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Is it true that an ideal $I$ in a commutative ring is primary iff $Rad(I)$ is prime?

If not, what are some nice counterexamples?

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Here is a simple example with $I$ not primary such that $Rad(I)$ is prime. Consider the ideal $(y^2,xy)\subset K[x,y]$. The radical of this ideal is $(y)$, i.e., it is prime.

At the same time, clearly $x\cdot y\in I$. Also, $y\notin I$. Nevertheless $x^n\notin I$ for any $n$. Hence $I$ is not primary!

Comment. As user26857 noted in a remark, one can even construct a prime ideal $p$ such that $p^2$ is not primary, though the example is a bit more complicated:

Is each power of a prime ideal a primary ideal?

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    @agleaner This example seems to be inordinately useful. I've seen it used and used it myself a couple different times over the past year! – rschwieb Mar 31 '14 at 12:52
  • If $xy\in I \subseteq \sqrt{I}$ and $\sqrt{I}=P$ is prime, then $x\in P$ or $y\in P$. Thus $x^n \in I$ or $y^m\in I$. Which fact am I missing here? – nkh99 Feb 09 '24 at 11:46
  • @nkh99 In our example, $x^n\notin I$ for all $n$ so you conclude $y^m\in I$ for some $m$. However, we do not get to conclude $y\in m$. The definition of primary is that $xy\in I\implies y\in I\text{ or }x\in\sqrt{I}$, or in other words the zero divisors of $R/I$ are precisely the nilpotents – FShrike Feb 24 '25 at 22:17