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Let $A$ be a commutative ring and $M$ an $A$-module. I realized recently that the property of $M$ having finite length is stronger than $M$ being finitely generated. Here is my reasoning: Suppose $M$ has finite length but it is not finitely-generated. Let $\left\{x_i\right\}_{i \in I}$ be an infinite set of generators. Let $J$ be a countable subset of $I$. Define $M_i=Ax_{a_1}+ \cdots +A_{a_i}, a_k \in J$. Then we have chains of length $i$ of the form $M \supset M_{i-1} \supset M_i \supset \cdots \supset M_1 \supset M_0=0$ and $i$ can grow arbitrarily large. This contradicts the finite length assumption. Hence $M$ must be finitely generated.

Can anyone think of a counterexample, where we have a module that is finitely generated but does not have finite length?

Manos
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1 Answers1

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A module has finite length if and only if it is Noetherian and Artinian. So any commutative ring that isn't Artinian does not have finite length over itself and is obviously finitely generated over itself.