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92.10 Morphisms of Schemes

The deformation theory of morphisms of schemes. Of course this is just an example of deformations of diagrams of schemes.

Example 92.10.1 (Morphisms of schemes). Let $\mathcal{F}$ be the category defined as follows

  1. an object is a pair $(A, X \to Y)$ consisting of an object $A$ of $\mathcal{C}_\Lambda $ and a morphism $X \to Y$ of schemes over $A$ with both $X$ and $Y$ flat over $A$, and

  2. a morphism $(f, g, h) : (A', X' \to Y') \to (A, X \to Y)$ consists of a morphism $f : A' \to A$ in $\mathcal{C}_\Lambda $ together with morphisms of schemes $g : X \to X'$ and $h : Y \to Y'$ such that

    \[ \xymatrix{ X \ar[r]_ g \ar[d] & X' \ar[d] \\ Y \ar[r]_ h \ar[d] & Y' \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r]^ f & \mathop{\mathrm{Spec}}(A') } \]

    is a commutative diagram of schemes where both squares are cartesian.

The functor $p : \mathcal{F} \to \mathcal{C}_\Lambda $ sends $(A, X \to Y)$ to $A$ and $(f, g, h)$ to $f$. It is clear that $p$ is cofibred in groupoids. Given a morphism of schemes $X \to Y$ over $k$, let $x_0 = (k, X \to Y)$ be the corresponding object of $\mathcal{F}(k)$. We set

\[ \mathcal{D}\! \mathit{ef}_{X \to Y} = \mathcal{F}_{x_0} \]

Lemma 92.10.2. Example 92.10.1 satisfies the Rim-Schlessinger condition (RS). In particular, $\mathcal{D}\! \mathit{ef}_{X \to Y}$ is a deformation category for any morphism of schemes $X \to Y$ over $k$.

Proof. Let $A_1 \to A$ and $A_2 \to A$ be morphisms of $\mathcal{C}_\Lambda $. Assume $A_2 \to A$ is surjective. According to Formal Deformation Theory, Lemma 89.16.4 it suffices to show that the functor $\mathcal{F}(A_1 \times _ A A_2) \to \mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)$ is an equivalence of categories. Observe that

\[ \xymatrix{ \mathop{\mathrm{Spec}}(A) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(A_2) \ar[d] \\ \mathop{\mathrm{Spec}}(A_1) \ar[r] & \mathop{\mathrm{Spec}}(A_1 \times _ A A_2) } \]

is a pushout diagram as in More on Morphisms, Lemma 37.14.3. Thus the lemma follows immediately from More on Morphisms, Lemma 37.14.6 as this describes the category of schemes flat over $A_1 \times _ A A_2$ as the fibre product of the category of schemes flat over $A_1$ with the category of schemes flat over $A_2$ over the category of schemes flat over $A$. $\square$

Lemma 92.10.3. In Example 92.9.1 let $f : X \to Y$ be a morphism of schemes over $k$. There is a canonical exact sequence of $k$-vector spaces

\[ \xymatrix{ 0 \ar[r] & \text{Inf}(\mathcal{D}\! \mathit{ef}_{X \to Y}) \ar[r] & \text{Inf}(\mathcal{D}\! \mathit{ef}_ X \times \mathcal{D}\! \mathit{ef}_ Y) \ar[r] & \text{Der}_ k(\mathcal{O}_ Y, f_*\mathcal{O}_ X) \ar[lld] \\ & T\mathcal{D}\! \mathit{ef}_{X \to Y} \ar[r] & T(\mathcal{D}\! \mathit{ef}_ X \times \mathcal{D}\! \mathit{ef}_ Y) \ar[r] & \text{Ext}^1_{\mathcal{O}_ X}(Lf^*\mathop{N\! L}\nolimits _{Y/k}, \mathcal{O}_ X) } \]

