \ar[ll]^{a_2'} \\
(B'_1, I_1) \ar[u]^{h'} & & (A'_1, J_1) \ar[ll]_{a_1'} \ar[u]_{f'}
}
}
\quad\text{and}\quad
\vcenter{
\xymatrix{
A'_2 & & R_2 \ar[ll] \\
A'_1 \ar[u] & & R_1 \ar[ll] \ar[u]
}
}
$$
The notation signifies that $I_1, I_2, J_1, J_2$ are ideals of square
zero and maps of pairs are ring maps sending ideals into ideals.
Set $A_1 = A'_1/J_1$, $A_2 = A'_2/J_2$, $B_1 = B'_1/I_1$, and
$B_2 = B'_2/I_2$. We are given that
$$
A_2 \otimes_{A_1} \Omega_{A_1/R_1} \longrightarrow \Omega_{A_2/R_2}
$$
is an isomorphism. Then
$\theta_1 : B_1 \otimes_{A_1} \Omega_{A_1/R_1} \to I_1$
is $B_1$-linear. This gives an $R_1$-derivation
$D_1 = \theta_1 \circ \text{d}_{A_1/R_1} : A_1 \to I_1$.
In a similar way we see that
$\theta_2 : B_2 \otimes_{A_2} \Omega_{A_2/R_2} \to I_2$
gives rise to a $R_2$-derivation
$D_2 = \theta_2 \circ \text{d}_{A_2/R_2} : A_2 \to I_2$.
The construction of $\theta_2$ implies the following compatibility between
$\theta_1$ and $\theta_2$: for every $x \in A_1$ we have
$$
h'(D_1(x)) = D_2(f'(x))
$$
as elements of $I_2$. We may view $D_1$ as a map $A'_1 \to B'_1$
using $A'_1 \to A_1 \xrightarrow{D_1} I_1 \to B_1$ similarly
we may view $D_2$ as a map $A'_2 \to B'_2$. Then the displayed
equality holds for $x \in A'_1$.
By the construction of the action in
Lemma \ref{lemma-action-by-derivations} and
Remark \ref{remark-action-by-derivations}
we know that $\theta_1 \cdot a_1'$ corresponds to the ring map
$a_1' + D_1 : A'_1 \to B'_1$ and $\theta_2 \cdot a_2'$ corresponds
to the ring map $a_2' + D_2 : A'_2 \to B'_2$. By the displayed equality
we obtain that
$h' \circ (a_1' + D_1) = (a_2' + D_2) \circ f'$
as desired.
\end{proof}
\begin{remark}
\label{remark-tiny-improvement}
Lemma \ref{lemma-action-by-derivations-etale-localization}
can be improved in the following way.
Suppose that we have commutative diagrams as in
Lemma \ref{lemma-action-by-derivations-etale-localization}
but we do not assume that $X_2 \to X_1$
and $S_2 \to S_1$ are \'etale. Next, suppose we have
$\theta_1 : a_1^*\Omega_{X_1/S_1} \to \mathcal{C}_{T_1/T'_1}$
and
$\theta_2 : a_2^*\Omega_{X_2/S_2} \to \mathcal{C}_{T_2/T'_2}$
such that
$$
\xymatrix{
f_*\mathcal{O}_{X_2} \ar[rr]_{f_*D_2} & &
f_*a_{2, *}\mathcal{C}_{T_2/T_2'} \\
\mathcal{O}_{X_1} \ar[rr]^{D_1} \ar[u]^{f^\sharp} & &
a_{1, *}\mathcal{C}_{T_1/T_1'} \ar[u]_{\text{induced by }(h')^\sharp}
}
$$
is commutative where $D_i$ corresponds to $\theta_i$ as in
Equation (\ref{equation-D}). Then we have the conclusion of
Lemma \ref{lemma-action-by-derivations-etale-localization}.
The importance of the condition that both $X_2 \to X_1$ and
$S_2 \to S_1$ are \'etale is that it allows us to construct a $\theta_2$
from $\theta_1$.
\end{remark}
\section{Infinitesimal deformations of schemes}
\label{section-deform}
\noindent
The following simple lemma is often a convenient tool to check whether
an infinitesimal deformation of a map is flat.
\begin{lemma}
\label{lemma-deform}
Let $(f, f') : (X \subset X') \to (S \subset S')$ be a morphism
of first order thickenings. Assume that $f$ is flat.
Then the following are equivalent
\begin{enumerate}
\item $f'$ is flat and $X = S \times_{S'} X'$, and
\item the canonical map $f^*\mathcal{C}_{S/S'} \to \mathcal{C}_{X/X'}$
is an isomorphism.
\end{enumerate}
\end{lemma}
\begin{proof}
As the problem is local on $X'$ we may assume that $X, X', S, S'$
are affine schemes. Say $S' = \Spec(A')$, $X' = \Spec(B')$,
$S = \Spec(A)$, $X = \Spec(B)$ with $A = A'/I$ and $B = B'/J$
for some square zero ideals. Then we obtain the following commutative
diagram
$$
\xymatrix{
0 \ar[r] &
J \ar[r] &
B' \ar[r] &
B \ar[r] & 0 \\
0 \ar[r] &
I \ar[r] \ar[u] &
A' \ar[r] \ar[u] &
A \ar[r] \ar[u] & 0
}
$$
with exact rows. The canonical map of the lemma is the map
$$
I \otimes_A B = I \otimes_{A'} B' \longrightarrow J.
$$
The assumption that $f$ is flat signifies that $A \to B$ is flat.
\medskip\noindent
Assume (1). Then $A' \to B'$ is flat and $J = IB'$. Flatness implies
$\text{Tor}_1^{A'}(B', A) = 0$ (see
Algebra, Lemma \ref{algebra-lemma-characterize-flat}).
This means $I \otimes_{A'} B' \to B'$ is injective (see
Algebra, Remark \ref{algebra-remark-Tor-ring-mod-ideal}).
Hence we see that $I \otimes_A B \to J$ is an isomorphism.
\medskip\noindent
Assume (2). Then it follows that $J = IB'$, so that $X = S \times_{S'} X'$.
Moreover, we get $\text{Tor}_1^{A'}(B', A'/I) = 0$ by reversing the
implications in the previous paragraph. Hence $B'$ is flat over $A'$ by
Algebra, Lemma \ref{algebra-lemma-what-does-it-mean}.
\end{proof}
\noindent
The following lemma is the ``nilpotent'' version of the
``crit\`ere de platitude par fibres'', see
Section \ref{section-criterion-flat-fibres}.
\begin{lemma}
\label{lemma-flatness-morphism-thickenings}
Consider a commutative diagram
$$
\xymatrix{
(X \subset X') \ar[rr]_{(f, f')} \ar[rd] & & (Y \subset Y') \ar[ld] \\
& (S \subset S')
}
$$
of thickenings. Assume
\begin{enumerate}
\item $X'$ is flat over $S'$,
\item $f$ is flat,
\item $S \subset S'$ is a finite order thickening, and
\item $X = S \times_{S'} X'$ and $Y = S \times_{S'} Y'$.
\end{enumerate}
Then $f'$ is flat and $Y'$ is flat over $S'$ at all points in
the image of $f'$.
\end{lemma}
\begin{proof}
Immediate consequence of
Algebra, Lemma \ref{algebra-lemma-criterion-flatness-fibre-nilpotent}.
