Situation 68.16.1. Here $S$ is a scheme and $X$ is a quasi-compact and quasi-separated algebraic space over $S$ with the following property: For every quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ we have $H^1(X, \mathcal{F}) = 0$. We set $A = \Gamma (X, \mathcal{O}_ X)$.

## 68.16 Vanishing of cohomology

In this section we show that a quasi-compact and quasi-separated algebraic space is affine if it has vanishing higher cohomology for all quasi-coherent sheaves. We do this in a sequence of lemmas all of which will become obsolete once we prove Proposition 68.16.7.

We would like to show that the canonical morphism

(see Properties of Spaces, Lemma 65.33.1) is an isomorphism. If $M$ is an $A$-module we denote $M \otimes _ A \mathcal{O}_ X$ the quasi-coherent module $p^*\tilde M$.

Lemma 68.16.2. In Situation 68.16.1 for an $A$-module $M$ we have $p_*(M \otimes _ A \mathcal{O}_ X) = \widetilde{M}$ and $\Gamma (X, M \otimes _ A \mathcal{O}_ X) = M$.

**Proof.**
The equality $p_*(M \otimes _ A \mathcal{O}_ X) = \widetilde{M}$ follows from the equality $\Gamma (X, M \otimes _ A \mathcal{O}_ X) = M$ as $p_*(M \otimes _ A \mathcal{O}_ X)$ is a quasi-coherent module on $\mathop{\mathrm{Spec}}(A)$ by Morphisms of Spaces, Lemma 66.11.2. Observe that $\Gamma (X, \bigoplus _{i \in I} \mathcal{O}_ X) = \bigoplus _{i \in I} A$ by Lemma 68.5.1. Hence the lemma holds for free modules. Choose a short exact sequence $F_1 \to F_0 \to M$ where $F_0, F_1$ are free $A$-modules. Since $H^1(X, -)$ is zero the global sections functor is right exact. Moreover the pullback $p^*$ is right exact as well. Hence we see that

is exact. The result follows. $\square$

The following lemma shows that Situation 68.16.1 is preserved by base change of $X \to \mathop{\mathrm{Spec}}(A)$ by $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$.

Lemma 68.16.3. In Situation 68.16.1.

Given an affine morphism $X' \to X$ of algebraic spaces, we have $H^1(X', \mathcal{F}') = 0$ for every quasi-coherent $\mathcal{O}_{X'}$-module $\mathcal{F}'$.

Given an $A$-algebra $A'$ setting $X' = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A')$ the morphism $X' \to X$ is affine and $\Gamma (X', \mathcal{O}_{X'}) = A'$.

**Proof.**
Part (1) follows from Lemma 68.8.2 and the Leray spectral sequence (Cohomology on Sites, Lemma 21.14.5). Let $A \to A'$ be as in (2). Then $X' \to X$ is affine because affine morphisms are preserved under base change (Morphisms of Spaces, Lemma 66.20.5) and the fact that a morphism of affine schemes is affine. The equality $\Gamma (X', \mathcal{O}_{X'}) = A'$ follows as $(X' \to X)_*\mathcal{O}_{X'} = A' \otimes _ A \mathcal{O}_ X$ by Lemma 68.11.1 and thus

by Lemma 68.16.2. $\square$

Lemma 68.16.4. In Situation 68.16.1. Let $Z_0, Z_1 \subset |X|$ be disjoint closed subsets. Then there exists an $a \in A$ such that $Z_0 \subset V(a)$ and $Z_1 \subset V(a - 1)$.

**Proof.**
We may and do endow $Z_0$, $Z_1$ with the reduced induced subspace structure (Properties of Spaces, Definition 65.12.5) and we denote $i_0 : Z_0 \to X$ and $i_1 : Z_1 \to X$ the corresponding closed immersions. Since $Z_0 \cap Z_1 = \emptyset $ we see that the canonical map of quasi-coherent $\mathcal{O}_ X$-modules

is surjective (look at stalks at geometric points). Since $H^1(X, -)$ is zero on the kernel of this map the induced map of global sections is surjective. Thus we can find $a \in A$ which maps to the global section $(0, 1)$ of the right hand side. $\square$

Lemma 68.16.5. In Situation 68.16.1 the morphism $p : X \to \mathop{\mathrm{Spec}}(A)$ is universally injective.

