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.
Best Answer
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.