Let $R$ be a right (and left) Noetherian ring and $T=R[x]$ its polynomial ring. It was shown by Stafford that the set $S=\{f\in T\:|\:f\text{ monic}\}$ is a right denominator set. So my question is, if $g\in S$, then is $G=\{g^a\:|\:a\geq 0\}$ a right denominator set also?

If $t\in T$ and $g^i$ then I want to show $tG\cap g^i T\neq\emptyset$. As $g^i\in S$ then I know there exists some $t^\prime\in T$ and $f\in S$ such that $g^i t^\prime=tf$. So the question then becomes, does $f=g^j$ for some $j\geq 0$?

Edit: I thought of using the following result:

If $A\subset B$ is a multiplicative subset of a ring $B$ such that $aB$ is a two-sided ideal for any $a\in A$, then $A$ is a right permutable set.

Now, for $g^i\in G$ it is clear that $g^iT$ is right ideal of $T$. So, let $t\in T$ then $tg^iS\cap g^iT\neq\emptyset$, i.e. for some $s\in S$ and $t^\prime\in T$ we have $tg^is=g^it^\prime$. This seems promising but then I am unsure of how to proceed. I need to show $tg^i\bar{t}\in g^iT$ but do we need $\bar{t}=s\bar{t}^\prime$?