The Stacks project

70.6 Weil divisors

This section is the analogue of Divisors, Section 31.26.

We will introduce Weil divisors and rational equivalence of Weil divisors for locally Noetherian integral algebraic spaces. Since we are not assuming our algebraic spaces are quasi-compact we have to be a little careful when defining Weil divisors. We have to allow infinite sums of prime divisors because a rational function may have infinitely many poles for example. In the quasi-compact case our Weil divisors are finite sums as usual. Here is a basic lemma we will often use to prove collections of closed subspaces are locally finite.

Lemma 70.6.1. Let $S$ be a scheme and let $X$ be a locally Noetherian algebraic space over $S$. If $T \subset |X|$ is a closed subset, then the collection of irreducible components of $T$ is locally finite.

Proof. The topological space $|X|$ is locally Noetherian (Properties of Spaces, Lemma 64.24.2). A Noetherian topological space has a finite number of irreducible components and a subspace of a Noetherian space is Noetherian (Topology, Lemma 5.9.2). Thus the lemma follows from the definition of locally finite (Topology, Definition 5.28.4). $\square$

Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $Z$ be an integral closed subspace of $X$ and let $\xi \in |Z|$ be the generic point. Then the codimension of $|Z|$ in $|X|$ is equal to the dimension of the local ring of $X$ at $\xi $ by Decent Spaces, Lemma 66.20.2. Recall that we also indicate this by saying that $\xi $ is a point of codimension $1$ on $X$, see Properties of Spaces, Definition 64.10.2.

Definition 70.6.2. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$.

  1. A prime divisor is an integral closed subspace $Z \subset X$ of codimension $1$, i.e., the generic point of $|Z|$ is a point of codimension $1$ on $X$.

  2. A Weil divisor is a formal sum $D = \sum n_ Z Z$ where the sum is over prime divisors of $X$ and the collection $\{ |Z| : n_ Z \not= 0\} $ is locally finite in $|X|$ (Topology, Definition 5.28.4).

The group of all Weil divisors on $X$ is denoted $\text{Div}(X)$.

Our next task is to define the Weil divisor associated to a rational function. In order to do this we need to define the order of vanishing of a rational function on a locally Noetherian integral algebraic space $X$ along a prime divisor $Z$. Let $\xi \in |Z|$ be the generic point. Here we run into the problem that the local ring $\mathcal{O}_{X, \xi }$ doesn't exist and the henselian local ring $\mathcal{O}_{X, \xi }^ h$ may not be a domain, see Example 70.6.11. To get around this we use the following lemma.

Lemma 70.6.3. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. Let $Z \subset X$ be a prime divisor and let $\xi \in |Z|$ be the generic point. Then the henselian local ring $\mathcal{O}_{X, \xi }^ h$ is a reduced $1$-dimensional Noetherian local ring and there is a canonical injective map

\[ R(X) \longrightarrow Q(\mathcal{O}_{X, \xi }^ h) \]

from the function field $R(X)$ of $X$ into the total ring of fractions.

Proof. We will use the results of Decent Spaces, Section 66.11. Let $(U, u) \to (X, \xi )$ be an elementary ├ętale neighbourhood. Observe that $U$ is locally Noetherian and reduced. Thus $\mathcal{O}_{U, u}$ is a $1$-dimensional (by our definition of prime divisors) reduced Noetherian ring. After replacing $U$ by an affine open neighbourhood of $u$ we may assume $U$ is Noetherian and affine. After replacing $U$ by a smaller open, we may assume every irreducible component of $U$ passes through $u$. Since $U \to X$ is open and $X$ irreducible, $U \to X$ is dominant. Hence we obtain a ring map $R(X) \to R(U)$ by composing rational maps, see Morphisms of Spaces, Section 65.47. Since $R(X)$ is a field, this map is injective. By our choice of $U$ we see that $R(U)$ is the total quotient ring $Q(\mathcal{O}_{U, u})$, see Morphisms, Lemma 29.48.5 and Algebra, Lemma 10.24.4.

