Non-Noetherian rings with an ideal not containing a product of prime ideals It is well-known that in every commutative Noetherian ring every ideal contains a product of prime ideals.
Are there examples of non-Noetherian rings with an ideal that does not contain any prime ideals? 
 A: Let the ring $R=\{ (a_n)_{n \in \mathbb{N}} \in \mathbb{Z}^{\mathbb{N}}\mid a_{n+1}=a_n\text{ for } n\text{ sufficiently large}\}$ and the ideal $I=(0)$.
For all $i\in \mathbb{N}$, let $e_i=(a_{i,n})_{n \in \mathbb{N}}$ with $a_{i,n}=1$ if $i=n$ and $a_{i,n}=0$ if $i \neq n$.
Let $P$ be a prime ideal.
If $i \neq j$, $e_i e_j=0$, so $e_i \in P$ or $e_j \in P$.
If there exists $i \in \mathbb{N}$ such that $e_i \notin P$, we have $e_j \in P$ for all $j \neq i$.
So $\bigoplus_{j \neq i}\mathbb{Z}e_j\subset P.$
If we choose a finite number of prime ideals $P_1,...,P_k$ with $\bigoplus_{j \neq i_m} \mathbb{Z} e_j \subset P_m$ for $m=1,...,k$,
we have $I=(0) \neq \bigoplus_{j \neq i_1,...,i_k} \mathbb{Z} e_j \subset P_1P_2\cdots P_k$.
So $I$ doesn't contain a product of prime ideals.
A: The ring of all algebraic integers $\overline{\mathbb{Z}}$ also supports a nice example. $\overline{\mathbb{Z}}$ is not Noetherian and every finitely generated ideal is principal. (This can be seen by the finiteness of class numbers of number fields. I think this is proved by Dedekind.)
We show that every non-zero principal ideal does not  contain a finite product of prime ideals.
If $(\alpha) \supset \mathfrak{p}_1\mathfrak{p}_2\cdots \mathfrak{p}_m$, one can choose a tower of field $K_n$ such that the number of prime factors (count multiplicity) of $\alpha \mathcal{O}_{n}$ increase strictly, where $\mathcal{O}_{n}$ is the ring of integers of $K_n$.  For example, let $K_n=\mathbb{Q}(\sqrt[n]{\alpha})$. Then the ramification of prime ideal factors of $\alpha \mathcal{O}_n$ will go to infinity when $n$ becomes large.
Since $\alpha \mathcal{O}_{n} =(\alpha) \cap \mathcal{O}_n \supset {\mathfrak{p}_1}_n{\mathfrak{p}_2}_n\cdots \mathfrak{p_m}_n$, we have $\alpha \mathcal{O}_{n} \mid {\mathfrak{p}_1}_n{\mathfrak{p}_2}_n\cdots \mathfrak{p_m}_n$, where ${\mathfrak{p}_i}_n=\mathfrak{p}_i \cap \mathcal{O}_n$ is a prime ideal of $\mathcal{O}_n$. We will get a contradiction when $n$ is large.
