## 31.29 Weil divisors on normal schemes

First we discuss properties of reflexive modules.

Lemma 31.29.1. Let $X$ be an integral locally Noetherian normal scheme. For $\mathcal{F}$ and $\mathcal{G}$ coherent reflexive $\mathcal{O}_ X$-modules the map

\[ (\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{O}_ X) \otimes _{\mathcal{O}_ X} \mathcal{G})^{**} \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) \]

is an isomorphism. The rule $\mathcal{F}, \mathcal{G} \mapsto (\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G})^{**}$ defines an abelian group law on the set of isomorphism classes of rank $1$ coherent reflexive $\mathcal{O}_ X$-modules.

**Proof.**
Although not strictly necessary, we recommend reading Remark 31.12.9 before proceeding with the proof. Choose an open subscheme $j : U \to X$ such that every irreducible component of $X \setminus U$ has codimension $\geq 2$ in $X$ and such that $j^*\mathcal{F}$ and $j^*\mathcal{G}$ are finite locally free, see Lemma 31.12.13. The map

\[ \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(j^*\mathcal{F}, \mathcal{O}_ U) \otimes _{\mathcal{O}_ U} j^*\mathcal{G} \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(j^*\mathcal{F}, j^*\mathcal{G}) \]

is an isomorphism, because we may check it locally and it is clear when the modules are finite free. Observe that $j^*$ applied to the displayed arrow of the lemma gives the arrow we've just shown is an isomorphism (small detail omitted). Since $j^*$ defines an equivalence between coherent reflexive modules on $U$ and coherent reflexive modules on $X$ (by Lemma 31.12.12 and Serre's criterion Properties, Lemma 28.12.5), we conclude that the arrow of the lemma is an isomorphism too. If $\mathcal{F}$ has rank $1$, then $j^*\mathcal{F}$ is an invertible $\mathcal{O}_ U$-module and the reflexive module $\mathcal{F}^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{F}, \mathcal{O}_ X)$ restricts to its inverse. It follows in the same manner as before that $(\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{F}^\vee )^{**} = \mathcal{O}_ X$. In this way we see that we have inverses for the group law given in the statement of the lemma.
$\square$

Lemma 31.29.2. Let $X$ be an integral locally Noetherian normal scheme. The group of rank $1$ coherent reflexive $\mathcal{O}_ X$-modules is isomorphic to the Weil divisor class group $\text{Cl}(X)$ of $X$.

**Proof.**
Let $\mathcal{F}$ be a rank $1$ coherent reflexive $\mathcal{O}_ X$-module. Choose an open $U \subset X$ such that every irreducible component of $X \setminus U$ has codimension $\geq 2$ in $X$ and such that $\mathcal{F}|_ U$ is invertible, see Lemma 31.12.13. Observe that $\text{Cl}(U) = \text{Cl}(X)$ as the Weil divisor class group of $X$ only depends on its field of rational functions and the points of codimension $1$ and their local rings. Thus we can define the Weil divisor class of $\mathcal{F}$ to be the Weil divisor class of $\mathcal{F}|_ U$ in $\text{Cl}(U)$. We omit the verification that this is independent of the choice of $U$.

Denote $\text{Cl}'(X)$ the set of isomorphism classes of rank $1$ coherent reflexive $\mathcal{O}_ X$-modules. The construction above gives a group homorphism

\[ \text{Cl}'(X) \longrightarrow \text{Cl}(X) \]

because for any pair $\mathcal{F}, \mathcal{G}$ of elements of $\text{Cl}'(X)$ we can choose a $U$ which works for both and the assignment (31.27.5.1) sending an invertible module to its Weil divisor class is a homorphism. If $\mathcal{F}$ is in the kernel of this map, then we find that $\mathcal{F}|_ U$ is trivial (Lemma 31.27.6) and hence $\mathcal{F}$ is trivial too by Lemma 31.12.12 and Serre's criterion Properties, Lemma 28.12.5. To finish the proof it suffices to check the map is surjective.

Let $D = \sum n_ Z Z$ be a Weil divisor on $X$. We claim that there is an open $U \subset X$ such that every irreducible component of $X \setminus U$ has codimension $\geq 2$ in $X$ and such that $Z|_ U$ is an effective Cartier divisor for $n_ Z \not= 0$. To prove the claim we may assume $X$ is affine. Then we may assume $D = n_1 Z_1 + \ldots + n_ r Z_ r$ is a finite sum with $Z_1, \ldots , Z_ r$ pairwise distinct. After throwing out $Z_ i \cap Z_ j$ for $i \not= j$ we may assume $Z_1, \ldots , Z_ r$ are pairwise disjoint. This reduces us to the case of a single prime divisor $Z$ on $X$. As $X$ is $(R_1)$ by Properties, Lemma 28.12.5 the local ring $\mathcal{O}_{X, \xi }$ at the generic point $\xi $ of $Z$ is a discrete valuation ring. Let $f \in \mathcal{O}_{X, \xi }$ be a uniformizer. Let $V \subset X$ be an open neighbourhood of $\xi $ such that $f$ is the image of an element $f \in \mathcal{O}_ X(V)$. After shrinking $V$ we may assume that $Z \cap V = V(f)$ scheme theoretically, since this is true in the local ring at $\xi $. In this case taking

