Lemma 10.63.18. Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$. Let $I \subset \mathfrak m$ be an ideal. Let $M$ be a finite $R$-module. The following are equivalent:

1. There exists an $x \in I$ which is not a zerodivisor on $M$.

2. We have $I \not\subset \mathfrak q$ for all $\mathfrak q \in \text{Ass}(M)$.

Proof. If there exists a nonzerodivisor $x$ in $I$, then $x$ clearly cannot be in any associated prime of $M$. Conversely, suppose $I \not\subset \mathfrak q$ for all $\mathfrak q \in \text{Ass}(M)$. In this case we can choose $x \in I$, $x \not\in \mathfrak q$ for all $\mathfrak q \in \text{Ass}(M)$ by Lemmas 10.63.5 and 10.15.2. By Lemma 10.63.9 the element $x$ is not a zerodivisor on $M$. $\square$

There are also:

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

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).