[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.])