Editing Primary decomposition (section)

== Primary decomposition from associated primes ==
Nowadays, it is common to do primary decomposition of ideals and modules within the theory of '''[[associated prime]]s'''. [[Nicolas Bourbaki|Bourbaki]]'s influential textbook ''Algèbre commutative'', in particular, takes this approach.

Let ''R'' be a ring and ''M'' a module over it. By definition, an '''associated prime''' is a prime ideal which is the [[annihilator (ring theory)|annihilator]] of a nonzero element of ''M''; that is, <math>\mathfrak{p} = \operatorname{Ann}(m)</math> for some <math>m\in M</math> (this implies <math>m \ne 0</math>). Equivalently, a prime ideal <math>\mathfrak{p}</math> is an associated prime of ''M'' if there is an injection of ''R''-modules <math>R/\mathfrak{p} \hookrightarrow M</math>.

A maximal element of the set of annihilators of nonzero elements of ''M'' can be shown to be a prime ideal and thus, when ''R'' is a Noetherian ring, there exists an associated prime of ''M'' if and only if ''M'' is nonzero.

The set of associated primes of ''M'' is denoted by <math>\operatorname{Ass}_R(M)</math> or <math>\operatorname{Ass}(M)</math>. Directly from the definition,
*If <math>M = \bigoplus_i M_i</math>, then <math>\operatorname{Ass}(M) = \bigcup_i \operatorname{Ass}(M_i)</math>.
*For an exact sequence <math>0 \to N \to M \to L \to 0</math>, <math>\operatorname{Ass}(N) \subset \operatorname{Ass}(M) \subset \operatorname{Ass}(N) \cup \operatorname{Ass}(L)</math>.<ref>{{harvnb|Bourbaki|loc=Ch. IV, § 1, no 1, Proposition 3.}}</ref>
*If ''R'' is a Noetherian ring, then <math>\operatorname{Ass}(M) \subset \operatorname{Supp}(M)</math> where <math>\operatorname{Supp}</math> refers to [[support of a module|support]].<ref name="Supp">{{harvnb|Bourbaki|loc=Ch. IV, § 1, no 3, Corollaire 1.}}</ref> Also, the set of minimal elements of <math>\operatorname{Ass}(M)</math> is the same as the set of minimal elements of <math>\operatorname{Supp}(M)</math>.<ref name="Supp" />

If ''M'' is a finitely generated module over ''R'', then there is a finite ascending sequence of submodules
: <math>0=M_0\subsetneq M_1\subsetneq\cdots\subsetneq M_{n-1}\subsetneq M_n=M\,</math>
such that each quotient ''M''<sub>''i''</sub> /''M''<sub>''i−1''</sub> is isomorphic to <math>R/\mathfrak{p}_i</math> for some prime ideals <math>\mathfrak{p}_i</math>, each of which is necessarily in the support of ''M''.<ref>{{harvnb|Bourbaki|loc=Ch. IV, § 1, no 4, Théorème 1.}}</ref> Moreover every associated prime of ''M'' occurs among the set of primes <math>\mathfrak{p}_i</math>; i.e.,
:<math>\operatorname{Ass}(M) \subset \{ \mathfrak{p}_1, \dots, \mathfrak{p}_n \} \subset \operatorname{Supp}(M)</math>.<ref>{{harvnb|Bourbaki|loc=Ch. IV, § 1, no 4, Théorème 2.}}</ref>
(In general, these inclusions are not the equalities.) In particular, <math>\operatorname{Ass}(M)</math> is a finite set when ''M'' is finitely generated.

Let <math>M</math> be a finitely generated module over a Noetherian ring ''R'' and ''N'' a submodule of ''M''. Given <math>\operatorname{Ass}(M/N) = \{ \mathfrak{p}_1, \dots, \mathfrak{p}_n \}</math>, the set of associated primes of <math>M/N</math>, there exist submodules <math>Q_i \subset M</math> such that <math>\operatorname{Ass}(M/Q_i) = \{ \mathfrak{p}_i \}</math> and
:<math>N = \bigcap_{i=1}^n Q_i.</math><ref>{{harvnb|Bourbaki|loc=Ch. IV, § 2, no. 2. Theorem 1.}}</ref><ref>Here is the proof of the existence of the decomposition (following Bourbaki). Let ''M'' be a finitely generated module over a Noetherian ring ''R'' and ''N'' a submodule. To show ''N'' admits a primary decomposition, by replacing ''M'' by <math>M/N</math>, it is enough to show that when <math>N = 0</math>. Now,
:<math>0 = \cap Q_i \iff \emptyset = \operatorname{Ass}(\cap Q_i) = \cap \operatorname{Ass}(Q_i)</math>
where <math>Q_i</math> are primary submodules of ''M''. In other words, 0 has a primary decomposition if, for each associated prime ''P'' of ''M'', there is a primary submodule ''Q'' such that <math>P \not\in \operatorname{Ass}(Q)</math>. Now, consider the set <math>\{ N \subseteq M | P \not\in \operatorname{Ass}(N) \}</math> (which is nonempty since zero is in it). The set has a maximal element ''Q'' since ''M'' is a Noetherian module. If ''Q'' is not ''P''-primary, say, <math>P' \ne P</math> is associated with <math>M/Q</math>, then <math>R/P' \simeq Q'/Q</math> for some submodule ''Q<nowiki>'</nowiki>'', contradicting the maximality. Thus, ''Q'' is primary and the proof is complete.

Remark: The same proof shows that if ''R'', ''M'', ''N'' are all graded, then <math>Q_i</math> in the decomposition may be taken to be graded as well.</ref>
A submodule ''N'' of ''M'' is called ''<math>\mathfrak{p}</math>-primary'' if <math>\operatorname{Ass}(M/N) = \{ \mathfrak{p} \}</math>. A submodule of the ''R''-module ''R'' is <math>\mathfrak{p}</math>-primary as a submodule if and only if it is a <math>\mathfrak{p}</math>-primary ideal; thus, when <math>M = R</math>, the above decomposition is precisely a primary decomposition of an ideal.

Taking <math>N = 0</math>, the above decomposition says the set of associated primes of a finitely generated module ''M'' is the same as <math>\{ \operatorname{Ass}(M/Q_i) | i \}</math> when <math>0 = \cap_1^n Q_i</math> (without finite generation, there can be infinitely many associated primes.)