# Unique factorization domain

In mathematics, a **unique factorization domain** (UFD) is, roughly speaking, a commutative ring in which every element 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.

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*=*p*_{1}*p*_{2}...*p*_{n}

and this representation is unique in the following sense: if *q*_{1},...,*q*_{m} are irreducible elements of *R* such that

*x*=*q*_{1}*q*_{2}...*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 UFD's:

- the integers. This is the fundamental theorem of arithmetic.
- any field; this includes the fields of rational numbers, real numbers, and complex numbers.
- Rings of polynomials with coefficients in a field.

Here are some more examples of UFDs:

- The Gaussian integers,
**Z**[*i*]. - The formal power series ring
*K*[[*X*_{1},...,*X*_{n}]] over a field*K*. - The ring of functions in
*n*complex variables holomorphic at the origin is a UFD.

Despite these examples, very few integral domains are UFDs. Here is a counterexample:

- The ring
**Z**[√ −5] of all complex numbers of the form*a*+*b*√ −5, where*a*and*b*are integers. Then 6 factors as both (2)(3) and as (1 + √ −5) (1 − √ −5). These truly are different factorizations, because the only units in this ring are 1 and −1; thus, none of 2, 3, 1 + √ −5, and 1 − √ −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
*R*be any commutative ring. Then*R*[*X*,*Y*,*Z*,*W*]/(*XY-ZW*) is not a UFD. It is clear that X, Y, Z, and W are all irreducibles, so the element XY=ZW has two factorizations into irreducible elements.

## Properties

Additional examples of UFDs can be constructed as follows:

- All principal ideal domains are UFDs.

- If
*R*is a UFD, then so is the polynomial ring*R*[*X*]. By induction, we can show that the polynomial rings**Z**[*X*_{1}, ...,*X*_{n}] as well as*K*[*X*_{1}, ...,*X*_{n}] (*K*a field) are UFD's. (Any polynomial ring with more than one variable is an example of a UFD that is not a principal ideal domain.)

Some concepts defined for integers can be generalized to UFDs:

- In UFD's, every irreducible element is prime. (In any integral domain, every prime element is irreducible, but the converse does not always hold.)

- 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.

## 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.

- A ring is a UFD if and only if its class group is zero.de:Faktorieller Ring