The Stacks project

80.13 Principal divisors

This section is the analogue of Chow Homology, Section 42.17. The following definition is the analogue of Spaces over Fields, Definition 70.6.7 in our current setup.

Definition 80.13.1. In Situation 80.2.1 let $X/B$ be good. Assume $X$ is integral with $\dim _\delta (X) = n$. Let $f \in R(X)^*$. The principal divisor associated to $f$ is the $(n - 1)$-cycle

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

defined in Spaces over Fields, Definition 70.6.7. This makes sense because prime divisors have $\delta $-dimension $n - 1$ by Lemma 80.12.1.

In the situation of the definition for $f, g \in R(X)^*$ we have

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

in $Z_{n - 1}(X)$. See Spaces over Fields, Lemma 70.6.8. The following lemma will allow us to reduce statements about principal divisors to the case of schemes.

Lemma 80.13.2. In Situation 80.2.1 let $f : X \to Y$ be an étale morphism of good algebraic spaces over $B$. Assume $Y$ is integral. Let $g \in R(Y)^*$. As cycles on $X$ we have

\[ f^*(\text{div}_ Y(g)) = \sum \nolimits _{X'} (X' \to X)_*\text{div}_{X'}(g \circ f|_{X'}) \]

where the sum is over the irreducible components of $X$ (Remark 80.5.1).

Proof. The map $|X| \to |Y|$ is open. The set of irreducible components of $|X|$ is locally finite in $|X|$. We conclude that $f|_{X'} : X' \to Y$ is dominant for every irreducible component $X' \subset X$. Thus $g \circ f|_{X'}$ is defined (Morphisms of Spaces, Section 65.47), hence $\text{div}_{X'}(g \circ f|_{X'})$ is defined. Moreover, the sum is locally finite and we find that the right hand side indeed is a cycle on $X$. The left hand side is defined by Definition 80.10.1 and the fact that an étale morphism is flat of relative dimension $0$.

Since $f$ is étale we see that $\delta _ X(x) = \delta _ y(f(x))$ for all $x \in |X|$. Thus if $\dim _\delta (Y) = n$, then $\dim _\delta (X') = n$ for every irreducible component $X'$ of $X$ (since generic points of $X$ map to the generic point of $Y$, see above). Thus both left and right hand side are $(n - 1)$-cycles.

Let $Z \subset X$ be an integral closed subspace with $\dim _\delta (Z) = n - 1$. To prove the equality, we need to show that the coefficients of $Z$ are the same. Let $Z' \subset Y$ be the integral closed subspace constructed in Lemma 80.7.1. Then $\dim _\delta (Z') = n - 1$ too. Let $\xi \in |Z|$ be the generic point. Then $\xi ' = f(\xi ) \in |Z'|$ is the generic point. Consider the commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }^ h) \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, \xi '}^ h) \ar[r] & Y } \]

of Decent Spaces, Remark 66.11.11. We have to be slightly careful as the reduced Noetherian local rings $\mathcal{O}_{X, \xi }^ h$ and $\mathcal{O}_{Y, \xi '}^ h$ need not be domains. Thus we work with total rings of fractions $Q(-)$ rather than fraction fields. By definition, to get the coefficient of $Z'$ in $\text{div}_ Y(g)$ we write the image of $g$ in $Q(\mathcal{O}_{Y, \xi '}^ h)$ as $a/b$ with $a, b \in \mathcal{O}_{Y, \xi '}^ h$ nonzerodivisors and we take

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

Observe that the coefficient of $Z$ in $f^*\text{div}_ Y(G)$ is the same integer, see Lemma 80.10.3. Suppose that $\xi \in X'$. Then we can consider the maps

\[ \mathcal{O}_{Y, \xi '}^ h \to \mathcal{O}_{X, \xi }^ h \to \mathcal{O}_{X', \xi }^ h \]

The first arrow is flat and the second arrow is a surjective map of reduced local Noetherian rings of dimension $1$. Therefore both these maps send nonzerodivisors to nonzerodivisors and we conclude the coefficient of $Z'$ in $\text{div}_{X'}(g \circ f|_{X'})$ is

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

by the same prescription as above. Thus it suffices to show

\[ \text{length}_{\mathcal{O}_{Y, \xi '}^ h} (\mathcal{O}_{Y, \xi '}^ h/a \mathcal{O}_{Y, \xi '}^ h) = \sum \nolimits _{\xi \in |X'|} \text{length}_{\mathcal{O}_{X', \xi }^ h} (\mathcal{O}_{X', \xi }^ h/a \mathcal{O}_{X', \xi }^ h) \]

First, since the ring map $\mathcal{O}_{Y, \xi '}^ h \to \mathcal{O}_{X, \xi }^ h$ is flat and unramified, we have

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

by Algebra, Lemma 10.51.13. Let $\mathfrak q_1, \ldots , \mathfrak q_ t$ be the nonmaximal primes of $\mathcal{O}_{X, \xi }^ h$ and set $R_ j = \mathcal{O}_{X, \xi }^ h/\mathfrak q_ j$. For $X'$ as above, denote $J(X') \subset \{ 1, \ldots , t\} $ the set of indices such that $\mathfrak q_ j$ corresponds to a point of $X'$, i.e., such that under the surjection $\mathcal{O}_{X, \xi }^ h \to \mathcal{O}_{X', \xi }$ the prime $\mathfrak q_ j$ corresponds to a prime of $\mathcal{O}_{X', \xi }$. By Chow Homology, Lemma 42.3.2 we get

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

and

\[ \text{length}_{\mathcal{O}_{X', \xi }^ h} (\mathcal{O}_{X', \xi }^ h/a \mathcal{O}_{X', \xi }^ h) = \sum \nolimits _{j \in J(X')} \text{length}_{R_ j}(R_ j/a R_ j) \]

Thus the result of the lemma holds because $\{ 1, \ldots , t\} $ is the disjoint union of the sets $J(X')$: each point of codimension $0$ on $X$ lies on a unique $X'$. $\square$


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