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In my course notes for an undergraduate course "Algebra I", I wrote at the point when I'm introducing the notion of divisibility in rings (in a section on unique factorization):

We want to study factorization in arbitrary integral domains$^1$,...

$^1$ (footnote) The study of factorization in arbitrary commutative rings that are not integral domains, is in principle possible, but leads to rather exotic behavior and is therefore less interesting.

I have to confess that I didn't think too seriously about this sentence when I wrote it, but a student asked me whether that doesn't make it more interesting instead of less interesting.

I couldn't really give a sensible answer, hence my question:

Have there been serious attempts of a study of divisibility and/or factorization in arbitrary commutative rings with zero divisors, or is that simply something that doesn't really make sense?

Of course there are immediate problems even with the smallest examples such as $\mathbb{Z}/6$, e.g. $$ 2 = 2 \cdot 4 = 2 \cdot 4 \cdot 4 = \dots $$ but I could imagine that there might be formal ways to overcome these obvious ambiguities when talking about uniqueness of factorization.

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    $\begingroup$ Do you mean factorization into prime elements or into irreducibles ? $\endgroup$
    – Ralph
    Jul 11, 2012 at 20:35
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    $\begingroup$ Anderson and Valdes-Leon's "Factorization in commutative rings with zero divisors" might be a place to start: projecteuclid.org/DPubS/Repository/1.0/… $\endgroup$
    – Guntram
    Jul 11, 2012 at 20:43
  • $\begingroup$ @Ralph: For integral domains, I would say "irreducibles", but that now seems to be part of the question... $\endgroup$
    – Tom De Medts
    Jul 11, 2012 at 20:48
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    $\begingroup$ Consider factorization in the ring $C(\mathbb{R})$ of continuous real-valued functions $\mathbb{R} \to \mathbb{R}$... $\endgroup$
    – Qiaochu Yuan
    Jul 11, 2012 at 23:47
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    $\begingroup$ Dear Tom, The ring $\mathbb Z/6$ is the product of two integral domains, in fact of two fields, i.e. $\mathbb Z/2 \times \mathbb Z/3$, so factorization in this ring is pretty easily understandable. In my experience, such questions are best understood by moving to the framework of algebraic geometry (by taking Spec of the ring in question), and using tools like Picard groups, which make sense in a great degree of generality. Regards, Matthew $\endgroup$
    – Emerton
    Jul 12, 2012 at 3:13

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The biggest problem complicating factorization when there are zero-divisors present in my experience stems from the fact that there are several ways to define "associate" that are no longer equivalent.

For instance, the reference given by Guntram provides a good introduction to the issues:

$a$ and $b$ are associates, written $a\sim b$ iff $(a)=(b)$

$a$ and $b$ are strong associates, written $a \approx b$ iff $a=\lambda b$ for some $\lambda \in U(R)$

$a$ and $b$ are very strong associates, written $a\cong b$ iff $(a)=(b)$ and if $a=rb$ for some $r\in R$, then $r$ is a unit.

Each gives rise to a different type of irreducible along the lines of if $a=a_1 \cdots a_n$ is a factorization then $a \sim a_i$, $a \approx a_i$, or $a \cong a_i$ for some $1 \leq i \leq n$, respectively $a$ is irreducible, strongly irreducible, or very strongly irreducible.

Examples are given to show how the different types of irreducibles are no longer equivalent.

This means when looking at something like how to define a Unique Factorization Ring (instead of domain) you need to pick what type of atoms you want your factorizations to be broken into and then when rearranging up to associate, what to pick for that.

Of course, this is only part of the story. Unique Factorization Rings are pretty nice, there are weaker properties that will fail by the existence of zero-divisors, idempotent elements and nilpotent elements.

Zero-divisors mean, you can factor zero: Say $ab=0$. Well so does $0=ab=ababab=abxyzw$ or anything after it has already become zero.

Or nilpotents: $x^n=0$ implies $0=(x^{n-1})^{i}$ for all $i \geq 2$.

So do you allow zero to be factored? Maybe just don't allow zero to appear as a factor? Even that is not enough to deal with it completely.

Suppose $e=e^2$, well then $e=e^i$ for all $i\geq 1$, so does that mean your ring is not even a Bounded Factorization Ring? This is what happened in your example $\mathbb{Z}/6 \mathbb{Z}$, $4$ is idempotent.

One approach that seems particularly nice was introduced by Fletcher in two papers in Proc. Cambridge Philos. Soc. in 1969 and 1970 called Unique Factorization Rings and The structure of Unique Factorization Rings. This was U-factorization where you have essential and inessential divisors to help deal with things like nilpotents and idempotents. M. Axtell has a couple of nice papers on it too in 2002 and 2003 U-factorizations in Commutative Rings with zero-divisors and Properties of U-factorizations.