\end{proof}
\noindent
Many properties of morphisms of schemes are preserved under flat
deformations.
\begin{lemma}
\label{lemma-deform-property}
Consider a commutative diagram
$$
\xymatrix{
(X \subset X') \ar[rr]_{(f, f')} \ar[rd] & & (Y \subset Y') \ar[ld] \\
& (S \subset S')
}
$$
of thickenings. Assume $S \subset S'$ is a finite order thickening,
$X'$ flat over $S'$, $X = S \times_{S'} X'$, and
$Y = S \times_{S'} Y'$. Then
\begin{enumerate}
\item $f$ is flat if and only if $f'$ is flat,
\label{item-flat}
\item $f$ is an isomorphism if and only if $f'$ is an isomorphism,
\label{item-isomorphism}
\item $f$ is an open immersion if and only if $f'$ is an open immersion,
\label{item-open-immersion}
\item $f$ is quasi-compact if and only if $f'$ is quasi-compact,
\label{item-quasi-compact}
\item $f$ is universally closed if and only if $f'$ is universally closed,
\label{item-universally-closed}
\item $f$ is (quasi-)separated if and only if $f'$ is (quasi-)separated,
\label{item-separated}
\item $f$ is a monomorphism if and only if $f'$ is a monomorphism,
\label{item-monomorphism}
\item $f$ is surjective if and only if $f'$ is surjective,
\label{item-surjective}
\item $f$ is universally injective if and only if $f'$ is universally injective,
\label{item-universally-injective}
\item $f$ is affine if and only if $f'$ is affine,
\label{item-affine}
\item
\label{item-finite-type}
$f$ is locally of finite type if and only if $f'$ is locally of finite type,
\item $f$ is locally quasi-finite if and only if $f'$ is locally quasi-finite,
\label{item-quasi-finite}
\item
\label{item-finite-presentation}
$f$ is locally of finite presentation if and only if $f'$ is locally of
finite presentation,
\item
\label{item-relative-dimension-d}
$f$ is locally of finite type of relative dimension $d$ if and only if
$f'$ is locally of finite type of relative dimension $d$,
\item $f$ is universally open if and only if $f'$ is universally open,
\label{item-universally-open}
\item $f$ is syntomic if and only if $f'$ is syntomic,
\label{item-syntomic}
\item $f$ is smooth if and only if $f'$ is smooth,
\label{item-smooth}
\item $f$ is unramified if and only if $f'$ is unramified,
\label{item-unramified}
\item $f$ is \'etale if and only if $f'$ is \'etale,
\label{item-etale}
\item $f$ is proper if and only if $f'$ is proper,
\label{item-proper}
\item $f$ is integral if and only if $f'$ is integral,
\label{item-integral}
\item $f$ is finite if and only if $f'$ is finite,
\label{item-finite}
\item
\label{item-finite-locally-free}
$f$ is finite locally free (of rank $d$) if and only if $f'$
is finite locally free (of rank $d$), and
\item add more here.
\end{enumerate}
\end{lemma}
\begin{proof}
The assumptions on $X$ and $Y$ mean that $f$ is the base change of
$f'$ by $X \to X'$.
The properties $\mathcal{P}$ listed in (1) -- (23) above are all stable
under base change, hence if $f'$ has property $\mathcal{P}$, then so
does $f$. See
Schemes, Lemmas \ref{schemes-lemma-base-change-immersion},
\ref{schemes-lemma-quasi-compact-preserved-base-change},
\ref{schemes-lemma-separated-permanence}, and
\ref{schemes-lemma-base-change-monomorphism}
and
Morphisms, Lemmas
\ref{morphisms-lemma-base-change-surjective},
\ref{morphisms-lemma-base-change-universally-injective},
\ref{morphisms-lemma-base-change-affine},
\ref{morphisms-lemma-base-change-finite-type},
\ref{morphisms-lemma-base-change-quasi-finite},
\ref{morphisms-lemma-base-change-finite-presentation},
\ref{morphisms-lemma-base-change-relative-dimension-d},
\ref{morphisms-lemma-base-change-syntomic},
\ref{morphisms-lemma-base-change-smooth},
\ref{morphisms-lemma-base-change-unramified},
\ref{morphisms-lemma-base-change-etale},
\ref{morphisms-lemma-base-change-proper},
\ref{morphisms-lemma-base-change-finite}, and
\ref{morphisms-lemma-base-change-finite-locally-free}.
\medskip\noindent
The interesting direction in each case is therefore to assume
that $f$ has the property and deduce that $f'$ has it too.
By induction on the order of the thickening we may
assume that $S \subset S'$ is a first order thickening, see
discussion immediately following
Definition \ref{definition-thickening}.
We make a couple of general remarks which we will use without further
mention in the arguments below.
(I) Let $W' \subset S'$ be an affine open and let $U' \subset X'$
and $V' \subset Y'$ be affine opens lying over $W'$ with $f'(U') \subset V'$.
Let $W' = \Spec(R')$ and denote $I \subset R'$ be the ideal
defining the closed subscheme $W' \cap S$. Say $U' = \Spec(B')$
and $V' = \Spec(A')$. Then we get a commutative diagram
$$
\xymatrix{
0 \ar[r] &
IB' \ar[r] &
B' \ar[r] &
B \ar[r] & 0 \\
0 \ar[r] &
IA' \ar[r] \ar[u] &
A' \ar[r] \ar[u] &
A \ar[r] \ar[u] & 0
}
$$
with exact rows. Moreover $IB' \cong I \otimes_R B$, see proof of
Lemma \ref{lemma-deform}.
(II) The morphisms $X \to X'$ and $Y \to Y'$ are universal homeomorphisms.
Hence the topology of the maps $f$ and $f'$ (after any base change)
is identical. (III) If $f$ is flat, then $f'$ is flat and
$Y' \to S'$ is flat at every point in the image of $f'$, see
Lemma \ref{lemma-flatness-morphism-thickenings}.
\medskip\noindent
Ad (\ref{item-flat}). This is general remark (III).
\medskip\noindent
Ad (\ref{item-isomorphism}). Assume $f$ is an isomorphism.
By (III) we see that $Y' \to S'$ is flat. Choose an
affine open $V' \subset Y'$ and set $U' = (f')^{-1}(V')$. Then
$V = Y \cap V'$ is affine which implies that
$V \cong f^{-1}(V) = U = Y \times_{Y'} U'$ is affine. By
Lemma \ref{lemma-thickening-affine-scheme}
we see that $U'$ is affine. Thus we have a diagram as in the
general remark (I) and moreover $IA \cong I \otimes_R A$ because
$R' \to A'$ is flat. Then
$IB' \cong I \otimes_R B \cong I \otimes_R A \cong IA'$
and $A \cong B$. By the exactness of the rows
in the diagram above we see that $A' \cong B'$,
i.e., $U' \cong V'$. Thus $f'$ is an isomorphism.
\medskip\noindent
Ad (\ref{item-open-immersion}). Assume $f$ is an open immersion.
Then $f$ is an isomorphism of $X$ with an open subscheme $V \subset Y$.
Let $V' \subset Y'$ be the open subscheme whose underlying topological
space is $V$. Then $f'$ is a map from $X'$ to $V'$ which is an isomorphism by
(\ref{item-isomorphism}). Hence $f'$ is an open immersion.
\medskip\noindent
Ad (\ref{item-quasi-compact}). Immediate from remark (II). See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-universally-closed}). Immediate from remark (II). See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-separated}). Note that
$X \times_Y X = Y \times_{Y'} (X' \times_{Y'} X')$ so that
$X' \times_{Y'} X'$ is a thickening of $X \times_Y X$.
Hence the topology of the maps $\Delta_{X/Y}$ and $\Delta_{X'/Y'}$
matches and we win. See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-monomorphism}). Assume $f$ is a monomorphism.
Consider the diagonal morphism $\Delta_{X'/Y'} : X' \to X' \times_{Y'} X'$.
The base change of $\Delta_{X'/Y'}$ by $S \to S'$ is $\Delta_{X/Y}$
which is an isomorphism by assumption. By (\ref{item-isomorphism})
we conclude that $\Delta_{X'/Y'}$ is an isomorphism.
\medskip\noindent
Ad (\ref{item-surjective}). This is clear. See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-universally-injective}). Immediate from remark (II). See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-affine}). Assume $f$ is affine. Choose an
affine open $V' \subset Y'$ and set $U' = (f')^{-1}(V')$.
Then $V = Y \cap V'$ is affine which implies that
$U = Y \times_{Y'} U'$ is affine. By
Lemma \ref{lemma-thickening-affine-scheme}
we see that $U'$ is affine. Hence $f'$ is affine. See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-finite-type}). Via remark (I) comes down to proving $A' \to B'$
is of finite type if $A \to B$ is of finite type. Suppose that
$x_1, \ldots, x_n \in B'$ are elements whose images in $B$ generate $B$
as an $A$-algebra. Then $A'[x_1, \ldots, x_n] \to B$ is surjective
as both $A'[x_1, \ldots, x_n] \to B$ is surjective and
$I \otimes_R A[x_1, \ldots, x_n] \to I \otimes_R B$ is surjective. See also
Lemma \ref{lemma-thicken-property-morphisms-cartesian}
for a more general statement.
