# Talk:Unique factorization

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 Definition:  Every positive integer can be expressed as a product of prime numbers in essentially only one way. [d] [e]
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## some clarification needed

However, and it did not apply to all whole numbers. - is and should be deleted? Also, to which numbers the Euclid result applied then? Andrey Khalyavin 08:49, 9 April 2008 (CDT)

Removed the extraneous 'and', and put in which whole numbers. Actually, Euclid didn't show that these numbers have unique factorization, but only that if a whole number is a product of distinct primes p_1 p_2 p_3...p_r, then the only primes which divide this number are the primes p_1, p_2, ..., p_r. It think saying this might be overkill, though.Barry R. Smith 12:14, 9 April 2008 (CDT)

## 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 ${\displaystyle \mathbb {Z} [i]}$) 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 ${\displaystyle {\sqrt {-3}}}$ 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 ${\displaystyle \mathbb {Z} [{\sqrt {-3}}]}$ is a little less natural a ring than ${\displaystyle \mathbb {Z} {\big [}(1+{\sqrt {-3}})/2{\big ]}}$. It's definitely true that unique factorization fails in ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$, the canonical example being ${\displaystyle 2\times 3=(1+{\sqrt {-5}})(1-{\sqrt {-5}})}$. - 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)