Let $R$ be a discrete integral domain. We say that an element $r\in R$ is a unit if it is invertible. A non-unit is called irreducible if it can not be represented as a product of two non-units. Note that then, $0$ is never irreducible, because it is a product of two non-units under the form $0 = 0.0$. It is often included in the definition of an irreducible element that it must be non-zero but this is in fact redudant.
An integral domain $R$ is a unique factorization domain (UFD for short) if every non-zero non-unit has a factorization $u = r_1 \cdots r_n$ (where $n \ge 1$) as product of irreducibles and this decomposition is unique up to renumbering and rescaling the irreducibles by units.
Put differently: $R$ is a UFD precisely when the multiplicative monoid of nonzero principal ideals of $R$ (which is isomorphic to the quotient monoid $Can(R)/R^\times$, where $Can(R) \coloneqq R \backslash \{0\}$ denotes the multiplicative subset of cancellative elements in $R$ and $R^\times$ denotes the group of units in $R$) is a commutative monoid freely generated by irreducible principal ideals. It follows that if $K$ is the field of fractions of $R$, then the quotient group $K^\times/R^\times$ is an abelian group that is freely generated by the set of cosets $f R^\times$ with $f$ ranging over irreducible elements. As a side remark, we observe that in this circumstance the exact sequence
splits and there is an isomorphism $K^\times \cong R^\times \oplus (K^\times/R^\times)$ of abelian groups.
The ring of (rational) integers $\mathbb{Z}$ is a UFD.
A principal ideal domain (PID) is a UFD. (In particular, a Euclidean domain is a UFD.) As a partial converse, a Dedekind domain that is a UFD is a PID.
If $R$ is a UFD, then its polynomial ring $R[x]$ is also a UFD.
If $R$ is a UFD, then so is any localization $S^{-1} R$.
A regular local ring (for example, a discrete valuation ring) is a UFD.
If $R$ is a UFD and all of its localizations at primes are regular local rings, then the ring of formal power series $R[ [x] ]$ is a UFD.
For $D$ a positive integer, the ring of integers in $\mathbb{Q}(\sqrt{-D})$ is a UFD iff $D$ is a Heegner number, namely one of the numbers $1, 2, 3, 7, 11, 19, 43, 67, 163$.
If $R$ is not integrally closed, then it is not a UFD.
As noted above, a UFD is necessarily integrally closed.
The lattice of principal ideals under the inclusion order is a distributive lattice.
There is a useful characterization of UFDs by Irving Kaplansky:
Let $R$ be an integral domain. Then, $R$ is a UFD iff every non-zero prime ideal contains a non-zero non-unit prime element.
Henri Lombardi, Claude Quitté (2010): Commutative algebra: Constructive methods (Finite projective modules) Translated by Tania K. Roblo, Springer (2015) (doi:10.1007/978-94-017-9944-7, pdf)
Irving Kaplansky, Commutative rings, rev. ed., the University of Chicago (1974), pdf
Last revised on January 23, 2023 at 17:56:38. See the history of this page for a list of all contributions to it.