Talk:Unique factorization domain

message: In http://userweb.cs.utexas.edu/users/EWD/transcriptions/EWD09xx/EWD993.html Dijkstra pleads for defining a product of zero factors to have the value one (or more generally the multiplicative identity element) in order to reduce the need for case analysis in proofs. HenningThielemann (talk) 19:49, 3 July 2010 (UTC)
thread_title:  Product of zero factors 

message: I corrected the example K[X,Y]/(Y^2 - X^2 + 1), which was described as a non-UFD. It is a UFD. With this correction, however, the exposition is awkward. It claims that most factor rings are not UFD's and then gives an example of one which is a UFD. Perhaps someone should write up a nice (and correct) counterexample.
:Gary Kennedy 18:18, 26 Apr 2005 (UTC)
thread_title: Fixing an example

message: Factorial ring redirects here, but there's no mention in the article. Could someone who knows about this either add a definition or remove the redirect?
:A factorial ring is a UFD.  I'll add this to the article if someone else has not already. Shawn M. O'Hare 15:34, 5 April 2006 (UTC)
replies: text: And I see someone already did. Shawn M. O'Hare 15:36, 5 April 2006 (UTC)
thread_title:  Factorial ring 

message: Let R be any commutative ring. Then R[X,Y,Z,W]/(XY-ZW) is not a UFD. It is clear that X, Y, Z, and W are all irreducibles, so the element XY=ZW has two factorizations into irreducible elements.
This proof is incorrect. (Though, the statement is true). X,Y,Z,W are irreducible in R[X,Y,Z,W],
but one has to prove that X+R[X,Y,Z,W](XY-ZW) is irreducible in R[X,Y,Z,W]/(XY-ZW), which is
more difficult.
:You're saying that this proof is lacking evidence at the statement "X,Y,Z,W are irreducible in R[X,Y,Z,W]"?--Rschwieb 02:16, 29 June 2006 (UTC)
thread_title: Fix this remark (TODO) 

message: Added the following two properties of UFDs (both of which are well-known):  Any UFD is integrally closed, and a domain R is a UFD if and only if every nonzero prime ideal of R has a nonzero prime element.192.236.44.130 00:50, 14 August 2007 (UTC)
thread_title: Added some properties

message: Fields have irreducibles? How can you write 2 as a product of irreducibles in, say, the rationals? --68.161.152.76 (talk) 07:19, 5 March 2008 (UTC)
replies: text: Fields don't have irreducibles. 2 is a unit of Q. Units in unique factorization domains can't be written as a product of irreducibles (except for 1, which could be considered as the empty product). --Zundark (talk) 08:24, 5 March 2008 (UTC)
replies: text: But they can (trivially) be written as a product of the irreducible 1 and a unit, as is allowed in the other products. Septentrionalis PMAnderson 23:31, 12 August 2008 (UTC)
replies: text: 1 isn't an irreducible, it's a unit. The statement of factorization into irreducibles either explicitly excludes units (as in the article) or allows units as associates of the empty product of irreducibles (1). Algebraist 23:33, 12 August 2008 (UTC)
thread_title:  "unique factorization domains ⊃ principal ideal domains ⊃ Euclidean domains ⊃ fields" 

