# Factorization in non-Noetherian integral domain

Let $A$ be an integral domain, suppose that $a\in A$ factors in a product of irreducible elements, $a=a_1a_2\cdots a_n$.

If $a$ factors also as $a=b_1b_2\cdots b_m$ can we conclude that every $b_i$ factors as a product of irreducibles?

I suspect that this is false, but I can't construct a simple counterexample and the standard examples of non-Noetherian integral domains don't seem to work

• Let $x$ be irreducible but not prime. Then there is a product $b_1b_2$ such that $x$ divides $b_1b_2$ but divides neither, i.e. $xy = b_1b_2$ for some $y$, but $x$ divides neither factor. Taking $a = xy$ might lead somewhere, maybe... – Dirk Apr 11 '17 at 13:03

To produce a counterexample, you can just force it.

Let $A$ be the ring defined by

$$A = \mathbb{Z}[x,y,t_1,t_2,t_3,...]/I$$

where

$$I = (xy - t_1^2,\,t_1 - t_2^2,\,t_2 - t_3^2,\,t_3-t_4^2,\,...)$$

and let $a = xy.$

Then $a$ is a product of the irreducible elements $x,y$.

But also $a = t_1^2,\,$ and $\,t_1$ is not a product of irreducibles.