Lemma 10.63.4. Let $R$ be a ring, and $M$ an $R$-module. Suppose 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 $R$. Then $\text{Ass}(M) \subset \{ \mathfrak p_1, \ldots , \mathfrak p_ n\} $.

**Proof.**
By induction on the length $n$ of the filtration $\{ M_ i \} $. Pick $m \in M$ whose annihilator is a prime $\mathfrak p$. If $m \in M_{n-1}$ we are done by induction. If not, then $m$ maps to a nonzero element of $M/M_{n-1} \cong R/\mathfrak p_ n$. Hence we have $\mathfrak p \subset \mathfrak p_ n$. If equality does not hold, then we can find $f \in \mathfrak p_ n$, $f \not\in \mathfrak p$. In this case the annihilator of $fm$ is still $\mathfrak p$ and $fm \in M_{n-1}$. Thus we win by induction.
$\square$

## Comments (5)

Comment #4945 by yogesh on

Comment #4946 by yogesh on

Comment #4947 by yogesh on

Comment #5967 by Maxim Mornev on

Comment #6146 by Johan on

There are also: