Proposition 3

Tag 04JD, permanent. Cite it as [Bourbaki, Tag 04JD].

Commutative Algebra, chapter IV, § 1. Prime ideals associated with a module

Let $A$ be a ring, $M$ an $A$ -module and $N$ a submodule of $N$ . Then

\operatorname{Ass}(N) \subset \operatorname{Ass}(M) \subset \operatorname{Ass}(N) \cup \operatorname{Ass}(M/N).

The inclusion $\operatorname{Ass}(N) \subset \operatorname{Ass}(M)$ is obvious. Let $p \in \operatorname{Ass}(M)$ , $E$ be a submodule of $M$ isomorphic to $A/p$ and $F = E \cap N$ . If $F = \{0\}$ , $E$ is isomorphic to a submodule of

M/N

, whence

p \in \operatorname{Ass}(M/N)

. If

F \neq \{0\}

, the annihilator of every element

\neq 0

F

p

(Proposition 1) and hence

p \in \operatorname{Ass}(F) \subset \operatorname{Ass}(N)

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Proposition 10, § 1
Corollary 1, § 1
Corollary 2, § 1
Proposition 4, § 1
Example, § 2
Proposition 2, § 2
Proposition 4, § 2
Proposition 8, § 2
Proposition 7, § 4

Cites

Proposition 1