Proof. The obvious map of deformation categories $\mathcal{D}\! \mathit{ef}_{X \to Y} \to \mathcal{D}\! \mathit{ef}_ X \times \mathcal{D}\! \mathit{ef}_ Y$ gives two of the arrows in the exact sequence of the lemma. Recall that $\text{Inf}(\mathcal{D}\! \mathit{ef}_{X \to Y})$ is the set of automorphisms of the trivial deformation

\[ f' : X' = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k[\epsilon ]) \xrightarrow {f \times \text{id}} Y' = Y \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k[\epsilon ]) \]

of $X \to Y$ to $k[\epsilon ]$ equal to the identity modulo $\epsilon $. This is clearly the same thing as pairs $(\alpha , \beta ) \in \text{Inf}(\mathcal{D}\! \mathit{ef}_ X \times \mathcal{D}\! \mathit{ef}_ Y)$ of infinitesimal automorphisms of $X$ and $Y$ compatible with $f'$, i.e., such that $f' \circ \alpha = \beta \circ f'$. By Deformation Theory, Lemma 90.7.1 for an arbitrary pair $(\alpha , \beta )$ the difference between the morphism $f' : X' \to Y'$ and the morphism $\beta ^{-1} \circ f' \circ \alpha : X' \to Y'$ defines an elment in

\[ \text{Der}_ k(\mathcal{O}_ Y, f_*\mathcal{O}_ X) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ Y}(\Omega _{Y/k}, f_*\mathcal{O}_ X) \]

Equality by More on Morphisms, Lemma 37.13.3. This defines the last top horizontal arrow and shows exactness in the first two places. For the map

\[ \text{Der}_ k(\mathcal{O}_ Y, f_*\mathcal{O}_ X) \to T\mathcal{D}\! \mathit{ef}_{X \to Y} \]

we interpret elements of the source as morphisms $f_\epsilon : X' \to Y'$ over $\mathop{\mathrm{Spec}}(k[\epsilon ])$ equal to $f$ modulo $\epsilon $ using Deformation Theory, Lemma 90.7.1. We send $f_\epsilon $ to the isomorphism class of $(f_\epsilon : X' \to Y')$ in $T\mathcal{D}\! \mathit{ef}_{X \to Y}$. Note that $(f_\epsilon : X' \to Y')$ is isomorphic to the trivial deformation $(f' : X' \to Y')$ exactly when $f_\epsilon = \beta ^{-1} \circ f \circ \alpha $ for some pair $(\alpha , \beta )$ which implies exactness in the third spot. Clearly, if some first order deformation $(f_\epsilon : X_\epsilon \to Y_\epsilon )$ maps to zero in $T(\mathcal{D}\! \mathit{ef}_ X \times \mathcal{D}\! \mathit{ef}_ Y)$, then we can choose isomorphisms $X' \to X_\epsilon $ and $Y' \to Y_\epsilon $ and we conclude we are in the image of the south-west arrow. Therefore we have exactness at the fourth spot. Finally, given two first order deformations $X_\epsilon $, $Y_\epsilon $ of $X$, $Y$ there is an obstruction in

\[ ob(X_\epsilon , Y_\epsilon ) \in \text{Ext}^1_{\mathcal{O}_ X}(Lf^*\mathop{N\! L}\nolimits _{Y/k}, \mathcal{O}_ X) \]

which vanishes if and only if $f : X \to Y$ lifts to $X_\epsilon \to Y_\epsilon $, see Deformation Theory, Lemma 90.7.1. This finishes the proof. $\square$

Lemma 92.10.4. In Lemma 92.10.3 if $X$ and $Y$ are both proper over $k$, then $\text{Inf}(\mathcal{D}\! \mathit{ef}_{X \to Y})$ and $T\mathcal{D}\! \mathit{ef}_{X \to Y}$ are finite dimensional.

Proof. Omitted. Hint: argue as in Lemma 92.9.4 and use the exact sequence of the lemma. $\square$

In Example 92.10.1 if $X \to Y$ is a morphism of proper schemes over $k$, then $\mathcal{D}\! \mathit{ef}_{X \to Y}$ admits a presentation by a smooth prorepresentable groupoid in functors over $\mathcal{C}_\Lambda $ and a fortiori has a (minimal) versal formal object. This follows from Lemmas 92.10.2 and 92.10.4 and the general discussion in Section 92.3.

Lemma 92.10.5. In Example 92.10.1 assume $X \to Y$ is a morphism of proper $k$-schemes. Assume $\Lambda $ is a complete local ring with residue field $k$ (the classical case). Then the functor

\[ F : \mathcal{C}_\Lambda \longrightarrow \textit{Sets},\quad A \longmapsto \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}\! \mathit{ef}_{X \to Y}(A))/\cong \]

of isomorphism classes of objects has a hull. If $\text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X) = \text{Der}_ k(\mathcal{O}_ Y, \mathcal{O}_ Y) = 0$, then $F$ is prorepresentable.