\[ U = X \setminus (Z \setminus V) = (X \setminus Z) \cup V \]

gives the desired open, thereby proving the claim.

In order to show that the divisor class of $D$ is in the image, we may write $D = \sum _{n_ Z < 0} n_ Z Z - \sum _{n_ Z > 0} (-n_ Z) Z$. By additivity of the map constructed above, we may and do assume $n_ Z \leq 0$ for all prime divisors $Z$ (this step may be avoided if the reader so desires). Let $U \subset X$ be as in the claim above. If $U$ is quasi-compact, then we write $D|_ U = -n_1 Z_1 - \ldots - n_ r Z_ r$ for pairwise distinct prime divisors $Z_ i$ and $n_ i > 0$ and we consider the invertible $\mathcal{O}_ U$-module

\[ \mathcal{L} = \mathcal{I}_1^{n_1} \ldots \mathcal{I}_ r^{n_ r} \subset \mathcal{O}_ U \]

where $\mathcal{I}_ i$ is the ideal sheaf of $Z_ i$. This is invertible by our choice of $U$ and Lemma 31.13.7. Also $\text{div}_\mathcal {L}(1) = D|_ U$. Since $\mathcal{L} = \mathcal{F}|_ U$ for some rank $1$ coherent reflexive $\mathcal{O}_ X$-module $\mathcal{F}$ by Lemma 31.12.12 we find that $D$ is in the image of our map.

If $U$ is not quasi-compact, then we define $\mathcal{L} \subset \mathcal{O}_ U$ locally by the displayed formula above. The reader shows that the construction glues and finishes the proof exactly as before. Details omitted.
$\square$

Lemma 31.29.3. Let $X$ be an integral locally Noetherian normal scheme. Let $\mathcal{F}$ be a rank 1 coherent reflexive $\mathcal{O}_ X$-module. Let $s \in \Gamma (X, \mathcal{F})$. Let

\[ U = \{ x \in X \mid s : \mathcal{O}_{X, x} \to \mathcal{F}_ x \text{ is an isomorphism}\} \]

Then $j : U \to X$ is an open subscheme of $X$ and

\[ j_*\mathcal{O}_ U = \mathop{\mathrm{colim}}\nolimits (\mathcal{O}_ X \xrightarrow {s} \mathcal{F} \xrightarrow {s} \mathcal{F}^{[2]} \xrightarrow {s} \mathcal{F}^{[3]} \xrightarrow {s} \ldots ) \]

where $\mathcal{F}^{[1]} = \mathcal{F}$ and inductively $\mathcal{F}^{[n + 1]} = (\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{F}^{[n]})^{**}$.

**Proof.**
The set $U$ is open by Modules, Lemmas 17.9.4 and 17.12.6. Observe that $j$ is quasi-compact by Properties, Lemma 28.5.3. To prove the final statement it suffices to show for every quasi-compact open $W \subset X$ there is an isomorphism

\[ \mathop{\mathrm{colim}}\nolimits \Gamma (W, \mathcal{F}^{[n]}) \longrightarrow \Gamma (U \cap W, \mathcal{O}_ U) \]

of $\mathcal{O}_ X(W)$-modules compatible with restriction maps. We will omit the verification of compatibilities. After replacing $X$ by $W$ and rewriting the above in terms of homs, we see that it suffices to construct an isomorphism

\[ \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{O}_ X, \mathcal{F}^{[n]}) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_ U, \mathcal{O}_ U) \]

Choose an open $V \subset X$ such that every irreducible component of $X \setminus V$ has codimension $\geq 2$ in $X$ and such that $\mathcal{F}|_ V$ is invertible, see Lemma 31.12.13. Then restriction defines an equivalence of categories between rank $1$ coherent reflexive modules on $X$ and $V$ and between rank $1$ coherent reflexive modules on $U$ and $V \cap U$. See Lemma 31.12.12 and Serre's criterion Properties, Lemma 28.12.5. Thus it suffices to construct an isomorphism

\[ \mathop{\mathrm{colim}}\nolimits \Gamma (V, (\mathcal{F}|_ V)^{\otimes n}) \longrightarrow \Gamma (V \cap U, \mathcal{O}_ U) \]

Since $\mathcal{F}|_ V$ is invertible and since $U \cap V$ is equal to the set of points where $s|_ V$ generates this invertible module, this is a special case of Properties, Lemma 28.17.2 (there is an explicit formula for the map as well).
$\square$

Lemma 31.29.4. Assumptions and notation as in Lemma 31.29.3. If $s$ is nonzero, then every irreducible component of $X \setminus U$ has codimension $1$ in $X$.

