The Stacks project

Proof. Immediate consquence of Properties, Proposition 28.26.13. $\square$

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 open covering $Y = V_1 \cup \ldots \cup V_ m$. Set $U_ j = f^{-1}(V_ j)$. By Properties, Proposition 28.26.13 for each $j$ we know there exists finitely many maps $s_{j, i} : \mathcal{L}^{\otimes n_{j, i}}|_{U_ j} \to \mathcal{F}|_{U_ j}$ which are jointly surjective. Consider the quasi-coherent $\mathcal{O}_ Y$-modules

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

are going to be jointly surjective and the lemma is proved. By Properties, Lemma 28.22.3 for each $i, j$ we can find a finite type quasi-coherent submodule $\mathcal{H}'_{i, j} \subset \mathcal{H}_{-n_{j, i}}$ which contains the section $s_{i, j}$ over $V_ j$. Thus the resolution property of $Y$ produces surjections $\mathcal{E}_{i, j} \to \mathcal{H}'_{j, i}$ and we conclude. $\square$

Proof. By Morphisms, Lemma 29.37.6 this is a special case of Lemma 36.36.3. $\square$

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 on quasi-coherent modules (Cohomology of Schemes, Lemma 30.2.3) we get a surjection

Taking duals we get a surjection

Since $f_*\mathcal{E}$ is finite locally free¹, we conclude. $\square$

Proof. One direction of the lemma is trivial. For the other, say $\mathcal{E}_ j \to \mathcal{I}_ j$ is a surjection with $\mathcal{E}_ j$ finite locally free. In the next paragraph, we reduce to the Noetherian case; we suggest the reader skip it.

The first observation is that $U_ j \cap U_{j'}$ is quasi-compact as the complement of the zero scheme of the quasi-coherent finite type ideal $\mathcal{I}_{j'}|{U_ j}$ on the affine scheme $U_ j$, see Properties, Lemma 28.24.1. Hence $X$ is quasi-compact and quasi-separated, see Schemes, Lemma 26.21.6. By Limits, Proposition 32.5.4 we can write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as the limit of a direct system of Noetherian schemes with affine transition morphisms. For each $j$ we can find an $i$ and a finite locally free $\mathcal{O}_{X_ i}$-module $\mathcal{E}_{i, j}$ pulling back to $\mathcal{E}_ j$, see Limits, Lemma 32.10.3. After increasing $i$ we may assume that the composition $\mathcal{E}_ j \to \mathcal{I}_ j \to \mathcal{O}_ X$ is the pullback of a map $\mathcal{E}_{i, j} \to \mathcal{O}_{X_ i}$, see Limits, Lemma 32.10.2. Denote $\mathcal{I}_{i, j} \subset \mathcal{O}_{X_ i}$ the image of this map; this is a quasi-coherent ideal sheaf on the Noetherian scheme $X_ i$ whose pullback to $X$ is $\mathcal{I}_ j$. Denoting $U_{i, j} \subset X_ i$ the complementary opens, we may assume these are affine for all $i, j$, see Limits, Lemma 32.4.13. If we can prove the lemma for the opens $U_{i, j}$ and the ideal sheaves $\mathcal{I}_{i, j}$ on $X_ i$ then $X$, being affine over $X_ i$, will have the resolution property by Lemma 36.36.4. In this way we reduce to the case of a Noetherian scheme.

Assume $X$ is Noetherian. For every coherent module $\mathcal{F}$ we can choose a finite list of sections $s_{jk} \in \mathcal{F}(U_ j)$, $k = 1, \ldots , e_ j$ which generate the restriction of $\mathcal{F}$ to $U_ j$. By Cohomology of Schemes, Lemma 30.10.5 we can extend $s_{jk}$ to a map $s'_{jk} : \mathcal{I}_ i^{n_{jk}} \to \mathcal{F}$ for some $n_{jk} \geq 1$. Then we can consider the compositions

to conclude. $\square$

Proof. Since $X$ is quasi-compact, there exists $n$ and pairs $(\mathcal{L}_ i, s_ i)$, $i = 1, \ldots , n$ where $\mathcal{L}_ i$ is an invertible $\mathcal{O}_ X$-module and $s_ i \in \Gamma (X, \mathcal{L}_ i)$ is a section such that the set of points $U_ i \subset X$ where $s_ i$ is nonvanishing is affine and $X = U_1 \cup \ldots \cup U_ n$. Let $\mathcal{I}_ i \subset \mathcal{O}_ X$ be the image of $s_ i : \mathcal{L}_ i^{\otimes -1} \to \mathcal{O}_ X$. Applying Lemma 36.36.6 we find that $X$ has the resolution property. $\square$

