Lemma 42.69.5. Let A be a Noetherian local ring. Let M be a finite A-module. Let a, b \in A. Assume
\dim (A) = 1,
both a and b are nonzerodivisors in A,
A has no embedded primes,
M has no embedded associated primes,
\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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