Editing Primary decomposition (section)

== Properties of associated primes ==
Let <math>R</math> be a Noetherian ring. Then
*The set of [[zero-divisor]]s on ''R'' is the same as the union of the associated primes of ''R'' (this is because the set of zerodivisors of ''R'' is the union of the set of annihilators of nonzero elements, the maximal elements of which are associated primes).<ref>{{harvnb|Bourbaki|loc=Ch. IV, § 1, Corollary 3.}}</ref>
* For the same reason, the union of the associated primes of an ''R''-module ''M'' is exactly the set of zero-divisors on ''M'', that is, an element ''r'' such that the endomorphism <math>m \mapsto rm, M \to M</math> is not injective.<ref>{{harvnb|Bourbaki|loc=Ch. IV, § 1, Corollary 2.}}</ref>
* Given a subset <math>\Phi \subset \operatorname{Ass}(M)</math>, ''M'' an ''R''-module , there exists a submodule <math>N \subset M</math> such that <math>\operatorname{Ass}(N) = \operatorname{Ass}(M) - \Phi</math> and <math>\operatorname{Ass}(M/N) = \Phi</math>.<ref>{{harvnb|Bourbaki|loc=Ch. IV, § 1, Proposition 4.}}</ref>
*Let <math>S \subset R</math> be a multiplicative subset, <math>M</math> an <math>R</math>-module and <math>\Phi</math> the set of all prime ideals of <math>R</math> not intersecting <math>S</math>. Then <math display="block">\mathfrak{p} \mapsto S^{-1}\mathfrak{p}, \, \operatorname{Ass}_R(M)\cap \Phi \to \operatorname{Ass}_{S^{-1}R}(S^{-1} M)</math> is a bijection.<ref>{{harvnb|Bourbaki|loc=Ch. IV, § 1, no. 2, Proposition 5.}}</ref> Also, <math>\operatorname{Ass}_R(M)\cap \Phi = \operatorname{Ass}_R(S^{-1}M)</math>.<ref>{{harvnb|Matsumura|1970|loc=7.C Lemma}}</ref>
* Any [[minimal prime ideal|prime ideal minimal]] with respect to containing an ideal ''J'' is in <math>\mathrm{Ass}_R(R/J).</math> These primes are precisely the isolated primes.
* A module ''M'' over ''R'' has [[finite length]] if and only if ''M'' is finitely generated and <math>\mathrm{Ass}(M)</math> consists of maximal ideals.<ref>{{citation|title=Basic Algebra|first=P. M.|last=Cohn|publisher=Springer|year=2003|isbn=9780857294289|at=Exercise&nbsp;10.9.7, p.&nbsp;391|url=https://books.google.com/books?id=bOElBQAAQBAJ&pg=PA391}}.</ref>
*Let <math>A \to B</math> be a ring homomorphism between Noetherian rings and ''F'' a ''B''-module that is [[flat module|flat]] over ''A''. Then, for each ''A''-module ''E'',
:<math>\operatorname{Ass}_B(E \otimes_A F) = \bigcup_{\mathfrak{p} \in \operatorname{Ass}(E)} \operatorname{Ass}_B(F/\mathfrak{p}F)</math>.<ref>{{harvnb|Bourbaki|loc=Ch. IV, § 2. Theorem 2.}}</ref>