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
.
Proof
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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