\medskip\noindent
Ad (\ref{item-quasi-finite}). Follows from (\ref{item-finite-type}) and that
quasi-finiteness of a morphism of finite type can be checked on fibres, see
Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-point-characterize}.
See also Lemma \ref{lemma-thicken-property-morphisms-cartesian}
for a more general statement.
\medskip\noindent
Ad (\ref{item-finite-presentation}). Via remark (I) comes down to proving
$A' \to B'$ is of finite presentation if $A \to B$ is of finite presentation.
We may assume that $B' = A'[x_1, \ldots, x_n]/K'$ for some ideal $K'$ by
(\ref{item-finite-type}). We get a short exact sequence
$$
0 \to K' \to A'[x_1, \ldots, x_n] \to B' \to 0
$$
As $B'$ is flat over $R'$ we see that $K' \otimes_{R'} R$ is the kernel of
the surjection $A[x_1, \ldots, x_n] \to B$. By assumption on $A \to B$ there
exist finitely many $f'_1, \ldots, f'_m \in K'$ whose images in
$A[x_1, \ldots, x_n]$ generate this kernel. Since $I$ is nilpotent we see
that $f'_1, \ldots, f'_m$ generate $K'$ by Nakayama's lemma, see
Algebra, Lemma \ref{algebra-lemma-NAK}.
\medskip\noindent
Ad (\ref{item-relative-dimension-d}). Follows from (\ref{item-finite-type})
and general remark (II). See also
Lemma \ref{lemma-thicken-property-morphisms-cartesian}
for a more general statement.
\medskip\noindent
Ad (\ref{item-universally-open}). Immediate from general remark (II). See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-syntomic}). Assume $f$ is syntomic. By
(\ref{item-finite-presentation}) $f'$ is locally of finite presentation,
by general remark (III) $f'$ is flat and the fibres of $f'$ are the fibres
of $f$. Hence $f'$ is syntomic by
Morphisms, Lemma \ref{morphisms-lemma-syntomic-flat-fibres}.
\medskip\noindent
Ad (\ref{item-smooth}). Assume $f$ is smooth. By
(\ref{item-finite-presentation}) $f'$ is locally of finite presentation,
by general remark (III) $f'$ is flat, and the fibres of $f'$ are the
fibres of $f$. Hence $f'$ is smooth by
Morphisms, Lemma \ref{morphisms-lemma-smooth-flat-smooth-fibres}.
\medskip\noindent
Ad (\ref{item-unramified}). Assume $f$ unramified. By
(\ref{item-finite-type}) $f'$ is locally of finite type
and the fibres of $f'$ are the fibres of $f$.
Hence $f'$ is unramified by
Morphisms, Lemma \ref{morphisms-lemma-unramified-etale-fibres}. See also
Lemma \ref{lemma-thicken-property-morphisms-cartesian}
for a more general statement.
\medskip\noindent
Ad (\ref{item-etale}). Assume $f$ \'etale. By
(\ref{item-finite-presentation}) $f'$ is locally of finite presentation,
by general remark (III) $f'$ is flat, and the fibres of $f'$ are the fibres
of $f$. Hence $f'$ is \'etale by
Morphisms, Lemma \ref{morphisms-lemma-etale-flat-etale-fibres}.
\medskip\noindent
Ad (\ref{item-proper}). This follows from a combination of
(\ref{item-separated}), (\ref{item-finite-type}), (\ref{item-quasi-compact}),
and (\ref{item-universally-closed}). See also
Lemma \ref{lemma-thicken-property-morphisms-cartesian}
for a more general statement.
\medskip\noindent
Ad (\ref{item-integral}). Combine (\ref{item-universally-closed}) and
(\ref{item-affine}) with
Morphisms, Lemma \ref{morphisms-lemma-integral-universally-closed}. See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-finite}). Combine (\ref{item-integral}),
and (\ref{item-finite-type}) with
Morphisms, Lemma \ref{morphisms-lemma-finite-integral}. See also
Lemma \ref{lemma-thicken-property-morphisms-cartesian}
for a more general statement.
\medskip\noindent
Ad (\ref{item-finite-locally-free}). Assume $f$ finite locally free. By
(\ref{item-finite}) we see that $f'$ is finite, by general remark (III)
$f'$ is flat, and by (\ref{item-finite-presentation}) $f'$ is locally of finite
presentation. Hence $f'$ is finite locally free by
Morphisms, Lemma \ref{morphisms-lemma-finite-flat}.
\end{proof}
\noindent
The following lemma is the ``locally nilpotent'' version of the
``crit\`ere de platitude par fibres'', see
Section \ref{section-criterion-flat-fibres}.
\begin{lemma}
\label{lemma-flatness-morphism-thickenings-fp-over-ft}
Consider a commutative diagram
$$
\xymatrix{
(X \subset X') \ar[rr]_{(f, f')} \ar[rd] & & (Y \subset Y') \ar[ld] \\
& (S \subset S')
}
$$
of thickenings. Assume
\begin{enumerate}
\item $Y' \to S'$ is locally of finite type,
\item $X' \to S'$ is flat and locally of finite presentation,
\item $f$ is flat, and
\item $X = S \times_{S'} X'$ and $Y = S \times_{S'} Y'$.
\end{enumerate}
Then $f'$ is flat and for all $y' \in Y'$ in the image of $f'$
the local ring $\mathcal{O}_{Y', y'}$ is
flat and essentially of finite presentation over $\mathcal{O}_{S', s'}$.
\end{lemma}
\begin{proof}
Immediate consequence of
Algebra, Lemma \ref{algebra-lemma-criterion-flatness-fibre-locally-nilpotent}.
\end{proof}
\noindent
Many properties of morphisms of schemes are preserved under flat
deformations as in the lemma above.
\begin{lemma}
\label{lemma-deform-property-fp-over-ft}
Consider a commutative diagram
$$
\xymatrix{
(X \subset X') \ar[rr]_{(f, f')} \ar[rd] & & (Y \subset Y') \ar[ld] \\
& (S \subset S')
}
$$
of thickenings. Assume $Y' \to S'$ locally of finite type,
$X' \to S'$ flat and locally of finite presentation,
$X = S \times_{S'} X'$, and $Y = S \times_{S'} Y'$. Then
\begin{enumerate}
\item $f$ is flat if and only if $f'$ is flat,
\label{item-flat-fp-over-ft}
\item $f$ is an isomorphism if and only if $f'$ is an isomorphism,
\label{item-isomorphism-fp-over-ft}
\item $f$ is an open immersion if and only if $f'$ is an open immersion,
\label{item-open-immersion-fp-over-ft}
\item $f$ is quasi-compact if and only if $f'$ is quasi-compact,
\label{item-quasi-compact-fp-over-ft}
\item $f$ is universally closed if and only if $f'$ is universally closed,
\label{item-universally-closed-fp-over-ft}
\item $f$ is (quasi-)separated if and only if $f'$ is (quasi-)separated,
\label{item-separated-fp-over-ft}
\item $f$ is a monomorphism if and only if $f'$ is a monomorphism,
\label{item-monomorphism-fp-over-ft}
\item $f$ is surjective if and only if $f'$ is surjective,
\label{item-surjective-fp-over-ft}
\item $f$ is universally injective if and only if $f'$ is universally injective,
\label{item-universally-injective-fp-over-ft}
\item $f$ is affine if and only if $f'$ is affine,
\label{item-affine-fp-over-ft}
\item $f$ is locally quasi-finite if and only if $f'$ is locally quasi-finite,
\label{item-quasi-finite-fp-over-ft}
\item
\label{item-relative-dimension-d-fp-over-ft}
$f$ is locally of finite type of relative dimension $d$ if and only if
$f'$ is locally of finite type of relative dimension $d$,
\item $f$ is universally open if and only if $f'$ is universally open,
\label{item-universally-open-fp-over-ft}
\item $f$ is syntomic if and only if $f'$ is syntomic,
\label{item-syntomic-fp-over-ft}
\item $f$ is smooth if and only if $f'$ is smooth,
\label{item-smooth-fp-over-ft}
\item $f$ is unramified if and only if $f'$ is unramified,
\label{item-unramified-fp-over-ft}
\item $f$ is \'etale if and only if $f'$ is \'etale,
\label{item-etale-fp-over-ft}
\item $f$ is proper if and only if $f'$ is proper,
\label{item-proper-fp-over-ft}
\item $f$ is finite if and only if $f'$ is finite,
\label{item-finite-fp-over-ft}
\item
\label{item-finite-locally-free-fp-over-ft}
$f$ is finite locally free (of rank $d$) if and only if $f'$
is finite locally free (of rank $d$), and
\item add more here.
