# Does every prime power generate a primary ideal?

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.

• $R=\mathbf{Z}[t]/(t^2,2t)$; $k=2$, $a=2$, $b=p=t$. – YCor Jun 10 '16 at 15:49
• @YCor Thanks! So even in Noetherian rings this is false (while I've found books that state the opposite...) – matthias.p Jun 10 '16 at 16:15
• And it's false also in artinian rings (replace $\mathbf{Z}$ with $\mathbf{Z}/4\mathbf{Z}$), and in reduced noetherian rings (replace $\mathbf{Z}[t]/(t^2,2t)$ with $\mathbf{Z}[t]/(t^2-2t)$. – YCor Jun 10 '16 at 16:16
• @YCor But in $(\mathbf Z/4\mathbf Z)[t]/(t^2,2t)$, $t$ is not a prime element, because $2\cdot2\in(t)$, but $2\notin(t)$. – matthias.p Jun 11 '16 at 10:46

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).