Infinite distinct factorizations into irreducibles for an element Consider the factorization into irreducibles of $6$ in $\mathbb{Z}[\sqrt{-5}]$. We have $6=2 \times 3$ and  $6=(1+\sqrt{-5}) \times (1-\sqrt{-5})$, i.e. $2$ distinct factorizations. And,
 $$6^2=3 \times 3\times2\times2$$
$$=(1+\sqrt{-5}) \times (1-\sqrt{-5}) \times (1+\sqrt{-5}) \times (1-\sqrt{-5})$$
$$=(1+\sqrt{-5}) \times (1-\sqrt{-5})\times3\times2.$$
More generally, $6^n$ will have $n+1$ distinct factorizations into irreducibles in $\mathbb{Z}[\sqrt{-5}]$ by a simple combinatorial argument. But, can we construct a ring in which there exists an element that has an infinite number of distinct factorizations into irreducibles? To make life harder, can we construct an extension of $\mathbb{Z}$ in which this happens? I have been thinking about this for a while and have managed to find no foothold.. Any help is appreciated.

If you are only interested in behaviour in the ring of integers of a number field (such as $\mathbb{Z}[\sqrt{-5}]$) then you will never get infinitely many different factorisations of an element.
These different factorisations come from reordering the (finitely many) prime ideals in the unique factorisation of the ideal generated by your element.
