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?


2 Answers 2


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.

  • 1
    $\begingroup$ Is there also a non-zero example of such an ideal? $\endgroup$
    – Pedro
    Jan 14, 2018 at 22:04

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .