Lemma 42.69.5. Let $A$ be a Noetherian local ring. Let $M$ be a finite $A$-module. Let $a, b \in A$. Assume

1. $\dim (A) = 1$,

2. both $a$ and $b$ are nonzerodivisors in $A$,

3. $A$ has no embedded primes,

4. $M$ has no embedded associated primes,

5. $\text{Supp}(M) = \mathop{\mathrm{Spec}}(A)$.

Let $I = \{ x \in A \mid x(a/b) \in A\}$. Let $\mathfrak q_1, \ldots , \mathfrak q_ t$ be the minimal primes of $A$. Then $(a/b)IM \subset M$ and

$\text{length}_ A(M/(a/b)IM) - \text{length}_ A(M/IM) = \sum \nolimits _ i \text{length}_{A_{\mathfrak q_ i}}(M_{\mathfrak q_ i}) \text{ord}_{A/\mathfrak q_ i}(a/b)$

Proof. Since $M$ has no embedded associated primes, and since the support of $M$ is $\mathop{\mathrm{Spec}}(A)$ we see that $\text{Ass}(M) = \{ \mathfrak q_1, \ldots , \mathfrak q_ t\}$. Hence $a$, $b$ are nonzerodivisors on $M$. Note that

\begin{align*} & \text{length}_ A(M/(a/b)IM) \\ & = \text{length}_ A(bM/aIM) \\ & = \text{length}_ A(M/aIM) - \text{length}_ A(M/bM) \\ & = \text{length}_ A(M/aM) + \text{length}_ A(aM/aIM) - \text{length}_ A(M/bM) \\ & = \text{length}_ A(M/aM) + \text{length}_ A(M/IM) - \text{length}_ A(M/bM) \end{align*}

as the injective map $b : M \to bM$ maps $(a/b)IM$ to $aIM$ and the injective map $a : M \to aM$ maps $IM$ to $aIM$. Hence the left hand side of the equation of the lemma is equal to

$\text{length}_ A(M/aM) - \text{length}_ A(M/bM).$

Applying the second formula of Lemma 42.3.2 with $x = a, b$ respectively and using Algebra, Definition 10.121.2 of the $\text{ord}$-functions we get the result. $\square$

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