At this point we have proved all the statements in the lemma with $\mathcal{O}_{U, u}$ in stead of $\mathcal{O}_{X, \xi }^ h$. However, $\mathcal{O}_{X, \xi }^ h$ is the henselization of $\mathcal{O}_{U, u}$. Thus $\mathcal{O}_{X, \xi }^ h$ is a $1$-dimensional reduced Noetherian ring, see More on Algebra, Lemmas 15.44.4, 15.44.7, and 15.44.3. Since $\mathcal{O}_{U, u} \to \mathcal{O}_{X, \xi }^ h$ is faithfully flat by More on Algebra, Lemma 15.44.1 it sends nonzerodivisors to nonzerodivisors. Therefore we obtain a canonical map $Q(\mathcal{O}_{U, u}) \to Q(\mathcal{O}_{X, \xi }^ h)$ and we obtain our map. We omit the verification that the map is independent of the choice of $(U, u) \to (X, x)$; a slightly better approach would be to first observe that $\mathop{\mathrm{colim}}\nolimits Q(\mathcal{O}_{U, u}) = Q(\mathcal{O}_{X, \xi }^ h)$. $\square$

Definition 70.6.4. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. Let $f \in R(X)^*$. For every prime divisor $Z \subset X$ we define the order of vanishing of $f$ along $Z$ as the integer

\[ \text{ord}_ Z(f) = \text{length}_{\mathcal{O}_{X, \xi }^ h} (\mathcal{O}_{X, \xi }^ h/a \mathcal{O}_{X, \xi }^ h) - \text{length}_{\mathcal{O}_{X, \xi }^ h} (\mathcal{O}_{X, \xi }^ h/b \mathcal{O}_{X, \xi }^ h) \]

where $a, b \in \mathcal{O}_{X, \xi }^ h$ are nonzerodivisors such that the image of $f$ in $Q(\mathcal{O}_{X, \xi }^ h)$ (Lemma 70.6.3) is equal to $a/b$. This is well defined by Algebra, Lemma 10.120.1.

If $\mathcal{O}_{X, \xi }^ h$ happens to be a domain, then we obtain

\[ \text{ord}_ Z(f) = \text{ord}_{\mathcal{O}_{X, \xi }^ h}(f) \]

where the right hand side is the notion of Algebra, Definition 10.120.2. Note that for $f, g \in R(X)^*$ we have

\[ \text{ord}_ Z(fg) = \text{ord}_ Z(f) + \text{ord}_ Z(g). \]

Of course it can happen that $\text{ord}_ Z(f) < 0$. In this case we say that $f$ has a pole along $Z$ and that $-\text{ord}_ Z(f) > 0$ is the order of pole of $f$ along $Z$. It is important to note that the condition $\text{ord}_ Z(f) \geq 0$ is not equivalent to the condition $f \in \mathcal{O}_{X, \xi }^ h$ unless the local ring $\mathcal{O}_{X, \xi }$ is a discrete valuation ring.

Lemma 70.6.5. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. Let $f \in R(X)^*$. If the prime divisor $Z \subset X$ meets the schematic locus of $X$, then the order of vanishing $\text{ord}_ Z(f)$ of Definition 70.6.4 agrees with the order of vanishing of Divisors, Definition 31.26.3.

Proof. After shrinking $X$ we may assume $X$ is an integral Noetherian scheme. If $\xi \in Z$ denotes the generic point, then we find that $\mathcal{O}_{X, \xi }^ h$ is the henselization of $\mathcal{O}_{X, \xi }$ (Decent Spaces, Lemma 66.11.8). To prove the lemma it suffices and is necessary to show that

\[ \text{length}_{\mathcal{O}_{X, \xi }} (\mathcal{O}_{X, \xi }/a \mathcal{O}_{X, \xi }) = \text{length}_{\mathcal{O}_{X, \xi }^ h} (\mathcal{O}_{X, \xi }^ h/a \mathcal{O}_{X, \xi }^ h) \]

This follows immediately from Algebra, Lemma 10.51.13 (and the fact that $\mathcal{O}_{X, \xi } \to \mathcal{O}_{X, \xi }^ h$ is a flat local ring homomorphism of local Noetherian rings). $\square$

Lemma 70.6.6. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. Let $f \in R(X)^*$. Then the collections

\[ \{ Z \subset X \mid Z\text{ a prime divisor with generic point }\xi \text{ and }f\text{ not in }\mathcal{O}_{X, \xi }\} \]


\[ \{ Z \subset X \mid Z \text{ a prime divisor and }\text{ord}_ Z(f) \not= 0\} \]

are locally finite in $X$.

