Can an element in a Noetherian domain have arbitrarily long factorizations?

Question

I tried to answer this question two days ago. Unfortunately, I said ring, rather than domain, which is what I wanted. So I try again.

Let $R$ be a Noetherian commutative domain and let $r\in R$. Is it possible that $r$ have arbitrarily long factorizations? That is, is it possible that for all $n>0$, there are elements $a_{1n}, a_{2n},\ldots,a_{nn}$, all non-units, such that $r=a_{1n}a_{2n}\cdots a_{nn}$?

Answer 1

No, it's not possible.

If $R$ is a noetherian domain then there are no non-zero elements having arbitrary long factorizations.

Let $r\in R$ be a non-zero non-unit, and $\mathfrak p_1,\dots,\mathfrak p_m$ the height one primes of $R$ containing $r$. Suppose that some non-unit $r'\in R$ divides $r$. Then there is $j\in\{1,\dots,m\}$ such that $r'\in\mathfrak p_j$. If $r=r_1\cdots r_n$ with $r_i$ non-units and $n$ arbitrarily large, then there is $j\in\{1,\dots,m\}$ such that $r\in\bigcap_{k\ge1}\mathfrak p_j^k=(0)$, a contradiction.