The Stacks project

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

  1. $Z_1 \to X_1$ is a closed immersion, and

  2. $Z_1 \to S_1$ 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

  1. $Y_ n \to X_ n$ is a finite morphism, and

  2. $Y_1 \to S_1$ 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

  1. $X \to S$ is proper, and

  2. $Y \to S$ 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

  1. $X_1 \to S_1$ is proper, and

  2. $\mathcal{L}_1$ 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)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0898. Beware of the difference between the letter 'O' and the digit '0'.



View Section 30.28 as pdf