# Is UFD an integral domain?

Let $$R$$ be a ring. Does part of the definition of $$R$$ being UFD contain that it is an integral domain, or does conditions of $$R$$ being a UFD forces $$R$$ being an integral domain?

If latter, why is it true?

• U.F.D. is the acronym of Unique Factorisation $\color{red}{\text{Domain}}$. – Bernard Nov 24 '20 at 23:20
• Yes, the D stands for domain – bounceback Nov 24 '20 at 23:20
• – lhf Nov 24 '20 at 23:33
• Something to think about: to answer this question one just had to look at the definition of "UFD". – David C. Ullrich Nov 24 '20 at 23:42

Being an integral domain is part of the definition of a UFD; this is what the "domain" part of "unique factorization domain" refers to. Consider for instance $$\mathbb{Z}\times\mathbb{Z}$$, which satisfies all the "conditions" to be a UFD except that it is not an integral domain. (To see this, note that the irreducible elements of $$\mathbb{Z}\times\mathbb{Z}$$ are precisely those of the form $$(p,1)$$ and $$(1,q)$$ for $$p,q\in\mathbb{Z}$$ prime.)
I put "conditions" in quotation marks because unique factorization into irreducibles in $$\mathbb{Z}\times\mathbb{Z}$$ will of course hold only for non-zero divisors, rather than all non-zero elements.
In fact, irreducible elements are themselves often defined only for integral domains, even though the definition carries over to more general cases. (The definition I use here is that an element $$x$$ in an arbitrary commutative ring $$R$$ is irreducible when $$x=yz$$ implies that exactly one of $$y,z$$ is a unit, for any $$y,z\in R$$. There are other reasonable definitions, too.)