Lemma 42.6.2. Let $(A, \mathfrak m)$ be a $2$-dimensional Noetherian local ring. Let $a, b \in A$ be nonzerodivisors. Then we have

$\sum \text{ord}_{A/\mathfrak q}(\partial _{A_{\mathfrak q}}(a, b)) = 0$

where the sum is over the height $1$ primes $\mathfrak q$ of $A$.

Proof. If $\mathfrak q$ is a height $1$ prime of $A$ such that $a, b$ map to a unit of $A_\mathfrak q$, then $\partial _{A_\mathfrak q}(a, b) = 1$. Thus the sum is finite. In fact, if $V(ab) = \{ \mathfrak m, \mathfrak q_1, \ldots , \mathfrak q_ r\}$ then the sum is over $i = 1, \ldots , r$. For each $i$ we pick an extension $A_{\mathfrak q_ i} \subset B_ i$ as in Lemma 42.4.4 for $a, b$. By Lemma 42.6.1 with $t = ab$ and the given list of primes we may assume we have a finite local extension $A \subset B$ with $B/A$ annihilated by a power of $ab$ and such that for each $i$ the $B_{\mathfrak q_ i} \cong B_ i$. Observe that if $\mathfrak q_{i, j}$ are the primes of $B$ lying over $\mathfrak q_ i$ then we have

$\text{ord}_{A/\mathfrak q_ i}(\partial _{A_{\mathfrak q_ i}}(a, b)) = \sum \nolimits _ j \text{ord}_{B/\mathfrak q_{i, j}}(\partial _{B_{\mathfrak q_{i, j}}}(a, b))$

by Lemma 42.5.3 and Algebra, Lemma 10.121.8. Thus we may replace $A$ by $B$ and reduce to the case discussed in the next paragraph.

Assume for each $i$ there is a nonzerodivisor $\pi _ i \in A_{\mathfrak q_ i}$ and units $u_ i, v_ i \in A_{\mathfrak q_ i}$ such that for some integers $e_ i, f_ i \geq 0$ we have

$a = u_ i \pi _ i^{e_ i},\quad b = v_ i \pi _ i^{f_ i}$

in $A_{\mathfrak q_ i}$. Setting $m_ i = \text{length}_{A_{\mathfrak q_ i}}(A_{\mathfrak q_ i}/\pi _ i)$ we have $\partial _{A_{\mathfrak q_ i}}(a, b) = ((-1)^{e_ if_ i}u_ i^{f_ i}v_ i^{-e_ i})^{m_ i}$ by definition. Since $a, b$ are nonzerodivisors the $(2, 1)$-periodic complex $(A/(ab), a, b)$ has vanishing cohomology. Denote $M_ i$ the image of $A/(ab)$ in $A_{\mathfrak q_ i}/(ab)$. Then we have a map

$A/(ab) \longrightarrow \bigoplus M_ i$

whose kernel and cokernel are supported in $\{ \mathfrak m\}$ and hence have finite length. Thus we see that

$\sum e_ A(M_ i, a, b) = 0$

by Lemma 42.2.5. Hence it suffices to show $e_ A(M_ i, a, b) = - \text{ord}_{A/\mathfrak q_ i}(\partial _{A_{\mathfrak q_ i}}(a, b))$.

Let us prove this first, in case $\pi _ i, u_ i, v_ i$ are the images of elements $\pi _ i, u_ i, v_ i \in A$ (using the same symbols should not cause any confusion). In this case we get

\begin{align*} e_ A(M_ i, a, b) & = e_ A(M_ i, u_ i\pi _ i^{e_ i}, v_ i\pi _ i^{f_ i}) \\ & = e_ A(M_ i, \pi _ i^{e_ i}, \pi _ i^{f_ i}) - e_ A(\pi _ i^{e_ i}M_ i, 0, u_ i) + e_ A(\pi _ i^{f_ i}M_ i, 0, v_ i) \\ & = 0 - f_ im_ i\text{ord}_{A/\mathfrak q_ i}(u_ i) + e_ im_ i\text{ord}_{A/\mathfrak q_ i}(v_ i) \\ & = -m_ i\text{ord}_{A/\mathfrak q_ i}(u_ i^{f_ i}v_ i^{-e_ i}) = -\text{ord}_{A/\mathfrak q_ i}(\partial _{A_{\mathfrak q_ i}}(a, b)) \end{align*}

The second equality holds by Lemma 42.3.4. Observe that $M_ i \subset (M_ i)_{\mathfrak q_ i} = A_{\mathfrak q_ i}/(\pi _ i^{e_ i + f_ i})$ and $(\pi _ i^{e_ i}M_ i)_{\mathfrak q_ i} \cong A_{\mathfrak q_ i}/\pi _ i^{f_ i}$ and $(\pi _ i^{f_ i}M_ i)_{\mathfrak q_ i} \cong A_{\mathfrak q_ i}/\pi _ i^{e_ i}$. The $0$ in the third equality comes from Lemma 42.3.3 and the other two terms come from Lemma 42.3.1. The last two equalities follow from multiplicativity of the order function and from the definition of our tame symbol.

In general, we may first choose $c \in A$, $c \not\in \mathfrak q_ i$ such that $c\pi _ i \in A$. After replacing $\pi _ i$ by $c\pi _ i$ and $u_ i$ by $c^{-e_ i}u_ i$ and $v_ i$ by $c^{-f_ i}v_ i$ we may and do assume $\pi _ i$ is in $A$. Next, choose an $c \in A$, $c \not\in \mathfrak q_ i$ with $cu_ i, cv_ i \in A$. Then we observe that

$e_ A(M_ i, ca, cb) = e_ A(M_ i, a, b) - e_ A(aM_ i, 0, c) + e_ A(bM_ i, 0, c)$

by Lemma 42.3.1. On the other hand, we have

$\partial _{A_{\mathfrak q_ i}}(ca, cb) = c^{m_ i(f_ i - e_ i)}\partial _{A_{\mathfrak q_ i}}(a, b)$

in $\kappa (\mathfrak q_ i)^*$ because $c$ is a unit in $A_{\mathfrak q_ i}$. The arguments in the previous paragraph show that $e_ A(M_ i, ca, cb) = - \text{ord}_{A/\mathfrak q_ i}(\partial _{A_{\mathfrak q_ i}}(ca, cb))$. Thus it suffices to prove

$e_ A(aM_ i, 0, c) = \text{ord}_{A/\mathfrak q_ i}(c^{m_ if_ i}) \quad \text{and}\quad e_ A(bM_ i, 0, c) = \text{ord}_{A/\mathfrak q_ i}(c^{m_ ie_ i})$

and this follows from Lemma 42.3.1 by the description (see above) of what happens when we localize at $\mathfrak q_ i$. $\square$

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