## 33.37 One dimensional Noetherian schemes

The main result of this section is that a Noetherian separated scheme of dimension $1$ has an ample invertible sheaf. See Proposition 33.37.12.

Lemma 33.37.1. Let $X$ be a scheme all of whose local rings are Noetherian of dimension $\leq 1$. Let $U \subset X$ be a retrocompact open. Denote $j : U \to X$ the inclusion morphism. Then $R^ pj_*\mathcal{F} = 0$, $p > 0$ for every quasi-coherent $\mathcal{O}_ U$-module $\mathcal{F}$.

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. $\square$

Lemma 33.37.2. Let $X$ be an affine scheme all of whose local rings are Noetherian of dimension $\leq 1$. Then any quasi-compact open $U \subset X$ is affine.

Proof. Denote $j : U \to X$ the inclusion morphism. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ U$-module. By Lemma 33.37.1 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. $\square$

Lemma 33.37.3. Let $X$ be a scheme. Let $U \subset X$ be an open. Assume

1. $U$ is a retrocompact open of $X$,

2. $X \setminus U$ is discrete, and

3. for $x \in X \setminus U$ the local ring $\mathcal{O}_{X, x}$ is Noetherian of dimension $\leq 1$.

Then (1) there exists an invertible $\mathcal{O}_ X$-module $\mathcal{L}$ and a section $s$ such that $U = X_ s$ and (2) the map $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (U)$ is surjective.

Proof. Let $X \setminus U = \{ x_ i; i \in I\}$. Choose affine opens $U_ i \subset X$ with $x_ i \in X$ 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

$\mathcal{U} : X = U \cup \bigcup U_ i$

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 $U_ i$). $\square$

Lemma 33.37.4. Let $X$ be an integral separated scheme. Let $U \subset X$ be a nonempty affine open such that $X \setminus U$ is a finite set of points $x_1, \ldots , x_ r$ with $\mathcal{O}_{X, x_ i}$ Noetherian of dimension $1$. Then there exists a globally generated invertible $\mathcal{O}_ X$-module $\mathcal{L}$ and a section $s$ such that $U = X_ s$.

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 33.36.8 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

$\mathcal{U} : X = U \cup \bigcup U_ i$

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. $\square$

Lemma 33.37.5. Let $X$ be a quasi-compact scheme. If for every $x \in X$ there exists a pair $(\mathcal{L}, s)$ consisting of a globally generated invertible sheaf $\mathcal{L}$ and a global section $s$ such that $x \in X_ s$ and $X_ s$ is affine, then $X$ has an ample invertible sheaf.

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 $X_{s_ i}$ 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 $t_{i, j}$, $j = 1, \ldots , m_ i$ such that $X = \bigcup X_{t_{i, j}}$. Set $t_{i, 0} = s_ i$. Consider the invertible sheaf

$\mathcal{L} = \mathcal{L}_1 \otimes _{\mathcal{O}_ X} \ldots \otimes _{\mathcal{O}_ X} \mathcal{L}_ n$

and the global sections

$\tau _ J = t_{1, j_1} \otimes \ldots \otimes t_{n, j_ n}$

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. $\square$

Lemma 33.37.6. Let $X$ be a Noetherian integral separated scheme of dimension $1$. Then $X$ has an ample invertible sheaf.

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 33.37.4 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 33.37.5. $\square$

Lemma 33.37.7. Let $f : X \to Y$ be a finite morphism of schemes. Assume there exists an open $V \subset Y$ such that $f^{-1}(V) \to V$ is an isomorphism and $Y \setminus V$ is a discrete space. Then every invertible $\mathcal{O}_ X$-module is the pullback of an invertible $\mathcal{O}_ Y$-module.

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

$H^1(X, \mathcal{O}_ X^*) \longrightarrow H^0(Y, R^1f_*\mathcal{O}_ X^*)$

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

$f^\sharp : \mathcal{O}_ Y^* \longrightarrow f_*\mathcal{O}_ X^*$

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

$0 \to \mathop{\mathrm{Ker}}(f^\sharp ) \to \mathcal{O}_ Y^* \to \mathop{\mathrm{Im}}(f^\sharp ) \to 0,\quad 0 \to \mathop{\mathrm{Im}}(f^\sharp ) \to f_*\mathcal{O}_ X^* \to \mathop{\mathrm{Coker}}(f^\sharp ) \to 0$

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. $\square$

Lemma 33.37.8. Let $X$ be a scheme. Let $Z_1, \ldots , Z_ n \subset X$ be closed subschemes. Let $\mathcal{L}_ i$ be an invertible sheaf on $Z_ i$. Assume that

1. $X$ is reduced,

2. $X = \bigcup Z_ i$ set theoretically, and

3. $Z_ i \cap Z_ j$ is a discrete topological space for $i \not= j$.

