TITLE: Infinite distinct factorizations into irreducibles for an element
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: Here's a construction similar to the one used by Math Gems, in the first
answer. Let $\mathbb Q$, $\mathbb R$ be the fields of rational and real numbers respectively and let $X$ be an indeterminate over $\mathbb R$. Consider the ring $S=\mathbb Q+X\mathbb R[X]=\{f\in \mathbb R[X]:f(0)\in \mathbb Q\}$. That such a monstrosity is an integral domain, can be easily checked, say from the reference given by Math Gems. Note that a typical element $f$ of $S$ can be written as $f=q_{0}+\sum_{i=1}^{n}r_{i}X^{i}$ where $q_{0}\in \mathbb Q$ and $r_{i}\in \mathbb R.$ So
coefficients of $X$ are real numbers. (Any (nonzero) real number can be a
coefficient of $X$!)
Notice that as $S$ is a subring of $\mathbb R[X]$ every unit of $S$ is a unit of $\mathbb
R[X]$. So $u\in S$ is a unit if and only if $u$ is of degree zero if and
only if $u\in \mathbb Q.$ Thus $f\in S$ is a nonunit if and only if $\deg (f)>0.$
Now this can be used to establish the following  
Observation A: $S$ satisfies ACC on principal ideals and so is atomic, i.e.
every nonzero nonunit is expressible as a product of at most a finite number
of irreducible elements.
Illustration: Let $f,g\in S$ such that $fS\subsetneq gS.$ Then $f=gh$ where $%
h$ is not a unit and so is of positive degree. But then in $S,$ $g$ properly
divides $f$ implies that $\deg (g)<\deg (f).$ Now as $\deg (f)$ is finite
for each $f\in S$ and so $fS\subsetneq f_{1}S\subsetneq \dots\subsetneq f_{n}S$
... must stop after a number of steps.
Having established that $S$ is an atomic domain we now proceed to show that 
Observation B: $S$ is not a UFD.
Illustration: Note that $X$ is an irreducible element or atom, being of
degree $1$, but then so are$\frac{1}{\sqrt{2}}X,$ $\sqrt{2}X$ and all  $X,$ 
$\frac{1}{\sqrt{2}}X,$ $\sqrt{2}X$ belong to $S.$ Note that $X$ and $\frac{1}{%
\sqrt{2}}X$ are not associates, since $X/\frac{1}{\sqrt{2}}X=\sqrt{2}$ which
is not in $S$ and $\frac{1}{\sqrt{2}}X/X=\frac{1}{\sqrt{2}}$ which again is
not in $S.$ In a similar fashion we show that $X$ and $\sqrt{2X}$ are not
associates.
Now note that $X^{2}=(\sqrt{2}X)(\frac{1}{\sqrt{2}}X).$ So $X\mid (\sqrt{2}%
X)(\frac{1}{\sqrt{2}}X)$ yet $X$ does not divide either of $(\sqrt{2}X),(%
\frac{1}{\sqrt{2}}X).$ Now recall the definition of a prime element it is a
nonzero nonunit $p$ such that $p\mid ab$ implies that $p\mid a$ \ or $p\mid
b.$ So, in $S,$ $X$ is not a prime. Now one of the criteria for a domain to
be a UFD is: $D$ is a UFD if and only if every nonzero nonunit of $D$ is
expressible as a product of at most a finite number of irreducible elements
and every irreducible element is a prime. As we have shown that the
irreducible element $X$ in $S$ is not a prime we have essentially
established that $S$ is not a UFD.
Observation C: In $S$, the element $X^{2}$ can be expressed in an infinite
number of ways as a product of two irreducible elements which are not
associates of $X.$
Illustration: (i) Our first choice is noting the following. There are
infinitely many distinct positive primes $p$ in $\mathbb Z,$ the ring of integers,
and for each $p$ $\sqrt{p}$ is irrational and so $X^{2}=(\sqrt{p}X)(\frac{1}{%
\sqrt{p}}X).$ Works fine using the same argument as we used in case of  $%
X^{2}=(\sqrt{2}X)(\frac{1}{\sqrt{2}}X).$ 
(ii) There is one snag with the factorization in (i) and that is,  we have
not been able to express $X^{2}$ as product of two non-associated atoms in
infinitely many ways. For, it so happens that $(\sqrt{p}X)$ and $(\frac{1}{%
\sqrt{p}}X)$ are associates as $(\sqrt{p}X=p(\frac{1}{\sqrt{p}}X)$ . \ The
problem arises from the fact that $(\sqrt{p})^{2}p$ a rational number. That
can be solved by selecting a set of distinct irrational numbers $\{\alpha
_{i}\}$ such that  that for each $i$ $\alpha _{i}^{2}$ is irrational, taking 
$\alpha _{i}=$ $\sqrt[3]{p}$ for instance where $p$ ranges over distinct
positive primes of $Z.$ In this case $X^{2}=$ $(\alpha _{i}X)(\frac{1}{%
\alpha _{i}}X)$ where  $(\alpha _{i}X)$ and $(\frac{1}{\alpha _{i}}X)$ are
irreducible elements that are not associated to each other and of course
neither of $(\alpha _{i}X)$ and $(\frac{1}{\alpha _{i}}X)$ is an associate
to $X.$
Observation C, hopefully, does answer the question. I have taken the longer
route to include the usual concepts. Now here's something that I would like
folks to think about:
Question: How can we express $X^{3}$as a product of three irreducible
elements, in $S,$ as $X^{3}=(\alpha _{1}X)(\alpha _{2}X)(\alpha _{3}X)$ in
infinitely many ways, where $X$ is not an associate of any of $\alpha _{i}X.$ One
answer is simple. Note that $pX^{3}=(\sqrt[3]{p}X)^{3}$ for any positive
prime, but the answer will be more interesting if $(\alpha _{1}X),(\alpha
_{2}X),(\alpha _{3}X)$ are mutually non associated.
OK guys I have probably overused your hospitality and probably made some mistakes too. So feel free to point out if you find any.