Is the radical of an irreducible ideal irreducible? 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?
EDIT : The answer below in the Noetherian case convinced me that this is harder than I thought.  I thus posted it to MathOverflow here
 A: Pham Hung Quy answered the question on MathOverflow.
(This is a community wiki post.)
(I understand nothing to Pham Hung Quy's answer, and think it would be great if more elementary explanations were given on MSE.)
[The part above was written by Pierre-Yves Gaillard.]

[Beginning of the part written by florian.]
Here is how Pham Hung Quy's counter-example with a non-noetherian ring works:
We start with the noetherian local ring $R = k[[x,y]]/(xy)\ $ (formal power series in $x$ and $y$ modulo the ideal $(xy)$). 
Note that $M := (x,y)$ is the unique maximal ideal in $R$. We will need this, and it is the reason we work with $k[[x,y]]$ instead of $k[x,y]$.
Note also that the zero ideal in $R$ is the same as $(xy)$ (since we factored $(xy)$ out), and since $(xy)=(x)\cap(y)$, we see that the zero ideal in $R$ is reducible. Moreover, $R$ doesn't have any nilpotent elements.
Now we construct an $R$-module $I$ with the following to properties:


*

*$I$ has a unique minimal submodule of dimension $1$, i.e. there is a non-zero submodule in $I$ which is contained in every other non-zero submodule.
Clearly, this shows that the zero submodule of $I$ cannot be written as an intersection of non-zero submodules of $I$, since this intersection again will contain the minimal submodule.

*$I$ is divisible. This means: Given $f\in I$ and $r\in R$ which isn't a zero divisor there is a $g\in I$ such that $f=r\cdot g$ (i.e. we can "divide elements of $I$ by elements of $R$ which aren't zero-divisors").


We will see that $I={\rm Hom}_{k}(R,k)$ does the job ($k$-linear maps from $R$ to $k$; this is an $R$-module by defining $(r\cdot f)(s) := f(r\cdot s)$ for $f\in I$ and $r,s\in R$). Let's now check that $I$ does indeed have the two desired properties:


*

*Suppose $J\subseteq I$ is a non-zero submodule. Then consider $J^\perp \subseteq R$ (those elements in $R$ annihilated by all elements in $J$; more generally, we have a non-degenerate bilinear pairing $I\times R \longrightarrow k: (g,r)\mapsto g(r)$, and we can always form perpendicular spaces with respect to this pairing). Since even in infinite dimensions we still have the well known equality from linear algebra $J^{\perp\perp}=J$, we must have $J^\perp \neq R$. Since one can show that $J^\perp$ is an ideal, and $M$ is the unique maximal ideal in $R$, it must follow $J^\perp \subseteq M$ and therefore $J=J^{\perp \perp} \supseteq M^\perp$. Thus it follows that $M^\perp$ (the set of all elements in $I$ which vanish on $M$) is the unique minimal submodule of $I$. It is also easy to see that $M^\perp$ is one-dimensional, since every element in $M^\perp$ must factor through the natural epimorphism $R\longrightarrow R/M\cong k$.

*Given $f\in I$ and $r\in R$ which isn't a zero divisor (this ensures well-definedness) we are looking for a $g\in I$ such that 
$(r\cdot g)(s)=f(s)$ for all $s\in R$, which by definition is the same
as $g(rs)=f(s)$ for all $s\in R$. So take $g$ to be the linear functional on
the $k$-vector space (even ideal) $rR$ which  is defined by the aforementioned equation, and extend it to $R$ (note that $g$ doesn't have to satisfy any structural properties other than being $k$-linear, and $k$-linear functionals can always  be extended to bigger spaces using the axiom of choice). 


Now we take the ring $R$ and the $R$-module $I$ and build a new ring
$$
     E := R\oplus I
$$ 
where addition is defined in the obvious way, i.e. $(r,i)+(r',i') := (r+r',i+i')$, and multiplication is defined by the formula $(r,i)\cdot (r',i') := (rr',ri'+ir')$.
Note that $I$ is an ideal in $E$, and in fact $I^2=(0)$ (in $E$.) Hence
$I$ is the radical of the zero ideal in $E$ (it is the entire radical since $E/I\cong R$ has no nilpotent elements). But $I= (I+(x))\cap (I+(y))$, and hence isn't irreducible (here it pays off that we made sure the zero ideal in $R$ wasn't irreducible).
Now all we have to check is that the zero ideal in $E$ is in fact irreducible. 
Let $f$ be a generator for the unique minimal submodule of $I$ and let 
$J\subseteq E$ be a non-zero ideal. We intend to show that $f\in J$, which will imply that $(0)$ cannot be the intersection of non-zero ideals, since each such intersection contains $f$. We have two possibilities:


*

*$J \subseteq I$: In this case, $J$ is a submodule of $I$, and we know it contains $f$ because $f$ generates the unique minimal submodule of $I$.

*$J \not\subseteq I$: In this case we may choose some $(r,h)\in J$ with 
$r\neq 0$. We can  choose a $j\in I$ such that
$r\cdot j \neq 0$ (since clearly we find a functional with $j(r)\neq 0$). 
Then $(r,h)\cdot(0,j)=(0,rj)\neq 0$, and hence $(0,rj)\cdot E \subset J$ is a sub-ideal of $J$ contained in $I$. This contains $f$ due to the first point.


(looks like we don't even need divisibility)
[End of the part written by florian.]
A: The answer is yes in a noetherian ring $A$. Let $I$ be a proper irreducible ideal of $A$. From AM-7.12, $I$ is a primary ideal. Hence $\sqrt{I}$ is prime (AM-4.1). Now, conclude because a prime ideal is irreducible (AM-1.11-ii).
One can reach the same conclusion if $I$ is supposed to be decomposable (i.e. $I$ is the intersection of a finite number of primary ideals). In a noetherian ring, every ideal is decomposable (AM-7.13). I don't know what happens if $I$ is not decomposable.
