Is the radical of an irreducible ideal irreducible? 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.
 A: I construct a counterexample for your question in the non-noetherian case:
(1) Let $A = k[[X,Y]]/(XY) = k[[x,y]]$, where $k$ be a field. Notice that $(0) = (x) \cap (y)$ so $(0)$ is reducible in $A$.
(2) We consider the injective hull $E(k)$ of $k$, and set $m \in E(k)$ be the element such that $mA \cong k$. Notice that every non-zero submodule of $E(k)$ contains $mA$ and $(0)$ is irreducible in $E(k)$
(3) Set $R = A \ltimes E(k)$ be the indealization. We have that $(0 \ltimes E(k))^2 = 0$ so $\sqrt{(0)R} = 0 \ltimes E(k)$ is reducible by (1).
(4) We can prove that for every non-zero element $(a,s)$ of $R$, we have $0 \ltimes mA \subseteq (a,s)R$. So the ideal $(0)$ is irreducible in $R$.
EDIT (13/02): It should be noted that this example is also a counterexample for a non-noetherian ring with an ideal is irreducible but not primary. Indeed, we have $(0)$ is irreducible as above. However
$$(x,0).(y,m) = (0,0) \in R,$$
and $(x,0)$ and $(y,m)$ are not nilpotent so $(0)$ is not primary in $R$
