Lemma 81.13.2. In Situation 81.2.1 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 81.5.1).

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 66.47), 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 81.10.1 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 81.7.1. 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 67.11.11. 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 81.10.3. 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 10.52.13. 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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