\end{enumerate}
\end{lemma}
\begin{proof}
The assumptions on $X$ and $Y$ mean that $f$ is the base change of
$f'$ by $X \to X'$.
The properties $\mathcal{P}$ listed in (1) -- (20) above are all stable
under base change, hence if $f'$ has property $\mathcal{P}$, then so
does $f$. See
Schemes, Lemmas \ref{schemes-lemma-base-change-immersion},
\ref{schemes-lemma-quasi-compact-preserved-base-change},
\ref{schemes-lemma-separated-permanence}, and
\ref{schemes-lemma-base-change-monomorphism}
and
Morphisms, Lemmas
\ref{morphisms-lemma-base-change-surjective},
\ref{morphisms-lemma-base-change-universally-injective},
\ref{morphisms-lemma-base-change-affine},
\ref{morphisms-lemma-base-change-quasi-finite},
\ref{morphisms-lemma-base-change-relative-dimension-d},
\ref{morphisms-lemma-base-change-syntomic},
\ref{morphisms-lemma-base-change-smooth},
\ref{morphisms-lemma-base-change-unramified},
\ref{morphisms-lemma-base-change-etale},
\ref{morphisms-lemma-base-change-proper},
\ref{morphisms-lemma-base-change-finite}, and
\ref{morphisms-lemma-base-change-finite-locally-free}.
\medskip\noindent
The interesting direction in each case is therefore to assume
that $f$ has the property and deduce that $f'$ has it too.
We make a couple of general remarks which we will use without further
mention in the arguments below.
(I) Let $W' \subset S'$ be an affine open and let $U' \subset X'$
and $V' \subset Y'$ be affine opens lying over $W'$ with $f'(U') \subset V'$.
Let $W' = \Spec(R')$ and denote $I \subset R'$ be the ideal
defining the closed subscheme $W' \cap S$. Say $U' = \Spec(B')$
and $V' = \Spec(A')$. Then we get a commutative diagram
$$
\xymatrix{
0 \ar[r] &
IB' \ar[r] &
B' \ar[r] &
B \ar[r] & 0 \\
0 \ar[r] &
IA' \ar[r] \ar[u] &
A' \ar[r] \ar[u] &
A \ar[r] \ar[u] & 0
}
$$
with exact rows.
(II) The morphisms $X \to X'$ and $Y \to Y'$ are universal homeomorphisms.
Hence the topology of the maps $f$ and $f'$ (after any base change) is
identical.
(III) If $f$ is flat, then $f'$ is flat and $Y' \to S'$ is flat at every
point in the image of $f'$, see
Lemma \ref{lemma-flatness-morphism-thickenings}.
\medskip\noindent
Ad (\ref{item-flat-fp-over-ft}). This is general remark (III).
\medskip\noindent
Ad (\ref{item-isomorphism-fp-over-ft}). Assume $f$ is an isomorphism.
Choose an affine open $V' \subset Y'$ and set $U' = (f')^{-1}(V')$.
Then $V = Y \cap V'$ is affine which implies that
$V \cong f^{-1}(V) = U = Y \times_{Y'} U'$ is affine. By
Lemma \ref{lemma-thickening-affine-scheme}
we see that $U'$ is affine. Thus we have a diagram as in the
general remark (I). By Algebra, Lemma
\ref{algebra-lemma-isomorphism-modulo-locally-nilpotent}
we see that $A' \to B'$ is an isomorphism, i.e., $U' \cong V'$.
Thus $f'$ is an isomorphism.
\medskip\noindent
Ad (\ref{item-open-immersion-fp-over-ft}). Assume $f$ is an open immersion.
Then $f$ is an isomorphism of $X$ with an open subscheme $V \subset Y$.
Let $V' \subset Y'$ be the open subscheme whose underlying topological
space is $V$. Then $f'$ is a map from $X'$ to $V'$ which is an isomorphism by
(\ref{item-isomorphism-fp-over-ft}). Hence $f'$ is an open immersion.
\medskip\noindent
Ad (\ref{item-quasi-compact-fp-over-ft}). Immediate from remark (II). See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-universally-closed-fp-over-ft}). Immediate from remark (II). See
also Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-separated-fp-over-ft}). Note that
$X \times_Y X = Y \times_{Y'} (X' \times_{Y'} X')$ so that
$X' \times_{Y'} X'$ is a thickening of $X \times_Y X$.
Hence the topology of the maps $\Delta_{X/Y}$ and $\Delta_{X'/Y'}$
matches and we win. See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-monomorphism-fp-over-ft}). Assume $f$ is a monomorphism.
Consider the diagonal morphism $\Delta_{X'/Y'} : X' \to X' \times_{Y'} X'$.
Observe that $X' \times_{Y'} X' \to S'$ is locally of finite type.
The base change of $\Delta_{X'/Y'}$ by $S \to S'$ is $\Delta_{X/Y}$
which is an isomorphism by assumption. By (\ref{item-isomorphism-fp-over-ft})
we conclude that $\Delta_{X'/Y'}$ is an isomorphism.
\medskip\noindent
Ad (\ref{item-surjective-fp-over-ft}). This is clear. See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-universally-injective-fp-over-ft}).
Immediate from remark (II). See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-affine-fp-over-ft}). Assume $f$ is affine. Choose an
affine open $V' \subset Y'$ and set $U' = (f')^{-1}(V')$.
Then $V = Y \cap V'$ is affine which implies that
$U = Y \times_{Y'} U'$ is affine. By
Lemma \ref{lemma-thickening-affine-scheme}
we see that $U'$ is affine. Hence $f'$ is affine. See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-quasi-finite-fp-over-ft}). Follows from the fact that $f'$
is locally of finite type
(by Morphisms, Lemma \ref{morphisms-lemma-permanence-finite-type}) and that
quasi-finiteness of a morphism of finite type can be checked on fibres, see
Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-point-characterize}.
\medskip\noindent
Ad (\ref{item-relative-dimension-d-fp-over-ft}).
Follows from general remark (II) and the fact that $f'$
is locally of finite type
(Morphisms, Lemma \ref{morphisms-lemma-permanence-finite-type}).
\medskip\noindent
Ad (\ref{item-universally-open-fp-over-ft}).
Immediate from general remark (II). See also
Lemma \ref{lemma-thicken-property-morphisms} for a more general statement.
\medskip\noindent
Ad (\ref{item-syntomic-fp-over-ft}). Assume $f$ is syntomic. By
Morphisms, Lemma \ref{morphisms-lemma-finite-presentation-permanence}
$f'$ is locally of finite presentation.
By general remark (III) $f'$ is flat. The fibres of $f'$ are the fibres
of $f$. Hence $f'$ is syntomic by
Morphisms, Lemma \ref{morphisms-lemma-syntomic-flat-fibres}.
\medskip\noindent
Ad (\ref{item-smooth-fp-over-ft}). Assume $f$ is smooth. By
Morphisms, Lemma \ref{morphisms-lemma-finite-presentation-permanence}
$f'$ is locally of finite presentation.
By general remark (III) $f'$ is flat. The fibres of $f'$ are the
fibres of $f$. Hence $f'$ is smooth by
Morphisms, Lemma \ref{morphisms-lemma-smooth-flat-smooth-fibres}.
\medskip\noindent
Ad (\ref{item-unramified-fp-over-ft}). Assume $f$ unramified. By
Morphisms, Lemma \ref{morphisms-lemma-permanence-finite-type}
$f'$ is locally of finite type. The fibres of $f'$ are the fibres of $f$.
Hence $f'$ is unramified by
Morphisms, Lemma \ref{morphisms-lemma-unramified-etale-fibres}.
\medskip\noindent
Ad (\ref{item-etale-fp-over-ft}). Assume $f$ \'etale. By
Morphisms, Lemma \ref{morphisms-lemma-finite-presentation-permanence}
$f'$ is locally of finite presentation.
By general remark (III) $f'$ is flat.
The fibres of $f'$ are the fibres of $f$. Hence $f'$ is \'etale by
Morphisms, Lemma \ref{morphisms-lemma-etale-flat-etale-fibres}.
