The Stacks project

Proof. We may check the vanishing of $R^ pj_*\mathcal{F}$ at stalks. Formation of $R^ qj_*$ commutes with flat base change, see Cohomology of Schemes, Lemma 30.5.2. Thus we may assume that $X$ is the spectrum of a Noetherian local ring of dimension $\leq 1$. In this case $X$ has a closed point $x$ and finitely many other points $x_1, \ldots , x_ n$ which specialize to $x$ but not each other (see Algebra, Lemma 10.31.6). If $x \in U$, then $U = X$ and the result is clear. If not, then $U = \{ x_1, \ldots , x_ r\} $ for some $r$ after possibly renumbering the points. Then $U$ is affine (Schemes, Lemma 26.11.8). Thus the result follows from Cohomology of Schemes, Lemma 30.2.3. $\square$

Proof. Denote $j : U \to X$ the inclusion morphism. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ U$-module. By Lemma 33.38.1 the higher direct images $R^ pj_*\mathcal{F}$ are zero. The $\mathcal{O}_ X$-module $j_*\mathcal{F}$ is quasi-coherent (Schemes, Lemma 26.24.1). Hence it has vanishing higher cohomology groups by Cohomology of Schemes, Lemma 30.2.2. By the Leray spectral sequence Cohomology, Lemma 20.13.6 we have $H^ p(U, \mathcal{F}) = 0$ for all $p > 0$. Thus $U$ is affine, for example by Cohomology of Schemes, Lemma 30.3.1. $\square$

Proof. Let $X \setminus U = \{ x_ i; i \in I\} $. Choose affine opens $U_ i \subset X$ with $x_ i \in U_ i$ and $x_ j \not\in U_ i$ for $j \not= i$. This is possible by condition (2). Say $U_ i = \mathop{\mathrm{Spec}}(A_ i)$. Let $\mathfrak m_ i \subset A_ i$ be the maximal ideal corresponding to $x_ i$. By our assumption on the local rings there are only a finite number of prime ideals $\mathfrak q \subset \mathfrak m_ i$, $\mathfrak q \not= \mathfrak m_ i$ (see Algebra, Lemma 10.31.6). Thus by prime avoidance (Algebra, Lemma 10.15.2) we can find $f_ i \in \mathfrak m_ i$ not contained in any of those primes. Then $V(f_ i) = \{ \mathfrak m_ i\} \amalg Z_ i$ for some closed subset $Z_ i \subset U_ i$ because $Z_ i$ is a retrocompact open subset of $V(f_ i)$ closed under specialization, see Algebra, Lemma 10.41.7. After shrinking $U_ i$ we may assume $V(f_ i) = \{ x_ i\} $. Then

is an open covering of $X$. Consider the $2$-cocycle with values in $\mathcal{O}_ X^*$ given by $f_ i$ on $U \cap U_ i$ and by $f_ i/f_ j$ on $U_ i \cap U_ j$. This defines a line bundle $\mathcal{L}$ such that the section $s$ defined by $1$ on $U$ and $f_ i$ on $U_ i$ is as in the statement of the lemma.

Let $\mathcal{N}$ be an invertible $\mathcal{O}_ U$-module. Let $N_ i$ be the invertible $(A_ i)_{f_ i}$ module such that $\mathcal{N}|_{U \cap U_ i}$ is equal to $\tilde N_ i$. Observe that $(A_{\mathfrak m_ i})_{f_ i}$ is an Artinian ring (as a dimension zero Noetherian ring, see Algebra, Lemma 10.60.5). Thus it is a product of local rings (Algebra, Lemma 10.53.6) and hence has trivial Picard group. Thus, after shrinking $U_ i$ (i.e., after replacing $A_ i$ by $(A_ i)_ g$ for some $g \in A_ i$, $g \not\in \mathfrak m_ i$) we can assume that $N_ i = (A_ i)_{f_ i}$, i.e., that $\mathcal{N}|_{U \cap U_ i}$ is trivial. In this case it is clear how to extend $\mathcal{N}$ to an invertible sheaf over $X$ (by extending it by a trivial invertible module over each $U_ i$). $\square$

