30.28 Grothendieck's algebraization theorem
Our first result is a translation of Grothendieck's existence theorem in terms of closed subschemes and finite morphisms.
Lemma 30.28.1. Let A be a Noetherian ring complete with respect to an ideal I. Write S = \mathop{\mathrm{Spec}}(A) and S_ n = \mathop{\mathrm{Spec}}(A/I^ n). Let X \to S be a separated morphism of finite type. For n \geq 1 we set X_ n = X \times _ S S_ n. Suppose given a commutative diagram
\xymatrix{ Z_1 \ar[r] \ar[d] & Z_2 \ar[r] \ar[d] & Z_3 \ar[r] \ar[d] & \ldots \\ X_1 \ar[r]^{i_1} & X_2 \ar[r]^{i_2} & X_3 \ar[r] & \ldots }
of schemes with cartesian squares. Assume that
Z_1 \to X_1 is a closed immersion, and
Z_1 \to S_1 is proper.
Then there exists a closed immersion of schemes Z \to X such that Z_ n = Z \times _ S S_ n. Moreover, Z is proper over S.
Proof.
Let's write j_ n : Z_ n \to X_ n for the vertical morphisms. As the squares in the statement are cartesian we see that the base change of j_ n to X_1 is j_1. Thus Morphisms, Lemma 29.45.7 shows that j_ n is a closed immersion. Set \mathcal{F}_ n = j_{n, *}\mathcal{O}_{Z_ n}, so that j_ n^\sharp is a surjection \mathcal{O}_{X_ n} \to \mathcal{F}_ n. Again using that the squares are cartesian we see that the pullback of \mathcal{F}_{n + 1} to X_ n is \mathcal{F}_ n. Hence Grothendieck's existence theorem, as reformulated in Remark 30.27.2, tells us there exists a map \mathcal{O}_ X \to \mathcal{F} of coherent \mathcal{O}_ X-modules whose restriction to X_ n recovers \mathcal{O}_{X_ n} \to \mathcal{F}_ n. Moreover, the support of \mathcal{F} is proper over S. As the completion functor is exact (Lemma 30.23.4) we see that the cokernel \mathcal{Q} of \mathcal{O}_ X \to \mathcal{F} has vanishing completion. Since \mathcal{F} has support proper over S and so does \mathcal{Q} this implies that \mathcal{Q} = 0 for example because the functor (30.27.0.1) is an equivalence by Grothendieck's existence theorem. Thus \mathcal{F} = \mathcal{O}_ X/\mathcal{J} for some quasi-coherent sheaf of ideals \mathcal{J}. Setting Z = V(\mathcal{J}) finishes the proof.
\square
In the following lemma it is actually enough to assume that Y_1 \to X_1 is finite as it will imply that Y_ n \to X_ n is finite too (see More on Morphisms, Lemma 37.3.3).
Lemma 30.28.2. Let A be a Noetherian ring complete with respect to an ideal I. Write S = \mathop{\mathrm{Spec}}(A) and S_ n = \mathop{\mathrm{Spec}}(A/I^ n). Let X \to S be a separated morphism of finite type. For n \geq 1 we set X_ n = X \times _ S S_ n. Suppose given a commutative diagram
\xymatrix{ Y_1 \ar[r] \ar[d] & Y_2 \ar[r] \ar[d] & Y_3 \ar[r] \ar[d] & \ldots \\ X_1 \ar[r]^{i_1} & X_2 \ar[r]^{i_2} & X_3 \ar[r] & \ldots }
of schemes with cartesian squares. Assume that
Y_ n \to X_ n is a finite morphism, and
Y_1 \to S_1 is proper.
Then there exists a finite morphism of schemes Y \to X such that Y_ n = Y \times _ S S_ n. Moreover, Y is proper over S.
Proof.
Let's write f_ n : Y_ n \to X_ n for the vertical morphisms. Set \mathcal{F}_ n = f_{n, *}\mathcal{O}_{Y_ n}. This is a coherent \mathcal{O}_{X_ n}-module as f_ n is finite (Lemma 30.9.9). Using that the squares are cartesian we see that the pullback of \mathcal{F}_{n + 1} to X_ n is \mathcal{F}_ n. Hence Grothendieck's existence theorem, as reformulated in Remark 30.27.2, tells us there exists a coherent \mathcal{O}_ X-module \mathcal{F} whose restriction to X_ n recovers \mathcal{F}_ n. Moreover, the support of \mathcal{F} is proper over S. As the completion functor is fully faithful (Theorem 30.27.1) we see that the multiplication maps \mathcal{F}_ n \otimes _{\mathcal{O}_{X_ n}} \mathcal{F}_ n \to \mathcal{F}_ n fit together to give an algebra structure on \mathcal{F}. Setting Y = \underline{\mathop{\mathrm{Spec}}}_ X(\mathcal{F}) finishes the proof.
