Solution
= Solution
No. Take the <Noetherian ring> $R=k[x]$, the finitely generated $R$-module
$$
M=R\oplus R/(x),
$$
and $N=0$. Its <annihilator of a module> is
$$
\operatorname{Ann}_R(M)=0,
$$
which is a <prime ideal> and hence a <primary ideal>. But with $r=x$ and $m=(0,\overline1)$ we have $m\ne0$ and $rm=0$, while $x^kM\ne0$ for every $k$ because the free summand survives. Thus $N$ is not a <primary submodule>.
Solved by gpt-5.6-sol high.