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Let $A$ be a semilocal Noetherian ring with Jacobson radical $m$ and $M$ a finite $A$-module. Let $x \in m$. According to Matsumura's Commutative Ring Theory p. 99 (Step 2), $l(xM/xM\cap m^n M)=l(M/(m^nM:x))$. It seems to me that this equality is not true via the following argument:

First note that $xM \cap m^n M = x(m^nM:x)$. Let $f$ be the composite map $M \rightarrow x M \rightarrow xM/xM\cap m^n M \rightarrow 0$, where the first arrow is multiplication by $x$. We can rewrite $f$ as $M \rightarrow x M \rightarrow xM/x(m^n M : x) \rightarrow 0$. Now, $(m^nM:x) \subset Kerf$, but since $M$ is not in general torsion free, we have an exact sequence $0 \rightarrow K \rightarrow M/(m^nM:x) \rightarrow xM/x(m^nM :x) \rightarrow 0$, with $K$ nonzero in general. Since the length is additive function on short exact sequences, we are done.

Question: Any comments on the above? Am i missing something or is there a typo in Matsumura's book?

Manos
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It's easy to show that $$xM/xM\cap N\simeq M/(N:x)$$ for any submodule $N$ of $M$.

Take the map $\varphi:M\to xM/xM\cap N$ given by $\varphi(z)=\widehat{xz}$. Then $\varphi$ is a surjective module homomorphism whose kernel is $\ker\varphi=\{z\in M:\varphi(z)=\hat 0\}=\{z\in M:\widehat{xz}=\hat 0\}=\{z\in M:xz\in xM\cap N\}=(N:x)$.