Lemma 42.3.2. Let $R$ be a Noetherian local ring. Let $x \in R$. If $M$ is a finite Cohen-Macaulay module over $R$ with $\dim (\text{Supp}(M)) = 1$ and $\dim (\text{Supp}(M/xM)) = 0$, then

$\text{length}_ R(M/xM) = \sum \nolimits _ i \text{length}_ R(R/(x, \mathfrak q_ i)) \text{length}_{R_{\mathfrak q_ i}}(M_{\mathfrak q_ i}).$

where $\mathfrak q_1, \ldots , \mathfrak q_ t$ are the minimal primes of the support of $M$. If $I \subset R$ is an ideal such that $x$ is a nonzerodivisor on $R/I$ and $\dim (R/I) = 1$, then

$\text{length}_ R(R/(x, I)) = \sum \nolimits _ i \text{length}_ R(R/(x, \mathfrak q_ i)) \text{length}_{R_{\mathfrak q_ i}}((R/I)_{\mathfrak q_ i})$

where $\mathfrak q_1, \ldots , \mathfrak q_ n$ are the minimal primes over $I$.

Proof. These are special cases of Lemma 42.3.1. $\square$

Comment #4980 by Kazuki Masugi on

Does this lemma hold in general $M$? (Is "Cohen-Macaulay module"-ness nessesary?)

Comment #5227 by on

@#4980: No, I think the lemma is wrong as soon as $M$ has depth $0$ (but everything else kept the same).

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