Lemma 42.26.1 (Key formula). In the situation above the cycle

is equal to the cycle

Lemma 42.26.1 (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.**
First, let us examine what happens if we replace $s_ i$ by $us_ i$ for some unit $u$ in $B_ i$. Then $f_ i$ gets replaced by $u^{-1} f_ i$. 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 Divisors, Lemma 31.27.3 and in the second expression we add

\[ \text{div}(\partial _{B_ i}(u^{-1}, g_ i)) \]

by bi-linearity of the tame symbol. These terms agree by property (6) of the tame symbol.

Let $Z \subset X$ be an irreducible closed with $\dim _\delta (Z) = n - 2$. To show that the coefficients of $Z$ of the two cycles of the lemma is the same, we may do a replacement $s_ i \mapsto us_ i$ as in the previous paragraph. In exactly the same way one shows that we may do a replacement $t_ i \mapsto vt_ i$ for some unit $v$ of $B_ i$.

Since we are proving the equality of cycles we may argue one coefficient at a time. Thus we choose an irreducible closed $Z \subset X$ with $\dim _\delta (Z) = n - 2$ and compare coefficients. Let $\xi \in Z$ be the generic point and set $A = \mathcal{O}_{X, \xi }$. This is a Noetherian local domain of dimension $2$. Choose generators $\sigma $ and $\tau $ for $\mathcal{L}_\xi $ and $\mathcal{N}_\xi $. After shrinking $X$, we may and do assume $\sigma $ and $\tau $ define trivializations of the invertible sheaves $\mathcal{L}$ and $\mathcal{N}$ over all of $X$. Because $Z_ i$ is locally finite after shrinking $X$ we may assume $Z \subset Z_ i$ for all $i \in I$ and that $I$ is finite. Then $\xi _ i$ corresponds to a prime $\mathfrak q_ i \subset A$ of height $1$. We may write $s_ i = a_ i \sigma $ and $t_ i = b_ i \tau $ for some $a_ i$ and $b_ i$ units in $A_{\mathfrak q_ i}$. By the remarks above, it suffices to prove the lemma when $a_ i = b_ i = 1$ for all $i$.

Assume $a_ i = b_ i = 1$ for all $i$. Then the first expression of the lemma is zero, because we choose $\sigma $ and $\tau $ to be trivializing sections. Write $s = f\sigma $ and $t = g \tau $ with $f$ and $g$ in the fraction field of $A$. By the previous paragraph we have reduced to the case $f_ i = f$ and $g_ i = g$ for all $i$. Moreover, for a height $1$ prime $\mathfrak q$ of $A$ which is not in $\{ \mathfrak q_ i\} $ we have that both $f$ and $g$ are units in $A_\mathfrak q$ (by our choice of the family $\{ Z_ i\} $ in the discussion preceding the lemma). Thus the coefficient of $Z$ in the second expression of the lemma is

\[ \sum \nolimits _ i \text{ord}_{A/\mathfrak q_ i}(\partial _{B_ i}(f, g)) \]

which is zero by the key Lemma 42.6.3. $\square$

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