Lemma 10.66.4. Let $R$ be a ring. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of $R$-modules. Then $\text{WeakAss}(M') \subset \text{WeakAss}(M)$ and $\text{WeakAss}(M) \subset \text{WeakAss}(M') \cup \text{WeakAss}(M'')$.

Proof. We will use the characterization of weakly associated primes of Lemma 10.66.2. Let $\mathfrak p$ be a prime of $R$. As localization is exact we obtain the short exact sequence $0 \to M'_{\mathfrak p} \to M_{\mathfrak p} \to M''_{\mathfrak p} \to 0$. Suppose that $m \in M_{\mathfrak p}$ is an element whose annihilator has radical $\mathfrak pR_{\mathfrak p}$. Then either the image $\overline{m}$ of $m$ in $M''_{\mathfrak p}$ is zero and $m \in M'_{\mathfrak p}$, or the radical of the annihilator of $\overline{m}$ is $\mathfrak pR_{\mathfrak p}$. This proves that $\text{WeakAss}(M) \subset \text{WeakAss}(M') \cup \text{WeakAss}(M'')$. The inclusion $\text{WeakAss}(M') \subset \text{WeakAss}(M)$ is immediate from the definitions. $\square$

Comment #4824 by Bogdan on

"or the annihilator of $\overline{m}$ is $\mathfrak p R_{\mathfrak p}$'' should read "or the radical of the annihilator of $\overline{m}$ is $\mathfrak p R_{\mathfrak p}$''.

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