Unique factorization domain

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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.

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

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

[edit] 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.

[edit] 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.)

[edit] Counterexamples

• The ring of all complex numbers of the form , where a and b are integers. Then 6 factors as both (2)(3) and as . These truly are different factorizations, because the only units in this ring are 1 and −1; thus, none of 2, 3, , and 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 αX + βY + γZ + δW and a degree zero element r. This gives X = rαX + rβY + rγZ + rδW. In R[X,Y,Z,W], then, the degree one element (rα − 1)X + rβY + rγZ + rδW must be an element of the ideal (XY − ZW), but the non-zero elements of that ideal are degree two and higher. Consequently, (rα − 1)X + rβY + rγZ + rδW must be zero in R[X,Y,Z,W]. That implies that rα = 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, i.e:

[edit] 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 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.

[edit] 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.