Lemma 42.3.1. Let $R$ be a Noetherian local ring. Let $M$ be a finite $R$-module. Let $x \in R$. Assume that

$\dim (\text{Supp}(M)) \leq 1$, and

$\dim (\text{Supp}(M/xM)) \leq 0$.

Write $\text{Supp}(M) = \{ \mathfrak m, \mathfrak q_1, \ldots , \mathfrak q_ t\} $. Then

\[ e_ R(M, 0, x) = \sum \nolimits _{i = 1, \ldots , t} \text{ord}_{R/\mathfrak q_ i}(x) \text{length}_{R_{\mathfrak q_ i}}(M_{\mathfrak q_ i}). \]

**Proof.**
We first make some preparatory remarks. The result of the lemma holds if $M$ has finite length, i.e., if $t = 0$, because both the left hand side and the right hand side are zero in this case, see Lemma 42.2.4. Also, if we have a short exact sequence $0 \to M \to M' \to M'' \to 0$ of modules satisfying (1) and (2), then lemma for 2 out of 3 of these implies the lemma for the third by the additivity of length (Algebra, Lemma 10.52.3) and additivty of multiplicities (Lemma 42.2.3).

Denote $M_ i$ the image of $M$ in $M_{\mathfrak q_ i}$, so $\text{Supp}(M_ i) = \{ \mathfrak m, \mathfrak q_ i\} $. The kernel and cokernel of the map $M \to \bigoplus M_ i$ have support $\{ \mathfrak m\} $ and hence have finite length. By our preparatory remarks, it follows that it suffices to prove the lemma for each $M_ i$. Thus we may assume that $\text{Supp}(M) = \{ \mathfrak m, \mathfrak q\} $. In this case we have a finite filtration $M \supset \mathfrak qM \supset \mathfrak q^2M \supset \ldots \supset \mathfrak q^ nM = 0$ by Algebra, Lemma 10.62.4. Again additivity shows that it suffices to prove the lemma in the case $M$ is annihilated by $\mathfrak q$. In this case we can view $M$ as a $R/\mathfrak q$-module, i.e., we may assume that $R$ is a Noetherian local domain of dimension $1$ with fraction field $K$. Dividing by the torsion submodule, i.e., by the kernel of $M \to M \otimes _ R K = V$ (the torsion has finite length hence is handled by our preliminary remarks) we may assume that $M \subset V$ is a lattice (Algebra, Definition 10.121.3). Then $x : M \to M$ is injective and $\text{length}_ R(M/xM) = d(M, xM)$ (Algebra, Definition 10.121.5). Since $\text{length}_ K(V) = \dim _ K(V)$ we see that $\det (x : V \to V) = x^{\dim _ K(V)}$ and $\text{ord}_ R(\det (x : V \to V)) = \dim _ K(V) \text{ord}_ R(x)$. Thus the desired equality follows from Algebra, Lemma 10.121.7 in this case.
$\square$

## Comments (2)

Comment #4979 by Kazuki Masugi on

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