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 $N$, and write $K_k:=\{y\in N:x^{k}y=0\}$. Since $R$ is commutative, multiplication by $x^k$ is $R$-linear, so $K_k$ and $x^{k}N$ are submodules. Also $K_1\mathrm{\ne }0$ and $K_k\subseteq K_{k+1}$. By [L4] this ascending sequence stabilizes; choose $n\ge 1$ beyond its stabilization index, so $K_n=K_{2n}$. If $z\in K_n\cap x^{n}N$, write $z=x^{n}y$. Then $x^{2n}y=x^{n}z=0$, hence $y\in K_{2n}=K_n$ and $z=0$. Thus $0=K_n\cap x^{n}N$. This uses only the choice-free implication of [L4].

Since $0\mathrm{\ne }{K}_1\subseteq K_n$, irreducibility of $0\subset N$ and $0=K_n\cap x^{n}N$ force $x^{n}N=0$. Consequently, if $xm\in Q$ and $m\mathrm{\notin }Q$, the nonzero coset $m+Q$ shows that $x$ is a zero divisor on $N$, so some $n\ge 1$ satisfies $x^{n}M\subseteq Q$. By [L3], $Q$ is primary.

Hence every irreducible submodule of a Noetherian module is primary, without an additional choice assumption.