When $A$ is an excellent ring this is [Proposition 1, Kato-Milnor-K].

Lemma 42.67.46 (Key Lemma). Let $A$ be a $2$-dimensional Noetherian local domain with fraction field $K$. Let $f, g \in K^*$. Let $\mathfrak q_1, \ldots , \mathfrak q_ t$ be the height $1$ primes $\mathfrak q$ of $A$ such that either $f$ or $g$ is not an element of $A^*_{\mathfrak q}$. Then we have

$\sum \nolimits _{i = 1, \ldots , t} \text{ord}_{A/\mathfrak q_ i}(d_{A_{\mathfrak q_ i}}(f, g)) = 0$

We can also write this as

$\sum \nolimits _{\text{height}(\mathfrak q) = 1} \text{ord}_{A/\mathfrak q}(d_{A_{\mathfrak q}}(f, g)) = 0$

since at any height one prime $\mathfrak q$ of $A$ where $f, g \in A^*_{\mathfrak q}$ we have $d_{A_{\mathfrak q}}(f, g) = 1$ by Lemma 42.67.33.

Proof. Since the tame symbols $d_{A_{\mathfrak q}}(f, g)$ are additive (Lemma 42.67.30) and the order functions $\text{ord}_{A/\mathfrak q}$ are additive (Algebra, Lemma 10.120.1) it suffices to prove the formula when $f = a \in A$ and $g = b \in A$. In this case we see that we have to show

$\sum \nolimits _{\text{height}(\mathfrak q) = 1} \text{ord}_{A/\mathfrak q}(\det \nolimits _\kappa (A_{\mathfrak q}/(ab), a, b)) = 0$

By Proposition 42.67.43 this is equivalent to showing that

$e_ A(A/(ab), a, b) = 0.$

Since the complex $A/(ab) \xrightarrow {a} A/(ab) \xrightarrow {b} A/(ab) \xrightarrow {a} A/(ab)$ is exact we win. $\square$

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