Proof. Say $U = \mathop{\mathrm{Spec}}(A)$ and let $K$ be the function field of $X$. Write $B_ i = \mathcal{O}_{X, x_ i}$ and $\mathfrak m_ i = \mathfrak m_{x_ i}$. Since $x_ i \not\in U$ we see that the open $U \times _ X \mathop{\mathrm{Spec}}(B_ i)$ of $\mathop{\mathrm{Spec}}(B_ i)$ has only one point, i.e., $U \times _ X \mathop{\mathrm{Spec}}(B_ i) = \mathop{\mathrm{Spec}}(K)$. Since $X$ is separated, we find that $\mathop{\mathrm{Spec}}(K)$ is a closed subscheme of $U \times \mathop{\mathrm{Spec}}(B_ i)$, i.e., the map $A \otimes B_ i \to K$ is a surjection. By Lemma 33.37.8 we can find a nonzero $f \in A$ such that $f^{-1} \in \mathfrak m_ i$ for $i = 1, \ldots , r$. Pick opens $x_ i \in U_ i \subset X$ such that $f^{-1} \in \mathcal{O}(U_ i)$. Then

is an open covering of $X$. Consider the $2$-cocycle with values in $\mathcal{O}_ X^*$ given by $f$ on $U \cap U_ i$ and by $1$ on $U_ i \cap U_ j$. This defines a line bundle $\mathcal{L}$ with two sections:

1. a section $s$ defined by $1$ on $U$ and $f^{-1}$ on $U_ i$ is as in the statement of the lemma, and

2. a section $t$ defined by $f$ on $U$ and $1$ on $U_ i$.

Note that $X_ t \supset U_1 \cup \ldots \cup U_ r$. Hence $s, t$ generate $\mathcal{L}$ and the lemma is proved. $\square$

Proof. Since $X$ is quasi-compact we can find a finite collection $(\mathcal{L}_ i, s_ i)$, $i = 1, \ldots , n$ of pairs such that $\mathcal{L}_ i$ is globally generated, $X_{s_ i}$ is affine and $X = \bigcup X_{s_ i}$. Again because $X$ is quasi-compact we can find, for each $i$, a finite collection of sections $t_{i, j}$ of $\mathcal{L}_ i$, $j = 1, \ldots , m_ i$ such that $X = \bigcup X_{t_{i, j}}$. Set $t_{i, 0} = s_ i$. Consider the invertible sheaf

and the global sections

By Properties, Lemma 28.26.4 the open $X_{\tau _ J}$ is affine as soon as $j_ i = 0$ for some $i$. It is a simple matter to see that these opens cover $X$. Hence $\mathcal{L}$ is ample by definition. $\square$

Proof. Choose an affine open covering $X = U_1 \cup \ldots \cup U_ n$. Since $X$ is Noetherian, each of the sets $X \setminus U_ i$ is finite. Thus by Lemma 33.38.4 we can find a pair $(\mathcal{L}_ i, s_ i)$ consisting of a globally generated invertible sheaf $\mathcal{L}_ i$ and a global section $s_ i$ such that $U_ i = X_{s_ i}$. We conclude that $X$ has an ample invertible sheaf by Lemma 33.38.5. $\square$

Proof. We will use that $\mathop{\mathrm{Pic}}\nolimits (X) = H^1(X, \mathcal{O}_ X^*)$, see Cohomology, Lemma 20.6.1. Consider the Leray spectral sequence for the abelian sheaf $\mathcal{O}_ X^*$ and $f$, see Cohomology, Lemma 20.13.4. Consider the induced map

Divisors, Lemma 31.17.1 says exactly that this map is zero. Hence Leray gives $H^1(X, \mathcal{O}_ X^*) = H^1(Y, f_*\mathcal{O}_ X^*)$. Next we consider the map

