Lemma 10.119.11. Let $R$ be a domain with fraction field $K$. Let $M$ be an $R$-submodule of $K^{\oplus r}$. Assume $R$ is Noetherian of dimension $1$. For any nonzero $x \in R$ we have $\text{length}_ R(M/xM) < \infty$.

Proof. Since $R$ has dimension $1$ we see that $x$ is contained in finitely many primes $\mathfrak m_ i$, $i = 1, \ldots , n$, each maximal. Since $R$ is Noetherian we see that $R/xR$ is Artinian and $R/xR = \prod _{i = 1, \ldots , n} (R/xR)_{\mathfrak m_ i}$ by Proposition 10.60.7 and Lemma 10.53.6. Hence $M/xM$ similarly decomposes as the product $M/xM = \prod (M/xM)_{\mathfrak m_ i}$ of its localizations at the $\mathfrak m_ i$. By Lemma 10.119.9 applied to $M_{\mathfrak m_ i}$ over $R_{\mathfrak m_ i}$ we see each $M_{\mathfrak m_ i}/xM_{\mathfrak m_ i} = (M/xM)_{\mathfrak m_ i}$ has finite length over $R_{\mathfrak m_ i}$. Thus $M/xM$ has finite length over $R$ as the above implies $M/xM$ has a finite filtration by $R$-submodules whose successive quotients are isomorphic to the residue fields $\kappa (\mathfrak m_ i)$. $\square$

Comment #5768 by Yoav on

Why $\prod (M/xM)_{\mathfrak m_ i} = \prod M/(\mathfrak m_ i^{e_ i}, x)M$? I also don't understand how it is easy to deduct that $M/xM$ has finite length over $R$.

Comment #5773 by on

OK, you don't need to use that $(M/xM)_{\mathfrak m_i} = M/(\mathfrak m_i^e, x)M$ to conclude the proof. All we need to use is that the product decomposition of $R/xR$ into its localizations from Proposition 10.60.7 induces a product decomposition of $M/xM$ into its localizations. Then Lemma 10.119.9 says that $(M/xM)_{\mathfrak m_i}$ has finite length over $(R/xR)_{\mathfrak m_i}$. Then to conclude you just have to show: if a ring $A = \prod A_i$ is a finite product of rings $A_i$ and a module $N$ over $A$ is correspondingly a product $N = \prod N_i$ then if the length of $N_i$ over $A_i$ is finite, then the length of $N$ over $A$ is finite.

So I should remove the stuff with the powers of the $\mathfrak m_i$ the next time I go through the comments.

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