# Primary decomposition of the zero submodule of a module $M$, and of the annihilator of $M$

Let $$A$$ be a commutative ring with one, $$M$$ an $$A$$-module and $$Q$$ a proper submodule of $$M$$. Using the terminology of Commutative Algebra by Zariski and Samuel, we say that $$Q$$ is primary if, for any $$a\in A$$, the map $$x\mapsto ax$$, $$M/Q\to M/Q$$, is injective or nilpotent. (Warning: some mathematicians, like Bourbaki, use another definition of a primary submodule. Note however that the definition of Zariski and Samuel is also used by Atiyah and MacDonald.)

Consider the following conditions on an $$A$$-module $$M$$:

(a) $$M$$ is faithful,

(b) $$M$$ is not a direct sum of two of its proper submodules,

(c) there are primary submodules $$Q_1,\dots,Q_n\subset M$$ such that $$Q_1\cap\dots\cap Q_n=0$$ is a reduced primary decomposition of the zero submodule of $$M$$.

Then it is well known, and easy to see, that the ideals $$\mathfrak q_i:=(Q_i:M)$$ are primary, and that the zero ideal $$(0)\subset A$$ is the intersection of the $$\mathfrak q_i$$.

Say that our commutative ring $$A$$ has Property (P) if for all $$A$$-module $$M$$ satisfying Conditions (a), (b) and (c) above, the equality $$\mathfrak q_1\cap\dots\cap\mathfrak q_n=(0)$$ is a reduced primary decomposition of the zero ideal $$(0)$$ of $$A$$.

Question. Do all commutative rings have Property (P)?

The ring $$A$$ has Property (P) if it satisfies at least one of the following conditions:

(d) the Krull dimension of $$A$$ is at most $$0$$,

(e) $$A$$ is isomorphic to a product of two nonzero rings,

(f) the zero ideal of $$A$$ admits no primary decomposition.

Sadly there is not a single ring $$A$$ which satisfies none of the three above conditions for which I know that it has Property (P).

In particular I don't know if $$\mathbb Z$$ has Property (P).

Note that a domain $$A$$ has Property (P) if and only if any module satisfying (a), (b) and (c) is torsion free.

EDIT 1. In fact principal ideal domains have Property (P). More generally, a domain $$A$$ such that any indecomposable $$A$$-module is either torsion or torsion-free has Property (P), and it is proved in Kaplansky's book Infinite abelian groups, p. 36, that principal ideal domains have this dichotomy property. I don't know if there are domains admitting indecomposable modules which are neither torsion nor torsion-free. This has been asked as a separate question. END OF EDIT 1.

EDIT 3. Kaplansky proved more generally that Dedekind domains have this dichotomy property: see Theorem 10 in Modules over Dedekind rings and valuation rings. END OF EDIT 3.

Let me end this post with a comment which can be safely skipped by the busy reader.

If we use Bourbaki's definition of a primary submodule, there is an analog to Property (P). Call it Property (P'). Then there are rings $$A$$ which don't have Property (P'). Here are more details.

To simplify, assume that $$A$$ is noetherian.

If $$M$$ is an $$A$$-module and $$Q$$ a proper submodule of $$M$$, Bourbaki says that $$Q$$ is primary if, for any $$a\in A$$, the map $$x\mapsto ax$$, $$M/Q\to M/Q$$, is injective or locally nilpotent. One can show that the set of $$a\in A$$ such that $$x\mapsto ax$$, $$M/Q\to M/Q$$, is locally nilpotent is a prime ideal $$\mathfrak p$$, and Bourbaki says that $$Q$$ is $$\mathfrak p$$-primary.

To avoid any misunderstanding let me write "primary" (with quotation marks) for primary in the sense of Bourbaki.

Consider the following conditions on an $$A$$-module $$M$$:

(a) $$M$$ is faithful,

(b) $$M$$ is not a direct sum of two of its proper submodules,

(c') for $$i=1,\dots,n$$ there are $$\mathfrak p_i$$-"primary" submodules $$Q_i\subset M$$ such that $$Q_1\cap\dots\cap Q_n=0$$ is a reduced "primary" decomposition of the zero submodule of $$M$$.

Say that our noetherian ring $$A$$ has Property (P') if for all $$A$$-module $$M$$ satisfying Conditions (a), (b) and (c') above, there is a reduced primary decomposition $$\mathfrak q_1\cap\dots\cap\mathfrak q_n=(0)$$ of the zero ideal $$(0)$$ of $$A$$, where each $$\mathfrak q_i$$ is $$\mathfrak p_i$$-primary.

Here is a noetherian ring $$A$$ which does not have Property (P'):

Let $$K$$ be a field of characteristic zero, let $$u$$ and $$v$$ be indeterminates, set $$A:=K[[u]]$$, $$M:=K[v]$$, and let $$\sum a_nu^n\in A$$ act on $$f\in M$$ by $$\left(\sum a_nu^n\right)f:=\sum a_nf^{(n)},$$ where $$f^{(n)}$$ is the $$n$$-th derivative of $$f$$. Clearly $$M$$ satisfies (a) and (b).

As $$0$$ is a $$(u)$$-"primary" submodule of $$M$$, Condition (c') is also satisfied. The zero ideal $$(0)$$ is "primary", but it is $$(0)$$-"primary" instead of being $$(u)$$-"primary", so that $$A$$ does not have Property (P').

Note that no submodule of $$M$$ is primary (without quotation marks), so that (c) is not satisfied. (Again, I don't know if $$K[[u]]$$ has Property (P).)

EDIT 2. In fact $$K[[u]]$$ has Property (P): see Edit 1 above.

• What do you mean by "reduced"? No embedded components? Sep 13, 2019 at 16:22
• @Youngsu - The primary decomposition $Q_1\cap\dots\cap Q_n=0$ is reduced if (1) the radicals of the $(Q_i:M)$ are distinct, and (2) for each $i$ the submodule $Q_i$ does not contain the intersection $\bigcap_{j\ne i}Q_j$ of the other primary submodules $Q_j$. One also says that the primary decomposition is "minimal", or "irredundant". Sep 13, 2019 at 16:53
• @Pierre-YvesGaillard Why this requirement: the module is not a direct sum of two proper sumbodules? Sep 13, 2019 at 20:35
• @user26857 - Otherwise you could take $A:=\mathbb Z,M:=\mathbb Z\oplus\mathbb Z/(2)$ and you would get in $\mathbb Z$ the primary decomposition $(0)=(0)\cap(2)$, which is not reduced. Sep 13, 2019 at 21:04

Mohan answered the present question in a comment to this question. For more details see this answer. Note that the module $$M$$ considered by Mohan, being finitely generated over a noetherian ring, satisfies the condition that its zero submodule has a primary decomposition.