Lemma 81.20.1 (Key formula). In the situation above the cycle

is equal to the cycle

Lemma 81.20.1 (Key formula). In the situation above the cycle

\[ \sum (Z_ i \to X)_*\left( \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{N}|_{Z_ i}}(t_ i|_{Z_ i}) - \text{ord}_{B_ i}(g_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i}) \right) \]

is equal to the cycle

\[ \sum (Z_ i \to X)_*\text{div}(\partial _{B_ i}(f_ i, g_ i)) \]

**Proof.**
The strategy of the proof will be: first reduce to the case where $\mathcal{L}$ and $\mathcal{N}$ are trivial invertible modules, then change our choices of local trivializations, and then finally use étale localization to reduce to the case of schemes^{1}.

First step. Let $q : T \to X$ be the morphism constructed in Lemma 81.17.3. We will use all properties stated in that lemma without further mention. In particular, it suffices to show that the cycles are equal after pulling back by $q$. Denote $s'$ and $t'$ the pullbacks of $s$ and $t$ to meromorphic sections of $q^*\mathcal{L}$ and $q^*\mathcal{N}$. Denote $Z'_ i = q^{-1}(Z_ i)$, denote $\xi '_ i \in |Z'_ i|$ the generic point, denote $B'_ i = \mathcal{O}_{T, \xi '_ i}^ h$, denote $\mathcal{L}_{\xi '_ i}$ and $\mathcal{N}_{\xi '_ i}$ the pullbacks of $\mathcal{L}$ and $\mathcal{N}$ to $\mathop{\mathrm{Spec}}(B'_ i)$. Recall that we have commutative diagrams

\[ \xymatrix{ \mathop{\mathrm{Spec}}(B'_ i) \ar[r]_-{c_{\xi '_ i}} \ar[d] & T \ar[d]^ q \\ \mathop{\mathrm{Spec}}(B_ i) \ar[r]^-{c_{\xi _ i}} & X } \]

see Decent Spaces, Remark 67.11.11. Denote $s'_ i$ and $t'_ i$ the pullbacks of $s_ i$ and $t_ i$ which are generators of $\mathcal{L}_{\xi '_ i}$ and $\mathcal{N}_{\xi '_ i}$. Then we have

\[ s' = f'_ i s'_ i \quad \text{and}\quad t' = g'_ i t'_ i \]

where $f'_ i$ and $g'_ i$ are the images of $f_ i, g_ i$ under the map $Q(B_ i) \to Q(B'_ i)$ induced by $B_ i \to B'_ i$. By Algebra, Lemma 10.52.13 we have

\[ \text{ord}_{B_ i}(f_ i) = \text{ord}_{B'_ i}(f'_ i) \quad \text{and}\quad \text{ord}_{B_ i}(g_ i) = \text{ord}_{B'_ i}(g'_ i) \]

By Lemma 81.19.1 applied to $q : Z'_ i \to Z_ i$ we have

\[ q^*\text{div}_{\mathcal{N}|_{Z_ i}}(t_ i|_{Z_ i}) = \text{div}_{q^*\mathcal{N}|_{Z'_ i}}(t'_ i|_{Z'_ i}) \quad \text{and}\quad q^*\text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i}) = \text{div}_{q^*\mathcal{L}|_{Z'_ i}}(s'_ i|_{Z'_ i}) \]

This already shows that the first cycle in the statement of the lemma pulls back to the corresponding cycle for $s', t', Z'_ i, s'_ i, t'_ i$. To see the same is true for the second, note that by Chow Homology, Lemma 42.5.4 we have

\[ \partial _{B_ i}(f_ i, g_ i) \mapsto \partial _{B'_ i}(f'_ i, g'_ i) \quad \text{via}\quad \kappa (\xi _ i) \to \kappa (\xi '_ i) \]

Hence the same lemma as before shows that

\[ q^*\text{div}(\partial _{B_ i}(f_ i, g_ i)) = \text{div}(\partial _{B'_ i}(f'_ i, g'_ i)) \]

Since $q^*\mathcal{L} \cong \mathcal{O}_ T$ we find that it suffices to prove the equality in case $\mathcal{L}$ is trivial. Exchanging the roles of $\mathcal{L}$ and $\mathcal{N}$ we see that we may similarly assume $\mathcal{N}$ is trivial. This finishes the proof of the first step.

