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more-morphisms.tex
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more-morphisms.tex
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\input{preamble}
% OK, start here.
%
\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, 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 (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-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
}