# 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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