# UFD

An integral domain $D$ satisfying

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

• If  $p_{1}p_{2}\cdots p_{r}$  and  $q_{1}q_{2}\cdots q_{s}$  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  $p_{1},\,p_{2},\,\ldots,\,p_{r}$  are called the of $a$.

Some of the classic results about UFDs:

• On a UFD, the concept of prime element and irreducible element coincide.

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

• 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.

• 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 2013-03-22 11:56:22 Last modified on 2013-03-22 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