Talk:Unique factorization

could be a good article

Right now this article only discusses unique factorization of integers - and that's definitely the most important sense of the words "unique factorization". However, there are other mathematical objects for which unique factorization is an important concept: polynomial rings and rings of integers in number fields (such as $\mathbb {Z} [i]$) come immediately to mind. The latter concept, which began to be investigated in relation to early attempts to prove Fermat's Last Theorem, eventually led to the definition of ideals in rings. Anyway, consider this an invitation to be bold and expand the current article. - Greg Martin 22:19, 29 April 2007 (CDT)

If I remember right, if you extend the integers by including ${\sqrt {-3}}$ you get a ring which does not have unique prime factorization. --Catherine Woodgold 20:42, 30 April 2007 (CDT)

I'm not sure whether that's true, because as it turns out $\mathbb {Z} [{\sqrt {-3}}]$ is a little less natural a ring than $\mathbb {Z} {\big [}(1+{\sqrt {-3}})/2{\big ]}$. It's definitely true that unique factorization fails in $\mathbb {Z} [{\sqrt {-5}}]$, the canonical example being $2\times 3=(1+{\sqrt {-5}})(1-{\sqrt {-5}})$. - Greg Martin 13:56, 8 May 2007 (CDT)

induction?

Note that the proof currently given is not formally a proof by induction: instead it uses the phrase "Continuing in this way". We should choose one or the other. - Greg Martin 22:21, 29 April 2007 (CDT)

Ambiguity

This first sentence looks ambiguous to me: "In mathematics, the unique factorization theorem, also known as the fundamental theorem of arithmetic states that every integer can be expressed as a product of prime numbers in essentially only one way." It could mean that no integer can be so expressed in (essentially) more than one way. Or, it could mean that every integer can be so expressed in one way and (essentially) only in that one way. I suggest rewording it so that it's very clear which of the two is meant. In other words: is one thing being proven here (that there aren't two significantly different prime factorizations for the same integer), or are two things being proven (what I just said, and also that there is at least one prime factorization for each integer)? I see similar ambiguity on this point on other web pages on this topic. Some might be clear about what they're stating and proving but not clear about which version is labelled the fundamental theorem of arithmetic, for example. Maybe someone could look up a primary source on the origin of the term "fundamental theorem of arithmetic". --Catherine Woodgold 08:00, 7 May 2007 (CDT)

There are indeed two important facts here: first, that every number can be expressed as a product of primes; second, that this expression is essentially (i.e., up to reordering) unique. The second statement is the more meaty of the two. In my mind, both parts are included under the idea "fundamental theorem of arithmetic", but I can't say whether that's historically accurate. (Notice that in the statement of the result, we should say "number" rather than "integer", since negative integers and 0 cannot be expressed as products of primes. As usual the number 1 is a sticky case, although an "empty product" with no factors equals 1 by convention.) - Greg Martin 14:01, 8 May 2007 (CDT)

The second is the more meaty. That's exactly why it's called the unique factorization theorem. It is also exactly why it is misleading to cite this theorem when existence rather than uniqueness is what matters. That's how the approved version of the prime number article is misleading.

I agree that this way of stating it is confusing and it should be rephrased.

I also don't think that it's mere convention that the empty product is 1, but not long ago I found out that some respectable mathematicians are confused about this. Michael Hardy 17:37, 8 May 2007 (CDT)

I had never thought about it, but I agree that citing "unique factorization" when existence is meant is misleading. I have always considered the two words in "unique factorization" to each designate one part of the result -- factorization entails having at least one factorization, and unique entails having only the one. I think this is also a valid interpretation, but I will try to always use fundamental theorem instead when I mean existence. Barry R. Smith 11:21, 6 April 2008 (CDT)

Proof is too advanced

An inductive proof is (tragically) far to difficult for the general audience -- even the "mathematical induction" page has had trouble trying to find an example that the general audience would understand. I moved the proof to the "advanced" subpage I just created. I did mention the "failure" of unique factorization in more advanced contexts in the history section. We could give an example on the advanced page, I suppose, but is it more appropriate just to include this in the "unique factorization domains" page?Barry R. Smith 11:24, 6 April 2008 (CDT)

I dunno about it being too difficult; I think the notation (with which I am unfamiliar) seemed like the only barrier. Of course, I'm not exactly a 'typical' reader!

I think an example of the failure might be appropriate on the /Advanced subpage - this is, after all, the article about "unique factorization". J. Noel Chiappa 09:17, 8 April 2008 (CDT)

I am mostly concerned by the use of mathematical induction. However, as written, the existence part of the proof doesn't make it too obvious that induction is used. On the one hand, this style is technically less precise, but probably more readable to the average reader. As it is currently on the advanced page, perhaps the more standard "induction" style of base case, induction hypothesis, induction step is preferable.Barry R. Smith 20:24, 8 April 2008 (CDT)