The Stacks project

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

Why ? I also don't understand how it is easy to deduct that has finite length over .

Comment #5773 by on

OK, you don't need to use that to conclude the proof. All we need to use is that the product decomposition of into its localizations from Proposition 10.60.7 induces a product decomposition of into its localizations. Then Lemma 10.119.9 says that has finite length over . Then to conclude you just have to show: if a ring is a finite product of rings and a module over is correspondingly a product then if the length of over is finite, then the length of over is finite.

So I should remove the stuff with the powers of the the next time I go through the comments.

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