Why exactly is the unique factorization of elements into irreducibles a natural thing to look for? Of course, it's true in $\mathbb{Z}$ and we'd like to see where else it is true; also, regardless of whether something is natural or not, studying it extends our knowledge of mathematics, which is always good. But the unique factorization of elements - being specifically a question of elements - seems completely counter to the category theory philosophy of characterizing structure via the maps between objects rather than their elements. Indeed, I feel like unique factorization of ideals into prime ideals is less a generalization of unique factorization of elements into irreducibles than the latter is a messier, unnatural special case of the former, a "purer" question (ideals, being the kernels of maps between rings, I feel meet my criteria for being a category-theoretically acceptable thing to look at). Certainly, the common theme in algebra (and most of mathematics) is to look at the decomposition of structures into simpler structures - but quite rarely at actual elements.

Now, for nice cases like rings of integers in number fields, we can characterize being a UFD in terms of the class group and other nice structures and not have to mess around with ring elements, but looking at the Wikipedia page on UFDs and the alternative characterizations they list for general rings, they all appear to depend on ring elements in some way (the link to "divisor theory" is broken, and I don't know what that is, so if someone could explain it and/or point me to some resources for it, it'd be much appreciated).

Sorry about the rambling question, but I was wondering if anyone had any thoughts or comments? Is "being a UFD" equivalent to any property which can be stated entirely without reference to ring elements? Should we care whether it is or not?

EDIT: Here's a more straightforward way of saying what I was trying to get at: The structure theorem for f.g. modules over a PID, the Artin-Wedderburn theorem, the Jordan-Holder theorem - these are structural decompositions. Unique factorization of elements is not, because elements are not a structure. My feeling is that this makes it a fundamentally less natural question, and I ask whether being a UFD can be characterized in purely structural terms, which would redeem the concept somewhat, I think.

  • $\begingroup$ You can view elements in a ring R as category-theoretic objects: they are maps of sets from the one-point set to the underlying set of R, and even as maps of abelian groups from the integers to (R, +). $\endgroup$ Dec 5, 2009 at 1:59
  • $\begingroup$ True, we can create structures in bijection with R, and put R's ring structure back on them (though my understanding is that we have to explicitly tell Hom_Sets({*},R) and Hom_Ab(Z,R) to have R's ring structure), which I would agree is a somewhat better situation because now we are using arrows instead of "elements". However, I don't think that just because we can do this without having to explicitly reference R's underlying set fundamentally changes the issue - $\endgroup$ Dec 5, 2009 at 2:42
  • $\begingroup$ now we effectively have a relabeling of R, and the question of whether a map - an element of the Hom-set - factors uniquely as a "product" (again, having to give the maps R's ring structure) of "irreducible" maps is still much more ungainly, in my opinion, than whether ideals factor as a product of prime ideals, which is in the proud tradition of other structural decomposition questions such as whether a ring is semi-simple. $\endgroup$ Dec 5, 2009 at 2:43
  • $\begingroup$ Another way of viewing elements of R category-theoretically is to use the identification of R with End_{R-mod}(R). This way multiplication in R agrees with category-theoretic composition. $\endgroup$ Dec 5, 2009 at 2:48
  • $\begingroup$ About divisor theory: a Google search turns up the PlanetMath page planetmath.org/encyclopedia/DivisorTheory.html and a book by Harold M. Edwards titled "Divisor Theory" books.google.com/books?id=TKRmT5L4CNcC $\endgroup$ Dec 5, 2009 at 2:52

3 Answers 3


This is sort of an anti-answer, but: my instinct is that ZC is taking the categorical perspective too far.

To start philosophically, I think it is quite appropriate to, when given a mathematical structure like a topological space or a ring -- i.e., a set with additional structure -- refrain from inquiring as to exactly what sort of object any element of the structure is. There is a famous essay "What numbers could not be" by Paul Benacerraf, in which he pokes fun at this idea by imagining two children who have been taught about the natural numbers by two different "militant logicists". Their education proceeds well until one day they get into an argument as to whether 3 is an element of 17. (The writing is very nice here and unusually witty for an essay on mathematical philosophy: the names of the children are Ernie and Johnny, an allusion to Zermelo and von Neumann, who had rival definitions of ordinal numbers.) The point of course is that it's a silly question, and a mathematically useless one: it won't help you to understand the structure of the natural numbers any better.

On the other hand, to deny that a set is an essential part of certain (indeed, many) mathematical structures seems to be carrying things too far. As far as I know, it is not one of the goals of category theory to eliminate sets (though one occasionally hears vague mutterings in this direction, I have never seen an explanation of this or, more critically, of the need for this).

Coming back to rings, it seems to me that very few properties of rings can be expressed without elements. You also seem to implicitly suggest that it is "more structural" to think about things in terms of ideals than elements. Can you explain this? It would seem that speaking of ideals involves more set theoretic machinery than speaking about elements: this is certainly true in model theory in the language of rings.

It seems wrong to say that unique factorization of ideals into primes is a "generalization" of unique factorization of elements, since neither property implies the other.

Finally a positive remark: it sounds like you might like the characterization of UFDs as Krull domains with trivial divisor class group.