Proof. The existence of a hull follows immediately from Lemmas 92.10.2 and 92.10.4 and Formal Deformation Theory, Lemma 89.16.6 and Remark 89.15.7.

Assume $\text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X) = \text{Der}_ k(\mathcal{O}_ Y, \mathcal{O}_ Y) = 0$. Then the exact sequence of Lemma 92.10.3 combined with Lemma 92.9.3 shows that $\text{Inf}(\mathcal{D}\! \mathit{ef}_{X \to Y}) = 0$. Then $\mathcal{D}\! \mathit{ef}_{X \to Y}$ and $F$ are equivalent by Formal Deformation Theory, Lemma 89.19.13. Hence $F$ is a deformation functor (because $\mathcal{D}\! \mathit{ef}_{X \to Y}$ is a deformation category) with finite tangent space and we can apply Formal Deformation Theory, Theorem 89.18.2. $\square$


Lemma 92.10.6. In Example 92.9.1 let $f : X \to Y$ be a morphism of schemes over $k$. If $f_*\mathcal{O}_ X = \mathcal{O}_ Y$ and $R^1f_*\mathcal{O}_ X = 0$, then the morphism of deformation categories

\[ \mathcal{D}\! \mathit{ef}_{X \to Y} \to \mathcal{D}\! \mathit{ef}_ X \]

is an equivalence.

Proof. We construct a quasi-inverse to the forgetful functor of the lemma. Namely, suppose that $(A, U)$ is an object of $\mathcal{D}\! \mathit{ef}_ X$. The given map $X \to U$ is a finite order thickening and we can use it to identify the underlying topological spaces of $U$ and $X$, see More on Morphisms, Section 37.2. Thus we may and do think of $\mathcal{O}_ U$ as a sheaf of $A$-algebras on $X$; moreover the fact that $U \to \mathop{\mathrm{Spec}}(A)$ is flat, means that $\mathcal{O}_ U$ is flat as a sheaf of $A$-modules. In particular, we have a filtration

\[ 0 = \mathfrak m_ A^ n\mathcal{O}_ U \subset \mathfrak m_ A^{n - 1}\mathcal{O}_ U \subset \ldots \subset \mathfrak m_ A^2\mathcal{O}_ U \subset \mathfrak m_ A\mathcal{O}_ U \subset \mathcal{O}_ U \]

with subquotients equal to $\mathcal{O}_ X \otimes _ k \mathfrak m_ A^ i/\mathfrak m_ A^{i + 1}$ by flatness, see More on Morphisms, Lemma 37.10.1 or the more general Deformation Theory, Lemma 90.5.2. Set

\[ \mathcal{O}_ V = f_*\mathcal{O}_ U \]

viewed as sheaf of $A$-algebras on $Y$. Since $R^1f_*\mathcal{O}_ X = 0$ we find by the description above that $R^1f_*(\mathfrak m_ A^ i\mathcal{O}_ U/\mathfrak m_ A^{i + 1}\mathcal{O}_ U) = 0$ for all $i$. This implies that the sequences

\[ 0 \to (f_*\mathcal{O}_ X) \otimes _ k \mathfrak m_ A^ i/\mathfrak m_ A^{i + 1} \to f_*(\mathcal{O}_ U/\mathfrak m_ A^{i + 1}\mathcal{O}_ U) \to f_*(\mathcal{O}_ U/\mathfrak m_ A^ i\mathcal{O}_ U) \to 0 \]

are exact for all $i$. Reading the references given above backwards (and using induction) we find that $\mathcal{O}_ V$ is a flat sheaf of $A$-algebras with $\mathcal{O}_ V/\mathfrak m_ A\mathcal{O}_ V = \mathcal{O}_ Y$. Using More on Morphisms, Lemma 37.2.2 we find that $(Y, \mathcal{O}_ V)$ is a scheme, call it $V$. The equality $\mathcal{O}_ V = f_*\mathcal{O}_ U$ defines a morphism of ringed spaces $U \to V$ which is easily seen to be a morphism of schemes. This finishes the proof by the flatness already established. $\square$

Comments (2)

Comment #3578 by on

In the statement of Lemma 0E3V, the morphism X --> Y should be called f (as f appears in the assertion), and then the second sentence of the proof can be omitted.

Comment #3702 by on

Thanks Ben! A rare change that shrinks the Stacks project... See changes here.

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