# Can a Unique Factorisation Domain be non-commutative?

The definition that our lecturer gave us for Unique Factorisation Domains is:

An integral domain $$R$$ is called a Unique Factorisation Domain (UFD) if every non-zero non-unit element of $$R$$ can be written as a product of irreducible elements and this product is unique up to order of the factors and multiplication by units.

If multiplication in this integral domain is non-commutative, then if $$x, a, b \in R$$ and $$x = ab = ba$$, do these count as different factorisations and mean that $$R$$ can't be a UFD?

## 3 Answers

In practice it's always safe to assume "integral domains" like UFD's are commutative, unless explicitly mentioned otherwise. (Commutative is the right word, not abelian. Abelian has other uses in ring theory.)

Of course, one can always go about trying to find an acceptable adaptation to the noncommutative case. One of the first papers I worked through in an undergrad seminar was this, which happens to be exactly the thing you are asking about: Cohn, P. M. "Noncommutative unique factorization domains." Transactions of the American Mathematical Society 109.2 (1963): 313-331.

You should check that out if you are interested in the subject. It outlines the changes which have to be made to make familiar arguments work out in the noncommutative case. According to Cohn's definition, those two factorizations would be considered identical.

• Oooh cool, thanks for the really informative answer. :) – ŠotiBriti Mar 29 '20 at 11:20

By the usual definition, an integral domain is commutative.

You can check out this survey. Smertnig did a great work in the developement of factorization theory of non-commutative rings. He also deals with the question of when factorizations should be considered different.