An irreducible submodule of a Noetherian module is primary

Let $N=M\mathrm{/}Q$. By [L1], the quotient module $N$ is Noetherian. The submodule $0\subset N$ is irreducible: if $0=A^{\mathrm{\prime }}\cap B^{\mathrm{\prime }}$ in $N$, then taking inverse images in $M$ gives $Q=A\cap B$ with $Q\subseteq A,B$, so the irreducibility of $Q$ forces $A=Q$ or $B=Q$, hence $A^{\mathrm{\prime }}=0$ or $B^{\mathrm{\prime }}=0$.

Let $x\in R$ be a zero divisor on $N$. Then $(0{:}_{N}x)\mathrm{\ne }0$. Because $N$ is Noetherian, the ascending chain $(0{:}_{N}x)\subseteq (0{:}_N{x}^2)\subseteq \cdots$ stabilizes; choose $n\ge 1$ with $(0{:}_N{x}^n)=(0{:}_N{x}^{n+1})=(0{:}_N{x}^{2n})$. If $z\in (0{:}_N{x}^n)\cap x^{n}N$, write $z=x^{n}y$. Then $x^{n}z=0$, so $x^{2n}y=0$ and hence $y\in (0{:}_N{x}^{2n})=(0{:}_N{x}^n)$. Therefore $z=x^{n}y=0$, and $0=(0{:}_N{x}^n)\cap x^{n}N\mathrm{.}$

Since $(0{:}_{N}x)\subseteq (0{:}_N{x}^n)$ is nonzero, the irreducibility of $0\subset N$ and the decomposition in step 2.1 force $x^{n}N=0$. Thus every zero divisor on $N$ acts nilpotently on $N$. By [L3], this means $Q$ is primary.

Hence every irreducible submodule of a Noetherian module is primary.