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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.

Some specific kinds of unique factorization domains are given with the following chain of set inclusions:

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

Contents 1 Definition 2 Examples 3 Counterexamples 4 Properties 5 Equivalent conditions for a ring to be a UFD

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 Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbb Z[\sqrt{-5}]

of all complex numbers of the form Failed to parse (Missing texvc executable; please see math/README to configure.): a+ib\sqrt{5}

, where a and b are integers. Then 6 factors as both (2)(3) and as Failed to parse (Missing texvc executable; please see math/README to configure.): \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, Failed to parse (Missing texvc executable; please see math/README to configure.): 1+i\sqrt{5} , and Failed to parse (Missing texvc executable; please see math/README to configure.): 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 Failed to parse (Missing texvc executable; please see math/README to configure.): R

be any commutative ring.  Then Failed to parse (Missing texvc executable; please see math/README to configure.): R[X,Y,Z,W]/(XY-ZW)
is not a UFD.  The proof is in two parts.

First we must show Failed to parse (Missing texvc executable; please see math/README to configure.): X

, Failed to parse (Missing texvc executable; please see math/README to configure.): Y , Failed to parse (Missing texvc executable; please see math/README to configure.): Z , and Failed to parse (Missing texvc executable; please see math/README to configure.): W

are all irreducible.  Grade Failed to parse (Missing texvc executable; please see math/README to configure.): R[X,Y,Z,W]/(XY-ZW)
by degree.  Assume for a contradiction that Failed to parse (Missing texvc executable; please see math/README to configure.): X
has a factorization into two non-zero non-units.  Since it is degree one, the two factors must be a degree one element Failed to parse (Missing texvc executable; please see math/README to configure.): \alpha X + \beta Y + \gamma Z + \delta W
and a degree zero element Failed to parse (Missing texvc executable; please see math/README to configure.): r

. This gives Failed to parse (Missing texvc executable; please see math/README to configure.): X = r\alpha X + r\beta Y + r\gamma Z + r\delta W . In Failed to parse (Missing texvc executable; please see math/README to configure.): R[X,Y,Z,W] , then, the degree one element Failed to parse (Missing texvc executable; please see math/README to configure.): (r\alpha-1) X + r\beta Y + r\gamma Z + r\delta W

must be an element of the ideal Failed to parse (Missing texvc executable; please see math/README to configure.): (XY-ZW)

, but the non-zero elements of that ideal are degree two and higher. Consequently, Failed to parse (Missing texvc executable; please see math/README to configure.): (r\alpha-1) X + r\beta Y + r\gamma Z + r\delta W

must be zero in Failed to parse (Missing texvc executable; please see math/README to configure.): R[X,Y,Z,W]

. That implies that Failed to parse (Missing texvc executable; please see math/README to configure.): r\alpha = 1 , so Failed to parse (Missing texvc executable; please see math/README to configure.): r

is a unit, which is a contradiction.  Failed to parse (Missing texvc executable; please see math/README to configure.): Y

, Failed to parse (Missing texvc executable; please see math/README to configure.): Z , and Failed to parse (Missing texvc executable; please see math/README to configure.): W

are irreducible by the same argument.

Next, the element Failed to parse (Missing texvc executable; please see math/README to configure.): XY

equals the element Failed to parse (Missing texvc executable; please see math/README to configure.): ZW
because of the relation Failed to parse (Missing texvc executable; please see math/README to configure.): XY - ZW = 0

. That means that Failed to parse (Missing texvc executable; please see math/README to configure.): XY

and Failed to parse (Missing texvc executable; please see math/README to configure.): ZW
are two different factorizations of the same element into irreducibles, so Failed to parse (Missing texvc executable; please see math/README to configure.): R[X,Y,Z,W]/(XY-ZW)
is not a UFD.

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 Failed to parse (Missing texvc executable; please see math/README to configure.): \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.de:Faktorieller Ring

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