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. $\square$

Comment #4945 by yogesh on

In the proof, I think the inclusion $\mathfrak{p} \subset \mathfrak{p}_n$ is going the wrong way and the rest of the proof is unnecessary. I think any nonzero element of $R/\mathfrak{p}_n$ is annihilated by exactly by $\mathfrak{p}_n$.

Comment #4946 by yogesh on

Oops, sorry disregard my previous comment. I got forgot we're talkng about the annihilator as an element of $M$, not $M/M_{n-1}$.

Comment #4947 by yogesh on

Oops, sorry disregard my previous comment. I got forgot we're talkng about the annihilator as an element of $M$, not $M/M_{n-1}$. I was thinking of using the previous result of associated primes of short exact sequences applied to $M_{n-1} \to M \to R/\mathfrak{p}_n$ so $Ass(M) \subseteq Ass(M_{n-1} \cup \{ \mathfrak{p}_n\}$ But the proof you give is basically the proof of that result, whose proof is omitted.

Comment #5967 by Maxim Mornev on

Here is an alternative argument. By Lemma 02M3 it is enough to show that $\operatorname{Ass}(R/\mathfrak p_i) = \\{\mathfrak p_i\\}$. By Lemma 05BY the set $\operatorname{Ass}_R(R/\mathfrak p_i)$ is the image of $\operatorname{Ass}_{R/\mathfrak p_i}(R/\mathfrak p_i)$ in $\operatorname{Spec} R$. The ring $R/\mathfrak p_i$ is a domain, so its only associated prime is $(0)$.

Comment #6146 by on

We can't use this argument as Lemma 10.63.14 comes later. The proof as we have it now is totally fine however.

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• 13 comment(s) on Section 10.63: Associated primes

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