The Stacks project

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

In the proof, I think the inclusion is going the wrong way and the rest of the proof is unnecessary. I think any nonzero element of is annihilated by exactly by .

Comment #4946 by yogesh on

Oops, sorry disregard my previous comment. I got forgot we're talkng about the annihilator as an element of , not .

Comment #4947 by yogesh on

Oops, sorry disregard my previous comment. I got forgot we're talkng about the annihilator as an element of , not . I was thinking of using the previous result of associated primes of short exact sequences applied to so 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 . By Lemma 05BY the set is the image of in . The ring is a domain, so its only associated prime is .

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.

There are also:

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

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