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.


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.


  • 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.)


  • 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.)