Proof. There exists a nonempty open subspace $U \subset X$ such that $f$ corresponds to a section of $\Gamma (U, \mathcal{O}_ X^*)$. Hence the prime divisors which can occur in the sets of the lemma all correspond to irreducible components of $|X| \setminus |U|$. Hence Lemma 70.6.1 gives the desired result. $\square$

This lemma allows us to make the following definition.

Definition 70.6.7. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. Let $f \in R(X)^*$. The principal Weil divisor associated to $f$ is the Weil divisor

\[ \text{div}(f) = \text{div}_ X(f) = \sum \text{ord}_ Z(f) [Z] \]

where the sum is over prime divisors and $\text{ord}_ Z(f)$ is as in Definition 70.6.4. This makes sense by Lemma 70.6.6.

Lemma 70.6.8. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. Let $f, g \in R(X)^*$. Then

\[ \text{div}_ X(fg) = \text{div}_ X(f) + \text{div}_ X(g) \]

as Weil divisors on $X$.

Proof. This is clear from the additivity of the $\text{ord}$ functions. $\square$

We see from the lemma above that the collection of principal Weil divisors form a subgroup of the group of all Weil divisors. This leads to the following definition.

Definition 70.6.9. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. The Weil divisor class group of $X$ is the quotient of the group of Weil divisors by the subgroup of principal Weil divisors. Notation: $\text{Cl}(X)$.

By construction we obtain an exact complex
\begin{equation} \label{spaces-over-fields-equation-Weil-divisor-class} R(X)^* \xrightarrow {\text{div}} \text{Div}(X) \to \text{Cl}(X) \to 0 \end{equation}

which we can think of as a presentation of $\text{Cl}(X)$. Our next task is to relate the Weil divisor class group to the Picard group.

Example 70.6.10. This is a continuation of Morphisms of Spaces, Example 65.53.3. Consider the algebraic space $X = \mathbf{A}^1_ k/\{ t \sim -t \mid t \not= 0\} $. This is a smooth algebraic space over the field $k$. There is a universal homeomorphism

\[ X \longrightarrow \mathbf{A}^1_ k = \mathop{\mathrm{Spec}}(k[t]) \]

which is an isomorphism over $\mathbf{A}^1_ k \setminus \{ 0\} $. We conclude that $X$ is Noetherian and integral. Since $\dim (X) = 1$, we see that the prime divisors of $X$ are the closed points of $X$. Consider the unique closed point $x \in |X|$ lying over $0 \in \mathbf{A}^1_ k$. Since $X \setminus \{ x\} $ maps isomorphically to $\mathbf{A}^1 \setminus \{ 0\} $ we see that the classes in $\text{Cl}(X)$ of closed points different from $x$ are zero. However, the divisor of $t$ on $X$ is $2[x]$. We conclude that $\text{Cl}(X) = \mathbf{Z}/2\mathbf{Z}$.

Example 70.6.11. Let $k$ be a field. Let

\[ U = \mathop{\mathrm{Spec}}(k[x, y]/(xy)) \]

be the union of the coordinate axes in $\mathbf{A}^2_ k$. Denote $\Delta : U \to U \times _ k U$ the diagonal and $\Delta ' : U \to U \times _ k U$ the map $u \mapsto (u, \sigma (u))$ where $\sigma : U \to U$, $(x, y) \mapsto (y, x)$ is the automorphism flipping the coordinate axes. Set

\[ R = \Delta (U) \amalg \Delta '(U \setminus \{ 0_ U\} ) \]

where $0_ U \in U$ is the origin. It is easy to see that $R$ is an ├ętale equivalence relation on $U$. The quotient $X = U/R$ is an algebraic space. The morphism $U \to \mathbf{A}^1_ k$, $(x, y) \mapsto x + y$ is $R$-invariant and hence defines a morphism

\[ X \longrightarrow \mathbf{A}^1_ k \]

This morphism is a universal homeomorphism and an isomorphism over $\mathbf{A}^1_ k \setminus \{ 0\} $. It follows that $X$ is integral and Noetherian. Exactly as in Example 70.6.10 the reader shows that $\text{Cl}(X) = \mathbf{Z}/2\mathbf{Z}$ with generator corresponding to the unique closed point $x \in |X|$ mapping to $0 \in \mathbf{A}^1_ k$. However, in this case the henselian local ring of $X$ at $x$ isn't a domain, as it is the henselization of $\mathcal{O}_{U, 0_ U}$.

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