Talk:Unique factorization domain
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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)
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[edit] 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)
[edit] Fixing an example
I corrected the example K[X,Y] / (Y2 − X2 + 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)
[edit] 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)
[edit] 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)
[edit] 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)