\medskip\noindent
Ad (\ref{item-proper-fp-over-ft}). This follows from a combination of
(\ref{item-separated-fp-over-ft}), the fact that $f$ is locally
of finite type (Morphisms, Lemma \ref{morphisms-lemma-permanence-finite-type}),
(\ref{item-quasi-compact-fp-over-ft}),
and (\ref{item-universally-closed-fp-over-ft}).
\medskip\noindent
Ad (\ref{item-finite-fp-over-ft}).
Combine (\ref{item-universally-closed-fp-over-ft}),
(\ref{item-affine-fp-over-ft}),
Morphisms, Lemma \ref{morphisms-lemma-integral-universally-closed},
the fact that $f$ is locally of finite type
(Morphisms, Lemma \ref{morphisms-lemma-permanence-finite-type}), and
Morphisms, Lemma \ref{morphisms-lemma-finite-integral}.
\medskip\noindent
Ad (\ref{item-finite-locally-free-fp-over-ft}).
Assume $f$ finite locally free. By
(\ref{item-finite-fp-over-ft}) we see that $f'$ is finite.
By general remark (III) $f'$ is flat.
By Morphisms, Lemma \ref{morphisms-lemma-finite-presentation-permanence}
$f'$ is locally of finite presentation. Hence $f'$ is finite locally free by
Morphisms, Lemma \ref{morphisms-lemma-finite-flat}.
\end{proof}
\begin{lemma}[Deformations of projective schemes]
\label{lemma-deform-projective}
Let $f : X \to S$ be a morphism of schemes which is proper, flat, and
of finite presentation. Let $\mathcal{L}$ be $f$-ample. Assume
$S$ is quasi-compact. There exists a $d_0 \geq 0$ such that
for every cartesian diagram
$$
\vcenter{
\xymatrix{
X \ar[r]_{i'} \ar[d]_f & X' \ar[d]^{f'} \\
S \ar[r]^i & S'
}
}
\quad\text{and}\quad
\begin{matrix}
\text{invertible }\mathcal{O}_{X'}\text{-module}\\
\mathcal{L}'\text{ with }\mathcal{L} \cong (i')^*\mathcal{L}'
\end{matrix}
$$
where $S \subset S'$ is a thickening and $f'$ is
proper, flat, of finite presentation we have
\begin{enumerate}
\item $R^p(f')_*(\mathcal{L}')^{\otimes d} = 0$
for all $p > 0$ and $d \geq d_0$,
\item $\mathcal{A}'_d = (f')_*(\mathcal{L}')^{\otimes d}$
is finite locally free for $d \geq d_0$,
\item $\mathcal{A}' =
\mathcal{O}_{S'} \oplus \bigoplus_{d \geq d_0} \mathcal{A}'_d$
is a quasi-coherent $\mathcal{O}_{S'}$-algebra of finite presentation,
\item there is a canonical isomorphism
$r' : X' \to \underline{\text{Proj}}_{S'}(\mathcal{A}')$, and
\item there is a canonical isomorphism
$\theta' : (r')^*\mathcal{O}_{\underline{\text{Proj}}_{S'}(\mathcal{A}')}(1)
\to \mathcal{L}'$.
\end{enumerate}
The construction of $\mathcal{A}'$, $r'$, $\theta'$
is functorial in the data $(X', S', i, i', f', \mathcal{L}')$.
\end{lemma}
\begin{proof}
We first describe the maps $r'$ and $\theta'$.
Observe that $\mathcal{L}'$ is $f'$-ample, see
Lemma \ref{lemma-thicken-property-relatively-ample}.
There is a canonical map of quasi-coherent graded
$\mathcal{O}_{S'}$-algebras
$\mathcal{A}' \to \bigoplus_{d \geq 0} (f')_*(\mathcal{L}')^{\otimes d}$
which is an isomorphism in degrees $\geq d_0$.
Hence this induces an isomorphism on relative Proj
compatible with the Serre twists of the structure sheaf, see
Constructions, Lemma
\ref{constructions-lemma-eventual-iso-graded-rings-map-relative-proj}.
Hence we get the morphism $r'$ by
Morphisms, Lemma \ref{morphisms-lemma-characterize-relatively-ample}
(which in turn appeals to the construction given in
Constructions, Lemma
\ref{constructions-lemma-invertible-map-into-relative-proj})
and it is an isomorphism by
Morphisms, Lemma \ref{morphisms-lemma-proper-ample-is-proj}.
We get the map $\theta'$ from Constructions, Lemma
\ref{constructions-lemma-invertible-map-into-relative-proj}.
By Properties, Lemma \ref{properties-lemma-ample-gcd-is-one}
we find that $\theta'$ is an isomorphism
(this also uses that the morphism $r'$ over affine
opens of $S'$ is the same as the morphism from
Properties, Lemma \ref{properties-lemma-map-into-proj}
as is explained in the proof
of Morphisms, Lemma \ref{morphisms-lemma-proper-ample-is-proj}).
\medskip\noindent
Assuming the vanishing and local freeness stated in parts
(1) and (2), the functoriality of the construction can be seen as follows.
Suppose that $h : T \to S'$ is a morphism of schemes, denote
$f_T : X'_T \to T$ the base change of $f'$ and
$\mathcal{L}_T$ the pullback of $\mathcal{L}$ to $X'_T$.
By cohomology and base change
(as formulated in Derived Categories of Schemes,
Lemma \ref{perfect-lemma-compare-base-change} for example)
we have the corresponding vanishing over $T$ and moreover
$h^*\mathcal{A}'_d = f_{T, *}\mathcal{L}_T^{\otimes d}$
(and thus the local freeness of pushforwards as well
as the finite generation of the corresponding graded
$\mathcal{O}_T$-algebra $\mathcal{A}_T$).
Hence the morphism
$r_T : X_T \to
\underline{\text{Proj}}_T(\bigoplus f_{T, *}\mathcal{L}_T^{\otimes d})$
is simply the base change of $r'$ to $T$ and the pullback of
$\theta'$ is the map $\theta_T$.
\medskip\noindent
Having said all of the above, we see that it suffices to prove
(1), (2), and (3). Pick $d_0$ such that
$R^pf_*\mathcal{L}^{\otimes d} = 0$ for all $d \geq d_0$ and $p > 0$, see
Cohomology of Schemes, Lemma \ref{coherent-lemma-coherent-proper-ample}.
We claim that $d_0$ works.
\medskip\noindent
By cohomology and base change
(Derived Categories of Schemes,
Lemma \ref{perfect-lemma-flat-proper-perfect-direct-image-general})
we see that $E'_d = Rf'_*(\mathcal{L}')^{\otimes d}$
is a perfect object of $D(\mathcal{O}_{S'})$
and its formation commutes with arbitrary base change.
In particular, $E_d = Li^*E'_d = Rf_*\mathcal{L}^{\otimes d}$.
By Derived Categories of Schemes, Lemma
\ref{perfect-lemma-vanishing-implies-locally-free}
we see that for $d \geq d_0$ the complex $E_d$ is isomorphic to
the finite locally free $\mathcal{O}_S$-module
$f_*\mathcal{L}^{\otimes d}$ placed in
cohomological degree $0$. Then by
Derived Categories of Schemes, Lemma
\ref{perfect-lemma-open-where-cohomology-in-degree-i-rank-r}
we conclude that $E'_d$ is isomorphic to a finite locally free
module placed in cohomological degree $0$.
Of course this means that $E'_d = \mathcal{A}'_d[0]$,
that $R^pf'_*(\mathcal{L}')^{\otimes d} = 0$ for $p > 0$,
and that $\mathcal{A}'_d$ is finite locally free.
This proves (1) and (2).
\medskip\noindent
The last thing we have to show is finite presentation of
$\mathcal{A}'$ as a sheaf of $\mathcal{O}_{S'}$-algebras
(this notion was introduced in Properties, Section
\ref{properties-section-extending-quasi-coherent-sheaves}).
Let $U' = \Spec(R') \subset S'$ be an affine open.
Then $A' = \mathcal{A}'(U')$ is a graded $R'$-algebra
whose graded parts are finite projective $R'$-modules.
We have to show that $A'$ is a finitely presented $R'$-algebra.
We will prove this by reduction to the Noetherian case.
