Just as in the title: I've seen the statement that Artinian rings are Noetherian several times (eg Commutative artinian ring is noetherian) but if we take $R=\oplus_\mathbb{N}\mathbb{Z}_2$, it seems it is artinian but not Noetherian:

Indeed, given any decreasing sequence $I_n$ of ideals, choose $k$ for which $I_1\subseteq\mathbb{Z}_2^k$. This is a finite ring, so the sequence stabilizes.

However, the increasing sequence of ideals $J_n=\mathbb{Z}_2^n$ does not stabilize.

Where is my mistake?

  • $\begingroup$ There are ideals not contained in $\mathbb Z_2^k$ for any $k$, for example $\mathrm{Ann}(J_n)=\{x\in R\mid x_i=0\text{ for }i\leq n\}$. $\endgroup$
    – stewbasic
    Oct 9, 2016 at 22:12

1 Answer 1


First of all, it is only true that all Artinian rings are Noetherian if "ring" means "unital ring". Since your ring $R$ is not unital, what you have written does not actually give a contradiction.

However, $R$ in fact is not Artinian. The reason is that there need not exist $k$ such that $I_1\subseteq \mathbb{Z}_2^k$. For instance, $I_1$ might be the ideal generated by $e_{2n}$ for each $n\in\mathbb{N}$, where $e_i$ is the element of $R$ whose $i$th coordinate is $1$ and all other coordinates are $0$. You can then get a descending chain of ideals by one-by-one removing one of the generators from this $I_1$.


You must log in to answer this question.

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