An example of radical ideal which is irreducible but not prime

Question

$\DeclareMathOperator\rad{rad}$I am searching an example of ideal $I$ of a ring $R$ such that $\rad(I)$ is irreducible but not prime ideal.

In case $R$ is Noetherian, the radical of $I$ being irreducible implies $\rad(I)$ is primary. Then it is straight forward to see that $\rad(I)$ is prime. So we want to look at non Noetherian rings. But I am not able to find such examples. Also, $I$ cannot be primary as $\rad(I)$ is prime if $I$ is primary.

Answer 1

There is no example.

Let $K=\textrm{rad}(I)$. $K$ is a radical ideal (i.e., $K=\textrm{rad}(K)$) which you assume to be irreducible. Assume that $a, b\notin K$, but $ab\in K$. Then $K+(a)$ and $K+(b)$ properly extend $K$, so by irreducibility $$L = (K+(a))\cap (K+(b))$$ properly extends $K$. But $L^2 \subseteq (K+(a))\cdot (K+(b)) \subseteq K^2 + (a)K + K(b) + (a)(b) \subseteq K$, contradicting the fact that $K$ is a radical ideal.