Namely, we can find a finite type $\mathbf{Z}$-subalgebra
$R'_0 \subset R'$ and a pair\footnote{With the same properties
as those enjoyed by $X' \to S'$ and $\mathcal{L}'$, i.e.,
$X'_0 \to \Spec(R'_0)$ is flat and proper and $\mathcal{L}'_0$
is ample.} $(X'_0, \mathcal{L}'_0)$ over $R'_0$
whose base change is $(X'_{U'}, \mathcal{L}'|_{X'_{U'}})$, see
Limits, Lemmas
\ref{limits-lemma-descend-modules-finite-presentation},
\ref{limits-lemma-descend-invertible-modules},
\ref{limits-lemma-eventually-proper},
\ref{limits-lemma-descend-flat-finite-presentation}, and
\ref{limits-lemma-limit-ample}.
Cohomology of Schemes, Lemma \ref{coherent-lemma-coherent-proper-ample}
implies
$A'_0 = \bigoplus_{d \geq 0} H^0(X'_0, (\mathcal{L}'_0)^{\otimes d})$
is a finitely generated graded $R'_0$-algebra and implies
there exists a $d'_0$ such that
$H^p(X'_0, (\mathcal{L}'_0)^{\otimes d}) = 0$, $p > 0$ for $d \geq d'_0$.
By the arguments given above applied to $X'_0 \to \Spec(R'_0)$ and
$\mathcal{L}'_0$ we see that $(A'_0)_d$ is a finite projective $R'_0$-module
and that
$$
A'_d = \mathcal{A}'_d(U') =
H^0(X'_{U'}, (\mathcal{L}')^{\otimes d}|_{X'_{U'}}) =
H^0(X'_0, (\mathcal{L}'_0)^{\otimes d}) \otimes_{R'_0} R' =
(A'_0)_d \otimes_{R'_0} R'
$$
for $d \geq d'_0$. Now a small twist in the argument is that we
don't know that we can choose $d'_0$ equal to $d_0$\footnote{Actually,
one can reduce to this case by doing more limit arguments.}. To
get around this we use the following sequence of arguments to finish
the proof:
\begin{enumerate}
\item[(a)] The algebra
$B = R'_0 \oplus \bigoplus_{d \geq \max(d_0, d'_0)} (A'_0)_d$
is an $R'_0$-algebra of finite type: apply
the Artin-Tate lemma to $B \subset A'_0$, see
Algebra, Lemma \ref{algebra-lemma-Artin-Tate}.
\item[(b)] As $R'_0$ is Noetherian we see that
$B$ is an $R'_0$-algebra of finite presentation.
\item[(c)] By right exactness of tensor product we see that
$B \otimes_{R'_0} R'$ is an $R'$-algebra of finite presentation.
\item[(d)] By the displayed equalities this exactly says that
$C = R' \oplus \bigoplus_{d \geq \max(d_0, d'_0)} A'_d$
is an $R'$-algebra of finite presentation.
\item[(e)] The quotient $A'/C$ is the direct sum of the finite
projective $R'$-modules $A'_d$, $d_0 \leq d \leq \max(d_0, d'_0)$,
hence finitely presented as $R'$-module.
\item[(f)] The quotient $A'/C$ is finitely presented
as a $C$-module by Algebra, Lemma
\ref{algebra-lemma-finitely-presented-over-subring}.
\item[(g)] Thus $A'$ is finitely presented as a $C$-module by
Algebra, Lemma \ref{algebra-lemma-extension}.
\item[(h)] By Algebra, Lemma \ref{algebra-lemma-finite-finite-type}
this implies $A'$ is finitely presented as a $C$-algebra.
\item[(i)] Finally, by
Algebra, Lemma \ref{algebra-lemma-compose-finite-type}
applied to $R' \to C \to A'$
this implies $A'$ is finitely presented as an $R'$-algebra.
\end{enumerate}
This finishes the proof.
\end{proof}
\section{Formally smooth morphisms}
\label{section-formally-smooth}
\noindent
Michael Artin's position on differential criteria of smoothness (e.g.,
Morphisms, Lemma \ref{morphisms-lemma-smooth-at-point}) is that they are
basically useless (in practice). In this section we introduce the
notion of a formally smooth morphism $X \to S$. Such a morphism is
characterized by the property that $T$-valued points of $X$ lift
to infinitesimal thickenings of $T$ provided $T$ is affine.
The main result is that a morphism which is formally smooth and
locally of finite presentation is smooth, see
Lemma \ref{lemma-smooth-formally-smooth}.
It turns out that this criterion is often easier to use than the
differential criteria mentioned above.
\medskip\noindent
Recall that a ring map $R \to A$ is called {\it formally smooth}
(see Algebra, Definition \ref{algebra-definition-formally-smooth})
if for every commutative solid diagram
$$
\xymatrix{
A \ar[r] \ar@{-->}[rd] & B/I \\
R \ar[r] \ar[u] & B \ar[u]
}
$$
where $I \subset B$ is an ideal of square zero, a dotted
arrow exists which makes the diagram commute. This motivates
the following analogue for morphisms of schemes.
\begin{definition}
\label{definition-formally-smooth}
Let $f : X \to S$ be a morphism of schemes.
We say $f$ is {\it formally smooth} if given any solid commutative diagram
$$
\xymatrix{
X \ar[d]_f & T \ar[d]^i \ar[l] \\
S & T' \ar[l] \ar@{-->}[lu]
}
$$
where $T \subset T'$ is a first order thickening of affine schemes over $S$
there exists a dotted arrow making the diagram commute.
\end{definition}
\noindent
In the cases of formally unramified and formally \'etale morphisms
the condition that $T'$ be affine could be dropped, see
Lemmas \ref{lemma-formally-unramified-not-affine} and
\ref{lemma-formally-etale-not-affine}.
This is no longer true in the case of formally smooth morphisms.
In fact, a slightly more natural condition would be that we should be
able to fill in the dotted arrow Zariski locally on $T'$.
In fact, analyzing the proof of
Lemma \ref{lemma-formally-smooth}
shows that this would be equivalent to the definition as it currently
stands. In particular, the being formally smooth is
Zariski local on the source (and in fact it is smooth local on the soure,
insert future reference here).
\begin{lemma}
\label{lemma-composition-formally-smooth}
A composition of formally smooth morphisms is formally smooth.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-base-change-formally-smooth}
A base change of a formally smooth morphism is formally smooth.
\end{lemma}
\begin{proof}
Omitted, but see Algebra, Lemma \ref{algebra-lemma-base-change-fs}
for the algebraic version.
\end{proof}
\begin{lemma}
\label{lemma-formally-etale-unramified-smooth}
Let $f : X \to S$ be a morphism of schemes.
Then $f$ is formally \'etale if and only if
$f$ is formally smooth and formally unramified.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-formally-smooth-on-opens}
Let $f : X \to S$ be a morphism of schemes.
Let $U \subset X$ and $V \subset S$ be open subschemes such that
$f(U) \subset V$. If $f$ is formally smooth, so is $f|_U : U \to V$.
\end{lemma}
\begin{proof}
Consider a solid diagram
$$
\xymatrix{
U \ar[d]_{f|_U} & T \ar[d]^i \ar[l]^a \\
V & T' \ar[l] \ar@{-->}[lu]
}
$$
as in Definition \ref{definition-formally-smooth}. If $f$ is formally
smooth, then there exists an $S$-morphism $a' : T' \to X$ such that
$a'|_T = a$. Since the underlying sets of $T$ and $T'$ are the same
we see that $a'$ is a morphism into $U$ (see Schemes, Section
\ref{schemes-section-open-immersion}). And it clearly is a $V$-morphism
as well. Hence the dotted arrow above as desired.
\end{proof}
\begin{lemma}
\label{lemma-affine-formally-smooth}
Let $f : X \to S$ be a morphism of schemes.
Assume $X$ and $S$ are affine.
Then $f$ is formally smooth if and only if
$\mathcal{O}_S(S) \to \mathcal{O}_X(X)$ is a formally smooth
ring map.
\end{lemma}
\begin{proof}
This is immediate from the definitions
(Definition \ref{definition-formally-smooth} and
Algebra, Definition \ref{algebra-definition-formally-smooth})
by the equivalence of categories of rings and affine schemes,
see
Schemes, Lemma \ref{schemes-lemma-category-affine-schemes}.
