Suppose $R=\mathbb{Q}[x]$ is a ring and define $T$ to be $T=\{f(x) \in R : f(0) \neq 0\}$.
Show that $T^{-1}R$ is a unique factorization domain.
We can show that $T^{-1}R$ is a principal ideal domain and therefore a unique factorization domain but I don't want to take the PID route.
So by definition we need to be able to write every nonzero non-unit element of $T^{-1}R$ as a product of irreducibles and this factorization needs to be unique up to associates.
We can check that all irreducibles in $T^{-1}R$ are associate to $\frac{x}{1}$ and that units are given by elements in $T^{-1}R$ with numerators with nonzero constant terms. So we need to prove that we can find unique factorizations (consisting of irreducibles) of fractions with numerators without a constant term.
I figured I could maybe apply the fundamental theorem of algebra but got stuck.
How do we show this?