Product of two Polynomials in a UFD

oblixps

Member
Let R be a UFD and K be its field of fractions. Let f(x) be in R[x] and f(x) = a(x)b(x) where a(x), b(x) are in K[x]. Show that there exists a c in K such that c*a(x) and c-1*b(x) are both in R[x] and such that f(x) = (c*a(x))(c-1*b(x)) in R[x].

I have been stuck on this for a while now as I am having trouble coming up with what possible c to use. Since c*a(x) must be in R[x], the "numerator" of c must contain the lcm of the denominators of a(x) (i will denote as m) so that multiplying c and a(x) will cancel out all the denominators. Similarly, since c-1*b(x) is in R[x] as well, i need the numerator of c-1 to contain the lcm of the denominators of b(x) (i denote as n). In other words, i need the denominator of c to contain the lcm of the denominators of b(x). So far i have that c = m/n.

This is where i start to have trouble. Looking at c*a(x), after canceling out the denominators of a(x) i still have the factor of n on the bottom and there is no reason to believe that n divides each of the numerators of a(x). So I "divide out" as much as possible and then i am left with n* on the bottom. i can't do anything else so i just shove a factor of n* to compensate on the numerator of c so i now have c = m(n*)/n. So multiplying by the new c will make c*a(x) be in R[x].

Now looking at c-1 = n/(m)(n*) we have that c-1*b(x) the n cancels out so we are left with (m)(n*) on the bottom. Just like before, there is no reason to believe that (m)(n*) can divide each of the numerators of b(x) so i have to add another factor to the numerator of c-1 to compensate for that.but then when i now multiply c*a(x) i may need to add a new compensating factor and so on. it is not clear to me that this process will end.

is this the best way to go about this problem? my argument feels awkward and I would appreciate it if someone could give me some hints on how to better approach this.