message: "The ring of functions in a fixed number of complex variables holomorphic at the origin is a UFD."
I suppose this is fairly straightforward, but do we have a reference with a proof?
I'm think about adding
"The ring of holomorphic functions in a single complex variable..."
but I'm not sure if this is an example or a counter-example!  We have the Weierstrass factorization theorem, but I think the factorization might not be unique.  Anybody know for sure?
Baccala@freesoft.org (talk) 05:01, 30 May 2008 (UTC)
replies: text: There's no factorisation into irreducibles here. The function sin has infinitely many pairwise non-associated irreducible factors (to wit (z-nπ)). Algebraist 13:08, 30 May 2008 (UTC)
replies: text: Good point, and I guess that's the answer to my question, but why does the factorization have to be finite?  I know that's the way the article is worded, and that's the standard definition in the literature (I just looked at Lang's book) but what's wrong with requiring just a bijection between the irreducibles, finite or not? Baccala@freesoft.org (talk) 19:51, 30 May 2008 (UTC)
replies: text: So you want unique factorization to mean that each element (up to associates) is uniquely determined by the powers of irreducibles that divide it? That's an interesting idea, and one I haven't seen before. (of course, it has no place on WP until someone publishes it) Algebraist 20:34, 30 May 2008 (UTC)
replies: text: Well, I'm adding this as a counter-example, using (basically) your explaination. Baccala@freesoft.org (talk) 02:16, 31 May 2008 (UTC)
replies: text: Infinite factorizations are not well-defined without convergence properties. Septentrionalis PMAnderson 23:32, 12 August 2008 (UTC)

text:  Is it correct, that the formal power series over a field (or even a PID) constitute a UFD? I mean, the holomorphic functions can be regarded as a subring of CX. Now, doesn't the same argument as with the sinus work somehow?  —Preceding unsigned comment added by 84.137.53.18 (talk) 10:43, 6 August 2010 (UTC)
thread_title:  Rings of holomorphic functions 

message: The phrase "not necessarily exhaustive" is obscure to me.  What is the intended meaning?  
Plclark (talk) 18:08, 6 July 2008 (UTC)
replies: text: I suppose it's supposed to mean there are interesting classes of rings it could contain but doesn't. For example, you could have Noetherian domains between integral domains and UFDs. Algebraist 23:26, 12 August 2008 (UTC)
replies: text: It's not quite right as it stands.  To say that UFDs are "characterized by the following (not necessarily exhaustive) chain" suggests that there is only one class of rings properly between integral domains and principal ideal domains and that that is the class of UFDs.  But that's quite untrue, consider for example the natural class of Dedekind domains.  Richard Pinch (talk) 20:49, 31 August 2008 (UTC)
thread_title:  Not necessarily exhaustive 

message: The article asserts, without explanation, that the ring can be "grade[d] by degree" such that, if X is reducible, then X is a product of two factors of the form listed, but this is not obvious. For example, take R=F_2[x]/(x^2), then X=(xX+1)(xX^2+X). Granted, xX+1 is a unit, so it doesn't prove the reducibility of X, but without clarification on what is meant by the degree of the element of R[X,Y,Z,W]/(XY-ZW), it's not clear whether this is a counterexample the the idea that the degree in mind has the desired properties.
Assuming that my suspicion that there is no such definition of degree that works in the general case is correct, the proof can be fixed without too much trouble (and if my suspicion is incorrect, I think it would be a good idea to clarify what is meant by degree): If R is not an integral domain, then R[X,Y,Z,W]/(XY-ZW) is not an integral domain either (the zero-divisors in R are sent to zero-divisors in R[X,Y,Z,W]/(XY-ZW) under the canonical homomorphism) and therefore is not a UFD. So assume R is an integral domain, in this case, we can define the degree of an element of R[X,Y,Z,W]/(XY-ZW) to be the minimal degree of all the elements in the corresponding equivalence class in R[X,Y,Z,W], then the degree of the product of any two elements is the sum of the degrees of the factors.
This last sentence I can see is true given that XY-ZW is homogeneous and prime in R[X,Y,Z,W], and given that R is an integral domain. I can see that XY-ZW is prime in R[X,Y,Z,W] since, if F is the factor field of R[Y,Z,W], then X-(ZW/Y) is prime in F[X], and Y is prime in R[X,Y,Z,W] so that if X-(ZW/Y) divides an element of R[X,Y,Z,W] in F[X], then XY-ZW divides it in R[X,Y,Z,W], though there may be a simpler way to show that XY-ZW is prime. 67.188.193.108 (talk) 02:09, 31 May 2011 (UTC)
thread_title:  R[X,Y,Z,W]/(XY-ZW) Counterexample 

message: Every finite domain is a field, which the polynomial ring over a finite field clearly is not. The article could be confusing UFDs with unique factorization rings. ᛭ LokiClock (talk) 10:11, 23 September 2012 (UTC)
replies: text: Nevermind, misunderstood free generation. ᛭ LokiClock (talk) 22:22, 25 September 2012 (UTC)
thread_title:  Polynomial rings over UFDs are not UFDs. 