\square
Lemma 30.28.3. Let A be a Noetherian ring complete with respect to an ideal I. Write S = \mathop{\mathrm{Spec}}(A) and S_ n = \mathop{\mathrm{Spec}}(A/I^ n). Let X, Y be schemes over S. For n \geq 1 we set X_ n = X \times _ S S_ n and Y_ n = Y \times _ S S_ n. Suppose given a compatible system of commutative diagrams
\xymatrix{ & & X_{n + 1} \ar[rd] \ar[rr]_{g_{n + 1}} & & Y_{n + 1} \ar[ld] \\ X_ n \ar[rru] \ar[rd] \ar[rr]_{g_ n} & & Y_ n \ar[rru] \ar[ld] & S_{n + 1} \\ & S_ n \ar[rru] }
Assume that
X \to S is proper, and
Y \to S is separated of finite type.
Then there exists a unique morphism of schemes g : X \to Y over S such that g_ n is the base change of g to S_ n.
Proof.
The morphisms (1, g_ n) : X_ n \to X_ n \times _ S Y_ n are closed immersions because Y_ n \to S_ n is separated (Schemes, Lemma 26.21.11). Thus by Lemma 30.28.1 there exists a closed subscheme Z \subset X \times _ S Y proper over S whose base change to S_ n recovers X_ n \subset X_ n \times _ S Y_ n. The first projection p : Z \to X is a proper morphism (as Z is proper over S, see Morphisms, Lemma 29.41.7) whose base change to S_ n is an isomorphism for all n. In particular, p : Z \to X is finite over an open neighbourhood of X_0 by Lemma 30.21.2. As X is proper over S this open neighbourhood is all of X and we conclude p : Z \to X is finite. Applying the equivalence of Proposition 30.25.4 we see that p_*\mathcal{O}_ Z = \mathcal{O}_ X as this is true modulo I^ n for all n. Hence p is an isomorphism and we obtain the morphism g as the composition X \cong Z \to Y. We omit the proof of uniqueness.
\square
In order to prove an “abstract” algebraization theorem we need to assume we have an ample invertible sheaf, as the result is false without such an assumption.
Theorem 30.28.4 (Grothendieck's algebraization theorem). Let A be a Noetherian ring complete with respect to an ideal I. Set S = \mathop{\mathrm{Spec}}(A) and S_ n = \mathop{\mathrm{Spec}}(A/I^ n). Consider a commutative diagram
\xymatrix{ X_1 \ar[r]_{i_1} \ar[d] & X_2 \ar[r]_{i_2} \ar[d] & X_3 \ar[r] \ar[d] & \ldots \\ S_1 \ar[r] & S_2 \ar[r] & S_3 \ar[r] & \ldots }
of schemes with cartesian squares. Suppose given (\mathcal{L}_ n, \varphi _ n) where each \mathcal{L}_ n is an invertible sheaf on X_ n and \varphi _ n : i_ n^*\mathcal{L}_{n + 1} \to \mathcal{L}_ n is an isomorphism. If
X_1 \to S_1 is proper, and
\mathcal{L}_1 is ample on X_1
then there exists a proper morphism of schemes X \to S and an ample invertible \mathcal{O}_ X-module \mathcal{L} and isomorphisms X_ n \cong X \times _ S S_ n and \mathcal{L}_ n \cong \mathcal{L}|_{X_ n} compatible with the morphisms i_ n and \varphi _ n.
Proof.
