Lemma 10.63.3. 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{Ass}(M') \subset \text{Ass}(M) and \text{Ass}(M) \subset \text{Ass}(M') \cup \text{Ass}(M''). Also \text{Ass}(M' \oplus M'') = \text{Ass}(M') \cup \text{Ass}(M'').
Proof. If m' \in M', then the annihilator of m' viewed as an element of M' is the same as the annihilator of m' viewed as an element of M. Hence the inclusion \text{Ass}(M') \subset \text{Ass}(M). Let m \in M be an element whose annihilator is a prime ideal \mathfrak p. If there exists a g \in R, g \not\in \mathfrak p such that m' = gm \in M' then the annihilator of m' is \mathfrak p. If there does not exist a g \in R, g \not\in \mathfrak p such that gm \in M', then the annilator of the image m'' \in M'' of m is \mathfrak p. This proves the inclusion \text{Ass}(M) \subset \text{Ass}(M') \cup \text{Ass}(M''). We omit the proof of the final statement. \square
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