## 75.43 Grothendieck's algebraization theorem

This section is the analogue of Cohomology of Schemes, Section 30.28. However, this section is missing the result on algebraization of deformations of proper algebraic spaces endowed with ample invertible sheaves, as a proper algebraic space which comes with an ample invertible sheaf is already a scheme. We do have an algebraization result on proper algebraic spaces of relative dimension $1$. Our first result is a translation of Grothendieck's existence theorem in terms of closed subschemes and finite morphisms.

Lemma 75.43.1. 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 morphism of algebraic spaces that is separated and 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 algebraic spaces 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 algebraic spaces $Z \to X$ such that $Z_ n = Z \times _ S S_ n$ for all $n \geq 1$. 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 Limits of Spaces, Lemma 69.15.5 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 75.42.12, 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 75.42.3) we see that $\mathcal{O}_ X \to \mathcal{F}$ is surjective. 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. $\square$

Lemma 75.43.2. 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 morphism of algebraic spaces that is separated and 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 algebraic spaces with cartesian squares. Assume that

1. $Y_1 \to X_1$ is a finite morphism, and

2. $Y_1 \to S_1$ is proper.

Then there exists a finite morphism of algebraic spaces $Y \to X$ such that $Y_ n = Y \times _ S S_ n$ for all $n \geq 1$. Moreover, $Y$ is proper over $S$.

Proof. Let's write $f_ n : Y_ n \to X_ n$ for the vertical morphisms. As the squares in the statement are cartesian we see that the base change of $f_ n$ to $X_1$ is $f_1$. Thus Lemma 75.10.2 shows that $f_ n$ is a finite morphism. Set $\mathcal{F}_ n = f_{n, *}\mathcal{O}_{Y_ n}$. 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 75.42.12, 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 75.42.11) 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 75.43.3. 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 algebraic spaces 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 algebraic spaces $g : X \to Y$ 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 (Morphisms of Spaces, Lemma 66.4.7). Thus by Lemma 75.43.1 there exists a closed subspace $Z \subset X \times _ S Y$ 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 of Spaces, Lemma 66.40.6) whose base change to $S_ n$ is an isomorphism for all $n$. In particular, $p : Z \to X$ is quasi-finite on an open subspace of $Z$ containing every point of $Z_0$ for example by Morphisms of Spaces, Lemma 66.34.7. As $Z$ is proper over $S$ this open neighbourhood is all of $Z$. We conclude that $p : Z \to X$ is finite by Zariski's main theorem (for example apply Lemma 75.34.3 and use properness of $Z$ over $X$ to see that the immersion is a closed immersion). Applying the equivalence of Theorem 75.42.11 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. $\square$

Remark 75.43.4. We can ask if in Grothendieck's algebraization theorem (in the form of Lemma 75.43.3), we can get by with weaker separation axioms on the target. Let us be more precise. Let $A$, $I$, $S$, $S_ n$, $X$, $Y$, $X_ n$, $Y_ n$, and $g_ n$ be as in the statement of Lemma 75.43.3 and assume that

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

2. $Y \to S$ is locally of finite type.

Does there exist a morphism of algebraic spaces $g : X \to Y$ over $S$ such that $g_ n$ is the base change of $g$ to $S_ n$? We don't know the answer in general; if you do please email stacks.project@gmail.com. If $Y \to S$ is separated, then the result holds by the lemma (there is an immediate reduction to the case where $X$ is finite type over $S$, by choosing a quasi-compact open containing the image of $g_1$). If we only assume $Y \to S$ is quasi-separated, then the result is true as well. First, as before we may assume $Y$ is quasi-compact as well as quasi-separated. Then we can use either or from to algebraize $(g_ n)$. Namely, to apply the first reference, we use

$D_{perf}(X) \to \mathop{\mathrm{lim}}\nolimits D_{perf}(X_ n) \xrightarrow {\mathop{\mathrm{lim}}\nolimits Lg_ n^*} \mathop{\mathrm{lim}}\nolimits D_{perf}(Y_ n) = D_{perf}(Y)$

where the last step uses a Grothendieck existence result for the derived category of the proper algebraic space $Y$ over $R$ (compare with Flatness on Spaces, Remark 76.13.7). The paper cited shows that this arrow determines a morphism $Y \to X$ as desired. To apply the second reference we use the same argument with coherent modules:

$\textit{Coh}(\mathcal{O}_ X) \to \mathop{\mathrm{lim}}\nolimits \textit{Coh}(\mathcal{O}_{X_ n}) \xrightarrow {\mathop{\mathrm{lim}}\nolimits g_ n^*} \mathop{\mathrm{lim}}\nolimits \textit{Coh}(\mathcal{O}_{Y_ n}) = \textit{Coh}(\mathcal{O}_ Y)$

where the final equality is a consequence of Grothendieck's existence theorem (Theorem 75.42.11). The second reference tells us that this functor corresponds to a morphism $Y \to X$ over $R$. If we ever need this generalization we will precisely state and carefully prove the result here.

Lemma 75.43.5. Let $(A, \mathfrak m, \kappa )$ be a complete local Noetherian ring. Set $S = \mathop{\mathrm{Spec}}(A)$ and $S_ n = \mathop{\mathrm{Spec}}(A/\mathfrak m^ n)$. 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 algebraic spaces with cartesian squares. If $\dim (X_1) \leq 1$, then there exists a projective morphism of schemes $X \to S$ and isomorphisms $X_ n \cong X \times _ S S_ n$ compatible with $i_ n$.

Proof. By Spaces over Fields, Lemma 71.9.3 the algebraic space $X_1$ is a scheme. Hence $X_1$ is a proper scheme of dimension $\leq 1$ over $\kappa$. By Varieties, Lemma 33.43.4 we see that $X_1$ is H-projective over $\kappa$. Let $\mathcal{L}_1$ be an ample invertible sheaf on $X_1$.

We are going to show that $\mathcal{L}_1$ lifts to a compatible system $\{ \mathcal{L}_ n\}$ of invertible sheaves on $\{ X_ n\}$. Observe that $X_ n$ is a scheme too by Lemma 75.9.5. Recall that $X_1 \to X_ n$ induces homeomorphisms of underlying topological spaces. In the rest of the proof we do not distinguish between sheaves on $X_ n$ and sheaves on $X_1$. Suppose, given a lift $\mathcal{L}_ n$ to $X_ n$. We consider the exact sequence

$1 \to (1 + \mathfrak m^ n\mathcal{O}_{X_{n + 1}})^* \to \mathcal{O}_{X_{n + 1}}^* \to \mathcal{O}_{X_ n}^* \to 1$

of sheaves on $X_{n + 1}$. The class of $\mathcal{L}_ n$ in $H^1(X_ n, \mathcal{O}_{X_ n}^*)$ (see Cohomology, Lemma 20.6.1) can be lifted to an element of $H^1(X_{n + 1}, \mathcal{O}_{X_{n + 1}}^*)$ if and only if the obstruction in $H^2(X_{n + 1}, (1 + \mathfrak m^ n\mathcal{O}_{X_{n + 1}})^*)$ is zero. As $X_1$ is a Noetherian scheme of dimension $\leq 1$ this cohomology group vanishes (Cohomology, Proposition 20.20.7).

By Grothendieck's algebraization theorem (Cohomology of Schemes, Theorem 30.28.4) we find a projective morphism of schemes $X \to S = \mathop{\mathrm{Spec}}(A)$ and a compatible system of isomorphisms $X_ n = S_ n \times _ S X$. $\square$

Lemma 75.43.6. Let $(A, \mathfrak m, \kappa )$ be a complete Noetherian local ring. Let $X$ be an algebraic space over $\mathop{\mathrm{Spec}}(A)$. If $X \to \mathop{\mathrm{Spec}}(A)$ is proper and $\dim (X_\kappa ) \leq 1$, then $X$ is a scheme projective over $A$.

Proof. Set $X_ n = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/\mathfrak m^ n)$. By Lemma 75.43.5 there exists a projective morphism $Y \to \mathop{\mathrm{Spec}}(A)$ and compatible isomorphisms $Y \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/\mathfrak m^ n) \cong X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/\mathfrak m^ n)$. By Lemma 75.43.3 we see that $X \cong Y$ and the proof is complete. $\square$

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