By assumption the kernel and cokernel of this map are supported on the closed subset $T = Y \setminus V$ of $Y$. Since $T$ is a discrete topological space by assumption the higher cohomology groups of any abelian sheaf on $Y$ supported on $T$ is zero (follows from Cohomology, Lemma 20.20.1, Modules, Lemma 17.6.1, and the fact that $H^ i(T, \mathcal{F}) = 0$ for any $i > 0$ and any abelian sheaf $\mathcal{F}$ on $T$). Breaking the displayed map into short exact sequences

we first conclude that $H^1(Y, \mathcal{O}_ Y^*) \to H^1(Y, \mathop{\mathrm{Im}}(f^\sharp ))$ is surjective and then that $H^1(Y, \mathop{\mathrm{Im}}(f^\sharp )) \to H^1(Y, f_*\mathcal{O}_ X^*)$ is surjective. Combining all the above we find that $H^1(Y, \mathcal{O}_ Y^*) \to H^1(X, \mathcal{O}_ X^*)$ is surjective as desired. $\square$

Proof. The existence of $\mathcal{L}$ can be deduced from Lemma 33.38.7 but we will also give a direct proof and we will use the direct proof to see the statement about sections is true. Set $T = \bigcup _{i \not= j} Z_ i \cap Z_ j$. As $X$ is reduced we have

as schemes. Assumption (3) implies $T$ is a discrete subset of $X$. Thus for each $t \in T$ we can find an open $U_ t \subset X$ with $t \in U_ t$ but $t' \not\in U_ t$ for $t' \in T$, $t' \not= t$. By shrinking $U_ t$ if necessary, we may assume that there exist isomorphisms $\varphi _{t, i} : \mathcal{L}_ i|_{U_ t \cap Z_ i} \to \mathcal{O}_{U_ t \cap Z_ i}$. Furthermore, for each $i$ choose an open covering

such that there exist isomorphisms $\varphi _{i, j} : \mathcal{L}_ i|_{U_{ij}} \cong \mathcal{O}_{U_{ij}}$. Observe that

is an open covering of $X$. We claim that we can use the isomorphisms $\varphi _{t, i}$ and $\varphi _{i, j}$ to define a $2$-cocycle with values in $\mathcal{O}_ X^*$ for this covering that defines $\mathcal{L}$ as in the statement of the lemma.

Namely, if $i \not= i'$, then $U_{i, j} \cap U_{i', j'} = \emptyset $ and there is nothing to do. For $U_{i, j} \cap U_{i, j'}$ we have $\mathcal{O}_ X(U_{i, j} \cap U_{i, j'}) = \mathcal{O}_{Z_ i}(U_{i, j} \cap U_{i, j'})$ by the first remark of the proof. Thus the transition function for $\mathcal{L}_ i$ (more precisely $\varphi _{i, j} \circ \varphi _{i, j'}^{-1}$) defines the value of our cocycle on this intersection. For $U_ t \cap U_{i, j}$ we can do the same thing. Finally, for $t \not= t'$ we have

and moreover the intersection $U_ t \cap U_{t'} \cap Z_ i$ is contained in $Z_ i \setminus T$. Hence by the same reasoning as before we see that

and we can use the transition functions for $\mathcal{L}_ i$ (more precisely $\varphi _{t, i} \circ \varphi _{t', i}^{-1}$) to define the value of our cocycle on $U_ t \cap U_{t'}$. This finishes the proof of existence of $\mathcal{L}$.

Given sections $s_ i$ as in the last assertion of the lemma, in the argument above, we choose $U_ t$ such that $s_ i|_{U_ t \cap Z_ i}$ is nonvanishing and we choose $\varphi _{t, i}$ such that $\varphi _{t, i}(s_ i|_{U_ t \cap Z_ i}) = 1$. Then using $1$ over $U_ t$ and $\varphi _{i, j}(s_ i|_{U_{i, j}})$ over $U_{i, j}$ will define a section of $\mathcal{L}$ which restricts to $s_ i$ over $Z_ i$. $\square$

Proof. Let $Z_ i$, $i = 1, \ldots , n$ be the irreducible components of $X$. We view these as reduced closed subschemes of $X$. By Lemma 33.38.6 there exist ample invertible sheaves $\mathcal{L}_ i$ on $Z_ i$. Set $T = \bigcup _{i \not= j} Z_ i \cap Z_ j$. As $X$ is Noetherian of dimension $1$, the set $T$ is finite and consists of closed points of $X$. For each $i$ we may, possibly after replacing $\mathcal{L}_ i$ by a power, choose $s_ i \in \Gamma (Z_ i, \mathcal{L}_ i)$ such that $(Z_ i)_{s_ i}$ is affine and contains $T \cap Z_ i$, see Properties, Lemma 28.29.6.

