Is an ideal generated by irreductible elements of a ring a radical ideal?

Question

I'm wondering if an ideal generated by irreductible elements of a ring is a radical ideal, I can't seem to prove it nor disprove it.

Answer 1

The answer is NO.

Let $R=\mathbb{Z}[\sqrt{-6}]$. It is easy to show that $3$ is irreducible. Now, let $I=3R.$

Clearly $\sqrt{-6}\in R\setminus I$ (it is not of the form $3a+3b\sqrt{-6}$ for some $a,b\in\mathbb{Z})$, but $(\sqrt{-6})^2=-6\in I$.