Every submodule of a finite module over a Noetherian ring has a minimal primary decomposition

The submodule $M$ has the empty primary decomposition, whose intersection is interpreted as $M$.

By [L1], the module $M$ is Noetherian. Suppose, toward contradiction, that some submodule of $M$ has no finite primary decomposition. Let $\mathrm{\Sigma }$ be the set of such submodules. By step 1.1, $M\mathrm{\notin }\mathrm{\Sigma }$. By [L2], choose a maximal element $Q$ of $\mathrm{\Sigma }$; then $Q⊊M$.

The submodule $Q$ is irreducible. Indeed, if $Q=A\cap B$ with $Q⊊A$ and $Q⊊B$, then the maximality of $Q$ in step 2.1 forces finite primary decompositions of $A$ and $B$. Intersecting those two finite decompositions gives a finite primary decomposition of $Q$, contrary to $Q\in \mathrm{\Sigma }$.

Because $Q$ is irreducible and proper and $M$ is Noetherian, [L3] shows that $Q$ is primary. But a primary submodule is already a one-term finite primary decomposition, again contradicting $Q\in \mathrm{\Sigma }$. Therefore $\mathrm{\Sigma }$ is empty: every submodule of $M$ has a finite primary decomposition.

If $N=M$, the empty decomposition from step 1.1 is irredundant and has pairwise distinct radicals vacuously, so it is minimal. If $N⊊M$, start from a finite primary decomposition supplied by step 4.1, apply [L4] to remove redundant components, and then apply [L5] to merge equal-radical blocks. The resulting decomposition is finite, has the same intersection, has no redundant component, and has pairwise distinct radicals; hence it is minimal.

Taking $M=R$ recovers the ideal case, because ideals are precisely the submodules of the regular module over a commutative ring.