Unique factorization domain
A unique factorization domain is an integral domain in which an analog of the fundamental theorem of arithmetic holds. More precisely an integral domain is a unique factorization domain if for any nonzero element which is not a unit:
- can be written in the form where are (not necessarily distinct) irreducible elements in .
- This representation is unique up to units and reordering, that is if where and are all irreducibles then and there is some permutation of such that for each there is a unit such that .
One of the most significant results about unique factorization domains is that any principal ideal domain (and hence any euclidean domain) is a unique factorization domain. This automatically implies that many well-known rings are unique factorization domains including:
- The ring of integers, (so in this sense, this result is a generalization of the fundamental theorem of arithmetic)
- The Gaussian integers,
- The polynomial ring over any field .
First we note that in any principal ideal domain, , the irreducible elements are precisely the prime elements. One implication (that any prime element is irreducible) is known to be true for any integral domain. For the other direction, let be irreducible. Then as is a principal ideal domain, must be a maximal ideal. But now in a commutative ring with unity maximal ideals are prime. Thus is prime, and hence is prime.
Now let be any principal domain. First we shall show that any nonzero non-unit in can be factored into irreducibles. Assume that this is not the case. Then let be the set of all non-units in which cannot be written as a product of irreducibles. Consider any . Clearly itself cannot be irreducible, so we may write for some nonzero , neither of which are units. Now if neither nor is in , then both and can be written as a product of irreducibles, and hence so can . This is a contradiction, so at least one of and must be in . WLOG let this be . Now , so and also as is not a unit. Thus and we have proved the following proposition:
- If then there is some element such that .
So now we can construct a sequence of elements of such that (simply let ). But now as is a principal ideal domain, and hence Noetherian, this contradicts the ascending chain condition and is therefore impossible. Thus every element of can indeed be written as the product of irreducibles.
It now remains to show that such representations are unique. For any nonzero , let be the smallest integer such that can be written as the product of irreducibles (this is guaranteed to exist by the previous work). We proceed by strong induction on .
If then is a unit. So now assume that has some other factorization (clearly or this factorization would not be different from the factorization ). Let (or if ). Then we have , which implies that is a unit, a contradiction. So the satement is true for .
Now assume that we have unique factorization for any with . Consider some with . Assume that for irreducibles and (note that by the definition of , ). Now as is irreducible by the above note it must also be prime. Hence as we must have for some . Renumbering the 's if necessary, we may assume WLOG that . So now , and so for some . But is irreducible and is not a unit, so must be a unit. Plugging this back into our expressions for , we get: Now as is an integral domain we get . Letting we get , so by the inductive hypothesis (since is clearly irreducible) there is a unique factorization for . Thus and there is a permutation of and units such that: Combining this with the fact that proves that the representation of is unique and finishes the induction.
Therefore factorization into irreducibles in is unique, and therefore is a unique factorization domain.
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