# Unique factorisation domain

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 \(\mathbb{C}\) 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 noteThat 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. (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. (Proof.)

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