R UFD if R[x] UFD - 65 Characters

[math_grl]

Homework Statement

Assuming R is an integral domain. If the polynomial ring of one variable, R[x], is a unique factorization domain, then R is a unique factorization domain.

The Attempt at a Solution

Should be straightforward...so much so that I don't know how to start...probably with a homomorphism...

 

[rasmhop]

An integral domain is a UFD iff every reducible element has a unique factorization. So consider an arbitrary reducible element $a \in R$. Then a is an element of R[x] of degree 0, and since R[x] is a UFD a must have a unique factorization in R[x]. Can you from this unique factorization conclude that it has a unique factorization in R? (HINT: If $a = a_1a_2\cdots a_n$ is $a_i \in R$ for all i?)

 

[icantadd]

There is an inclusion of R into R[x] (in fact, and isomorphic copy of R in R[x]), so define a property on R[x] s..t p(y) iff y has a unique factorization. Then p(z) for all z in R[x], (including the irreducibles, since in this case the factorization of x = x and is unique by irreducibility). This includes those z which are in R. But all r in R are in R[x].