Talk:Unique factorization ___domain
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 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.
- Also, please add comments to the bottom of the page, not the top.
- 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 , 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.
![{\displaystyle K[X,Y]/(Y^{2}-X^{2}+1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b41afdaff80de6f667f21574f5494642298c598b)
- 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)
question about quadratic ring extensions
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 some properties
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)
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?
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?