Lemma 42.68.36. Let $A$ be a Noetherian local ring. Let $M$ be a finite $A$-module with $\dim (\text{Supp}(M)) = 1$. Let $a, b \in A$ nonzerodivisors on $M$. Let $\mathfrak q_1, \ldots , \mathfrak q_ t$ be the minimal primes in the support of $M$. Then

$d_ M(a, b) = \prod \nolimits _{i = 1, \ldots , t} d_{A/\mathfrak q_ i}(a, b)^{ \text{length}_{A_{\mathfrak q_ i}}(M_{\mathfrak q_ i})}$

as elements of $\kappa ^*$.

Proof. Choose a filtration by $A$-submodules

$0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M$

such that each quotient $M_ j/M_{j - 1}$ is isomorphic to $A/\mathfrak p_ j$ for some prime ideal $\mathfrak p_ j$ of $A$. See Algebra, Lemma 10.62.1. For each $j$ we have either $\mathfrak p_ j = \mathfrak q_ i$ for some $i$, or $\mathfrak p_ j = \mathfrak m_ A$. Moreover, for a fixed $i$, the number of $j$ such that $\mathfrak p_ j = \mathfrak q_ i$ is equal to $\text{length}_{A_{\mathfrak q_ i}}(M_{\mathfrak q_ i})$ by Algebra, Lemma 10.62.5. Hence $d_{M_ j}(a, b)$ is defined for each $j$ and

$d_{M_ j}(a, b) = \left\{ \begin{matrix} d_{M_{j - 1}}(a, b) d_{A/\mathfrak q_ i}(a, b) & \text{if} & \mathfrak p_ j = \mathfrak q_ i \\ d_{M_{j - 1}}(a, b) & \text{if} & \mathfrak p_ j = \mathfrak m_ A \end{matrix} \right.$

by Lemma 42.68.34 in the first instance and Lemma 42.68.35 in the second. Hence the lemma. $\square$

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