stacks/stacks-project

 \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, \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