[summary(Technical): An Integral domain $~$R$~$ is said to be a unique factorisation domain if every nonzero non-unit element of $~$R$~$ may be written as a product of irreducibles, and moreover this product is unique up to reordering and multiplying by units.]

[summary: A ring $~$R$~$ is a unique factorisation domain if an analog of the Fundamental Theorem of Arithmetic holds in it. The condition is as follows: every nonzero element, which does not have a multiplicative inverse, must be expressible as a product of irreducibles, and the expression must be unique if we do not care about ordering or about multiplying by elements which have multiplicative inverses.]

Ring theory is the art of extracting properties from the integers and working out how they interact with each other. From this point of view, a unique factorisation domain is a ring in which the integers' Fundamental Theorem of Arithmetic holds.

There have been various incorrect "proofs" of Fermat's Last Theorem throughout the ages; it turns out that if we assume the false "fact" that all subrings of [4zw $~$\mathbb{C}$~$] are unique factorisation domains, then FLT is not particularly difficult to prove. This is an example of where abstraction really is helpful: having a name for the concept of a UFD, and a few examples, makes it clear that it is not a trivial property and that it does need to be checked whenever we try and use it.

Formal statement

Let $~$R$~$ be an Integral domain. Then $~$R$~$ is a unique factorisation domain if every nonzero non-unit element of $~$R$~$ may be expressed as a product of irreducibles, and moreover this expression is unique up to reordering and multiplying by units.

Why ignore units?

We must set things up so that we don't care about units in the factorisations we discover. Indeed, if $~$u$~$ is a unit %%note:That is, it has a multiplicative inverse $~$u^{-1}$~$.%%, then $~$p \times q$~$ is always equal to $~$(p \times u) \times (q \times u^{-1})$~$, and this "looks like" a different factorisation into irreducibles. ($~$p \times u$~$ is irreducible if $~$p$~$ is irreducible and $~$u$~$ is a unit.) The best we could possibly hope for is that the factorisation would be unique if we ignored multiplying by invertible elements, because those we may always forget about.

Example

In $~$\mathbb{Z}$~$, the units are precisely $~$1$~$ and $~$-1$~$. We have that $~$-10 = -1 \times 5 \times 2$~$ or $~$-5 \times 2$~$ or $~$5 \times -2$~$; we need these to be "the same" somehow.

The way we make them be "the same" is to insist that the $~$5$~$ and $~$-5$~$ are "the same" and the $~$2$~$ and $~$-2$~$ are "the same" (because they only differ by multiplication of the unit $~$-1$~$), and to note that $~$-1$~$ is not irreducible (because irreducibles are specifically defined to be non-unit) so $~$-1 \times 5 \times 2$~$ is not actually a factorisation into irreducibles.

That way, $~$-1 \times 5 \times 2$~$ is not a valid decomposition anyway, and $~$-5 \times 2$~$ is just the same as $~$5 \times -2$~$ because each of the irreducibles is the same up to multiplication by units.

Examples

• Every Principal ideal domain is a unique factorisation domain. ([pid_implies_ufd Proof.]) This fact is not trivial! Therefore $~$\mathbb{Z}$~$ is a UFD, though we can also prove this directly; this is the Fundamental Theorem of Arithmetic.
• $~$\mathbb{Z}[-\sqrt{3}]$~$ is not a UFD. Indeed, $~$4 = 2 \times 2$~$ but also $~$(1+\sqrt{-3})(1-\sqrt{-3})$~$; these are both decompositions into irreducible elements. (See the page on irreducibles for a proof that $~$2$~$ is irreducible; the same proof can be adapted to show that $~$1 \pm \sqrt{-3}$~$ are both irreducible.)

Properties

• If it is hard to test for uniqueness up to reordering and multiplying by units, there is an easier but equivalent condition to check: an integral domain is a unique factorisation domain if and only if every element can be written (not necessarily uniquely) as a product of irreducibles, and all irreducibles are prime. ([alternative_condition_for_ufd Proof.])