Lemma 10.121.8. Let A \to B be a ring map. Assume
A is a Noetherian local domain of dimension 1,
A \subset B is a finite extension of domains.
Let L/K be the corresponding finite extension of fraction fields. Let y \in L^* and x = \text{Nm}_{L/K}(y). In this situation B is semi-local. Let \mathfrak m_ i, i = 1, \ldots , n be the maximal ideals of B. Then
\text{ord}_ A(x) = \sum \nolimits _ i [\kappa (\mathfrak m_ i) : \kappa (\mathfrak m_ A)] \text{ord}_{B_{\mathfrak m_ i}}(y)
where \text{ord} is defined as in Definition 10.121.2.
Proof.
The ring B is semi-local by Lemma 10.113.2. Write y = b/b' for some b, b' \in B. By the additivity of \text{ord} and multiplicativity of \text{Nm} it suffices to prove the lemma for y = b or y = b'. In other words we may assume y \in B. In this case the right hand side of the formula is
\sum [\kappa (\mathfrak m_ i) : \kappa (\mathfrak m_ A)] \text{length}_{B_{\mathfrak m_ i}}((B/yB)_{\mathfrak m_ i})
By Lemma 10.52.12 this is equal to \text{length}_ A(B/yB). By Lemma 10.121.7 we have
\text{length}_ A(B/yB) = d(B, yB) = \text{ord}_ A(\det \nolimits _ K(L \xrightarrow {y} L)).
Since x = \text{Nm}_{L/K}(y) = \det \nolimits _ K(L \xrightarrow {y} L) by definition the lemma is proved.
\square
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