This makes something like $\mathbb{Z}/6 \mathbb{Z}$ a U-UFR, U-HFR, U-BFR, etc, but not a UFR, HFR, or a BFR.

Interesting stuff in my opinion.

EDIT: I just wanted to add a few mathematical comments to respond to your mention of Galovich's paper.

This was something that I found very, very confusing when I first began studying factorization in rings with zero-divisors. Nearly every author uses the term "irreducible" and "associate"; however, the problem is they mean different things in each case. As far as I can tell D.D. Anderson and Valdez-Leon's method of 3 choices for associate, and several choices of irreducible (irreducible, strongly irreducible, m-irreducible, very strongly irreducible) is the most general and encompasses the others in the following way:

In Galovich's paper he chooses for associate the relation $\approx$. That is $a$ and $b$ are "associate" in the Galovich sense if $a=\lambda b$ for some unit $\lambda$. His choice for "irreducible", is what Anderson and Valdez-Leon termed "very strongly irreducible."

In Bouvier's papers we have "associate" is $\sim$ as in Anderson and Valdez-Leon $a$ and $b$ are associate if $(a)=(b)$. However, the choice for "irreducible" is what is called m-irreducible, which comes from the fact that $a$ is m-irreducible if it is maximal among all principal ideals. For non-zero elements, the strength of this type of irreducible falls between strong irreducible and very strong irreducible. For zero, $0$ is m-irreducible if and only if $R$ is a field. For zero to be irreducible, strongly irreducible, or very strongly irreducible is equivalent to $R$ being a domain.

There is yet another type of atomic referred to as i-atomic which is in terms of factorization into products of principal ideals, and I believe this is equivalent to the U-factorizations studied by Axtell, and Fletcher.

As you can see this is perhaps why most people skip the factorization with zero-divisors present. Things get complicated rather quickly, but as the student alluded, perhaps this is what makes it more interesting in some ways.

This might be way more information than you want, but I just came across this site and it seems really exciting and useful and I saw your question was about what I study, and I may have gone a little overboard.

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    $\begingroup$ Thanks for your answer and for the references! While Googling for the papers, I bumped into another related paper "Unique Factorization Rings with Zero Divisors" by S. Galovich (Mathematics Magazine 51, No. 5, 1978) that also seems interesting. $\endgroup$
    – Tom De Medts
    Aug 16, 2012 at 7:57
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    $\begingroup$ This is a very good paper as well! Bouvier and Allard also have written a lot of papers on this subject; however, I have only been able to find French versions of their papers, which might pose some problems. D.D. Anderson and Frazier's paper linked by Guntram has a pretty extensive bibliography. This is unfortunately why it gets a bit complicated. Many authors use different terminology for similar content. I have run out of characters, so I will edit my answer above to add a few comments there about this. (I am new, not sure what protocol is here, sorry if this is bad form to do) $\endgroup$
    – CPM
    Aug 16, 2012 at 9:29
  • $\begingroup$ This is an excellent answer! I just took a look at Anderson and Valdes-Leon's paper and there are a couple of "gotchas" I wanted to point out in case it is helpful for future readers. First, AV define $a \cong b$ slightly differently than in this answer: they add the fact that $0 \cong 0$. But this seems like a mistake in definitions; it means their "$a$ is very strongly irreducible" only agrees with the usual def. of irreducible when $a \ne 0$. And this mistake shows up later in Theorem 2.13 (very strong irreducible implies (contd) $\endgroup$
    – Caleb Stanford
    Feb 19 at 3:43
  • $\begingroup$ m-irreducible), which holds only for $a \ne 0$. With the definition in this answer, things work out better, very strong irreducible implies m-irreducible for all $a \in R$. Second, there seems to be an error in Theorem 2.13: they say that $p$ is strongly irreducible implies that $a$ is both weakly irreducible and prime. But this is false for integral domains. That is, it is well-known that there are examples of irreducible (hence very strongly irreducible) elements that are not prime. Other than this, I think the theorem is correct. $\endgroup$
    – Caleb Stanford
    Feb 19 at 3:45
  • $\begingroup$ For reference, the theorem is that: "$a$ very strongly irreducible $\implies$ $a$ is $m$-reducible $\implies$ $a$ is strongly irreducible $\implies$ $a$ is weakly irreducible. Moreover, none of these implications can be reversed." I've posted a related answer on mathSE. $\endgroup$
    – Caleb Stanford
    Feb 19 at 3:46

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