I originally posted this to math.stackexchange.com here. I got a partial answer, but I now suspect that the complete answer is much harder than I thought, so I'm posting it here.

Fix a commutative ring $R$. Recall that an ideal $I$ of $R$ is **irreducible** if $I = J_1 \cap J_2$ for ideals $J_1$ and $J_2$ only when either $I = J_1$ or $I = J_2$.

Question : Assume that $I$ is an irreducible ideal. Must the radical of $I$ be an irreducible ideal?

On math.stackchange.com, I learned that the answer is "yes" if $R$ is Noetherian. My guess is that there is a counterexample if $R$ is not assumed to be Noetherian, but I have no idea how to construct it.