**Proof.**
Let $A \to k$ be a ring homomorphism where $k$ is a field. It suffices to show that $\mathop{\mathrm{Spec}}(k) \times _{\mathop{\mathrm{Spec}}(A)} X$ has at most one point (see Morphisms of Spaces, Lemma 66.19.6). Using Lemma 68.16.3 we may assume that $A$ is a field and we have to show that $|X|$ has at most one point.

Let's think of $X$ as an algebraic space over $\mathop{\mathrm{Spec}}(k)$ and let's use the notation $X(K)$ to denote $K$-valued points of $X$ for any extension $K/k$, see Morphisms of Spaces, Section 66.24. If $K/k$ is an algebraically closed field extension of large transcendence degree, then we see that $X(K) \to |X|$ is surjective, see Morphisms of Spaces, Lemma 66.24.2. Hence, after replacing $k$ by $K$, we see that it suffices to prove that $X(k)$ is a singleton (in the case $A = k)$.

Let $x, x' \in X(k)$. By Decent Spaces, Lemma 67.14.4 we see that $x$ and $x'$ are closed points of $|X|$. Hence $x$ and $x'$ map to distinct points of $\mathop{\mathrm{Spec}}(k)$ if $x \not= x'$ by Lemma 68.16.4. We conclude that $x = x'$ as desired. $\square$

Lemma 68.16.6. In Situation 68.16.1 the morphism $p : X \to \mathop{\mathrm{Spec}}(A)$ is separated.

**Proof.**
By Decent Spaces, Lemma 67.9.2 we can find a scheme $Y$ and a surjective integral morphism $Y \to X$. Since an integral morphism is affine, we can apply Lemma 68.16.3 to see that $H^1(Y, \mathcal{G}) = 0$ for every quasi-coherent $\mathcal{O}_ Y$-module $\mathcal{G}$. Since $Y \to X$ is quasi-compact and $X$ is quasi-compact, we see that $Y$ is quasi-compact. Since $Y$ is a scheme, we may apply Cohomology of Schemes, Lemma 30.3.1 to see that $Y$ is affine. Hence $Y$ is separated. Note that an integral morphism is affine and universally closed, see Morphisms of Spaces, Lemma 66.45.7. By Morphisms of Spaces, Lemma 66.9.8 we see that $X$ is a separated algebraic space.
$\square$

Proposition 68.16.7. A quasi-compact and quasi-separated algebraic space is affine if and only if all higher cohomology groups of quasi-coherent sheaves vanish. More precisely, any algebraic space as in Situation 68.16.1 is an affine scheme.

**Proof.**
Choose an affine scheme $U = \mathop{\mathrm{Spec}}(B)$ and a surjective étale morphism $\varphi : U \to X$. Set $R = U \times _ X U$. As $p$ is separated (Lemma 68.16.6) we see that $R$ is a closed subscheme of $U \times _{\mathop{\mathrm{Spec}}(A)} U = \mathop{\mathrm{Spec}}(B \otimes _ A B)$. Hence $R = \mathop{\mathrm{Spec}}(C)$ is affine too and the ring map

is surjective. Let us denote the two maps $s, t : B \to C$ as usual. Pick $g_1, \ldots , g_ m \in B$ such that $s(g_1), \ldots , s(g_ m)$ generate $C$ over $t : B \to C$ (which is possible as $t : B \to C$ is of finite presentation and the displayed map is surjective). Then $g_1, \ldots , g_ m$ give global sections of $\varphi _*\mathcal{O}_ U$ and the map

is surjective: you can check this by restricting to $U$. Namely, $\varphi ^*\varphi _*\mathcal{O}_ U = t_*\mathcal{O}_ R$ (by Lemma 68.11.2) hence you get exactly the condition that $s(g_ i)$ generate $C$ over $t : B \to C$. By the vanishing of $H^1$ of the kernel we see that

is surjective. Thus we conclude that $B$ is a finite type $A$-algebra. Hence $X \to \mathop{\mathrm{Spec}}(A)$ is of finite type and separated. By Lemma 68.16.5 and Morphisms of Spaces, Lemma 66.27.5 it is also locally quasi-finite. Hence $X \to \mathop{\mathrm{Spec}}(A)$ is representable by Morphisms of Spaces, Lemma 66.51.1 and $X$ is a scheme. Finally $X$ is affine, hence equal to $\mathop{\mathrm{Spec}}(A)$, by an application of Cohomology of Schemes, Lemma 30.3.1. $\square$

