TITLE: Can an element in a Noetherian ring have arbitrarily long factorizations?
QUESTION [9 upvotes]: Suppose $R$ is a Noetherian ring.  Is it possible that an element $r\in R$ have arbitrarily long factorizations? That is, for all $n>0$, is there a factorization $r=a_{1n}a_{2n}\cdots a_{nn}$ such that each $a_{in}$ is a non-unit? If this is not possible, what about the weaker hypothesis of ascending chain condition on principal ideals? It is evident that this cannot happen in a UFD.

REPLY [6 votes]: You can even have arbitrarily long factorizations with a finite ring.
Take the field of two elements $F$, form $R=F\times F$, and let $e=(1,0)$. Then $e$ is not a unit and $e=e^n$ for any positive integer.
Of course, if you want to rule out idempotents like this you can by looking at domains and local rings. As other posters have explained, your luck will definitely better with Noetherian domains.
Addendum: there was a comment and a solution that I was referring to that no longer appear here. The solution for domains was moved here.