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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\} $. $\square$

Comments (3)

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 is not needed, so "a finite set of generators" would be more to the point. Also, it would be worth pointing out that the 's are finite modules, and perhaps that the last quotient is finitely presented if is.

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

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