Lemma 10.62.1. Let $R$ be a Noetherian ring, and let $M$ be a finite $R$-module. There exists a filtration by $R$-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 $R/\mathfrak p_ i$ for some prime ideal $\mathfrak p_ i$ of $R$.

First proof. By Lemma 10.5.4 it suffices to do the case $M = R/I$ for some ideal $I$. Consider the set $S$ of ideals $J$ such that the lemma does not hold for the module $R/J$, and order it by inclusion. To arrive at a contradiction, assume that $S$ is not empty. Because $R$ is Noetherian, $S$ has a maximal element $J$. By definition of $S$, the ideal $J$ cannot be prime. Pick $a, b\in R$ such that $ab \in J$, but neither $a \in J$ nor $b\in J$. Consider the filtration $0 \subset aR/(J \cap aR) \subset R/J$. Note that $aR/(J \cap aR)$ is a quotient of $R/(J + bR)$ and the second quotient equals $R/(aR + J)$. Hence by maximality of $J$, each of these has a filtration as above and hence so does $R/J$. Contradiction. $\square$

Second proof. For an $R$-module $M$ we say $P(M)$ holds if there exists a filtration as in the statement of the lemma. Observe that $P$ is stable under extensions and holds for $0$. By Lemma 10.5.4 it suffices to prove $P(R/I)$ holds for every ideal $I$. If not then because $R$ is Noetherian, there is a maximal counter example $J$. By Example 10.28.7 and Proposition 10.28.8 the ideal $J$ is prime which is a contradiction. $\square$

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