Unique factorization ___domain

In mathematics, a unique factorization ___domain (UFD) is, roughly speaking, a commutative ring in which every element, with special exceptions, can be uniquely written as a product of prime elements, analogous to the fundamental theorem of arithmetic for the integers. UFDs are sometimes called factorial rings, following the terminology of Bourbaki.

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:

• integral domains ⊃ unique factorization domains ⊃ principal ideal domains ⊃ Euclidean domains ⊃ fields

Definition

Formally, a unique factorization ___domain is defined to be an integral ___domain R in which every non-zero non-unit x of R can be written as a product of irreducible elements of R:

x = p₁ p₂ ... p_n

and this representation is unique in the following sense: if q₁,...,q_m are irreducible elements of R such that

x = q₁ q₂ ... q_m,

then m = n and there exists a bijective map φ : {1,...,n} -> {1,...,n} such that p_i is associated to q_φ(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:

• All principal ideal domains, hence all Euclidean domains, are UFDs. In particular, the integers (also see fundamental theorem of arithmetic), the Gaussian integers and the Eisenstein integers are UFDs.
• Any field is trivially a UFD, since every non-zero element is a unit. Examples of fields include rational numbers, real numbers, and complex numbers.
• If R is a UFD, then so is R[x], 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.

Further examples of UFDs are:

• The formal power series ring K[[X₁,...,X_n]] over a field K.
• 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 Z[X₁, ..., X_n] as well as K[X₁, ..., X_n] (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 ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$  of all complex numbers of the form ${\displaystyle a+ib{\sqrt {5}}}$ , where a and b are integers. Then 6 factors as both (2)(3) and as ${\displaystyle \left(1+i{\sqrt {5}}\right)\left(1-i{\sqrt {5}}\right)}$ . These truly are different factorizations, because the only units in this ring are 1 and −1; thus, none of 2, 3, ${\displaystyle 1+i{\sqrt {5}}}$ , and ${\displaystyle 1-i{\sqrt {5}}}$  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 ${\displaystyle R}$  be any commutative ring. Then ${\displaystyle R[X,Y,Z,W]/(XY-ZW)}$  is not a UFD. The proof is in two parts.

First we must show ${\displaystyle X}$ , ${\displaystyle Y}$ , ${\displaystyle Z}$ , and ${\displaystyle W}$  are all irreducible. Grade ${\displaystyle R[X,Y,Z,W]/(XY-ZW)}$  by degree. Assume for a contradiction that ${\displaystyle X}$  has a factorization into two non-zero non-units. Since it is degree one, the two factors must be a degree one element ${\displaystyle \alpha X+\beta Y+\gamma Z+\delta W}$  and a degree zero element ${\displaystyle r}$ . This gives ${\displaystyle X=r\alpha X+r\beta Y+r\gamma Z+r\delta W}$ . In ${\displaystyle R[X,Y,Z,W]}$ , then, the degree one element ${\displaystyle (r\alpha -1)X+r\beta Y+r\gamma Z+r\delta W}$  must be an element of the ideal ${\displaystyle (XY-ZW)}$ , but the non-zero elements of that ideal are degree two and higher. Consequently, ${\displaystyle (r\alpha -1)X+r\beta Y+r\gamma Z+r\delta W}$  must be zero in ${\displaystyle R[X,Y,Z,W]}$ . That implies that ${\displaystyle r\alpha =1}$ , so ${\displaystyle r}$  is a unit, which is a contradiction. ${\displaystyle Y}$ , ${\displaystyle Z}$ , and ${\displaystyle W}$  are irreducible by the same argument.

Next, the element ${\displaystyle XY}$  equals the element ${\displaystyle ZW}$  because of the relation ${\displaystyle XY-ZW=0}$ . That means that ${\displaystyle XY}$  and ${\displaystyle ZW}$  are two different factorizations of the same element into irreducibles, so ${\displaystyle R[X,Y,Z,W]/(XY-ZW)}$  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, i.e: ${\displaystyle \sin \pi z=\pi z\prod _{n=1}^{\infty }(1-{{z^{2}} \over {n^{2}}})}$ 

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 ${\displaystyle \Rightarrow }$  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.

• Any UFD is integrally closed. In other words, if R is an integral ___domain with quotient field K, and if an element k in K is a root of a monic polynomial with coefficients in R, then k is an element of R.

Equivalent conditions for a ring to be a UFD

Under some circumstances, it is possible to give equivalent conditions for a ring to be a UFD.

• A Noetherian integral ___domain is a UFD if and only if every height 1 prime ideal is principal.

• 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.

• There is a nice ideal-theoretic characterization of UFDs, due to Kaplansky. If R is an integral ___domain, then R is a UFD if and only if every nonzero prime ideal of R has a nonzero prime element.