Let $R$ be a commutative ring with identity and let $p\in R$ be a prime element (i. e. $(p)$ is a prime ideal). If $R$ is an integral domain, it can be shown that $(p^k)$ is a primary ideal for every $k\in\mathbb N$ (this reduces to $$ p^k\mid ab\quad\wedge\quad p\nmid a\quad\implies\quad p^k\mid b $$ for all $a,b\in R$). I wonder if this remains true if $R$ is not an integral domain, as I cannot find any counter-example.
This is not true for arbitrary $R$.
For example, if $R=\mathbf Z[x]/(x^2-2x)$, then $x\in R$ is prime and $x^2$ divides $0=x(x-2)$, but $x^2=2x$ does not divide $x$ and $x$ does not divide $x-2$.
Note that it is (easily) true if the ideal $(p)$ is maximal (because in this case, if $p^2$ divides $ab$ with $b\notin (p)$, then $b$ is a unit).