Is it true that if some power of an ideal is primary, then the ideal itself is also primary?

Question

Is it true that if some power of an ideal $I$ is primary, then $I$ itself is also a primary ideal?

I do not know whether the above statement is true or there is a counterexample. If one wants to prove it, then the natural way is as follows. Suppose the $n$-th power of $I$ is primary for some $n\geq 2$. Let $ab\in I$ and $a\notin I$. Then one should show that $b\in \sqrt{I}$. Since $ab\in I$, we have $a^nb^n={(ab)}^n\in I^n$ (we are in a commutative ring). If $a^n\notin I^n$, then we get $b^n\in \sqrt{I^n}=\sqrt{I}$, since $I^n$ is primary by assumption. So, in this case, we get $b\in \sqrt{I}$, since $\sqrt{I}$ is a prime ideal as well as $\sqrt{I^n}$, and hence we are done. But, if $a^n\in I^n$, what can we say? (Indeed, the assumption $a\notin I$ does not necessarily imply that $a^n\notin I^n$). So, maybe one can find a counterexample, or somehow finish the proof.

Answer 1

No, it's not.

Let $R=I_{\mathbb Z}(\mathbb Q)$ be the idealization of the $\mathbb Z$-module $\mathbb Q$.
The primary ideals of $R$ are of the following form: $p^n\mathbb Z\times\mathbb Q$ with $p$ prime and $n\ge 1$, $\{0\}\times\mathbb Q$ or $\{0\}\times\{0\}$ (why?).
Set $I=\{0\}\times H$ with $H$ a proper subgroup of $\mathbb Q$. Then $I^2=\{0\}\times\{0\}$ is primary, but $I$ is not.