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" = "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,...,"m"} 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""X"₁,...,"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 ["X"₁, ..., "X"_"n"] as well as "K" ["X"₁, ..., "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 $mathbb\; Z\; [sqrt\{-5\}]$ of all complex numbers of the form $a+ibsqrt\{5\}$, where "a" and "b" are integers. Then 6 factors as both (2)(3) and as $left(1+isqrt\{5\}\; ight)left(1-isqrt\{5\}\; ight)$. These truly are different factorizations, because the only units in this ring are 1 and −1; thus, none of 2, 3, $1+isqrt\{5\}$, and $1-isqrt\{5\}$ 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. The proof is in two parts.

:First we must show $X$, $Y$, $Z$, and $W$ are all irreducible. Grade $R\; [X,Y,Z,W]\; /(XY-ZW)$ 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 and a degree zero element $r$. This gives . In $R\; [X,Y,Z,W]$, then, the degree one element must be an element of the ideal $(XY-ZW)$, but the non-zero elements of that ideal are degree two and higher. Consequently, must be zero in $R\; [X,Y,Z,W]$. That implies that $ralpha\; =\; 1$, 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 $R\; [X,Y,Z,W]\; /(XY-ZW)$ 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, e.g.:::$sin\; pi\; z\; =\; pi\; z\; prod\_\{n=1\}^\{infty\}\; left(1-$z^2}over{n^2 ight).

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 $Rightarrow$ 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.

* 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.

References

* Chap. 4.
* Chap.II.5.
*