Since the squares in the diagram are cartesian and since the morphisms S_ n \to S_{n + 1} are closed immersions, we see that the morphisms i_ n are closed immersions too. In particular we may think of X_ m as a closed subscheme of X_ n for m < n. In fact X_ m is the closed subscheme cut out by the quasi-coherent sheaf of ideals I^ m\mathcal{O}_{X_ n}. Moreover, the underlying topological spaces of the schemes X_1, X_2, X_3, \ldots are all identified, hence we may (and do) think of sheaves \mathcal{O}_{X_ n} as living on the same underlying topological space; similarly for coherent \mathcal{O}_{X_ n}-modules. Set
\mathcal{F}_ n = \mathop{\mathrm{Ker}}(\mathcal{O}_{X_{n + 1}} \to \mathcal{O}_{X_ n})
so that we obtain short exact sequences
0 \to \mathcal{F}_ n \to \mathcal{O}_{X_{n + 1}} \to \mathcal{O}_{X_ n} \to 0
By the above we have \mathcal{F}_ n = I^ n\mathcal{O}_{X_{n + 1}}. It follows \mathcal{F}_ n is a coherent sheaf on X_{n + 1} annihilated by I, hence we may (and do) think of it as a coherent module \mathcal{O}_{X_1}-module. Observe that for m > n the sheaf
I^ n\mathcal{O}_{X_ m}/I^{n + 1}\mathcal{O}_{X_ m}
maps isomorphically to \mathcal{F}_ n under the map \mathcal{O}_{X_ m} \to \mathcal{O}_{X_{n + 1}}. Hence given n_1, n_2 \geq 0 we can pick an m > n_1 + n_2 and consider the multiplication map
I^{n_1}\mathcal{O}_{X_ m} \times I^{n_2}\mathcal{O}_{X_ m} \longrightarrow I^{n_1 + n_2}\mathcal{O}_{X_ m} \to \mathcal{F}_{n_1 + n_2}
This induces an \mathcal{O}_{X_1}-bilinear map
\mathcal{F}_{n_1} \times \mathcal{F}_{n_2} \longrightarrow \mathcal{F}_{n_1 + n_2}
which in turn defines the structure of a graded \mathcal{O}_{X_1}-algebra on \mathcal{F} = \bigoplus _{n \geq 0} \mathcal{F}_ n.
Set B = \bigoplus I^ n/I^{n + 1}; this is a finitely generated graded A/I-algebra. Set \mathcal{B} = (X_1 \to S_1)^*\widetilde{B}. The discussion above provides us with a canonical surjection
\mathcal{B} \longrightarrow \mathcal{F}
of graded \mathcal{O}_{X_1}-algebras. In particular we see that \mathcal{F} is a finite type quasi-coherent graded \mathcal{B}-module. By Lemma 30.19.3 we can find an integer d_0 such that H^1(X_1, \mathcal{F} \otimes \mathcal{L}^{\otimes d}) = 0 for all d \geq d_0. Pick a d \geq d_0 such that there exist sections s_{0, 1}, \ldots , s_{N, 1} \in \Gamma (X_1, \mathcal{L}_1^{\otimes d}) which induce an immersion
\psi _1 : X_1 \to \mathbf{P}^ N_{S_1}
over S_1, see Morphisms, Lemma 29.39.4. As X_1 is proper over S_1 we see that \psi _1 is a closed immersion, see Morphisms, Lemma 29.41.7 and Schemes, Lemma 26.10.4. We are going to “lift” \psi _1 to a compatible system of closed immersions of X_ n into \mathbf{P}^ N.
Upon tensoring the short exact sequences of the first paragraph of the proof by \mathcal{L}_{n + 1}^{\otimes d} we obtain short exact sequences
0 \to \mathcal{F}_ n \otimes \mathcal{L}_{n + 1}^{\otimes d} \to \mathcal{L}_{n + 1}^{\otimes d} \to \mathcal{L}_{n + 1}^{\otimes d} \to 0
Using the isomorphisms \varphi _ n we obtain isomorphisms \mathcal{L}_{n + 1} \otimes \mathcal{O}_{X_ l} = \mathcal{L}_ l for l \leq n. Whence the sequence above becomes
0 \to \mathcal{F}_ n \otimes \mathcal{L}_1^{\otimes d} \to \mathcal{L}_{n + 1}^{\otimes d} \to \mathcal{L}_ n^{\otimes d} \to 0
The vanishing of H^1(X, \mathcal{F}_ n \otimes \mathcal{L}_1^{\otimes d}) implies we can inductively lift s_{0, 1}, \ldots , s_{N, 1} \in \Gamma (X_1, \mathcal{L}_1^{\otimes d}) to sections s_{0, n}, \ldots , s_{N, n} \in \Gamma (X_ n, \mathcal{L}_ n^{\otimes d}). Thus we obtain a commutative diagram
\xymatrix{ X_1 \ar[r]_{i_1} \ar[d]_{\psi _1} & X_2 \ar[r]_{i_2} \ar[d]_{\psi _2} & X_3 \ar[r] \ar[d]_{\psi _3} & \ldots \\ \mathbf{P}^ N_{S_1} \ar[r] & \mathbf{P}^ N_{S_2} \ar[r] & \mathbf{P}^ N_{S_3} \ar[r] & \ldots }
where \psi _ n = \varphi _{(\mathcal{L}_ n, (s_{0, n}, \ldots , s_{N, n}))} in the notation of Constructions, Section 27.13. As the squares in the statement of the theorem are cartesian we see that the squares in the above diagram are cartesian. We win by applying Lemma 30.28.1.
\square
Comments (0)