# Unique factorization domain

(Redirected from Unique factorisation domain)

In mathematics, a unique factorization domain (UFD) is a commutative ring, which is an integral domain, and in which every non-zero non-unit element can be written as a product of prime elements (or irreducible elements), uniquely up to order and units, analogous to the fundamental theorem of arithmetic for the integers. UFDs are sometimes called factorial rings, following the terminology of Bourbaki.

Unique factorization domains appear in the following chain of class inclusions:

commutative ringsintegral domainsintegrally closed domainsGCD domainsunique factorization domainsprincipal ideal domainsEuclidean domainsfieldsfinite fields

## Definition

Formally, a unique factorization domain is defined to be an integral domain R in which every non-zero element x of R can be written as a product (an empty product if x is a unit) of irreducible elements pi of R and a unit u:

x = u p1 p2 ... pn with n ≥ 0

and this representation is unique in the following sense: If q1,...,qm are irreducible elements of R and w is a unit such that

x = w q1 q2 ... qm with m ≥ 0,

then m = n, and there exists a bijective map φ : {1,...,n} → {1,...,m} such that pi is associated to qφ(i) for i ∈ {1, ..., n}.

The uniqueness part is usually 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 element can be written as a product of a unit and 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.
• ${\displaystyle Z[e^{\frac {2\pi i}{n}}]}$ is a UFD for all integers 1 ≤ n ≤ 22, but not for n = 23.
• If R is a UFD, then so is R[X], the ring of polynomials with coefficients in R. Unless R is a field, R[X] is not a principal ideal domain. By iteration, a polynomial ring in any number of variables over any UFD (and in particular over a field) is a UFD.
• The Auslander–Buchsbaum theorem states that every regular local ring is a UFD.
• The formal power series ring K[[X1,...,Xn]] over a field K (or more generally over a regular UFD such as a PID) is a UFD. On the other hand, the formal power series ring over a UFD need not be a UFD, even if the UFD is local. For example, if R is the localization of k[x,y,z]/(x2 + y3 + z7) at the prime ideal (x,y,z) then R is a local ring that is a UFD, but the formal power series ring R[[X]] over R is not a UFD.
• Mori showed that if the completion of a Zariski ring, such as a Noetherian local ring, is a UFD, then the ring is a UFD.[1] The converse of this is not true: there are Noetherian local rings that are UFDs but whose completions are not. The question of when this happens is rather subtle: for example, for the localization of k[x,y,z]/(x2 + y3 + z5) at the prime ideal (x,y,z), both the local ring and its completion are UFDs, but in the apparently similar example of the localization of k[x,y,z]/(x2 + y3 + z7) at the prime ideal (x,y,z) the local ring is a UFD but its completion is not.
• Let ${\displaystyle R}$ be any field of characteristic not 2. Klein and Nagata showed that the ring R[X1,...,Xn]/Q is a UFD whenever Q is a nonsingular quadratic form in the X's and n is at least 5. When n=4 the ring need not be a UFD. For example, ${\displaystyle R[X,Y,Z,W]/(XY-ZW)}$ is not a UFD, because the element ${\displaystyle XY}$ equals the element ${\displaystyle ZW}$ so that ${\displaystyle XY}$ and ${\displaystyle ZW}$ are two different factorizations of the same element into irreducibles.
• The ring of formal power series over the complex numbers is a UFD, but the subring of those that converge everywhere, in other words the ring of entire functions in a single complex variable is not a UFD, since there exist entire functions with an infinity of zeros, and thus an infinity of irreducible factors, while a UFD factorization must be finite, e.g.:
${\displaystyle \sin \pi z=\pi z\prod _{n=1}^{\infty }\left(1-{{z^{2}} \over {n^{2}}}\right).}$
• The ring Q[x,y]/(x2 + 2y2 + 1) is a UFD, but the ring Q(i)[x,y]/(x2 + 2y2 + 1) is not. On the other hand, The ring Q[x,y]/(x2 + y2 – 1) is not a UFD, but the ring Q(i)[x,y]/(x2 + y2 – 1) is (Samuel 1964, p.35). Similarly the coordinate ring R[X,Y,Z]/(X2 + Y2 + Z2 − 1) of the 2-dimensional real sphere is a UFD, but the coordinate ring C[X,Y,Z]/(X2 + Y2 + Z2 − 1) of the complex sphere is not.
• Suppose that the variables Xi are given weights wi, and F(X1,...,Xn) is a homogeneous polynomial of weight w. Then if c is coprime to w and R is a UFD and either every finitely generated projective module over R is free or c is 1 mod w, the ring R[X1,...,Xn,Z]/(Zc − F(X1,...,Xn)) is a UFD (Samuel 1964, p.31).

Non-example:

• The quadratic integer ring ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$ of all complex numbers of the form ${\displaystyle a+b{\sqrt {-5}}}$, where a and b are integers, is not a UFD because 6 factors as both (2)(3) and as ${\displaystyle \left(1+{\sqrt {-5}}\right)\left(1-{\sqrt {-5}}\right)}$. These truly are different factorizations, because the only units in this ring are 1 and −1; thus, none of 2, 3, ${\displaystyle 1+{\sqrt {-5}}}$, and ${\displaystyle 1-{\sqrt {-5}}}$ are associate. It is not hard to show that all four factors are irreducible as well, though this may not be obvious.[2] See also algebraic integer.
• For a square-free positive integer n, the quadratic integer ring ${\displaystyle \mathbb {Z} [{\sqrt {-n}}]}$ will fail to be a UFD unless n is a Heegner number.

## 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. For instance, the element ${\displaystyle z\in K[x,y,z]/(z^{2}-xy)}$ is irreducible, but not prime.) Note that this has a partial converse: a domain satisfying the ACCP is a UFD if and only if every irreducible element is prime.
• 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 a UFD 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.
• Let S be a multiplicatively closed subset of a UFD A. Then the localization ${\displaystyle S^{-1}A}$ is a UFD. A partial converse to this also holds; see below.

## 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 (a proof is given below). Also, a Dedekind domain is a UFD if and only if its ideal class group is trivial. In this case it is in fact a principal ideal domain.

There are also equivalent conditions for non-noetherian integral domains. Let A be an integral domain. Then the following are equivalent.

1. A is a UFD.
2. Every nonzero prime ideal of A contains a prime element. (Kaplansky)
3. A satisfies ascending chain condition on principal ideals (ACCP), and the localization S−1A is a UFD, where S is a multiplicatively closed subset of A generated by prime elements. (Nagata criterion)
4. A satisfies ACCP and every irreducible is prime.
5. A is atomic and every irreducible is prime.
6. A is a GCD domain (i.e., any two elements have a greatest common divisor) satisfying (ACCP).
7. A is a Schreier domain,[3] and atomic.
8. A is a pre-Schreier domain and atomic.
9. A has a divisor theory in which every divisor is principal.
10. A is a Krull domain in which every divisorial ideal is principal (in fact, this is the definition of UFD in Bourbaki.)
11. A is a Krull domain and every prime ideal of height 1 is principal.[4]

In practice, (2) and (3) are the most useful conditions to check. For example, it follows immediately from (2) that a PID is a UFD, since, in a PID, every prime ideal is generated by a prime element.

For another example, consider a Noetherian integral domain in which every height one prime ideal is principal. Since every prime ideal has finite height, it contains height one prime ideal (induction on height), which is principal. By (2), the ring is a UFD.