
Lemma 10.120.8. Let $A \to B$ be a ring map. Assume

1. $A$ is a Noetherian local domain of dimension $1$,

2. $A \subset B$ 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 10.120.2.

Proof. The ring $B$ is semi-local by Lemma 10.112.2. 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 10.51.12 this is equal to $\text{length}_ A(B/yB)$. By Lemma 10.120.7 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. $\square$

There are also:

• 2 comment(s) on Section 10.120: Orders of vanishing

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).