\end{proof}
\noindent
The following lemma is the main result of this section. It is a victory of the
functorial point of view in that it implies (combined with
Limits,
Proposition \ref{limits-proposition-characterize-locally-finite-presentation})
that we can recognize whether a morphism $f : X \to S$ is smooth in terms of
``simple'' properties of the functor $h_X : \Sch/S \to \textit{Sets}$.
\begin{lemma}[Infinitesimal lifting criterion]
\label{lemma-smooth-formally-smooth}
Let $f : X \to S$ be a morphism of schemes.
The following are equivalent:
\begin{enumerate}
\item The morphism $f$ is smooth, and
\item the morphism $f$ is locally of finite presentation and
formally smooth.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume $f : X \to S$ is locally of finite presentation and formally smooth.
Consider a pair of affine opens $\Spec(A) = U \subset X$ and
$\Spec(R) = V \subset S$
such that $f(U) \subset V$. By Lemma \ref{lemma-formally-smooth-on-opens}
we see that $U \to V$ is formally smooth. By Lemma
\ref{lemma-affine-formally-smooth} we see that $R \to A$ is formally
smooth. By
Morphisms, Lemma \ref{morphisms-lemma-locally-finite-presentation-characterize}
we see that $R \to A$ is of finite presentation.
By Algebra, Proposition \ref{algebra-proposition-smooth-formally-smooth}
we see that $R \to A$ is smooth.
Hence by the definition of a smooth morphism we see that $X \to S$ is smooth.
\medskip\noindent
Conversely, assume that $f : X \to S$ is smooth. Consider a solid commutative
diagram
$$
\xymatrix{
X \ar[d]_f & T \ar[d]^i \ar[l]^a \\
S & T' \ar[l] \ar@{-->}[lu]
}
$$
as in Definition \ref{definition-formally-smooth}.
We will show the dotted arrow exists thereby
proving that $f$ is formally smooth.
\medskip\noindent
Let $\mathcal{F}$ be the sheaf of sets on $T'$ of Lemma \ref{lemma-sheaf}
in the special case discussed in Remark \ref{remark-special-case}.
Let
$$
\mathcal{H} =
\SheafHom_{\mathcal{O}_T}(a^*\Omega_{X/S}, \mathcal{C}_{T/T'})
$$
be the sheaf of $\mathcal{O}_T$-modules with action
$\mathcal{H} \times \mathcal{F} \to \mathcal{F}$ as in
Lemma \ref{lemma-action-sheaf}. Our goal is simply
to show that $\mathcal{F}(T) \not = \emptyset$. In other words we
are trying to show that $\mathcal{F}$ is a trivial $\mathcal{H}$-torsor
on $T$ (see Cohomology, Section \ref{cohomology-section-h1-torsors}).
There are two steps: (I) To show that $\mathcal{F}$ is a torsor
we have to show that $\mathcal{F}_t \not = \emptyset$ for all $t \in T$ (see
Cohomology, Definition \ref{cohomology-definition-torsor}).
(II) To show that $\mathcal{F}$ is the trivial torsor it suffices
to show that $H^1(T, \mathcal{H}) = 0$ (see
Cohomology, Lemma \ref{cohomology-lemma-torsors-h1} --
we may use either cohomology
of $\mathcal{H}$ as an abelian sheaf or as an $\mathcal{O}_T$-module,
see Cohomology, Lemma \ref{cohomology-lemma-modules-abelian}).
\medskip\noindent
First we prove (I). To see this, for every $t \in T$ we can
choose an affine open $U \subset T$ neighbourhood of $t$
such that $a(U)$ is contained
in an affine open $\Spec(A) = W \subset X$
which maps to an affine open $\Spec(R) = V \subset S$.
By Morphisms, Lemma \ref{morphisms-lemma-smooth-characterize}
the ring map $R \to A$ is smooth.
Hence by Algebra, Proposition \ref{algebra-proposition-smooth-formally-smooth}
the ring map $R \to A$ is formally smooth.
Lemma \ref{lemma-affine-formally-smooth}
in turn implies that $W \to V$ is formally smooth.
Hence we can lift $a|_U : U \to W$ to a $V$-morphism
$a' : U' \to W \subset X$ showing that $\mathcal{F}(U) \not = \emptyset$.
\medskip\noindent
Finally we prove (II).
By Morphisms, Lemma \ref{morphisms-lemma-finite-presentation-differentials}
we see that $\Omega_{X/S}$ is of finite presentation
(it is even finite locally free by
Morphisms, Lemma \ref{morphisms-lemma-smooth-omega-finite-locally-free}).
Hence $a^*\Omega_{X/S}$ is of finite presentation (see
Modules, Lemma \ref{modules-lemma-pullback-finite-presentation}).
Hence the sheaf
$\mathcal{H} =
\SheafHom_{\mathcal{O}_T}(a^*\Omega_{X/S}, \mathcal{C}_{T/T'})$
is quasi-coherent by the discussion in
Schemes, Section \ref{schemes-section-quasi-coherent}.
Thus by Cohomology of Schemes, Lemma
\ref{coherent-lemma-quasi-coherent-affine-cohomology-zero}
we have $H^1(T, \mathcal{H}) = 0$ as desired.
\end{proof}
\noindent
Locally projective quasi-coherent modules are defined in
Properties, Section \ref{properties-section-locally-projective}.
\begin{lemma}
\label{lemma-formally-smooth-sheaf-differentials}
Let $f : X \to Y$ be a formally smooth morphism of schemes.
Then $\Omega_{X/Y}$ is locally projective on $X$.
\end{lemma}
\begin{proof}
Choose $U \subset X$ and $V \subset Y$ affine open such that
$f(U) \subset V$. By
Lemma \ref{lemma-formally-smooth-on-opens}
$f|_U : U \to V$ is formally smooth. Hence
$\Gamma(V, \mathcal{O}_V) \to \Gamma(U, \mathcal{O}_U)$ is
a formally smooth ring map, see
Lemma \ref{lemma-affine-formally-smooth}.
Hence by
Algebra, Lemma \ref{algebra-lemma-characterize-formally-smooth-again}
the $\Gamma(U, \mathcal{O}_U)$-module
$\Omega_{\Gamma(U, \mathcal{O}_U)/\Gamma(V, \mathcal{O}_V)}$
is projective. Hence $\Omega_{U/V}$ is locally projective, see
Properties, Section \ref{properties-section-locally-projective}.
\end{proof}
\begin{lemma}
\label{lemma-h1-is-zero}
Let $T$ be an affine scheme. Let $\mathcal{F}$, $\mathcal{G}$ be quasi-coherent
$\mathcal{O}_T$-modules. Consider
$\mathcal{H} = \SheafHom_{\mathcal{O}_T}(\mathcal{F}, \mathcal{G})$.
If $\mathcal{F}$ is locally projective, then $H^1(T, \mathcal{H}) = 0$.
\end{lemma}
\begin{proof}
By the definition of a locally projective sheaf on a scheme (see
Properties, Definition \ref{properties-definition-locally-projective})
we see that $\mathcal{F}$ is a direct summand of a free
$\mathcal{O}_T$-module. Hence we may assume that
$\mathcal{F} = \bigoplus_{i \in I} \mathcal{O}_T$ is a free module.
In this case $\mathcal{H} = \prod_{i \in I} \mathcal{G}$ is
a product of quasi-coherent modules. By
Cohomology, Lemma \ref{cohomology-lemma-cohomology-products}
we conclude that $H^1 = 0$ because the cohomology of a quasi-coherent sheaf
on an affine scheme is zero, see Cohomology of Schemes, Lemma
\ref{coherent-lemma-quasi-coherent-affine-cohomology-zero}.
\end{proof}
\begin{lemma}
\label{lemma-formally-smooth}
Let $f : X \to Y$ be a morphism of schemes. The following are equivalent:
\begin{enumerate}
\item $f$ is formally smooth,
\item for every $x \in X$ there exist opens $x \in U \subset X$ and
$f(x) \in V \subset Y$ with $f(U) \subset V$ such that
$f|_U : U \to V$ is formally smooth,
\item for every pair of affine opens $U \subset X$ and $V \subset Y$
with $f(U) \subset V$ the ring map $\mathcal{O}_Y(V) \to \mathcal{O}_X(U)$
is formally smooth, and
\item there exists an affine open covering $Y = \bigcup V_j$ and
for each $j$ an affine open covering $f^{-1}(V_j) = \bigcup U_{ji}$
such that $\mathcal{O}_Y(V) \to \mathcal{O}_X(U)$ is a formally smooth
ring map for all $j$ and $i$.
