A unique factorization domain is a specific type of integral domain, and can be characterized by the following (not necessarily exhaustive) chain of class inclusions:
Formally, a unique factorization domain is defined to be an integral domainR in which every non-zero non-unitx of R can be written as a product of irreducible elements of R:
x = p1p2 ... pn
and this representation is unique in the following sense: if q1,...,qm are irreducible elements of R such that
x = q1q2 ... qm,
then m = n and there exists a bijective map φ : {1,...,n} -> {1,...,n} such that pi is associated to qφ(i) for i = 1, ..., n.
The uniqueness part is sometimes hard to verify, which is why the following equivalent definition is useful: a unique factorization domain is an integral domain R in which every non-zero non-unit can be written as a product of prime elements of R.
Examples
Most rings familiar from elementary mathematics are UFDs:
If R is a UFD, then so is Rx, the ring of polynomials with coefficients in R. A special case of this, due to the above, is that the polynomial ring over any field is a UFD.
The ring of functions in a fixed number of complex variables holomorphic at the origin is a UFD.
By induction one can show that the polynomial rings ZX1, ..., Xn as well as KX1, ..., Xn (K a field) are UFDs. (Any polynomial ring with more than one variable is an example of a UFD that is not a principal ideal domain.)
Counterexamples
The ring of all complex numbers of the form , where a and b are integers. Then 6 factors as both (2)(3) and as . These truly are different factorizations, because the only units in this ring are 1 and −1; thus, none of 2, 3, , and are associate. It is not hard to show that all four factors are irreducible as well, though this may not be obvious. See also algebraic integer.
Most factor rings of a polynomial ring are not UFDs. Here is an example:
Let R be any commutative ring. Then RX,Y,Z,W / (XY − ZW) is not a UFD. The proof is in two parts.
First we must show X, Y, Z, and W are all irreducible. Grade RX,Y,Z,W / (XY − ZW) by degree. Assume for a contradiction that X has a factorization into two non-zero non-units. Since it is degree one, the two factors must be a degree one element αX + βY + γZ + δW and a degree zero element r. This gives X = rαX + rβY + rγZ + rδW. In RX,Y,Z,W, then, the degree one element (rα − 1)X + rβY + rγZ + rδW must be an element of the ideal (XY − ZW), but the non-zero elements of that ideal are degree two and higher. Consequently, (rα − 1)X + rβY + rγZ + rδW must be zero in RX,Y,Z,W. That implies that rα = 1, so r is a unit, which is a contradiction. Y, Z, and W are irreducible by the same argument.
Next, the element XY equals the element ZW because of the relation XY − ZW = 0. That means that XY and ZW are two different factorizations of the same element into irreducibles, so RX,Y,Z,W / (XY − ZW) is not a UFD.
The ring of holomorphic functions in a single complex variable is not a UFD, since there exist holomorphic functions with an infinity of zeros, and thus an infinity of irreducible factors, while a UFD factorization must be finite, e.g.:
Properties
Some concepts defined for integers can be generalized to UFDs:
In UFDs, every irreducible element is prime. (In any integral domain, every prime element is irreducible, but the converse does not always hold.) Note that this has a partial converse: any Noetherian domain is a UFD iff every irreducible element is prime (this is one proof of the implication PID UFD).
Any two (or finitely many) elements of a UFD have a greatest common divisor and a least common multiple. Here, a greatest common divisor of a and b is an element d which divides both a and b, and such that every other common divisor of a and b divides d. All greatest common divisors of a and b are associated.
An integral domain is a UFD if and only if the ascending chain condition holds for principal ideals, and any two elements of A have a least common multiple.