The Stacks project

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

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

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

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

What does “In this case we can filter by powers of " mean? (Should I use lemma 10.61.1(Tag 00L0)?)

Comment #5226 by on

@#4979: No it just means to use the finite filtration . Of course, you can also use Lemma 10.62.1 if you wish. Thanks for pointing this out. Fixed here.

There are also:

  • 2 comment(s) on Section 42.3: Calculation of some multiplicities

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02QF. Beware of the difference between the letter 'O' and the digit '0'.