\end{enumerate}
\end{lemma}
\begin{proof}
The implications (1) $\Rightarrow$ (2),
(1) $\Rightarrow$ (3), and (2) $\Rightarrow$ (4) follow from
Lemma \ref{lemma-formally-smooth-on-opens}.
The implication (3) $\Rightarrow$ (4) is immediate.
\medskip\noindent
Assume (4). The proof that $f$ is formally smooth is the same
as the second part of the proof of Lemma \ref{lemma-smooth-formally-smooth}.
Consider a solid commutative diagram
$$
\xymatrix{
X \ar[d]_f & T \ar[d]^i \ar[l]^a \\
Y & T' \ar[l] \ar@{-->}[lu]
}
$$
as in Definition \ref{definition-formally-smooth}.
We will show the dotted arrow exists thereby
proving that $f$ is formally smooth.
Let $\mathcal{F}$ be the sheaf of sets on $T'$ of
Lemma \ref{lemma-sheaf} as in the special case discussed in
Remark \ref{remark-special-case}.
Let
$$
\mathcal{H} =
\SheafHom_{\mathcal{O}_T}(a^*\Omega_{X/Y}, \mathcal{C}_{T/T'})
$$
be the sheaf of $\mathcal{O}_T$-modules on $T$
with action $\mathcal{H} \times \mathcal{F} \to \mathcal{F}$ as in
Lemma \ref{lemma-action-sheaf}.
The action $\mathcal{H} \times \mathcal{F} \to \mathcal{F}$
turns $\mathcal{F}$ into a pseudo $\mathcal{H}$-torsor, see
Cohomology, Definition \ref{cohomology-definition-torsor}.
Our goal is to show that $\mathcal{F}$ is a trivial $\mathcal{H}$-torsor.
There are two steps: (I) To show that $\mathcal{F}$ is a torsor
we have to show that $\mathcal{F}$ locally has a
section. (II) To show that $\mathcal{F}$ is the trivial torsor
it suffices to show that $H^1(T, \mathcal{H}) = 0$, see
Cohomology, Lemma \ref{cohomology-lemma-torsors-h1}.
\medskip\noindent
First we prove (I). To see this, for every $t \in T$ we can
choose an affine open $W \subset T$ neighbourhood of $t$
such that $a(W)$ is contained in $U_{ji}$ for some $i, j$.
Let $W' \subset T'$ be the corresponding open subscheme.
By assumption (4) we can lift $a|_W : W \to U_{ji}$
to a $V_j$-morphism $a' : W' \to U_{ji}$ showing that
$\mathcal{F}(W')$ is nonempty.
\medskip\noindent
Finally we prove (II). By
Lemma \ref{lemma-formally-smooth-sheaf-differentials}
we see that $\Omega_{U_{ji}/V_j}$ locally projective.
Hence $\Omega_{X/Y}$ is locally projective, see
Properties, Lemma \ref{properties-lemma-locally-projective}.
Hence $a^*\Omega_{X/Y}$ is locally projective, see
Properties, Lemma \ref{properties-lemma-locally-projective-pullback}.
Hence
$$
H^1(T, \mathcal{H}) =
H^1(T, \SheafHom_{\mathcal{O}_T}(a^*\Omega_{X/Y}, \mathcal{C}_{T/T'}) = 0
$$
by
Lemma \ref{lemma-h1-is-zero}
as desired.
\end{proof}
\begin{lemma}
\label{lemma-triangle-differentials-formally-smooth}
Let $f : X \to Y$, $g : Y \to S$ be morphisms of schemes.
Assume $f$ is formally smooth. Then
$$
0 \to f^*\Omega_{Y/S} \to \Omega_{X/S} \to \Omega_{X/Y} \to 0
$$
(see
Morphisms, Lemma \ref{morphisms-lemma-triangle-differentials})
is short exact.
\end{lemma}
\begin{proof}
The algebraic version of this lemma is the following:
Given ring maps $A \to B \to C$ with $B \to C$ formally smooth, then
the sequence
$$
0 \to C \otimes_B \Omega_{B/A} \to \Omega_{C/A} \to \Omega_{C/B} \to 0
$$
of
Algebra, Lemma \ref{algebra-lemma-exact-sequence-differentials}
is exact. This is
Algebra, Lemma \ref{algebra-lemma-ses-formally-smooth}.
\end{proof}
\begin{lemma}
\label{lemma-differentials-formally-unramified-formally-smooth}
Let $h : Z \to X$ be a formally unramified morphism of schemes over $S$.
Assume that $Z$ is formally smooth over $S$. Then the
canonical exact sequence
$$
0 \to \mathcal{C}_{Z/X} \to i^*\Omega_{X/S} \to \Omega_{Z/S} \to 0
$$
of
Lemma \ref{lemma-universally-unramified-differentials-sequence}
is short exact.
\end{lemma}
\begin{proof}
Let $Z \to Z'$ be the universal first order thickening of $Z$ over $X$.
From the proof of
Lemma \ref{lemma-universally-unramified-differentials-sequence}
we see that our sequence is identified with the sequence
$$
\mathcal{C}_{Z/Z'} \to \Omega_{Z'/S} \otimes \mathcal{O}_Z \to
\Omega_{Z/S} \to 0.
$$
Since $Z \to S$ is formally smooth we can locally on $Z'$ find
a left inverse $Z' \to Z$ over $S$ to the inclusion map $Z \to Z'$.
Thus the sequence is locally split, see
Morphisms, Lemma \ref{morphisms-lemma-differentials-relative-immersion-section}.
\end{proof}
\begin{lemma}
\label{lemma-two-unramified-morphisms-formally-smooth}
Let
$$
\xymatrix{
Z \ar[r]_i \ar[rd]_j & X \ar[d]^f \\
& Y
}
$$
be a commutative diagram of schemes where $i$ and $j$ are formally
unramified and $f$ is formally smooth. Then the canonical exact sequence
$$
0 \to
\mathcal{C}_{Z/Y} \to
\mathcal{C}_{Z/X} \to
i^*\Omega_{X/Y} \to 0
$$
of
Lemma \ref{lemma-two-unramified-morphisms}
is exact and locally split.
\end{lemma}
\begin{proof}
Denote $Z \to Z'$ the universal first order thickening of $Z$ over $X$.
Denote $Z \to Z''$ the universal first order thickening of $Z$ over $Y$.
By
Lemma \ref{lemma-universally-unramified-differentials-sequence}
here is a canonical morphism $Z' \to Z''$ so that we have a commutative
diagram
$$
\xymatrix{
Z \ar[r]_{i'} \ar[rd]_{j'} & Z' \ar[r]_a \ar[d]^k & X \ar[d]^f \\
& Z'' \ar[r]^b & Y
}
$$
In the proof of
Lemma \ref{lemma-two-unramified-morphisms}
we identified the sequence above with the sequence
$$
\mathcal{C}_{Z/Z''} \to
\mathcal{C}_{Z/Z'} \to
(i')^*\Omega_{Z'/Z''} \to 0
$$
Let $U'' \subset Z''$ be an affine open. Denote $U \subset Z$ and
$U' \subset Z'$ the corresponding affine open subschemes.
As $f$ is formally smooth there exists a morphism $h : U'' \to X$
which agrees with $i$ on $U$ and such that $f \circ h$ equals $b|_{U''}$.
Since $Z'$ is the universal first order thickening we obtain a unique
morphism $g : U'' \to Z'$ such that $g = a \circ h$. The universal
property of $Z''$ implies that $k \circ g$ is the inclusion map
$U'' \to Z''$. Hence $g$ is a left inverse to $k$. Picture
$$
\xymatrix{
U \ar[d] \ar[r] & Z' \ar[d]^k \\
U'' \ar[r] \ar[ru]^g & Z''
}
$$
Thus $g$ induces a map $\mathcal{C}_{Z/Z'}|_U \to \mathcal{C}_{Z/Z''}|_U$
which is a left inverse to the map
$\mathcal{C}_{Z/Z''} \to \mathcal{C}_{Z/Z'}$ over $U$.
\end{proof}
\section{Smoothness over a Noetherian base}
\label{section-smooth-Noetherian}
\noindent