Finite modules have filtrations such that successive quotients are cyclic modules.

Lemma 10.5.4. Let $R$ be a ring, and let $M$ be a finite $R$-module. There exists a filtration by finite $R$-submodules

$0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M$

such that each quotient $M_ i/M_{i - 1}$ is isomorphic to $R/I_ i$ for some ideal $I_ i$ of $R$.

Proof. By induction on the number of generators of $M$. Let $x_1, \ldots , x_ r \in M$ be generators. Let $M' = Rx_1 \subset M$. Then $M/M'$ has $r - 1$ generators and the induction hypothesis applies. And clearly $M' \cong R/I_1$ with $I_1 = \{ f \in R \mid fx_1 = 0\}$. $\square$

Comment #1107 by Evan Warner on

Suggested slogan: Finite modules have filtrations such that successive quotients are quotients of the underlying ring.

Comment #8470 by Laurent Moret-Bailly on

In the proof, the minimality of $n$ is not needed, so "a finite set of generators" would be more to the point. Also, it would be worth pointing out that the $M_i$'s are finite modules, and perhaps that the last quotient $M/M_{n-1}$ is finitely presented if $M$ is.

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• 11 comment(s) on Section 10.5: Finite modules and finitely presented modules

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