Second step. Assume $\mathcal{L} = \mathcal{O}_ X$ and $\mathcal{N} = \mathcal{O}_ X$. Denote $1$ the trivializing section of $\mathcal{L}$. Then $s_ i = u \cdot 1$ for some unit $u \in B_ i$. Let us examine what happens if we replace $s_ i$ by $1$. Then $f_ i$ gets replaced by $u f_ i$. Thus the first part of the first expression of the lemma is unchanged and in the second part we add

\[ \text{ord}_{B_ i}(g_ i)\text{div}(u|_{Z_ i}) \]

where $u|_{Z_ i}$ is the image of $u$ in the residue field by Spaces over Fields, Lemma 71.7.3 and in the second expression we add

\[ \text{div}(\partial _{B_ i}(u, g_ i)) \]

by bi-linearity of the tame symbol. These terms agree by the property of the tame symbol given in Chow Homology, Equation (6).

Let $Y \subset X$ be an integral closed subspace with $\dim _\delta (Y) = n - 2$. To show that the coefficients of $Y$ of the two cycles of the lemma is the same, we may do a replacement of $s_ i$ by $1$ as in the previous paragraph. In exactly the same way one shows that we may do a replacement of $t_ i$ by $1$. Since there are only a finite number of $Z_ i$ such that $Y \subset Z_ i$ we may assume $s_ i = 1$ and $t_ i = 1$ for all these $Z_ i$.

Third step. Here we prove the coefficients of $Y$ in the cycles of the lemma agree for an integral closed subspace $Y$ with $\dim _\delta (Y) = n - 2$ such that moreover $\mathcal{L} = \mathcal{O}_ X$ and $\mathcal{N} = \mathcal{O}_ X$ and $s_ i = 1$ and $t_ i = 1$ for all $Z_ i$ such that $Y \subset Z_ i$. After replacing $X$ by a smaller open subspace we may in fact assume that $s_ i$ and $t_ i$ are equal to $1$ for all $i$. In this case the first cycle is zero. Our task is to show that the coefficient of $Y$ in the second cycle is zero as well.

First, since $\mathcal{L} = \mathcal{O}_ X$ and $\mathcal{N} = \mathcal{O}_ X$ we may and do think of $s, t$ as rational functions $f, g$ on $X$. Since $s_ i$ and $t_ i$ are equal to $1$ we find that $f_ i$, resp. $g_ i$ is the image of $f$, resp. $g$ in $Q(B_ i)$ for all $i$. Let $\zeta \in |Y|$ be the generic point. Choose an étale neighbourhood

\[ (U, u) \longrightarrow (X, \zeta ) \]

and denote $Y' = \overline{\{ u\} } \subset U$. Since an étale morphism is flat, we can pullback $f$ and $g$ to regular meromorphic functions on $U$ which we will also denote $f$ and $g$. For every prime divisor $Y \subset Z \subset X$ the scheme $Z \times _ X U$ is a union of prime divisors of $U$. Conversely, given a prime divisor $Y' \subset Z' \subset U$, there is a prime divisor $Y \subset Z \subset X$ such that $Z'$ is a component of $Z \times _ X U$. Given such a pair $(Z, Z')$ the ring map

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

is étale (in fact it is finite étale). Hence we find that

\[ \partial _{\mathcal{O}_{X, \xi }^ h}(f, g) \mapsto \partial _{\mathcal{O}_{U, \xi '}^ h}(f, g) \quad \text{via}\quad \kappa (\xi ) \to \kappa (\xi ') \]

by Chow Homology, Lemma 42.5.4. Thus Lemma 81.13.2 applies to show

\[ (Z \times _ X U \to Z)^*\text{div}_ Z(\partial _{\mathcal{O}_{X, \xi }^ h}(f, g)) = \sum \nolimits _{Z' \subset Z \times _ X U} \text{div}_{Z'}(\partial _{\mathcal{O}_{U, \xi '}^ h}(f, g)) \]

Since flat pullback commutes with pushforward along closed immersions (Lemma 81.11.1) we see that it suffices to prove that the coefficient of $Y'$ in

\[ \sum \nolimits _{Z' \subset U} (Z' \to U)_*\text{div}_{Z'}(\partial _{\mathcal{O}_{U, \xi '}^ h}(f, g)) \]

is zero.

Let $A = \mathcal{O}_{U, u}$. Then $f, g \in Q(A)^*$. Thus we can write $f = a/b$ and $g = c/d$ with $a, b, c, d \in A$ nonzerodivisors. The coefficient of $Y'$ in the expression above is

\[ \sum \nolimits _{\mathfrak q \subset A\text{ height }1} \text{ord}_{A/\mathfrak q}(\partial _{A_\mathfrak q}(f, g)) \]

By bilinearity of $\partial _ A$ it suffices to prove

\[ \sum \nolimits _{\mathfrak q \subset A\text{ height }1} \text{ord}_{A/\mathfrak q}(\partial _{A_\mathfrak q}(a, c)) \]

is zero and similarly for the other pairs $(a, d)$, $(b, c)$, and $(b, d)$. This is true by Chow Homology, Lemma 42.6.2. $\square$

[1] It is possible that a shorter proof can be given by immediately applying étale localization.

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