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R[x] UFD, then R UFD

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data

    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.

    3. The attempt at a solution

    Should be straightforward...so much so that I don't know how to start...probably with a homomorphism...
     
  2. jcsd
  3. Nov 23, 2009 #2
    An integral domain is a UFD iff every reducible element has a unique factorization. So consider an arbitrary reducible element [itex]a \in R[/itex]. 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 [itex]a = a_1a_2\cdots a_n[/itex] is [itex]a_i \in R[/itex] for all i?)
     
  4. Nov 23, 2009 #3
    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].
     
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