TITLE: Ideal in a Noetherian ring which is not equal to a product of prime ideals
QUESTION [6 upvotes]: In a Noetherian ring, it is known that every ideal contains a product of prime ideals. 

Is there any example of a Noetherian ring in which an ideal is not equal to any product of prime ideals?

This is a natural question,which I didn't find even as an exercise in common references where the above property is stated. 
My intuitive answer: consider a suitable subring of the ring of integers in a  number field. For example, the ideal $(2,1+\sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$; am I right?

REPLY [5 votes]: I belong to the Geometric Faith but for those of the Arithmetic Persuasion here is a number-theoretic example.
Let $A=\mathbb Z[i\sqrt 3$] and consider the principal ideal $I=\langle 1+i\sqrt 3\rangle\subset A$.
I claim that $I$ is not a product of prime ideals.
Indeed any prime ideal $\mathfrak p$ containing $I$ must contain $ (1+i\sqrt 3)(1-i\sqrt 3)=4$ and thus $\mathfrak p$ must also contain $2$ so that 
 $\langle 1+i\sqrt 3,2\rangle\subset \mathfrak p$.
However $\langle 1+i\sqrt 3,2\rangle$ is a maximal ideal and we conclude that $ \mathfrak p=\langle 1+i\sqrt 3,2\rangle$.
This proves that the only way $I$ could be a product of primeideal would be that $I=\mathfrak p^n$.
But this is impossible because $$\mathfrak p=\langle 1+i\sqrt 3,2\rangle\supsetneq I=\langle 1+i\sqrt 3\rangle\supsetneq \mathfrak p^2=\langle 2+2i\sqrt 3,4\rangle \supsetneq\cdots\supsetneq \mathfrak p^n\supsetneq\cdots$$ 
NB
The ring $A=\mathbb Z[i\sqrt 3]$ is of course not  Dedekind: the main theorem about Dedekind rings is that each ideal of such a ring is a product of prime ideals !