message: Take a unique factorization domain R such that the only unit in R is 1 and assume a fixed total ordering ≤ on the set of primes in R. Then the factorization into primes (put in order using ≤) is unique on the nose.
In general, define a category C whose objects are the elements of R and whose morphisms are given by divisibility (a preorder). (0 does not uniquely divide itself, so one can also define C to have the endomorphisms of 0 correspond to the elements of R so that C is no longer a preorder.) Isomorphic objects of C correspond to associate elements of R, and C is skeletal if and only if the unit group of R is trivial.
A domain with trivial unit group necessarily has characteristic 2, because –1 is a unit. GeoffreyT2000 (talk) 23:57, 24 May 2015 (UTC)
thread_title:  Trivial unit group 

message: Somebody (not myself) has noted that there is a mistake in the 'Non-examples' section concerning quadratic integer rings and Heegner numbers, however they noted this in parentheses in the article. I have added a clarification tag and am also mentioning it here on the talk page. If anybody in the know could resolve this point, it would be appreciated. Joel Brennan (talk) 17:11, 14 May 2019 (UTC)
thread_title:  Mistake in Non-examples section 

message: I've recently had some trouble with edit warring myself, so this is me trying to do a good deed and help out here.  I agree with Joel's edit removing "non-zero".  It's the lead of a different article and is just serving to remind people what an integral domain is; we don't need to get pedantic about edge cases.  If you really have to point this out, "nontrivial" would probably be best.  The latest "1≠0" (needs spaces anyway) is unfortunately probably worse since it will just look utterly baffling to someone who doesn't understand what the shorthand refers to. –Deacon Vorbis (carbon • videos) 16:49, 16 April 2020 (UTC)
replies: text: Dammit, fixing ping to . –Deacon Vorbis (carbon • videos) 16:50, 16 April 2020 (UTC)
replies: text:  Agreed. Feel free to make the edit you suggested. Joel Brennan (talk) 16:53, 16 April 2020 (UTC)
replies: text: Well, mathematically spending, I view the “nonzero ring” requirement is a key assumption; it is equivalent to saying the zero ideal is proper and “proper” is an important part of the definition of a prime ideal. Unfortunately (or fortunately), the leading paragraph is often quoted like by Google and such; thus, the accuracy is rather important: when Wikipedia is accused for an inaccuracy, that accusation often refers to the leading section. I think “nontrivial” can work here (will make that change). —- Taku (talk) 17:12, 16 April 2020 (UTC)

text: I have also rephrased the definition to avoid ambiguity that a product there might be infinite. (Saying “a product of some things” always leaves a possibility of infinite product). Mathematically, a good definition is “a commutative ring in which a product of a finite number of nonzero elements is nonzero”. But I can use this nice definition only if I am writing a textbook on algebra... —- Taku (talk) 17:28, 16 April 2020 (UTC)
replies: text:  Infinite products do not make sense (there is no way to define them) unless there is a notion of limit (e.g. metric spaces/topological spaces/topological groups etc.). In particular, groups and rings do not have a notion of limit, so infinite products do not make sense, and so there is no need to say that products are finite. Joel Brennan (talk) 12:07, 17 April 2020 (UTC)
replies: text: An infinite product may make sense in a topological ring, for instant; so it’s usually a better writing style not to leave such an “ambiguity”. —- Taku (talk) 12:49, 17 April 2020 (UTC)
thread_title:  Non-zero