Lemma 10.119.11 (00PF)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.119: Around Krull-Akizuki
Lemma 10.119.11 (cite)

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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 $\kappa (\mathfrak m_ i)$ . $\square$

Comments (3)

Comment #5768 by Yoav on November 17, 2020 at 05:43

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 Johan on November 17, 2020 at 12:03

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.

Comment #5774 by Johan on November 17, 2020 at 13:54

@#5768: thanks! I've fixed this as explained above in this commit.

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