Then there exists an invertible sheaf $\mathcal{L}$ on $X$ whose restriction to $Z_ i$ is $\mathcal{L}_ i$. Moreover, if we are given sections $s_ i \in \Gamma (Z_ i, \mathcal{L}_ i)$ which are nonvanishing at the points of $Z_ i \cap Z_ j$, then we can choose $\mathcal{L}$ such that there exists a $s \in \Gamma (X, \mathcal{L})$ with $s|_{Z_ i} = s_ i$ for all $i$.

Proof. The existence of $\mathcal{L}$ can be deduced from Lemma 33.37.7 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

$X \setminus T = \bigcup (Z_ i \setminus T)$

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

$Z_ i \setminus T = \bigcup \nolimits _ j U_{ij}$

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

$\mathcal{U} : X = \bigcup U_ t \cup \bigcup U_{ij}$

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

$U_ t \cap U_{t'} = \coprod (U_ t \cap U_{t'}) \cap Z_ i$

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

$\mathcal{O}_ X(U_ t \cap U_{t'}) = \prod \mathcal{O}_{Z_ i}(U_ t \cap U_{t'} \cap Z_ i)$

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 $Z_ i$. $\square$

Remark 33.37.9. Let $A$ be a reduced ring. Let $I, J$ be ideals of $A$ such that $V(I) \cup V(J) = \mathop{\mathrm{Spec}}(A)$. Set $B = A/J$. Then $I \to IB$ is an isomorphism of $A$-modules. Namely, we have $IB = I + J/J = I/(I \cap J)$ and $I \cap J$ is zero because $A$ is reduced and $\mathop{\mathrm{Spec}}(A) = V(I) \cup V(J) = V(I \cap J)$. Thus for any projective $A$-module $P$ we also have $IP = I(P/JP)$.

Lemma 33.37.10. Let $X$ be a Noetherian reduced separated scheme of dimension $1$. Then $X$ has an ample invertible sheaf.

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 33.37.6 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 33.37.8 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

$\Gamma (X, \mathcal{I}\mathcal{L}^{\otimes m}) = \Gamma (Z_1, \mathcal{I}_1 \mathcal{L}_1^{\otimes m})$

see Remark 33.37.9. 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. $\square$

Lemma 33.37.11. Let $i : Z \to X$ be a closed immersion of schemes. If the underlying topological space of $X$ is Noetherian and $\dim (X) \leq 1$, then $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (Z)$ is surjective.

Proof. Consider the short exact sequence

$0 \to (1 + \mathcal{I}) \cap \mathcal{O}_ X^* \to \mathcal{O}^*_ X \to i_*\mathcal{O}^*_ Z \to 0$

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. $\square$

Proposition 33.37.12. Let $X$ be a Noetherian separated scheme of dimension $1$. Then $X$ has an ample invertible sheaf.

Proof. Let $Z \subset X$ be the reduction of $X$. By Lemma 33.37.10 the scheme $Z$ has an ample invertible sheaf. Thus by Lemma 33.37.11 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. $\square$

Remark 33.37.13. In fact, if $X$ is a scheme whose reduction is a Noetherian separated scheme of dimension $1$, then $X$ has an ample invertible sheaf. The argument to prove this is the same as the proof of Proposition 33.37.12 except one uses Limits, Lemma 32.11.4 instead of Cohomology of Schemes, Lemma 30.17.5.

The following lemma actually holds for quasi-finite separated morphisms as the reader can see by using Zariski's main theorem (More on Morphisms, Lemma 37.39.3) and Lemma 33.37.3.

Lemma 33.37.14. Let $f : X \to Y$ be a morphism of schemes. Assume $Y$ is Noetherian of dimension $\leq 1$, $f$ is finite, and there exists a dense open $V \subset Y$ such that $f^{-1}(V) \to V$ is a closed immersion. Then every invertible $\mathcal{O}_ X$-module is the pullback of an invertible $\mathcal{O}_ Y$-module.

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 33.37.7 applies. On the other hand, Lemma 33.37.11 applies to $Z \to Y$. Some details omitted. $\square$

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