UFD
An integral domain^{} $D$ satisfying

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Every nonzero element of $D$ that is not a unit can be factored into a product of a finite number of irreducibles^{},

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If ${p}_{1}{p}_{2}\mathrm{\cdots}{p}_{r}$ and ${q}_{1}{q}_{2}\mathrm{\cdots}{q}_{s}$ are two factorizations of the same element $a$ into irreducibles, then $r=s$ and we can reorder the ${q}_{j}$’s in a way that ${q}_{j}$ is an associate^{} element of ${p}_{j}$ for all $j$
is called a unique factorization domain^{} (UFD), also a factorial ring.
The factors ${p}_{1},{p}_{2},\mathrm{\dots},{p}_{r}$ are called the prime factors^{} of $a$.
Some of the classic results about UFDs:

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On a UFD, the concept of prime element^{} and irreducible element coincide.

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If $F$ is a field, then $F[x]$ is a UFD.

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If $D$ is a UFD, then $D[x]$ (the ring of polynomials on the variable $x$ over $D$) is also a UFD.
Since $R[x,y]\cong R[x][y]$, these results can be extended to rings of polynomials with a finite number of variables.

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If $D$ is a principal ideal domain^{}, then it is also a UFD.
The converse is, however, not true. Let $F$ a field and consider the UFD $F[x,y]$. Let $I$ the ideal consisting of all the elements of $F[x,y]$ whose constant term is $0$. Then it can be proved that $I$ is not a principal ideal^{}. Therefore not every UFD is a PID.
Title  UFD 
Canonical name  UFD 
Date of creation  20130322 11:56:22 
Last modified on  20130322 11:56:22 
Owner  drini (3) 
Last modified by  drini (3) 
Numerical id  19 
Author  drini (3) 
Entry type  Definition 
Classification  msc 13G05 
Synonym  unique factorization domain 
Related topic  IntegralDomain 
Related topic  Irreducible 
Related topic  EuclideanRing 
Related topic  EuclideanValuation 
Related topic  ProofThatAnEuclideanDomainIsAPID 
Related topic  WhyEuclideanDomains 
Related topic  Y2X32 
Related topic  PID 
Related topic  PIDsAreUFDs 
Related topic  FundamentalTheoremOfArithmetic 
Defines  factorial ring 
Defines  prime factor 
Defines  UFD 
Defines  unique factorization 