The Stacks project

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 67.11.2. Observe that $\Gamma (X, \bigoplus _{i \in I} \mathcal{O}_ X) = \bigoplus _{i \in I} A$ by Lemma 69.5.1. 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. $\square$

Proof. Part (1) follows from Lemma 69.8.2 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 67.20.5) 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 69.11.1 and thus

by Lemma 69.16.2. $\square$

Proof. We may and do endow $Z_0$, $Z_1$ with the reduced induced subspace structure (Properties of Spaces, Definition 66.12.5) 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. $\square$

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 67.19.6). Using Lemma 69.16.3 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 67.24. 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 67.24.2. 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 68.14.4 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 69.16.4. We conclude that $x = x'$ as desired. $\square$

Proof. By Decent Spaces, Lemma 68.9.2 we can find a scheme $Y$ and a surjective integral morphism $Y \to X$. Since an integral morphism is affine, we can apply Lemma 69.16.3 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 67.45.7. By Morphisms of Spaces, Lemma 67.9.8 we see that $X$ is a separated algebraic space. $\square$

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 69.16.6) 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 69.11.2) 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 69.16.5 and Morphisms of Spaces, Lemma 67.27.5 it is also locally quasi-finite. Hence $X \to \mathop{\mathrm{Spec}}(A)$ is representable by Morphisms of Spaces, Lemma 67.51.1 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. $\square$

Proof. The assumption implies that $H^1(X, \mathcal{F}) = 0$ for every quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ by Lemmas 69.15.1 and 69.5.1. Then $X$ is affine by Proposition 69.16.7. $\square$

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 69.16.7. 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 67.20.3). Hence we have

by Lemma 69.8.2 (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 69.15.1. Then

by Lemma 69.5.1. 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 68.14.6 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 68.6). 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 66.13.1) 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 66.24.3. Finally, if $X$ is a scheme, then we can apply Cohomology of Schemes, Lemma 30.3.3 to conclude that $\mathcal{L}$ is ample. $\square$