Lemma 68.16.8. Let $S$ be a scheme. Let $X$ be a Noetherian algebraic space over $S$. Assume that for every coherent $\mathcal{O}_ X$-module $\mathcal{F}$ we have $H^1(X, \mathcal{F}) = 0$. Then $X$ is an affine scheme.

**Proof.**
The assumption implies that $H^1(X, \mathcal{F}) = 0$ for every quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ by Lemmas 68.15.1 and 68.5.1. Then $X$ is affine by Proposition 68.16.7.
$\square$

Lemma 68.16.9. Let $S$ be a scheme. Let $X$ be a Noetherian algebraic space over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Assume that for every coherent $\mathcal{O}_ X$-module $\mathcal{F}$ there exists an $n \geq 1$ such that $H^1(X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) = 0$. Then $X$ is a scheme and $\mathcal{L}$ is ample on $X$.

**Proof.**
Let $s \in H^0(X, \mathcal{L}^{\otimes d})$ be a global section. Let $U \subset X$ be the open subspace over which $s$ is a generator of $\mathcal{L}^{\otimes d}$. In particular we have $\mathcal{L}^{\otimes d}|_ U \cong \mathcal{O}_ U$. We claim that $U$ is affine.

Proof of the claim. We will show that $H^1(U, \mathcal{F}) = 0$ for every quasi-coherent $\mathcal{O}_ U$-module $\mathcal{F}$. This will prove the claim by Proposition 68.16.7. Denote $j : U \to X$ the inclusion morphism. Since étale locally the morphism $j$ is affine (by Morphisms, Lemma 29.11.10) we see that $j$ is affine (Morphisms of Spaces, Lemma 66.20.3). Hence we have

by Lemma 68.8.2 (and Cohomology on Sites, Lemma 21.14.6). Write $j_*\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ as a filtered colimit of coherent $\mathcal{O}_ X$-modules, see Lemma 68.15.1. Then

by Lemma 68.5.1. Thus it suffices to show that $H^1(X, \mathcal{F}_ i)$ maps to zero in $H^1(U, j^*\mathcal{F}_ i)$. By assumption there exists an $n \geq 1$ such that

Hence there exists an $a \geq 0$ such that $H^1(X, \mathcal{F}_ i \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes ad}) = 0$. On the other hand, the map

is an isomorphism after restriction to $U$. Contemplating the commutative diagram

we conclude that the map $H^1(X, \mathcal{F}_ i) \to H^1(U, j^*\mathcal{F}_ i)$ is zero and the claim holds.

Let $x \in |X|$ be a closed point. By Decent Spaces, Lemma 67.14.6 we can represent $x$ by a closed immersion $i : \mathop{\mathrm{Spec}}(k) \to X$ (this also uses that a quasi-separated algebraic space is decent, see Decent Spaces, Section 67.6). Thus $\mathcal{O}_ X \to i_*\mathcal{O}_{\mathop{\mathrm{Spec}}(k)}$ is surjective. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the kernel and choose $d \geq 1$ such that $H^1(X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d}) = 0$. Then

is surjective by the long exact cohomology sequence. Hence there exists an $s \in H^0(X, \mathcal{L}^{\otimes d})$ such that $x \in U$ where $U$ is the open subspace corresponding to $s$ as above. Thus $x$ is in the schematic locus (see Properties of Spaces, Lemma 65.13.1) of $X$ by our claim.

To conclude that $X$ is a scheme, it suffices to show that any open subset of $|X|$ which contains all the closed points is equal to $|X|$. This follows from the fact that $|X|$ is a Noetherian topological space, see Properties of Spaces, Lemma 65.24.3. Finally, if $X$ is a scheme, then we can apply Cohomology of Schemes, Lemma 30.3.3 to conclude that $\mathcal{L}$ is ample. $\square$

## 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.

## Comments (0)