Proof. Combine Divisors, Lemma 31.16.8 and the above Lemma 36.36.7. $\square$

Proof. If $X_ i$ has the resolution property, then $X$ does by Lemma 36.36.4. Assume $X$ has the resolution property. Choose $i \in I$. Choose a finite affine open covering $X_ i = U_{i, 1} \cup \ldots \cup U_{i, m}$. For each $j$ choose a finite type quasi-coherent sheaf of ideals $\mathcal{I}_{i, j} \subset \mathcal{O}_{X_ i}$ such that $X_ i \setminus V(\mathcal{I}_{i, j}) = U_{i, j}$, see Properties, Lemma 28.24.1. Denote $U_ j \subset X$ the inverse image of $U_{i, j}$ and denote $\mathcal{I}_ j \subset \mathcal{O}_ X$ the pullback of $\mathcal{I}_{i, j}$. Since $X$ has the resolution property, we may choose finite locally free $\mathcal{O}_ X$-modules $\mathcal{E}_ j$ and surjections $\mathcal{E}_ j \to \mathcal{I}_ j$. By Limits, Lemmas 32.10.3 and 32.10.2 after increasing $i$ we can find finite locally free $\mathcal{O}_{X_ i}$-modules $\mathcal{E}_{i, j}$ and maps $\mathcal{E}_{i, j} \to \mathcal{O}_{X_ i}$ whose base changes to $X$ recover the compositions $\mathcal{E}_ j \to \mathcal{I}_ j \to \mathcal{O}_ X$, $j = 1, \ldots , m$. The pullbacks of the finitely presented $\mathcal{O}_{X_ i}$-modules $\mathop{\mathrm{Coker}}(\mathcal{E}_{i, j} \to \mathcal{O}_{X_ i})$ and $\mathcal{O}_{X_ i}/\mathcal{I}_{i, j}$ to $X$ agree as quotients of $\mathcal{O}_ X$. Hence by Limits, Lemma 32.10.2 we may assume that these agree, in other words that the image of $\mathcal{E}_{i, j} \to \mathcal{O}_{X_ i}$ is equal to $\mathcal{I}_{i, j}$. Then we conclude that $X_ i$ has the resolution property by Lemma 36.36.6. $\square$

Proof. Combining Limits, Proposition 32.5.4 and Lemma 36.36.9 this reduces to the case where $X$ is Noetherian (small detail omitted). Assume $X$ is Noetherian. Recall that $X \times X$ is covered by the affine opens $U \times V$ for affine opens $U$, $V$ of $X$, see Schemes, Section 26.17. Hence to show that the diagonal $\Delta : X \to X \times X$ is affine, it suffices to show that $U \cap V = \Delta ^{-1}(U \times V)$ is affine for all affine opens $U$, $V$ of $X$, see Morphisms, Lemma 29.11.3. In particular, it suffices to show that the inclusion morphism $j : U \to X$ is affine if $U$ is an affine open of $X$. By Cohomology of Schemes, Lemma 30.3.4 it suffices to show that $R^1j_*\mathcal{G} = 0$ for any quasi-coherent $\mathcal{O}_ U$-module $\mathcal{G}$. By Proposition 36.8.3 (this is where we use that we've reduced to the Noetherian case) we can represent $Rj_*\mathcal{G}$ by a complex $\mathcal{H}^\bullet $ of quasi-coherent $\mathcal{O}_ X$-modules. Assume

is nonzero in order to get a contradiction. Then we can find a coherent $\mathcal{O}_ X$-module $\mathcal{F}$ and a map

such that the composition with the projection onto $H^1(\mathcal{H}^\bullet )$ is nonzero. Namely, we can write $\mathop{\mathrm{Ker}}(\mathcal{H}^1 \to \mathcal{H}^2)$ as the filtered union of its coherent submodules by Properties, Lemma 28.22.3 and then one of these will do the job. Next, we choose a finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$ and a surjection $\mathcal{E} \to \mathcal{F}$ using the resolution property of $X$. This produces a map in the derived category

which is nonzero on cohomology sheaves and hence nonzero in $D(\mathcal{O}_ X)$. By adjunction, this is the same thing as a map

nonzero in $D(\mathcal{O}_ U)$. Since $\mathcal{E}$ is finite locally free this is the same thing as a nonzero element of

where $\mathcal{E}^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}, \mathcal{O}_ X)$ is the dual finite locally free module. However, this group is zero by Cohomology of Schemes, Lemma 30.2.2 which is the desired contradiction. (If in doubt about the step using duals, please see the more general Cohomology, Lemma 20.50.5.) $\square$