# 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.)

Parents: