Talk:Unique factorization ___domain: Difference between revisions

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== Product of zero factors ==

In http://userweb.cs.utexas.edu/users/EWD/transcriptions/EWD09xx/EWD993.html Dijkstra pleads for defining a [[empty product|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. [[User:HenningThielemann|HenningThielemann]] ([[User talk: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.

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==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. <small><span class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Tejolson|Tejolson]] ([[User talk:Tejolson|talk]] • [[Special:Contributions/Tejolson|contribs]]) 19:33, 24 March 2013 (UTC)</span></small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->

Why does the [[prime element]]s link go to integral ___domain and not [[prime number]]?

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: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). --[[User:Zundark|Zundark]] ([[User talk: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. [[User:Pmanderson|Septentrionalis]] <small>[[User talk:Pmanderson|PMAnderson]]</small> 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). [[User talk:Algebraist|Algebraist]] 23:33, 12 August 2008 (UTC)

== Rings of holomorphic functions ==

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[[User:[e-mail]|[e-mail]]] ([[User talk:[e-mail]|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)). [[User talk:Algebraist|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? [[User:[e-mail]|[e-mail]]] ([[User talk:[e-mail]|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) [[User talk:Algebraist|Algebraist]] 20:34, 30 May 2008 (UTC)

::::Well, I'm adding this as a counter-example, using (basically) your explaination. [[User:[e-mail]|[e-mail]]] ([[User talk:[e-mail]|talk]]) 02:16, 31 May 2008 (UTC)

:::::Infinite factorizations are not well-defined without convergence properties. [[User:Pmanderson|Septentrionalis]] <small>[[User talk:Pmanderson|PMAnderson]]</small> 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 C[[X]]. Now, doesn't the same argument as with the sinus work somehow? <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/84.137.53.18|84.137.53.18]] ([[User talk:84.137.53.18|talk]]) 10:43, 6 August 2010 (UTC)</span><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

== Not necessarily exhaustive ==

The phrase "not necessarily exhaustive" is obscure to me. What is the intended meaning?

[[User:Plclark|Plclark]] ([[User talk: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. [[User talk:Algebraist|Algebraist]] 23:26, 12 August 2008 (UTC)

::It's not quite right as it stands. To say that UFDs are "<u>characterized</u> 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 ___domain]]s. [[User:Richard Pinch|Richard Pinch]] ([[User talk: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 <math>X</math> is reducible, then <math>X</math> is a product of two factors of the form listed, but this is not obvious. For example, take <math>R=F_2[x]/(x^2)</math>, then <math>X=(xX+1)(xX^2+X)</math>. Granted, <math>xX+1</math> 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 <math>R[X,Y,Z,W]/(XY-ZW)</math>, 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 <math>R</math> is not an integral ___domain, then <math>R[X,Y,Z,W]/(XY-ZW)</math> is not an integral ___domain either (the zero-divisors in <math>R</math> are sent to zero-divisors in <math>R[X,Y,Z,W]/(XY-ZW)</math> under the canonical homomorphism) and therefore is not a UFD. So assume <math>R</math> is an integral ___domain, in this case, we can define the degree of an element of <math>R[X,Y,Z,W]/(XY-ZW)</math> to be the minimal degree of all the elements in the corresponding equivalence class in <math>R[X,Y,Z,W]</math>, 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 <math>XY-ZW</math> is homogeneous and prime in <math>R[X,Y,Z,W]</math>, and given that <math>R</math> is an integral ___domain. I can see that <math>XY-ZW</math> is prime in <math>R[X,Y,Z,W]</math> since, if <math>F</math> is the factor field of <math>R[Y,Z,W]</math>, then <math>X-(ZW/Y)</math> is prime in <math>F[X]</math>, and <math>Y</math> is prime in <math>R[X,Y,Z,W]</math> so that if <math>X-(ZW/Y)</math> divides an element of <math>R[X,Y,Z,W]</math> in <math>F[X]</math>, then <math>XY-ZW</math> divides it in <math>R[X,Y,Z,W]</math>, though there may be a simpler way to show that <math>XY-ZW</math> is prime. [[Special:Contributions/67.188.193.108|67.188.193.108]] ([[User talk: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. ᛭ [[User:LokiClock|LokiClock]] ([[User talk:LokiClock|talk]]) 10:11, 23 September 2012 (UTC)

: Nevermind, misunderstood free generation. ᛭ [[User:LokiClock|LokiClock]] ([[User talk: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.[[Special:Contributions/173.218.108.174|173.218.108.174]] ([[User talk: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. [[User:GeoffreyT2000|GeoffreyT2000]] ([[User talk:GeoffreyT2000|talk]]) 23:57, 24 May 2015 (UTC)

== Mistake in Non-examples section ==

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. [[User:Joel Brennan|Joel Brennan]] ([[User talk:Joel Brennan|talk]]) 17:11, 14 May 2019 (UTC)

== Non-zero ==

{{re|Joel Brennan|TakuyaMurata}} 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. &ndash;[[User:Deacon Vorbis|Deacon Vorbis]]&nbsp;([[User Talk:Deacon Vorbis|carbon]]&nbsp;&bull;&nbsp;[[Special:Contributions/Deacon Vorbis|videos]]) 16:49, 16 April 2020 (UTC)

:<small>Dammit, fixing ping to {{u|TakuyaMurata}}. &ndash;[[User:Deacon Vorbis|Deacon Vorbis]]&nbsp;([[User Talk:Deacon Vorbis|carbon]]&nbsp;&bull;&nbsp;[[Special:Contributions/Deacon Vorbis|videos]]) 16:50, 16 April 2020 (UTC)</small>

:: Agreed. Feel free to make the edit you suggested. [[User:Joel Brennan|Joel Brennan]] ([[User talk:Joel Brennan|talk]]) 16:53, 16 April 2020 (UTC)

:::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). —- [[User:TakuyaMurata|Taku]] ([[User talk:TakuyaMurata|talk]]) 17:12, 16 April 2020 (UTC)

:::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... —- [[User:TakuyaMurata|Taku]] ([[User talk:TakuyaMurata|talk]]) 17:28, 16 April 2020 (UTC)

:::: 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. [[User:Joel Brennan|Joel Brennan]] ([[User talk:Joel Brennan|talk]]) 12:07, 17 April 2020 (UTC)

:::::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”. —- [[User:TakuyaMurata|Taku]] ([[User talk:TakuyaMurata|talk]]) 12:49, 17 April 2020 (UTC)

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