# Show that the following localization is a unique factorization domain

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?

• Commented Nov 10, 2016 at 17:53
• @Watson I would still need to show that $\mathbb{Q}[x]$ is a UFD. And the only way I know how to show that is to prove that if a field is a UFD then $F[x]$ is a UFD as well. That would become too lengthy/messy for something that seems pretty straightforward. Commented Nov 10, 2016 at 18:30
• @Jeffrey I think you're thinking about the result $D$ is a UFD $\implies$ $D[X]$ is a UFD. Instead, the result when $D$ is a field is very simple because in that case $D[X]$ is a PID, hence a UFD.
– Xam
Commented Mar 17, 2017 at 23:56

As pointed out in the comments, it suffices to show that $$\mathbb{Q}[x]$$ is a unique factorization domain. Every euclidean domain is a principal ideal domain and every principal ideal domain is a unique factorization domain (as pointed out in the comments as well). I believe that in this case the easiest property to check is being an euclidean domain:
Let $$k$$ be a field. Then $$k[x]$$ is an euclidean domain with euclidean function $$\deg{f}$$ for all $$f\neq 0\in k[x]$$.