# 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?

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## 1 Answer

Let the ring $R=\mathbb{Z}^{\mathbb{(N)}}$ and the ideal $I=(0)$.

$R= \{ (a_n)_{n \in \mathbb{N}} \in \mathbb{Z}^{\mathbb{N}}| a_n=0$ except for a finite number of $n\}$.

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 $P$ 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 an $i \in \mathbb{N}$ such that $e_i \notin P$, we have $e_i e_j=0$ if $j \neq i$. So for all $j \neq i$, $e_j \in P$.

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 $I=(0) \neq \bigoplus_{j \neq i_1,...,i_k} \mathbb{Z} e_j \subset P_1P_2...P_k$.

So $I$ doesn't contain a product of prime ideals.

EDIT: $R= \{ (a_n)_{n \in \mathbb{N}} \in \mathbb{Z}^{\mathbb{N}}| a_{n+1}=a_n$ for $n$ sufficiently large $\}$.

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Technical detail: Your ring doesn't have a $1$, which might be considered as an implicit requirement. But this can easily be fixed. +1! –  Pierre-Yves Gaillard Dec 7 '11 at 18:32
I have edited the answer. Thanks, Pierre-Yves. –  francis-jamet Dec 7 '11 at 19:00