The Stacks project

Lemma 10.66.7. Let $R$ be a ring. Let $M$ be an $R$-module. The union $\bigcup _{\mathfrak q \in \text{WeakAss}(M)} \mathfrak q$ is the set of elements of $R$ which are zerodivisors on $M$.

Proof. Suppose $f \in \mathfrak q \in \text{WeakAss}(M)$. Then there exists an element $m \in M$ such that $\mathfrak q$ is minimal over $I = \{ x \in R \mid xm = 0\} $. Hence there exists a $g \in R$, $g \not\in \mathfrak q$ and $n > 0$ such that $f^ ngm = 0$. Note that $gm \not= 0$ as $g \not\in I$. If we take $n$ minimal as above, then $f (f^{n - 1}gm) = 0$ and $f^{n - 1}gm \not= 0$, so $f$ is a zerodivisor on $M$. Conversely, suppose $f \in R$ is a zerodivisor on $M$. Consider the submodule $N = \{ m \in M \mid fm = 0\} $. Since $N$ is not zero it has a weakly associated prime $\mathfrak q$ by Lemma 10.66.5. Clearly $f \in \mathfrak q$ and by Lemma 10.66.4 $\mathfrak q$ is a weakly associated prime of $M$. $\square$

Comments (3)

Comment #4088 by Matthieu Romagny on

Word omitted in statement of the Lemma: "The union (...) is the set of elements..."

Comment #7208 by Matthieu Romagny on

Missing word in statement of lemma : The union ⋃q∈WeakAss(M) q is the set of elements...

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  • 3 comment(s) on Section 10.66: Weakly associated primes

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