Ideal in a Noetherian ring which is not equal to a product of prime ideals

In the polynomial ring $k[X,Y]$ over a field $k$ I claim that the ideal $I=\langle X^2,Y^2\rangle$ cannot be written as a product $I=\mathfrak p_1\cdot\cdots\mathfrak p_n$ where the $\mathfrak p_i$'s are prime ideals, not necessarily distinct.
Indeed, if this were the case we would have $I\subset \mathfrak p_i$ for every $i$ so that $$ \sqrt I= \langle X,Y\rangle\subset \sqrt {\mathfrak p_i}=\mathfrak p_i$$ Since $\mathfrak m :=\langle X,Y\rangle$ is maximal this forces $\mathfrak p_i=\mathfrak m$ for all $i$ and thus we would have $I=\mathfrak m^n$.
But this is false for any $n$ since $$ \mathfrak m=\langle X,Y\rangle\supsetneq \mathfrak m^2=\langle X,Y\rangle^2=\langle X^2,XY,Y^2\rangle\supsetneq I=\langle X^2,Y^2\rangle\supsetneq \mathfrak m^3\supsetneq \cdots \supsetneq \mathfrak m^n\supsetneq\cdots $$

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 !