Lemma 42.6.3 (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

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

We can also write this as

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

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

**Proof.**
Since the tame symbols $\partial _{A_{\mathfrak q}}(f, g)$ are bilinear and the order functions $\text{ord}_{A/\mathfrak q}$ are additive it suffices to prove the formula when $f$ and $g$ are elements of $A$. This case is proven in Lemma 42.6.2.
$\square$

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