# Talk:Unique factorization domain

## Product of zero factors

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)

## Edit of 30 November 2002

I removed the non-example, because the previous one wasn't a ring, so there is no hope that it could ever be a UFD.

I also reverted the statement about the uniqueness of the expression, since it is simply wrong that the p's are a permutation of the q's: 2*4 = (-2) * (-4).

I also removed the repetion of the fact that in UFD's, irreducible implies prime. AxelBoldt 23:56 Nov 30, 2002 (UTC)

## Couple Qs

Why is this significant? \sin \left( \pi \,z \right) =\sin \left( \pi \,z \right) significant? It seems to me when a zero gets in the product the whole this will become zero. The zero will happen when z and n are equal, 1-z^2/z^2. — Preceding unsigned comment added by Tejolson (talkcontribs) 19:33, 24 March 2013 (UTC)

Why does the prime elements link go to integral domain and not prime number?

Why say prime elements rather than prime numbers? Are prime elements prime numbers? Cyclotronwiki 27 April 01:33 Taipei

Because this is an algebra page, defining everything in commutative ring terms.
Please note also that we don't use computer notation * for product, ^ for exponent. You need to write
2 × 52
for that, or use TeX.
I am reverting your edits, since they really relate to something different, the fundamental theorem of arithmetic.
Charles Matthews 18:18, 26 Apr 2005 (UTC)

I'm sorry but I still feel the redirection of UF to UFD without explaining the simple concept of UF (which the user was expecting when they clicked on/searched for UF) is an error.

Does your algebra page reasoning explain why prime elements is being used instead of prime numbers or why prime elements links to integral domain? Cyclotronwiki 27 April 02:51 Taipei

Sorry, this post is about two months late, but I can answer for completeness of this section. In the integral domain article under the heading "Divisibility, prime and irreducible elements" you can find a detailed definition of what a prime element is. 'Prime numbers' and 'prime elements of an integral domain' are not the same thing. You can think of prime elements as a generalization of the idea of prime numbers to any integral domain. Indeed, prime numbers are prime elements of the ring of integers.
--Rschwieb 00:51, 27 June 2006 (UTC)

## Fixing an example

I corrected the example ${\displaystyle K[X,Y]/(Y^{2}-X^{2}+1)}$, 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)

## Factorial ring

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)
And I see someone already did. Shawn M. O'Hare 15:36, 5 April 2006 (UTC)

## Fix this remark (TODO)

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)

It's some time I have been out of school, but I heard there is only a finite number of rings of the form { a+b sqrt(d) | a,b integers} (real quadratic ring extensions I think is the term), which are also unique factorization domains. Is it true? --Samohyl Jan 19:25, 7 July 2006 (UTC)

Actually, I believe it is unknown whether there there are infinitely many quadratic rings of algebraic integers which are UFDs. It is known that there are only infinitely many such rings for d negative. In fact, there are only finitely may such negative d for which the ring of integers is half factorial--i.e. given two irreducible factorizations of any element, say x_1 x_2...x_n = y_1 y_2...y_m, then n=m (Carlitz's theorem states that any ring of algebraic integers is an HFD if and only if the class number is at most 2).
However, Gauss conjectured that there are infinitely many quadratic rings of integers that are UFDs with d>0. I haven't heard of anyone resolving this conjecture.192.236.44.130 00:21, 14 August 2007 (UTC)
It is unresolved. See class number problem ('R has class number 1' is equivalent to 'R is a UFD'). Algebraist 15:41, 21 May 2008 (UTC)

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)

## "unique factorization domains ⊃ principal ideal domains ⊃ Euclidean domains ⊃ fields"

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)

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)
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)
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)

## Rings of holomorphic functions

"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?

"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)

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)
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)
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)
Well, I'm adding this as a counter-example, using (basically) your explaination. Baccala@freesoft.org (talk) 02:16, 31 May 2008 (UTC)
Infinite factorizations are not well-defined without convergence properties. Septentrionalis PMAnderson 23:32, 12 August 2008 (UTC)
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)

## Not necessarily exhaustive

The phrase "not necessarily exhaustive" is obscure to me. What is the intended meaning? Plclark (talk) 18:08, 6 July 2008 (UTC)

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)
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)

## R[X,Y,Z,W]/(XY-ZW) Counterexample

The article asserts, without explanation, that the ring can be "grade[d] by degree" such that, if ${\displaystyle X}$ is reducible, then ${\displaystyle X}$ is a product of two factors of the form listed, but this is not obvious. For example, take ${\displaystyle R=F_{2}[x]/(x^{2})}$, then ${\displaystyle X=(xX+1)(xX^{2}+X)}$. Granted, ${\displaystyle xX+1}$ 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 ${\displaystyle R[X,Y,Z,W]/(XY-ZW)}$, 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 ${\displaystyle R}$ is not an integral domain, then ${\displaystyle R[X,Y,Z,W]/(XY-ZW)}$ is not an integral domain either (the zero-divisors in ${\displaystyle R}$ are sent to zero-divisors in ${\displaystyle R[X,Y,Z,W]/(XY-ZW)}$ under the canonical homomorphism) and therefore is not a UFD. So assume ${\displaystyle R}$ is an integral domain, in this case, we can define the degree of an element of ${\displaystyle R[X,Y,Z,W]/(XY-ZW)}$ to be the minimal degree of all the elements in the corresponding equivalence class in ${\displaystyle R[X,Y,Z,W]}$, 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 ${\displaystyle XY-ZW}$ is homogeneous and prime in ${\displaystyle R[X,Y,Z,W]}$, and given that ${\displaystyle R}$ is an integral domain. I can see that ${\displaystyle XY-ZW}$ is prime in ${\displaystyle R[X,Y,Z,W]}$ since, if ${\displaystyle F}$ is the factor field of ${\displaystyle R[Y,Z,W]}$, then ${\displaystyle X-(ZW/Y)}$ is prime in ${\displaystyle F[X]}$, and ${\displaystyle Y}$ is prime in ${\displaystyle R[X,Y,Z,W]}$ so that if ${\displaystyle X-(ZW/Y)}$ divides an element of ${\displaystyle R[X,Y,Z,W]}$ in ${\displaystyle F[X]}$, then ${\displaystyle XY-ZW}$ divides it in ${\displaystyle R[X,Y,Z,W]}$, though there may be a simpler way to show that ${\displaystyle XY-ZW}$ is prime. 67.188.193.108 (talk) 02:09, 31 May 2011 (UTC)

## Polynomial rings over UFDs are not UFDs.

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)

Nevermind, misunderstood free generation. ᛭ LokiClock (talk) 22:22, 25 September 2012 (UTC)

(2014, Oct 9): Request to add GCD Domains in the inclusion chain. Even a greedier request would be a digraph with nodes, each with a phrase - {I, I[X} is {ID, Intergrally closed Domain, GCD, UFD, Noetherian, PID, Field} the arrows of this Digraph being one-way or two-way implications, thanks.173.218.108.174 (talk) 00:27, 10 October 2014 (UTC)Svatan

## Trivial unit group

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)