# R is UFD iff R[x] is UFD.

In this proof of "$R$ is UFD $\Rightarrow$$R[x]$ is UFD", the author showed that any nonzero $f \in R[x]$ can be factorized into primes. Then how does $R[x]$'s UFD property follow?

I believe we also need:

If $R$ is integral domain: $R$ is UFD $\Leftrightarrow$ every irreducible is prime, every non unit factors into irreducible $\Leftrightarrow$every non unit factors into primes.

Or is the above unnecessary?

• @G.Sassatelli I tried to undo my edit but the site would not allow it – Arbuja Apr 29 '17 at 1:02

• In a domain, primes are always irreducible: suppose $p=ab$. By primality, $p$ divides one of them, say $a$. Then $p=p\alpha b$, i.e. $(1-\alpha b)p=0$, i.e. $1=\alpha b$. So $b$ is invertible, q.e.d.
• If all non-units except $0$ factor into primes, then irreducible elements are primes. In fact, let $a$ be irreducible. By definition, it is neither zero nor a unit, therefore there is a prime $q$ and something else $b$ such that $a=qb$. By irreducibility, $b$ must be a unit. Therefore, $a$ is actually a conjugate of $q$.
So you can see that for a domain $R$ the condition "every non-zero, non-unit element can be written as a product of primes" is equivalent to "every irreducible is prime and every non-zero, non-unit can be written as a product of irreducible elements".