Example of non-decomposable ideal An ideal $I$ of a commutative unital ring $R$ is called decomposable if it has a primary decomposition.

Can you give an example of an ideal that is not decomposable? 

All the examples I can think of are decomposable. Thanks.
 A: Zero ideal in $C[0,1]$ is not decomposable.
More generally,

If $X$ is an infinite compact Hausdorff space then the zero ideal of $C(X)$, the ring of real valued continuous functions on $X$, is not decomposable.

Proof : First let us note that every maximal ideal of $C(X)$ is of the form $M_x=\lbrace f \in C(X) : f(x)=0\rbrace$, for some $x \in X$. Note that, if the zero ideal of $C(X)$ were decomposable, then there
would be only finitely many minimal prime ideals of $C(X)$. This certainly
sounds strange since every maximal ideal $M_x$ of $C(X)$ contains a minimal
prime ideal as every maximal ideal is also prime ideal. Hence to show that the zero ideal of $C(X)$ is not decomposable it is enough to show that if $x\not= y\in X$ then any two minimal prime ideals
$P_1\subseteq M_x, P_2\subseteq M_y$ are different: Since $X$ is Hausdorff and normal, there is an open set $U$
such that $x\in U$ and $y\not\in\overline{U}$. By Urysohn's lemma there are$ f,g\in C(X)$
such that $f(U) = 0, f(y) = 1, g(x) = 1,$ and $g(X \setminus U) = 0.$ So $fg = 0$. Therefore $fg \in P_1$ but $g \not\in P_1$, because $g(x)\not=0$, hence $f \in P_1$. But $f \not\in P_2$, because $f(y)\not=0$. Hence $P_1\not=P_2$. 