By Lemma 33.38.8 we can find an invertible sheaf $\mathcal{L}$ on $X$ and $s \in \Gamma (X, \mathcal{L})$ such that $(\mathcal{L}, s)|_{Z_ i} = (\mathcal{L}_ i, s_ i)$. Observe that $X_ s$ contains $T$ and is set theoretically equal to the affine closed subschemes $(Z_ i)_{s_ i}$. Thus it is affine by Limits, Lemma 32.11.3. To finish the proof, it suffices to find for every $x \in X$, $x \not\in T$ an integer $m > 0$ and a section $t \in \Gamma (X, \mathcal{L}^{\otimes m})$ such that $X_ t$ is affine and $x \in X_ t$. Since $x \not\in T$ we see that $x \in Z_ i$ for some unique $i$, say $i = 1$. Let $Z \subset X$ be the reduced closed subscheme whose underlying topological space is $Z_2 \cup \ldots \cup Z_ n$. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the ideal sheaf of $Z$. Denote that $\mathcal{I}_1 \subset \mathcal{O}_{Z_1}$ the inverse image of this ideal sheaf under the inclusion morphism $Z_1 \to X$. Observe that

see Remark 33.38.9. Thus it suffices to find $m > 0$ and $t \in \Gamma (Z_1, \mathcal{I}_1 \mathcal{L}_1^{\otimes m})$ with $x \in (Z_1)_ t$ affine. Since $\mathcal{L}_1$ is ample and since $x$ is not in $Z_1 \cap T = V(\mathcal{I}_1)$ we can find a section $t_1 \in \Gamma (Z_1, \mathcal{I}_1 \mathcal{L}_1^{\otimes m_1})$ with $x \in (Z_1)_{t_1}$, see Properties, Proposition 28.26.13. Since $\mathcal{L}_1$ is ample we can find a section $t_2 \in \Gamma (Z_1, \mathcal{L}_1^{\otimes m_2})$ with $x \in (Z_1)_{t_2}$ and $(Z_1)_{t_2}$ affine, see Properties, Definition 28.26.1. Set $m = m_1 + m_2$ and $t = t_1 t_2$. Then $t \in \Gamma (Z_1, \mathcal{I}_1 \mathcal{L}_1^{\otimes m})$ with $x \in (Z_1)_ t$ by construction and $(Z_1)_ t$ is affine by Properties, Lemma 28.26.4. $\square$

Proof. Consider the short exact sequence

of sheaves of abelian groups on $X$ where $\mathcal{I}$ is the quasi-coherent sheaf of ideals corresponding to $Z$. Since $\dim (X) \leq 1$ we see that $H^2(X, \mathcal{F}) = 0$ for any abelian sheaf $\mathcal{F}$, see Cohomology, Proposition 20.20.7. Hence the map $H^1(X, \mathcal{O}^*_ X) \to H^1(X, i_*\mathcal{O}_ Z^*)$ is surjective. By Cohomology, Lemma 20.20.1 we have $H^1(X, i_*\mathcal{O}_ Z^*) = H^1(Z, \mathcal{O}_ Z^*)$. This proves the lemma by Cohomology, Lemma 20.6.1. $\square$

Proof. Let $Z \subset X$ be the reduction of $X$. By Lemma 33.38.10 the scheme $Z$ has an ample invertible sheaf. Thus by Lemma 33.38.11 there exists an invertible $\mathcal{O}_ X$-module $\mathcal{L}$ on $X$ whose restriction to $Z$ is ample. Then $\mathcal{L}$ is ample by an application of Cohomology of Schemes, Lemma 30.17.5. $\square$

Proof. We factor $f$ as $X \to Z \to Y$ where $Z$ is the scheme theoretic image of $f$. Then $X \to Z$ is an isomorphism over $V \cap Z$ and Lemma 33.38.7 applies. On the other hand, Lemma 33.38.11 applies to $Z \to Y$. Some details omitted. $\square$