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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?

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    $\begingroup$ What about this definition? $\endgroup$ – azif00 Nov 24 '20 at 23:20
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    $\begingroup$ U.F.D. is the acronym of Unique Factorisation $\color{red}{\text{Domain}}$. $\endgroup$ – Bernard Nov 24 '20 at 23:20
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    $\begingroup$ Yes, the D stands for domain $\endgroup$ – bounceback Nov 24 '20 at 23:20
  • $\begingroup$ See also Unique Factorization Rings with Zero Divisors. $\endgroup$ – lhf Nov 24 '20 at 23:33
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    $\begingroup$ Something to think about: to answer this question one just had to look at the definition of "UFD". $\endgroup$ – David C. Ullrich Nov 24 '20 at 23:42
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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.)

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The domain part of Unique factorization domain refers to the ring being an integral domain. (atleast, this is the way I understood it, however, it is the case that being an integral domain is part of the definition of a UFD)

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