**Proof.**
Let $\xi \in X$ be a generic point of an irreducible component $Z$ of $X \setminus U$. After replacing $X$ by an open neighbourhood of $\xi $ we may assume that $Z = X \setminus U$ is irreducible. Since $s : \mathcal{O}_ U \to \mathcal{F}|_ U$ is an isomorphism, if the codimension of $Z$ in $X$ is $\geq 2$, then $s : \mathcal{O}_ X \to \mathcal{F}$ is an isomorphism by Lemma 31.12.12 and Serre's criterion Properties, Lemma 28.12.5. This would mean that $Z = \emptyset $, a contradiction.
$\square$

Lemma 31.29.6. Assumptions and notation as in Lemma 31.29.3. The following are equivalent

the inclusion morphism $j : U \to X$ is affine, and

for every $x \in X \setminus U$ there is an $n > 0$ such that $s^ n \in \mathfrak m_ x \mathcal{F}^{[n]}_ x$.

**Proof.**
Assume (1). Then for $x \in X \setminus U$ the inverse image $U_ x$ of $U$ under the canonical morphism $f_ x : \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \to X$ is affine and does not contain $x$. Thus $\mathfrak m_ x \Gamma (U_ x, \mathcal{O}_{U_ x})$ is the unit ideal. In particular, we see that we can write

\[ 1 = \sum f_ i g_ i \]

with $f_ i \in \mathfrak m_ x$ and $g_ i \in \Gamma (U_ x, \mathcal{O}_{U_ x})$. By Lemma 31.29.3 we have $\Gamma (U_ x, \mathcal{O}_{U_ x}) = \mathop{\mathrm{colim}}\nolimits \mathcal{F}^{[n]}_ x$ with transition maps given by multiplication by $s$. Hence for some $n > 0$ we have

\[ s^ n = \sum f_ i t_ i \]

for some $t_ i = s^ ng_ i \in \mathcal{F}^{[n]}_ x$. Thus (2) holds.

Conversely, assume that (2) holds. To prove $j$ is affine is local on $X$, see Morphisms, Lemma 29.11.3. Thus we may and do assume that $X$ is affine. Our goal is to show that $U$ is affine. By Cohomology of Schemes, Lemma 30.17.8 it suffices to show that $H^ p(U, \mathcal{O}_ U) = 0$ for $p > 0$. Since $H^ p(U, \mathcal{O}_ U) = H^0(X, R^ pj_*\mathcal{O}_ U)$ (Cohomology of Schemes, Lemma 30.4.6) and since $R^ pj_*\mathcal{O}_ U$ is quasi-coherent (Cohomology of Schemes, Lemma 30.4.5) it is enough to show the stalk $(R^ pj_*\mathcal{O}_ U)_ x$ at a point $x \in X$ is zero. Consider the base change diagram

\[ \xymatrix{ U_ x \ar[d]_{j_ x} \ar[r] & U \ar[d]^ j \\ \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \ar[r] & X } \]

By Cohomology of Schemes, Lemma 30.5.2 we have $(R^ pj_*\mathcal{O}_ U)_ x = R^ pj_{x, *}\mathcal{O}_{U_ x}$. Hence we may assume $X$ is local with closed point $x$ and we have to show $U$ is affine (because this is equivalent to the desired vanishing by the reference given above). In particular $d = \dim (X)$ is finite (Algebra, Proposition 10.60.9). If $x \in U$, then $U = X$ and the result is clear. If $d = 0$ and $x \not\in U$, then $U = \emptyset $ and the result is clear. Now assume $d > 0$ and $x \not\in U$. Since $j_*\mathcal{O}_ U = \mathop{\mathrm{colim}}\nolimits \mathcal{F}^{[n]}$ our assumption means that we can write

\[ 1 = \sum f_ i g_ i \]

for some $n > 0$, $f_ i \in \mathfrak m_ x$, and $g_ i \in \mathcal{O}(U)$. By induction on $d$ we know that $D(f_ i) \cap U$ is affine for all $i$: going through the whole argument just given with $X$ replaced by $D(f_ i)$ we end up with Noetherian local rings whose dimension is strictly smaller than $d$. Hence $U$ is affine by Properties, Lemma 28.27.3 as desired.
$\square$

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