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 $p_1p_2\cdots p_r$ , and $q_1q_2\cdots q_s$ , 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 $j$

The factors $p_1,\,p_2,\,\ldots,\,p_r$ , are called the prime factors 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.