  • $\begingroup$ I think your positive remark is the perfect answer in the spirit of the question. I must confess I don't know whats the definition of divisor class group of a Krull domain; I assume that it has to be analog to the class group of a Dedekind domain but I'd like to see if there is a short explanation of its definition. $\endgroup$ Dec 5, 2009 at 6:03
  • $\begingroup$ All right, then: see math.uga.edu/~pete/classgroup.pdf (It is one of the two ways of generalizing the class group of a Dedekind domain. The other is via the Picard group.) $\endgroup$ Dec 5, 2009 at 6:12
  • $\begingroup$ At least for commutative rings, we can study them entirely without looking directly at the elements of the ring by looking at the category of modules over the ring. There's a nice result about how how we can recover R from R-mod up to isomorphism. $\endgroup$ Dec 5, 2009 at 6:36
  • $\begingroup$ I really appreciate your insightful and constructive criticism, and of course your provided characterization of UFDs - I'm not at a level where I can understand it (I'm still working through the first half of both Atiyah-Macdonald and Marcus's Number Fields), but it's precisely the kind of answer I was hoping for. People have often told me that ideals factoring into prime ideals is "the correct generalization" of unique factorization of elements, though now that I see that in general neither implies the other, I suppose they probably meant that only for rings of integers in number fields. $\endgroup$ Dec 5, 2009 at 6:57
  • $\begingroup$ For ideals being more "structural", I was thinking along the lines of ideals being modules over their ring, as well as being kernels of homomorpisms, which allows us to specify them with arrows instead of having to name a set. Also, though I've never done any model theory or universal algebra, from what I'd heard they prefer structures which can in one way or another be characterized completely by equations and do not require axioms for the existence of identities, negatives, etc. $\endgroup$ Dec 5, 2009 at 6:59

First of all, the ubiquity of category theory in algebra is fairly recent, at least given how long people have been working on algebra (not even including elementary number theory). Much of algebraic number theory was developed in the mid-19th century in attempts to prove Fermat's Last Theorem. Since category theory would not show up for another century, mathematicians like Kummer and Dedekind had little reason to think in those terms. The notion of class group showed up as an obstruction to Kummer's attempted proof of Fermat's Last Theorem, which assumed that all the cyclotomic fields $\mathbb{Q}(\zeta_p)$ had class number 1 (or at least prime to $p$). It's hard to see even what form Kummer's arguments would take if phrased in the language of factorization of ideals. I think that when flaws in Kummer's arguments were exposed, mathematicians realized that factorization of ideals behaves much better than factorization of elements. But for a mid-19th century mathematician, it must have felt a lot more natural to try to factor elements than ideals; they only studied the latter because the former usually fails. Now we understand that being a UFD (class number 1) is simply the nicest case, and the class group in general is the obstruction.

  • $\begingroup$ My understanding is that Kummer was much more concerned about higher-order reciprocity laws than he was about Fermat's Last Theorem. $\endgroup$ Dec 5, 2009 at 2:49
  • $\begingroup$ Indeed. It was Kummer who pointed out the flaw in Lame's attempted proof, which assumed the cyclotomic integers where a UFD $\endgroup$
    – Ben Webster
    Dec 5, 2009 at 5:17

Not an answer, actually it is more of a question.

Since we are talking about UFdomains I'll assume that my rings are domains. Let $R$ be a commutative domain with $1$. Notice that the notion of principal ideal can be defined without talking about the elements in $R$.

It is the following true?

$R$ is a UFD if and only if $R$ satisfies ACC for principal ideals and every prime ideal $P$ with $ht(P)=1$ is principal.

if the answer to the above is yes, this gives a characterization of UFDs that does not talk about elements.

  • $\begingroup$ I'm not sure about the characterization that you suggest (although I probably should know...) I do know that there is a similar famous characterization due to Kaplansky: a domain is a UFD iff each nonzero prime ideal contains a principal prime ideal. I disagree though that my characterization (and yours, if it's true) "does not talk about elements": what does it mean for an ideal to be principal?? $\endgroup$ Dec 5, 2009 at 5:20
  • $\begingroup$ @pete: $J \leq R$ is principal iff it is $R$-isomorphic to $R$. (thats why I insisted in domains) I also suspect of my "characterization", but I remember thinking about this a couple of years ago, and I could not came up with an example of a non-UFD domain with ACC on principal ideals and such that primes of height are principal. I'm sure that this can be found somewhere in Matsumura. $\endgroup$ Dec 5, 2009 at 5:33
  • $\begingroup$ @GM: Oh, so we're allowed to talk about modules? I still don't understand the rules of the game. By the way, let R be a commutative domain with WHAT? $\endgroup$ Dec 5, 2009 at 5:59
  • $\begingroup$ @pete: Sorry did not see the rules you mention about not talking about modules in commutative algebra questions about rings(it's going to be hard to give answers with that rule in hand but I'll keep it in mind) When you ask "commutative domain with WHAT?" is it to make the point that I'm using elements of the ring in my answer? Just in case that was not the reason what I mean is a commutative ring with unity. $\endgroup$ Dec 5, 2009 at 6:19
  • $\begingroup$ Yes, it was a rhetorical question. :) $\endgroup$ Dec 5, 2009 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.