Lemma 10.63.4 (00LB)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.63: Associated primes
Lemma 10.63.4 (cite)

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 (7)

Comment #4945 by yogesh on February 22, 2020 at 20:34

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 February 22, 2020 at 20:53

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 February 22, 2020 at 20:58

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 March 10, 2021 at 19:57

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 Johan on April 30, 2021 at 22:42

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

Comment #9032 by Elías Guisado on April 29, 2024 at 03:48

I think one needs to treat the case $n=1$ separately. Something like "the case $n=1$ is the observation that a non-zero element of $R/\mathfrak{p}$ has annihilator equal to $\mathfrak{p}$ ."

Comment #9187 by Stacks project on May 14, 2024 at 15:47

I think the induction can start with $n = 0$ . Going to leave as is.

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