The Stacks project

75.28 The resolution property

This section is the analogue of Derived Categories of Schemes, Section 36.36 for algebraic spaces; please read that section first. It is currently not known if a smooth proper algebraic space over a field always has the resolution property or if this is false. If you know the answer to this question, please email stacks.project@gmail.com.

We can make the following definition although it scarcely makes sense to consider it for general algebraic spaces.

Definition 75.28.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. We say $X$ has the resolution property if every quasi-coherent $\mathcal{O}_ X$-module of finite type is the quotient of a finite locally free $\mathcal{O}_ X$-module.

If $X$ is a quasi-compact and quasi-separated algebraic space, then it suffices to check every $\mathcal{O}_ X$-module module of finite presentation (automatically quasi-coherent) is the quotient of a finite locally free $\mathcal{O}_ X$-module, see Limits of Spaces, Lemma 70.9.3. If $X$ is a Noetherian algebraic space, then finite type quasi-coherent modules are exactly the coherent $\mathcal{O}_ X$-modules, see Cohomology of Spaces, Lemma 69.12.2.

Lemma 75.28.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume

  1. $Y$ is quasi-compact and quasi-separated and has the resolution property,

  2. there exists an $f$-ample invertible module on $X$ (Divisors on Spaces, Definition 71.14.1).

Then $X$ has the resolution property.

Proof. Let $\mathcal{F}$ be a finite type quasi-coherent $\mathcal{O}_ X$-module. Let $\mathcal{L}$ be an $f$-ample invertible module. Choose an affine scheme $V$ and a surjective étale morphism $V \to Y$. Set $U = V \times _ Y X$. Then $\mathcal{L}|_ U$ is ample on $U$. By Properties, Proposition 28.26.13 we know there exists finitely many maps $s_ i : \mathcal{L}^{\otimes n_ i}|_ U \to \mathcal{F}|_ U$ which are jointly surjective. Consider the quasi-coherent $\mathcal{O}_ Y$-modules

\[ \mathcal{H}_ n = f_*(\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) \]

We may think of $s_ i$ as a section over $V$ of the sheaf $\mathcal{H}_{-n_ i}$. Suppose we can find finite locally free $\mathcal{O}_ Y$-modules $\mathcal{E}_ i$ and maps $\mathcal{E}_ i \to \mathcal{H}_{-n_ i}$ such that $s_ i$ is in the image. Then the corresponding maps

\[ f^*\mathcal{E}_ i \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n_ i} \longrightarrow \mathcal{F} \]

are going to be jointly surjective and the lemma is proved. By Limits of Spaces, Lemma 70.9.2 for each $i$ we can find a finite type quasi-coherent submodule $\mathcal{H}'_ i \subset \mathcal{H}_{-n_ i}$ which contains the section $s_ i$ over $V$. Thus the resolution property of $Y$ produces surjections $\mathcal{E}_ i \to \mathcal{H}'_ i$ and we conclude. $\square$

Lemma 75.28.3. Let $S$ be a scheme. Let $f : X \to Y$ be an affine or quasi-affine morphism of algebraic spaces over $S$ with $Y$ quasi-compact and quasi-separated. If $Y$ has the resolution property, so does $X$.

Proof. By Divisors on Spaces, Lemma 71.14.7 this is a special case of Lemma 75.28.2. $\square$

Here is a case where one can prove the resolution property goes down.

Lemma 75.28.4. Let $S$ be a scheme. Let $f : X \to Y$ be a surjective finite locally free morphism of algebraic spaces over $S$. If $X$ has the resolution property, so does $Y$.

Proof. The condition means that $f$ is affine and that $f_*\mathcal{O}_ X$ is a finite locally free $\mathcal{O}_ Y$-module of positive rank. Let $\mathcal{G}$ be a quasi-coherent $\mathcal{O}_ Y$-module of finite type. By assumption there exists a surjection $\mathcal{E} \to f^*\mathcal{G}$ for some finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$. Since $f_*$ is exact (Cohomology of Spaces, Section 69.4) we get a surjection

\[ f_*\mathcal{E} \longrightarrow f_*f^*\mathcal{G} = \mathcal{G} \otimes _{\mathcal{O}_ Y} f_*\mathcal{O}_ X \]

Taking duals we get a surjection

\[ f_*\mathcal{E} \otimes _{\mathcal{O}_ Y} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(f_*\mathcal{O}_ X, \mathcal{O}_ Y) \longrightarrow \mathcal{G} \]

Since $f_*\mathcal{E}$ is finite locally free, we conclude. $\square$

For more on the resolution property of algebraic spaces, please see More on Morphisms of Spaces, Section 76.56.


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